EST+Superbook+pw(jma,pece)-Review

To the woman who continually electrifies my once boring life… To my students who compelled me how to learn for the sake of teaching… To my baby who always regenerates my strength whenever I’m heavily attenuated… And to Him that gave us Maxwell, Hertz, Faraday, Marconi, Tesla, Gauss, Fessenden, etc… To Him is the kingdom, the power, and the glory forever…

“He that walketh with wise men shall be wise” Proverbs 13:20

Self-Sufficient Guide to Electronic Communications Engineering by Jason M. Ampoloquio copyright 2005 All rights reserved.

No part of this book may be reproduced in any form or by any means, electronic, or mechanical including photocopying, recording, mimeographing, or by any information and retrieval system, without the written permission from the copyright holder.

ISBN 971-92592-7-2

Printed in the Philippines

Important Legal Information: Warning and Disclaimer

This review book presents the fundamentals of Communications Engineering using modified concepts, principles, and examples from common textbook and reference materials in the field of Communications Engineering. While the author/compiler believes that the concepts and data in this book are accurate, correct and meticulously prepared, the materials in this book is intended solely as a quick reference aid, and is not represented to be an appropriate or safe solution to pass the ECE licensure exam. For this reason, the author makes no warranties, express or implied, that the concepts, examples, and data contained in this book are free from errors. I would, however, appreciate readers bringing to my attention any errors that may appear in the printed version of this book for reasons beyond my control. These errors and any other comments can be communicated to me by e-mail addressed to: [email protected]

The Author Jason M. Ampoloquio, PECE PECE consultant – Beta Electric Corporation Senior System Analyst, Department of Finance (Insurance Commission) Mobile/Web Application Developer Faculty Member – University of Santo Tomas President, Powerful Review Center MSECE Major in DSP-De La Salle University (units earned) BSECE-Central Colleges of the Philippines HR Reyes Scholar Coach, IECEP Quizzers Champion: 1. ECE Quiz Show (1999) 2. 1st Brain Encounter (1998) 3. Physics Quiz Show (1996) 4. Mathematics Wizard (1996) 5. Inter Engineering Quiz Show (1995) Battle of the Brain School Representative (RPN-9) Quizzer-19th and 20th IECEP Quiz Show Author: 1. Electronics Engineering SUPERBook 2. EST Refresher 3. Mathematics Refresher 4. GEAS Refresher 5. Electronics Refresher EST Review Director Former Faculty: TUP-M, CCP Resource Speaker, Various Topics in Communications In-house reviewer, Various Colleges and Universities Sought after reviewer in Communications Engineering

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Although it is a relatively easy process to select a book from a catalog or shelf in a bookstore, the actual effort involved in its preparation and publication can be quite complex and can extend over a significant period of time (for this book I personally experienced almost 2 years of hundreds of revision, an average of eight hrs every night for searching, researching, reading, comparing, computing, analyzing, and typing—3 hrs of which is allocated for drawings and illustrations). As a beginner, I realized that the creation of a written work is a team effort requiring the cooperation and assistance of many persons. Thus, I would be remiss if I did not thank those who made this book possible. First & foremost, to God for His indescribable gift. To my parents Jayme Ampoloquio and Teresita Ampoloquio, to my sisters and brothers namely Elizabeth, Ma. Teresa, Dennis, and Emmanuel, to whom I owe one of my deepest appreciations. To all the people who in invaluable ways help to mold my character as a teacher and a reviewer namely Dean Cynthia Llanes & Carlos Llanes, and Leonardo Samaniego. To my fellows, Engrs. Anthony Lopos, Gerard Ang, Rufino Orpia, Reginaldo Marinay, Dennis Paus. And to the following: Vic Buenconsejo (Colegio de San Agustin Bacolod) and Dean Christopher Taclobos (UNO-R Bacolod) for their continuous trust in my potentials and warmth accommodation during my stay in their respective places. To Engr. Reynante Abuyan: Thank you for lending me your awesome printer and editing the first few chapters of this superb guidebook. To all ECE reviewees: Thank you for your patronage and trust! And may all your dreams be electric! To Engrs. Randel Espina of Ateneo de Davao, Neil Abalajon of TUP-Visayas, Marlon Lagulos of Univ. of Southeastern Philippines, Carlos Sison of PLM, Nieves Elarco of SLPC, Dean Jose Magleo of Collegio de Dagupan, Evelyn Raguindin of Adamson University, Dean Roman Palo of Holy Angel University, Engr Oliver Mariano of Bulacan State University, and to those people whom I cannot individually write their names. I sincerely thank you all… jma/pece

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or as much as many have tried to create a good ECE guide book, it seems good to me also, having had a fine understanding to finally craft a review book intended not to compete to those which are existing but in my own little way and with God’s help, to properly guide ECE students on their quest to the elusive licensure examination.

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he Self-sufficient Guide to Electronic Communications Engineering (ECE) that I preferred to call SUPERBook, is a comprehensive and meticulous compilation of modified concepts in every aspect of communications engineering discipline. The SUPERBook is carefully scrutinized to cover important details that are imperative to board exam preparation but also taken into account minute pieces of information commonly neglected by reviewees that lead to occasionally guessing the board exam questions.

S

ince it is also a proven fact that 50% of learning process comes from visual aids and illustration, and past ECE board questions require the knowledge of block diagram to be answered correctly, the author/compiler painstakingly prepared lavish illustration that will scintillate the interest of readers.

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lso the SUPERBook flagships is its minefield of 250+ scrupulously solved and 2100+ multiple choice practice problems aimed to help ECE reviewees to keep abreast to the recent trend of ECE board examination. he answer to even-numbered questions are intentionally withheld by the author/compiler but will be discuss to EXPERTS reviewees during review, refresher and coaching program for the reason that the author is anticipating ECE school teachers and co-reviewers will use some of the question as a safe and quick exam references since the answers are withheld and encourage readers to test their knowledge in the field of communications engineering.

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nd finally, the author/compiler firmly believed that knowledge is cumulative. It is not based on each person’s re-inventing that is known, but rather accumulating what has been learned from the past, synthesizing new ideas from old formulas and principles, and creating completely new insights, I hope elements of all three can be found in these pages, and I want to acknowledge the contribution of all those works (textbooks, journals, manuals, magazine, and website tutorials) has added to my understanding of the various topics of this lavish guidebook. Read it till it Hertz… Happy reading… c”,) Jason M. Ampoloquio, PECE

Contents Title page Dedication Acknowledgments Preface CHAPTER 1: Introduction to Electronic Communication Section 1. Basic Principles of Communications Engineering Section 2. Amplitude Modulation Section 3. Angle Modulation Section 4. Noise Analysis and dB Calculations Section 5. Transmitters and Receivers

1-xx 1-1 1-38 1-68 1-94 1-125

CHAPTER 2: Acoustics & Broadcasting Section 6. Acoustics Fundamentals Section 7. Television Fundamentals Section 8. AM, FM AND TV Broadcasting Standards Section 9. Microphones & Loudspeakers

2-xx 2-1 2-28 2-69 2-89

CHAPTER 3: Wire Communications Section 10. Transmission Lines and Waveguides Section 11. Fiber Optics Communications Section 12. Telephone Networks and System Section 13. Facsimile Transmission

3-xx 3-1 3-50 3-106 3-159

CHAPTER 4: Computer Communications Section 14. Pulse Modulation Section 15. Digital Communications Section 16. Data Communications

4-xx 4-1 4-19 4-55

CHAPTER 5: Wireless Communications Section 17. Antenna Fundamentals Section 18. Radio-Wave Propagation Section 19. Microwave Engineering Section 20. Satellite Communications Section 21. Cellular Communication System

5-xx 5-1 5-42 5-62 5-86 5-113

CHAPTER 6: Navigational Aids Section 22. Navigation System Section 23. Radar Fundamentals

6-xx 6-1 6-29

CHAPTER 7: Miscellaneous Topics Section 24. Laws & Ethics Section 25. Basic Signals & System

7-xx 7-1 7-41

PRC Examinee’s Guide Answers to Odd-Numbered Questions Study Tips Major References Author’s Page

8-xx 9-xx

Section 1 Basic Principles of Communications Engineering

PLSB =

m 2Pc 4

PUSB =

PLSB =

m 2Pc 4

PUSB = 0

PLSB =

m 2Pc 4

PUSB = 0

m 2Pc 4

Section 2 Amplitude Modulation Section 3 Angle Modulation Section 4 Noise Analysis and dB Calculations Section 5 Transmitters & Receivers

Introduction to Electronics Communication

Loading ECE SUPERBook

Self-Sufficient Guide to ECE by JASON AMPOLOQUIO

Section

Basic Principles of

1

Communications Engineering

1-1

Read it till it Hertz!

Electronic Communication is the transmission, reception, and processing of information between two or more locations using electronic circuits.

DEFINITION.

Allocation: Entry in the Table of Frequency Allocations of a given frequency band for the purpose of its use by one or more terrestrial or space radio communication services or the radio astronomy service under specified conditions.

DEFINITION.

Allotment: Entry of a designated frequency channel in the agreed plan, adopted by the ITU, for use by one or more nations for a terrestrial or space radio communication services in one or more identified countries or geographic areas and under specified conditions.

DEFINITION.

DEFINITION. Assignment: Authorization given by a nation for a radio station to use a radio-frequency channel under specified conditions.

HISTORICAL PERSPECTIVE 1820

Hans Christian Oersted discovered the relation between electricity and magnetism, later known as electromagnetism.

1821

Andre Marie Ampere already observed momentarily the phenomenon we now call electromagnetic induction and hypothesized the existence of magnetic field around a current-carrying conductor.

1822

Michael Faraday discovered electromagnetic induction, the reverse of Oersted discovery.

1830

An American, Joseph Henry, demonstrated telecommand by sending an electronic current over one mile of wire to activate an electromagnet which caused a bell to strike, thus wire telegraphy was born. Samuel F.B. Morse, successfully exploited Henry’s invention commercially.

1866

James Clerk Maxwell put together the principles of Oersted, Faraday and hypothesized the existence of electromagnetic waves.


BASIC PRINCIPLES OF COMMUNICATIONS Engineering

1-2

1886

German physicist, Heinrich Hertz performed an experiment on spark gap transmission verifying Maxwell statement. Hertz experimentally showed the existence of such waves which he called radio waves that paved the way for wireless communication.

1896

An Italian, Guglielmo Marconi developed the first wireless telegraph and successfully sent a message over a distance of few kilometers using a spark gap transmitter.

1900

Reginald Aubrey Fessenden, invented AM and successfully transmits a few words using spark gap transmitter.

1936

Major Edwin Armstrong developed the first successful FM radio system.

A. .NOMENCLATURE OF RADIO FREQUENCY BAND.

Based on ITU-R Recommendations V.431-6 Frequency Range 0.03 0.3 3 30 300 3 30 300 3 30 300 3 30 300 3 30 300

to to to to to to to to to to to to to to to to to

0.3 3 30 300 3400 30 300 3000 30 300 3000 30 300 3000 30 300 3000

Hz Hz Hz Hz Hz kHz kHz kHz MHz MHz MHz GHz GHz GHz THz THz THz

Metric Subdivision

Adjectival Designation

Gigametric Hectomegametric Decamegametric Megametric Hectokilometric Myriametric kilometric Hectometric Decametric metric decimetric centimetric millimetric decimillimetric centimillimetric Micrometric Decimicrometric

ELF ELF ELF ELF ULF (Voice) VLF LF MF HF VHF UHF SHF EHF EHF EHF EHF EHF


1-3

B. .ELEMENTS OF BASIC COMMUNICATIONS SYSTEM.

1.

Transmitter A collection of electronic components and circuits designed to convert the information or intelligence into a signal suitable for transmission over a given communication medium.

2.

Channel The medium by which the electronic/electromagnetic signal is sent from one place to another. 2 General Categories i.

Wire Medium The signal is confined within the proximity of the channel or medium.

ii.

Wireless Medium The signal is not subjected to limits, boundaries, or channel restrictions.

a.k.a. Bounded or Guided medium

a.k.a. Unbounded or Unguided Medium Noise Noise is a random, undesirable electrical energy that enters the communications system and interferes with the transmitted message. 3.

Receiver The receiver is another collection of electronic components and circuits that accept the transmitted message from the channel and convert it back into a form understandable by humans.



1-4

Read it till it Hertz…jma Noise SPECTRUM ª

White Noise - White noise is defined as a noise that has equal amount of energy per frequency.

This means that if you could measure the amount of white noise energy between 100 Hz and 200 Hz it would equal the amount of energy between 1000 Hz and 1100 Hz. ª

Pink Noise - Pink noise is noise that has an equal amount of energy

per octave.

This means that pink noise would have equal power in the frequency range from 40 to 60 Hz as in the band from 4000 to 6000 Hz. ª

Brown noise - Brown noise is similar to pink noise, but with a power

density decrease of 6 dB per octave with increasing frequency (density proportional to 1/f2) over a frequency range which does not include DC.

ª

Blue Noise - Blue noise is noise that is the opposite of pink noise in that it doubles the amount of energy each time you go up 1 octave.

ª

Purple noise - Purple noise's power density increases 6 dB per octave with increasing frequency (density proportional to f2) over a finite frequency range. It is also known as differentiated white noise or violet noise.

ª

Orange noise - Orange noise is quasi-stationary noise with a finite

power spectrum with a finite number of small bands of zero energy dispersed throughout a continuous spectrum.

ª

Black Noise - Noise that has a frequency spectrum of predominately zero power level over all frequencies except for a few narrow bands or spikes.


1-5

C. .MODULATION. Modulation is the process of varying a carrier signal, typically a sinusoidal signal, in order to use that signal to convey information. The three key parameters of a sinusoid are its amplitude, its phase and its frequency, all of which can be modified in accordance with an information signal to obtain the modulated signal.

Three General Subdivisions 1.

Analog modulation

2.

3.

Amplitude modulation (AM)

Double-sideband Full Carrier (A3E) Single-sideband Full Carrier (H3E) Single-sideband Suppressed Carrier (J3E) Single-sideband Reduced Carrier (R3E) Vestigial-sideband Modulation (C3F)

Angle Modulation

Frequency modulation (FM) Phase modulation (PM)

Digital Modulation

Amplitude Shift Keying (ASK)

Frequency Shift Keying (FSK)

Phase Shift Keying (PSK) Quadrature Amplitude Modulation (QAM) Trellis Code Modulation (TCM)

On Off Keying

Audio FSK (AFSK) Continuous Phase FSK (CP-FSK) i. Minimum-shift keying (MSK) ii. Gaussian minimum-shift keying (GMSK) iii. Very minimum-shift keying (VMSK)

Hybrid Modulation (combined analog digital techniques)

Pulse modulation

Pulse-amplitude modulation (PAM) Pulse-code modulation (PCM) i. Differential PCM (DPCM) ii. Delta Modulation (DM) iii. Adaptive DM (ADM) iv. Continuously Variable Slope Delta (CVSD) v. Sigma-Delta Modulation (∑Δ)



1-6

Pulse Frequency Modulation (PFM) Pulse Time Modulation (PTM) i. Pulse-width modulation (PWM) ii. Pulse-position modulation (PPM) D. .WAVEFORM REPRESENTATION. 1.

Time Domain Representation

A standard oscilloscope is used to display the amplitude versus time representation of the input signal.

i.

Frequency (f) The number of times a particular phenomenon occurs in a given period of time expressed in Hertz.

ii.

Wavelength (λ) Wavelength is the distance between two points of similar cycles of a periodic wave or the distance traveled by an electromagnetic wave during the time of one cycle typically expressed in meters.

iii. Period (T) The time required for one complete cycle of a repetitive system, or simply the reciprocal of frequency. Relation between Wavelength, Frequency, and Period

λ=

c f

where: λ = wavelength in meters c = speed of light = 3 x 108 m/s f = frequency in Hertz T = period in sec

f=

c λ

T=

1 f


1-7

ECE Board Exam: APRIL 2005

Determine the wavelength of radio waves propagated using a frequency of 30 MHz.

Solution: Wavelength : λ=

c 3 x 108 = f 30 x 106

= 10 m

2.

Frequency Domain Representation

A spectrum analyzer is used to display the amplitude versus frequency representation of the input signal.

The time and frequency domain representation of three sine waves A A

t f



1-8

DOPPLER EFFECT

A perceived change in the frequency of a wave as the distance between the source and the observer changes.

Doppler Frequency of Sound Waves

ν ± νo fo = fs ν ∓ νs

fo = Observed frequency in Hz ν o = Velocity of observer in m/s ν s = Velocity of source in m/s fs = Source frequency in Hz

The top sign apply if the source and/or object are moving toward each other and the bottom sign apply if they move away from each other.

Sample Problem:

An ambulance travels down a highway at a speed of 75.0 mi/h with its siren emitting a sound with a frequency of 400 Hz. What frequency is heard (a) by someone standing still when the ambulance approaches? (b) by a passenger in a car traveling at 55 mi/h in the opposite direction as it approaches the ambulance? (c) by a passenger in a car traveling at 55 mi/h in the opposite direction as it moves away from the ambulance? 75 mi/h = 33.5 m/s, 55 mi/h = 24.6 m/s.

Solution:

(a) ν ≈ 345 m s , ν s = 33.5 m s, ν o = 0 (observer is not moving) fo = fs

ν + νo ⎛ 345 + 0 ⎞ = 400⎜ ⎟ = 443 Hz ν − νs ⎝ 345 − 33.5 ⎠

(b) ν ≈ 345 m s , ν s = 33.5 m s , ν o = 24.6 m s fo = fs

ν + νo ⎛ 345 + 24.6 ⎞ = 400⎜ ⎟ = 475 Hz ν − νs ⎝ 345 − 33.5 ⎠

(c) ν ≈ 345 m s , ν s = 33.5 m s , ν o = 24.6 m s fo = fs

ν − νo ⎛ 345 − 24.6 ⎞ = 400⎜ ⎟ = 339 Hz ν + νs ⎝ 345 + 33.5 ⎠


1-9

Doppler Frequency of Electromagnetic Waves

fo = fs

c ± νr c ∓ νr

fo = Observed frequency in Hz c = Speed of light = 3 x 10 8 m / s νr = Velocity of source relative to observer in m/s fs = Source frequency in Hz


Sample Problem: A LEO communications satellite is orbiting the earth at 27,000 kph (7,500 m/s). Calculate the frequency received by a mobile station antenna due to Doppler shift 450 km below if the satellite is operating at 1.28 GHz. Also compute the Doppler shift. (Assume the satellite is moving away from the subscriber) Solution: fo = fs

c − νr c + νr

= 1.28 GHz

27,000 kph c − 7500 m s

c + 7500 m s

450 km

= 1.279968 GHz For the Doppler Shift : fD = fs − fo = 1.28 GHz − 1.279968 GHz = 31.99 kHz

E. .ENGINEERING DEFINITIONS OF BANDWIDTH. 1.

Absolute bandwidth Absolute bandwidth is the difference between the upper and lower frequency limits (f2-f1), where the spectrum is zero outside the interval f1


1-10

2.

-3dB bandwidth (Half-power BW) -3dB bandwidth is f2-f1, where for frequencies inside the f1
3.

Null-to-null bandwidth (zero-crossing BW) Null-to-null bandwidth is f2-f1, where f2 is the first null in the envelope of the magnitude spectrum above f0 and, for bandpass system, f1 is the first null in the enveloped below f0, where f0 is the frequency where the magnitude spectrum is maximum.

4.

Bounded spectrum bandwidth Bounded spectrum bandwidth is f2-f1 such that outside the band f1
5.

Power bandwidth Power bandwidth is f2-f1, where f1
Read it till it Hertz…jma …2 LEGAL DEFINITIONS OF BANDWIDTH ª

FCC bandwidth FCC bandwidth is an authorized bandwidth parameter assigned by the FCC to specify the spectrum allowed in communication systems.

ª

ITU Necessary Bandwidth Necessary bandwidth for a given class of emission is defined as the width of the frequency band that is just sufficient to ensure the transmission of information at the rate and with the quality required under specified conditions.

ITU Necessary Bandwidth ª ª ª ª

Between Between Between Between

0.001 and 999 Hz shall be expressed in hertz 1.00 and 999 kHz shall be expressed in kilohertz 1.00 and 999 MHz shall be expressed in megahertz 1.00 and 999 GHz shall be expressed in gigahertz

H K M G


1-11

Example: What is the necessary bandwidth designation for 180.5 kHz? A. 181K B. 180K5 C. 181K5 D. 180.5K Answer. A .HInt. The necessary bandwidth shall be expressed by 3 numerals and one letter. The letter occupies the position of the decimal point and represents the unit of bandwidth.

Read it till it Hertz…jma ª

The ITU divides the world into three regions, with each region having its own allocations although there is much commonality between the regions.

ª

Philippines frequency allocations belong to Region 3 of the ITU Radio Regulations.

F. .ITU EMISSION DESIGNATION. According to ITU (Article 4; Radio Regulations, Geneva, 1982 revised 1985), emission shall be designated according to their necessary bandwidth and their classification.

DEFINITION.

Classification of the signal is given by three alphanumeric symbols.

FIRST SYMBOL Type of Modulation of the Main Carrier (1.1) (1.2)

Emission of unmodulated carrier Emission in which the main carrier is AMPLITUDE MODULATED (1.2.1) Double Sideband (1.2.2) Independent Sideband (1.2.3) Vestigial Sideband (1.2.4) Single Sideband, full carrier (1.2.5) Single Sideband, reduced or variable-level carrier (1.2.6) Single Sideband, suppressed carrier


N A B C H R J

1-12 (1.3)

(1.4) (1.5)

BASIC PRINCIPLES OF COMMUNICATIONS Engineering Emission in which the main carrier is ANGLE-MODULATED (1.3.1) Frequency Modulation (1.3.2) Phase Modulation Emission in which the main carrier is amplitude and angle-modulated either simultaneously or in a pre-established sequence. Emission of PULSES (1.5.1) Sequence of unmodulated pulses (1.5.2) A sequence of pulse (1.5.2.1) PAM (1.5.2.2) PDM/PWM (1.5.2.3) PPM (1.5.2.4) in which the carrier is angle-modulated during the period of the pulse. (1.5.2.5) which is a combination of the foregoing or

is produced by other means

F G D

P K L M Q

V

SECOND SYMBOL Nature of Signal(s) Modulating the Main (2.1) (2.2) (2.3) (2.4) (2.5) (2.6) (2.7) (2.8)

No modulating signal A single channel containing quantized or digital information WITHOUT the use of a modulating subcarrier A single channel containing quantized or digital information WITH the use of a modulating subcarrier A single channel containing analog information Two or more channels containing quantized or digital information Two or more channels containing analog information Composite system with one or more channels containing quantized or digital information, together with one or more channels containing analog information Case not otherwise covered

0 1 2 3 7 8 9 X

THIRD SYMBOL Type of Information(s) to be transmitted (3.1) (3.2) (3.3) (3.4) (3.5) (3.6) (3.7) (3.8) (3.9)

No information transmitted Telegraphy (manual) Telegraphy (automatic) Facsimile Data transmission, Telemetry, Telecommand Telephony (including sound broadcasting) Television (video) Combination of the above Case not otherwise covered

N A B C D E F W X


1-13


Two or more optional symbols can be added to the basic characteristics described above for a more complete description of an emission (Appendix 6, Radio Regulation, ITU; Geneva, 1982).

T FOURTH SYMBOL FIFTH SYMBOL

N

T

D

N

Details of Signal(s) Nature of Multiplexing

FOURTH SYMBOL Details of Signal(s) (4.1) (4.2) (4.3) (4.4) (4.5) (4.6) (4.7) (4.8) (4.9) (4.10) (4.11) (4.12) (4.13) (4.14) (4.15)

Two-condition code with elements of differing numbers and/or conditions Two-condition code with elements of the same number and duration without error correction Two-condition code with elements of the same number and duration with error correction Four-condition code in which each condition represents a signal element (of one or more bits) Multi-condition code in which each condition represents a signal element (of one or more bits) Multi-condition code in which each condition or combinations of conditions represent a character. Sound broadcasting quality (monophonic) Sound broadcasting quality (stereophonic or quadraphonic) Sound of commercial quality Sound commercial quality with the use of frequency inversion or band splitting Sound commercial quality with separate frequency-modulated signals to control the levels of demodulated signal Monochrome Color Combination of the above Cases not otherwise covered


A B C D E F G H J K L M N W X


1-14

FIFTH SYMBOL Nature of Multiplexing (5.1) (5.2) (5.3) (5.4) (5.5) (5.6)

N C F T W

None Code Division Multiplex Frequency Division Multiplex Time Division Multiplex Combination of frequency division multiplex and time division multiplex Other types of multiplexing

X

G. .FCC EMISSION DESIGNATION. FIRST SYMBOL Type of Modulation of the Main Carrier (1.1) (1.2) (1.3)

A F P

Amplitude Frequency or Phase Pulse

SECOND SYMBOL Type of Transmission (2.1) (2.2) (2.3) (2.4) (2.5) (2.6) (2.7) (2.8) (2.9)

Absence of any modulation intended to carry information Telegraphy without the use of a modulating audio frequency Telegraphy by the on-off keying of a modulating audio frequency, or by the on-off keying of the modulated emission (special case: an unkeyed modulated emission) Telephony (including sound broadcasting) Facsimile (with modulation of main carrier directly or by a Frequency-modulated subcarrier) Television (visual only) Four-frequency duplex telegraphy Multichannel voice-frequency telegraphy Cases not otherwise covered

0 1 2 3 4 5 6 7 9

THIRD SYMBOL Supplementary Characteristics (3.1) (3.2)

(3.3) (3.4)

Double sideband Single sideband (3.2.1) Reduced carrier (3.2.2) Full carrier (3.2.3) Suppressed carrier Two independent sideband Vestigial sideband

A H J B C

Self-Sufficient Guide to ECE by JASON AMPOLOQUIO (3.5)

Pulse

(3.5.1) (3.5.2) (3.5.3) (3.5.4) (3.6)

Amplitude modulated Width (or duration) modulated Phase (or position) modulated Code modulated Digital modulations

H. .FORMER DESIGNATION. 1.

Amplitude Modulated (AM) A0 A1 A2 A3 A3A A3J A3H A3B A3Y A4 A5C A9B A9Y

2.

Frequency or Phase–Modulated (FM/PM) F1 F2 F3 F3Y F9Y F4 F5 F6

3.

No modulation Telegraphy; on-off ; no other modulation Telegraphy; on-off ; amplitude-modulated tone Telephony; DSBFC Telephony; SSBRC Telephony; SSBSC Telephony; SSBFC Telephony; ISB Digital voice modulation Facsimile Television with vestigial sideband Telephony or telegraphy with ISB Nonvoice digital modulation

Telegraphy; FSK Telegraphy; on-off ; frequency-modulated tone Telephony; FM or PM Digital voice modulation Nonvoice digital modulation Facsimile Television Telegraphy; four-frequency duplex

Pulse–Modulated P0 P1D P2D P2E P2F P3D P3E P3F

RADAR Telegraphy; ASK Telegraphy; pulse-carrier tone-modulated Telegraphy; pulse-width tone-modulated Telegraphy; phase or position tone-modulated Telephony; amplitude-modulated pulses Telephony; pulse-width modulated Telephony; pulse phase or position-modulated


1-15

D E F G Y


1-16

I.

J.

.TRANSMISSION MODES. 1.

Simplex Transmissions can occur only in one direction. Sometimes called oneway-only, receive-only, or transmit-only.

2.

Half-Duplex Transmissions can occur in both directions, but not at the same time. Sometimes called two-way-alternate, either-way, or over-and-out systems.

3.

Full Duplex Transmissions can occur in both directions at the same time. Sometimes called two-way-simultaneous, duplex, or both-way line.

4.

Full/Full Duplex Possible to transmit and receive simultaneously, but not necessary between the same two locations (i.e. one station can transmit to a second station and receive from a third station at the same time).

.CIRCUIT ARRANGEMENTS. 1.

Two-Wire Transmission Two-wire circuits are those that carry information signals in both directions over the same physical link or path. Typically, such a circuit is provisioned through the use of a single twisted pair, copper wire connection.

2.

Four-Wire Transmission

Four-wire circuits are those that carry information signals in both directions over separate physical links or paths, and in support of simultaneous, two way transmission.


1-17

K. .MULTIPLEXING TECHNIQUE. Multiplexing is a technique used in communications and input/output operations for transmitting a number of separate signals simultaneously over a single channel or line. 1.

Frequency Division Multiplexing (FDM)

A unique band of frequencies within the wideband frequency spectrum of the medium is allotted to each communication channel on a continuous time basis. 2.

Time Division Multiplexing (TDM) Each communication channel is allotted a fixed time slot within a sampling frame, occupying essentially the entire wideband frequency spectrum for the allocated time.



1-18

Statistical Time Division Multiplexing (STDM) Statistical Time Division Multiplexing (STDM) is much improved over TDM, as the MUXs are intelligent. STDMs can allocate bandwidth, in the form of time slots, in consideration of the transmission requirements of individual devices serving specific applications.

3.

Wavelength Division Multiplexing (WDM)

WDM resembles FDM in that the idea is to send information signals that occupy the same frequency band of frequencies through the same fiber at the same time without them interfering with each other.

L.

.TRANSMISSION FACILITIES.

In terms of bandwidth, and in contemporary digital context, transmission facilities can be categorized as narrowband, wideband or broadband. 1.

Narrowband A single channel (64 Kbps) or some number of 64 Kbps channels (N × 64 Kbps), but less than wideband.

2.

Wideband Wideband is multi-channel capacity that is between 1.544 Mbps and 45 Mbps according to U.S. standards (2.048 Mbps-34 Mbps according to European/international standards.)

Self-Sufficient Guide to ECE by JASON AMPOLOQUIO 3.

1-19

Broadband Broadband is multi-channel capacity which is 45 Mbps according to U.S. standards and 34 Mbps according to European/international standards.

M. .TRANSMISSION IMPAIRMENTS. Signals travel through transmission media, which are not perfect. The imperfection causes signal impairment. This means that the signal at the beginning of the medium is not the same as the signal at the end of the medium. What is sent is not what is received. Three causes of impairment are attenuation, distortion, and noise. 1.

Attenuation

A type of transmission impairment in which the signal loses strength due to the resistance and length of the transmission medium.

TRANSMISSION MEDIUM

2.

A

Distortion The alteration of information in which the original proportions are changed, resulting from a defect in communication system.

CHANNEL SOURCE

RECEIVER



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3.

Noise

A type of transmission impairment in which an outside source such as crosstalk corrupts a signal.

SOURCE

RECEIVER

N. .CLASSIFICATION OF COMMUNICATIONS. 1.

2.

Distress A mobile station in distress is in need of immediate assistance.

Communication

Distress Signal

Radiotelegraph

…---… or SOS sent as one character with no spacing between letters.

Radiotelephone

MAYDAY (from the French m’aider), usually transmitted three or more times to attract attention.

Urgency Radio messages with an urgency classification refer to a situation that requires immediate attention and might conceivably become distress in nature.

Communication

Distress Signal

Radiotelegraph

XXX

Radiotelephone

PAN PAN


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Safety Radio communications with a safety classification refer to meteorological information, particularly about storms, hurricanes, etc.

Communication

Distress Signal

Radiotelegraph

TTT

Radiotelephone

SECURITY

O. .MESSAGE PRIORITIES. 1. 2. 3. 4. 5. 6.

Distress calls, messages, and traffic Communications preceded by the urgency signal Communications preceded by the safety signal Communications relative to radio direction finding Message relative to navigation of aircraft Message relative to navigation, movements, and needs of ships and official weather-observation messages

P. .OPERATIONAL WORDS. Code

Meaning

Roger

I received your message.

Over

I have completed transmitting and await your reply.

Go ahead

Same as over.

Out

I have completed my communication and do not expect to transmit again.

Clear

I have no further traffic. (Sometimes used in place of Out)

Stand by

Wait for another instructions.

Break

I am changing from one part of the message to another. (Also used to request the received operator to indicate if he has received the portion of the message transmitted thus far.)

Copy

Respond to Break

call

or

further



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Q. .INTERNATIONAL PHONETIC ALPHABET. A B C D E F G H I

Alpha Bravo Charlie Delta Echo Foxtrot Golf Hotel India

J K L M N O P Q R

Juliet Kilo Lima Mike November Oscar Papa Quebec Romeo

S T U V W X Y Z

Sierra Tango Uniform Victor Whiskey X-ray Yankee Zulu ---


I

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H

1.

A 5-frame TDM multiplexer has a total frame period of 625 μs. If the timeslot per channel is 5.208 μsec, determine the number of channels per frame and the total number of digital channels. A. 60, 12 B. 120, 24 C. 12, 60 D. 24, 120

2.

If the measured wavelength is 6 m, calculate the frequency in MHz. A. 25 MHz B. 10 MHz C. 50 MHz D. 100 MHz

3.

If several musical instruments are playing the same note, you should be able to distinguish one instrument from another because of which of the following characteristics of sound? A. Frequency B. Intensity C. Overtones D. Quality

4.

A large volume of light radiating in a given direction is referred to as a A. pencil B. beam C. ray D. shaft

5.

What are the primary colors of light? A. Red, blue, and green C. Blue, green, and violet

B. D.

Red, violet, and indigo Red, blue, and yellow

6.

____ noise is noise that has an equal amount of energy per octave. A. Pink B. Yellow C. White D. Blue

7.

_______ is multi-channel capacity that is between 1.544 Mbps and 45 Mbps according to U.S. standards (2.048 Mbps-34 Mbps according to European/international standards.) A. Wideband B. Baseband C. Narrowband D. Broadband

8.

Based on ITU-R Recommendations V.431-6, the adjectival designation of 30 to 300 kHz is _____. A. ELF B. LF C. ULF D. VLF


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9.

BASIC PRINCIPLES OF COMMUNICATIONS Engineering The world's first operational packet switching network, and the progenitor of the global Internet. A. DECNET B. ARPANET C. ISDN D. NMT

10. ______ would have equal power in the frequency range from 40 to 60 Hz as in the band from 4000 to 6000 Hz. A. pink noise B. blue noise C. white noise D. green noise 11. He hypothesized the existence of magnetic field around a current-carrying conductor. A. Michael Faraday B. Nikola Tesla C. Andre Marie Ampere D. Karl Gauss 12. In FCC former designation, F3Y means ______. A. Telegraphy; four-frequency duplex B. Digital voice modulation C. Telegraphy; FSK D. Telephony; amplitude-modulated pulses 13. Based on ITU-R Recommendations V.431-6, the metric subdivision of 0.030 to 0.300 Hz is _____. A. Decamegametric B. Gigametric C. Myriametric D. Hectomegametric 14. ______ would have equal power in the frequency range from 40 to 60 Hz as in the band from 4000 to 4020 Hz. A. pink noise B. blue noise C. white noise D. green noise 15. What is the wavelength for a color in the middle of the visible light band if the visible light frequency range is from 0.39 PHz (red) to 0.79 PHz (violet)? A. 5000 nm B. 50 nm C. 5 D. 500 nm 16. The person who sent the first radio signal across the Atlantic ocean was: A. Marconi B. Bell C. Maxwell D. Hertz 17. What percentage of the VHF band does a 6-MHz TV signal occupy? A. 2.2% B. 8.8% C. 4.4% D. 1.1% 18. How many AM stations can be accommodated in a 0.15 MHz BW if the allocated BW per station is 10 KHz? A. 100 B. 50 C. 150 D. 15 19. How many international commercial AM broadcast channels (assume BW=10 KHz) can fit into the bandwidth occupied by a commercial FM broadcast channel (assume BW=200 KHz)? A. 20 B. 200 C. 10 D. 5


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20. Determine the BW of sonic frequency range if the infrasonic frequency ends at 20 Hz while ultrasonic frequency begins at 20, 000 Hz. A. 4 kHz C. 10 kHz B. 8 kHz D. 19.98 kHz 21. What percentage of the audio passband does a high fidelity CD quality music occupy? A. 66.67% B. 43.25% C. 74.82% D. 50% 22. A standard TV channel occupies 6 MHz of BW. If a standard commercial FM broadcast channels is 200 KHz wide, how many of this channel can fit into the bandwidth occupied by a commercial TV station? A. 200 B. 30 C. 20 D. 300 23. In FCC former designation, A4 means ______. A. Telegraphy; on-off ; amplitude-modulated tone B. Facsimile C. Nonvoice digital modulation D. Digital voice modulation 24. Based on ITU-R Recommendations V.431-6, the adjectival designation of 0.003 to 0.030 GHz is _____. A. VHF B. MF C. SHF D. EHF 25. In AT&T FDM hierarchy the group BW is 48 KHz, determine the number of voice channel that a supergroup can handled if it needs 5 groups to form a supergroup. A. 12 B. 300 B. 60 D. 600 26. Another name for amplitude shift keying A. variable angle modulation B. C. continuous wave modulation D.

minimum shift keying beat frequency modulation

27. AM radio's main limitation is its susceptibility to A. ignition noise B. Rayleigh fading C. transit time noise D. atmospheric interference 28. A quasi-stationary noise with a finite power spectrum with a finite number of small bands of zero energy dispersed throughout a continuous spectrum A. white noise B. orange noise C. violet noise D. blue noise 29. ______ is an arrangement of conductors designed to radiate (transmit) an electromagnetic field in response to an applied alternating voltage and the associated alternating electric current A. transmission lines B. antenna C. counterpoise D. stub


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30. The transmission of radio waves was first done by: A. Marconi B. Bell C. Maxwell D. Hertz 31. A complete communication system must include: A. a transmitter and receiver B. a transmitter, a receiver, and a channel C. a transmitter, a receiver, and a spectrum analyzer D. a multiplexer, a demultiplexer, and a channel 32. In FCC former designation, P0 means ______. A. Telephony; pulse-width modulated B. Telephony; amplitude-modulated pulses C. Digital voice modulation D. RADAR 33. Radians per second is equal to: A. 2πf C. 2πλ 34. The A. B. C. D.

B. D.

f/2π 2π/λ

bandwidth required for a modulated carrier depends on: the carrier frequency the signal-to-noise ratio the signal-plus-noise to noise ratio the baseband frequency range

35. When two or more signals share a common channel, it is called: A. sub-channeling B. signal switching C. SINAD D. Multiplexing 36. The A. B. C. D.

wavelength of a radio signal is: equal to f /c equal to c /2πf the distance a wave travels in one period how far the signal can travel without distortion

37. In FCC former designation, A9B means ______. A. Facsimile B. Television with vestigial sideband C. Telephony or telegraphy with ISB D. Telephony; SSBRC 38. Distortion is caused by: A. creation of harmonics of baseband frequencies B. baseband frequencies "mixing" with each other C. shift in phase relationships between baseband frequencies D. all of the above 39. The power density of "flicker" noise is: A. the same at all frequencies B. C. greater at low frequencies D.

greater at high frequencies the same as "white" noise

Self-Sufficient Guide to ECE by JASON AMPOLOQUIO 40. So called "1/f" noise is also called: A. random noise C. white noise 41. The A. B. C. D.

B. D.

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pink noise partition noise

part, or parts, of a sinusoidal carrier that can be modulated are: its amplitude its amplitude and frequency its amplitude, frequency, and direction its amplitude, frequency, and phase angle

42. In FCC former designation, F5 means ______. A. Telephony; SSBFC B. Facsimile C. Television D. Telephony; FM or PM 43. It is also known as differentiated white noise or violet noise A. pink noise B. green noise C. blue noise D. purple noise 44. In 1820, he discovered the relation between electricity and magnetism, later known as electromagnetism. A. Michael Faraday B. Hans Christian Oersted C. Karl Gauss D. Nikolai Tesla 45. In 1886, a German physicist performed an experiment on spark gap transmission verifying Maxwell statement experimentally showed the existence of such waves which he called radio waves that paved the way for wireless communication. A. Guglielmo Marconi B. Dr. Hidetsugi Yagi C. Heinrich Hertz D. Isaac Asimov 46. A transmission technique where each communication channel is allotted an epoch or time slot within a sampling frame, occupying essentially the entire wideband frequency spectrum for the allocated time. A. Time Division Multiplexing (TDM) B. Wavelength Division Multiplexing (WDM) C. Frequency Division Multiplexing (FDM) D. Statistical Time Division Multiplexing (STDM) 47. He developed the first wireless telegraph and successfully sent a message over a distance of few kilometers using a spark gap transmitter. A. James Clerk Maxwell B. Edwin Arsmtrong C. Guglielmo Marconi D. Heinrich Hertz 48. Based on ITU-R Recommendations V.431-6, the metric subdivision of 0.030 to 0.300 THz is _____. A. micrometric B. kilometric C. millimetric D. decimicrometric 49. In FCC former designation, F6 means ______. A. Telephony; FM or PM B. Television C. Telegraphy; four-frequency duplex D. Telegraphy; FSK


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50. The theory of radio waves was originated by: A. Marconi B. Bell C. Maxwell D. Hertz 51. Based on ITU-R Recommendations V.431-6, the metric subdivision of 3.00 to 30.0 kHz is _____. A. Hectomegametric B. Hectokilometric C. Myriametric D. Decamegametric 52. Defined as the noise that has equal amount of energy per frequency. A. Pink Noise B. Transit-time noise C. Blue Noise D. White Noise 53. In FCC former designation, P3D means ______. A. Telegraphy; pulse-width tone-modulated B. Telephony; amplitude-modulated pulses C. Telegraphy; phase or position tone-modulated D. Telephony; pulse-width modulated 54. Radio waves travel at what speed? A. Speed of the Earth's rotation B. Speed of light C. Speed of the Earth's orbit around the sun D. Speed of sound 55. Which of the following types of energy cannot be seen, heard, or felt? A. Heat waves B. Sound waves C. Light waves D. Radio waves 56. The operational word “I have completed transmitting and await your reply” means A. Break B. Over C. Roger D. Copy 57. Hectomegametric occupies the frequency band of ______. A. 0.3 to 3 Hz B. 0.030 to 0.300 kHz C. 300 to 3000 MHz D. 30 to 300 kHz 58. In FCC former designation, P2E means ______. A. Telegraphy; pulse-carrier tone-modulated B. Telephony; amplitude-modulated pulses C. Telegraphy; pulse-width tone-modulated D. Telephony; pulse-width modulated 59. A sound wave that moves back and forth in the direction of propagation is an example of which of the following types of wave motion? A. Longitudinal B. Composite C. Concentric D. Transverse 60. What wave propagation principle accounts for the apparent increase in frequency as a train whistle approaches and the apparent decrease in frequency as it moves away? A. Reflection B. Diffraction C. Refraction D. Doppler effect


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61. What is the urgency signal in radiotelegraphy? A. MAYDAY B. XXX C. SOS D. PANPAN 62. What are the three audible frequency ranges? A. Infrasonic, sonic, and ultrasonic B. Infrasonic, subsonic, and ultrasonic C. Infrasonic, subsonic, and supersonic D. Subsonic, sonic, and supersonic 63. The operational word “I have completed my communication and do not expect to transmit again” means A. Over B. Copy C. Out D. Roger 64. Noise that is the opposite of pink noise in that it doubles the amount of energy each time you go up 1 octave. A. Yellow B. White C. Blue D. Magenta 65. The number of times a particular phenomenon occurs in a given period of time expressed in Hertz. A. Period B. Crest C. Frequency D. Wavelength 66. ______ is the distance between two points of similar cycles of a periodic wave or the distance traveled by an electromagnetic wave during the time of one cycle typically expressed in meters. A. Peak-to-peak value B. Crest C. Wavelength D. Period 67. ______ is f2-f1, where for frequencies inside the f1
B. D.

Electric and magnetic fields Noise and data

71. Electromagnetic waves produced primarily by heat are called A. Microwaves B. Shortwaves C. X-rays D. Infrared rays


1-30


72. Noise whose power density increases 3 dB per octave with increasing frequency over a finite frequency range. A. pink noise B. blue noise C. white noise D. green noise 73. In FCC former designation, P3F means ______. A. Telephony; amplitude-modulated pulses B. Telephony; pulse phase or position-modulated C. Telegraphy; phase or position tone-modulated D. Telephony; pulse-width modulated 74. Another name for signals in the HF range is A. Shortwaves B. Microwaves C. Millimeter waves D. RF waves 75. A micron is A. One-millionth of a foot C. One ten-thousandth of an inch

B. D.

One-thousandth of a meter One-millionth of a meter

76. Radio-frequency waves cannot be seen for which of the following reasons? A. Because the human eye detects only magnetic energy B. Because radio-frequency waves are below the sensitivity range of the human eye C. Because radio-frequency waves are above the sensitivity range of the human eye D. Because radio-frequency energy is low powered 77. In FCC former designation, A3J means ______. A. Telephony; SSBSC B. Telephony; SSBRC C. Telephony; SSBFC D. Telephony; DSBFC 78. In FCC former designation, P2D means ______. A. Telegraphy; pulse-carrier tone-modulated B. Telephony; pulse-width modulated C. Telephony; amplitude-modulated pulses D. Telegraphy; pulse-width tone-modulated 79. ______ is the difference between the where the spectrum is zero outside frequency axis. A. half-power BW C. -3dB bandwidth

upper and lower frequency limits (f2-f1), the interval f1
zero-crossing BW Absolute bandwidth

80. In FCC former designation, P2F means ______. A. Telephony; amplitude-modulated pulses B. Telegraphy; pulse-carrier tone-modulated C. Telegraphy; phase or position tone-modulated D. Telegraphy; pulse-width tone-modulated 81. The process of transmitting two or more information signals simultaneously over the same channel is called A. Mixing B. Telemetry C. Modulation D. Multiplexing


1-31

82. Recovering information from a carrier is known as A. Modulation B. Detection C. Demultiplexing D. Carrier recovery 83. In FCC former designation, P3E means ______. A. Telephony; pulse-width modulated B. Telegraphy; phase or position tone-modulated C. Telephony; amplitude-modulated pulses D. Telegraphy; pulse-width tone-modulated 84. ______ is a technique used in communications and input/output operations for transmitting a number of separate signals simultaneously over a single channel or line. A. Accessing B. Polling C. Multiplexing D. Contention 85. Centimetric wave occupies the frequency band of ______. A. 0.3 to 3 GHz B. 0.030 to 0.300 MHz C. 3 to 30 GHz D. 30 to 300 MHz 86. A unique band of frequencies within the wideband frequency spectrum of the medium is allotted to each communication channel on a continuous time basis. A. Time Division Duplexing (TDD) B. Frequency Division Duplexing (FDD) C. Frequency Division Multiplexing (FDM) D. Time Division Multiplexing (TDM) 87. In FCC former designation, A3H means ______. A. Telephony; DSBFC B. Telegraphy; SSBSC C. Telegraphy; FSK D. Telephony; SSBFC 88. _____ can allocate bandwidth, in the form of time slots, in consideration of the transmission requirements of individual devices serving specific applications. A. Statistical Time Division Multiplexing (STDM) B. Passive Time Division Multiplexing (TDM) C. Frequency Division Multiplexing (FDM) D. Wavelength Division Multiplexing (WDM) 89. A single channel (64 Kbps) or some number of 64 Kbps channels. A. Wideband B. Baseband C. Narrowband D. Broadband 90. _______ is multi-channel capacity which is 45 Mbps according to U.S. standards and 34 Mbps according to European/international standards. A. Wideband B. Baseband C. Narrowband D. Broadband 91. What is the distress signal in radiotelegraphy? A. SOS B. PANPAN C. MAYDAY D. XXX



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92. Based on ITU-R Recommendations V.431-6, the adjectival designation of 30 to 300 Hz is _____. A. ULF B. VLF C. ELF D. LF 93. The operational word “I am changing from one part of the message to another” means A. Roger B. Copy C. Break D. Over 94. Noise whose power density increases 6 dB per octave with increasing frequency over a finite frequency range. A. pink noise B. green noise C. blue noise D. purple noise 95. ______ for a given class of emission is defined as the width of the frequency band that is just sufficient to ensure the transmission of information at the rate and with the quality required under specified conditions. A. -3dB bandwidth B. Necessary Bandwidth C. Absolute bandwidth D. half-power BW 96. In FCC former designation, A3 means ______. A. Telegraphy; pulse-carrier tone-modulated B. Telephony; DSBFC C. Telephony; ISB D. Telephony; SSBRC 97. In 1822, he discovered electromagnetic induction, the reverse of Oersted discovery. A. Michael Faraday B. Joseph Henry C. James Clerk Maxwell D. Wilhelm Weber 98. In 1866, put together the principles of Oersted and Faraday and hypothesized the existence of electromagnetic waves. A. Guglielmo Marconi B. Arthur Clarke C. Heinrich Hertz D. James Clerk Maxwell 99. decimillimetric occupies the frequency band of ______. A. 30 to 300 kHz B. 300 to 3000 GHz C. 0.3 to 3 GHz D. 0.030 to 0.300 THz

A. C.

What is the urgency signal in radiotelephony? PANPAN B. XXX MAYDAY D. SECURITY

A. C.

In a time-domain plot, the horizontal axis is a measure of ________. signal amplitude B. frequency phase D. time

100.

101.

102.

If the bandwidth of a signal is 5 kHz and the lowest frequency is 52 kHz, what is the highest frequency? A. 57 KHz B. 47 KHz C. 10 KHz D. 5 KHz


A. C.

Electronic communications was discovered in which century? 16th century B. 19th century th 18 century D. 20th century

A. C.

One-way communications is called Half duplex Monocomm

103.

104.

105.

A. C.

Television broadcasting occurs in which ranges? HF B. EHF VHF D. SHF

A. C.

The frequency range of infrared rays is approximately 30 to 300 GHz B. 4000 to 8000 Ǻ 1000 to 10,000 A D. 0.7 to 100 μm

107.

Given two sine waves A and B, if the frequency of A is twice that of B, then the period of B is ________ that of A. A. one-half B. twice C. the same as D. indeterminate from

109.

A. C. 110.

Full duplex Simplex

________ is a type of transmission impairment in which the signal loses strength due to the different propagation speeds of each frequency that makes up the signal. A. Attenuation B. Distortion C. Noise D. Decibel

106.

108.

B. D.

1-33

_________ can impair a signal. Attenuation Distortion

B. D.

Noise All of the above

________ is a type of transmission impairment in which the signal loses strength due to the resistance of the transmission medium. A. Attenuation B. Distortion C. Noise D. Decibel

111.

A. C.

The communications medium causes the signal to be Amplified B. Modulated Attenuated D. Interfered with

112.

Which of the following is not a source of noise? A. Another communications signal B. Atmospheric effects C. Manufactured electrical systems D. Thermal agitation in electronic components

113.

When propagation speed is multiplied by propagation time, we get the ________. A. throughput B. wavelength of the signal C. distortion factor D. distance a signal or bit has traveled



1-34

114.

A. C.

The _________ product defines the number of bits that can fill the link. bandwidth-period B. frequency-amplitude bandwidth-delay D. delay-amplitude

115.

Baseband transmission of a digital signal is possible only if we have a ____ channel. A. low-pass B. low rate C. bandpass D. high rate

116.

In the United States, the electromagnetic spectrum is regulated and managed by A. Business and industry B. ITU C. FCC D. The United Nations

117.

For a given bandwidth signal, more channel space is available for signals in the range of A. VHF B. UHF C. SHF D. EHF

118.

A. C. 119.

B. D.

proportional to each other B and C

A _________ sine wave is not useful in data communications; we need to send a _______ signal. A. composite; single-frequency B. single-frequency; composite C. single-frequency; double-frequency D. complex, composite A. C.

________is the rate of change with respect to time. Amplitude B. Frequency Time D. Voltage

A. C.

Calculate the energy related to the frequency occupied by VLF. 124 μeV to 1.24 μeV B. 1.24 peV to 12.4 peV 12.4 peV to 124 peV D. 124 meV to 1.24 eV

A. C.

Communication is the process of Keeping in touch Exchanging information

120.

121.

122.

123.

Frequency and period are ______. inverse of each other the same

B. D.

Broadcasting Entertainment by electronics

________ is a type of transmission impairment in which an outside source such as crosstalk corrupts a signal. A. Attenuation B. Distortion C. Noise D. Decibel

124.

A. C.

As frequency increases, the period ________. decreases B. increases remains the same D. doubles


A. C.

The approximate wavelength of red light is 1000 μm B. 7000 Ǻ 3500 A D. 4000 A

A. C.

Which of the following is not used for communications? X-rays B. Millimeter waves Infrared D. Microwaves

125.

126.

127.

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A sine wave in the ______ domain can be represented by one single spike in the _____ domain. A. time; frequency B. frequency; time C. time; phase D. phase; time A. C.

Receiving electromagnetic emissions from stars is called Astrology B. Optical astronomy Radio astronomy D. Space surveillance

A. C.

A person communications hobby for individuals is Ham radio B. Electronic bulletin board CB radio D. Cellular radio

128.

129.

130.

For a ______ channel, the Nyquist bit rate formula defines the theoretical maximum bit rate. A. noisy B. noiseless C. bandpass D. low-pass

131.

If the available channel is a ____ channel, we cannot send a digital signal directly to the channel. A. low-pass B. bandpass C. low rate D. high rate

132.

A. C. 133.

The process of modifying a high-frequency carrier with the information to be transmitted is called A. Multiplexing B. Telemetry C. Modulation D. Detection A. C.

A periodic signal completes one cycle in 0.001 s. What is the frequency? 1 MHz B. 1 kHz 100 Hz D. 1 Hz

A. C.

In a frequency-domain plot, the horizontal axis measures the ________. peak amplitude B. frequency phase D. slope

134.

135.

136.

The original electrical information signal to be transmitted is called the Modulating signal B. Carrier Baseband signal D. Source signal

The _____ of a composite signal is the difference between the highest and the lowest frequencies contained in that signal. A. frequency B. period C. bandwidth D. amplitude



1-36

A. C.

_______ describes the position of the waveform relative to time 0. Frequency B. Phase Amplitude D. Voltage

A. C.

Calculate the energy related to the frequency occupied by ELF. 124 feV to 1.24 peV B. 1.24 peV to 12.4 peV 12.4 peV to 124 peV D. 124 peV to 1.24 neV

A. C.

Simultaneous two-way communications is called Half duplex B. Full duplex Bicomm D. Simplex

A. C.

Continuous voice or video signals are referred to as being Baseband B. Analog Digital D. Continuous waves

137.

138.

139.

140.

141.

Transmission of graphical information over the telephone network is accomplished by A. Television B. CATV C. Videotext D. Facsimile

142.

A. C.

A sine wave is ________. periodic and continuous periodic and discrete

B. D.

aperiodic and continuous aperiodic and discrete

143.

If the maximum amplitude of a sine wave is 2 V, the minimum amplitude is ________ V. A. 2 B. 1 C. -2 D. between -2 and 2

144.

Measuring physical conditions at some remote location and transmitting this data for analysis is the process of A. Telemetry B. Instrumentation C. Modulation D. Multiplexing

145.

Radar is bases upon A. Microwaves B. A water medium C. The directional nature of radio signals D. Reflected radio signals A. C.

A frequency of 27 MHz has a wavelength of approximately 11 m B. 27 m 30 m D. 81 m

A. C.

The voice frequency range is 30 to 300 Hz 20 Hz to 20 kHz

146.

147.

148.

B. D.

300 to 3000 Hz 0 Hz to 15 kHz

A signal occupies the spectrum space from 1.115 to 1.122 GHz. The bandwidth is A. 0.007 MHz B. 7 MHz C. 237 MHz D. 700 MHz


1-37

149.

A and A. C.

signal is measured at two different points. The power is P1 at the first point P2 at the second point. The dB is 0. This means ________. P2 is zero B. P2 equals P1 P2 is much larger than P1 D. P2 is much smaller than P1

150.

For a ______ channel, we need to use the Shannon capacity to find the maximum bit rate. A. noisy B. noiseless C. bandpass D. low-pass

--End of Text--


AMPLITUDE MODULATION

1-38

Section

2

= Amplitude

Modulation


DEFINITION. Modulation is the process of impressing or imparting a lowfrequency source information (voice, audio, video, or data) onto a highfrequency bandpass signal with a carrier frequency fc by the introduction of amplitude, frequency or phase perturbation. Demodulation is the reverse process where the received signals are transformed back to their original form.

DEFINITION.

Amplitude Modulation is an analog modulation scheme in which the amplitude of a relatively high-frequency carrier signal is varied in accordance with the instantaneous amplitude of an information signal.

DEFINITION.

A. .THE AM WAVEFORM.

1.

Instantaneous Amplitude of the Modulated Wave

ν AM ( t ) = ( Vc + em ) sin ωc t

The modulating signal is more often an arbitrary waveform, such as audio signal. However, an analysis of sine wave modulation is very useful, since Fourier analysis often allows complex signals to be express as a series of sinusoids.


Hence,

ν AM (t) = ( Vc + Vm sin ωmt ) sin ωc t

By applying little trigonometry,

ν AM (t) = Vc sin ωc t −

mVc mVc cos ( ωc + ωm ) t + cos ( ωc − ωm ) t 2 2

where : ν AM (t) = AM modulated wave Vc = peak amplitude of the carrier in V em = instantaneous amplitude of the modulating signal in V ωc = angular frequency of the carrier wave in rad/s ωm = angular frequency of the modulating signal in rad/s m = modulation index


First, note that the amplitude of the carrier after modulation is the same as it was before modulation

(In short, AM is a bit of a misnomer). ª

Second, the amplitude of the upper and lower side frequencies is a function of both the carrier amplitude and the modulation index.

ª

Third, the carrier component is a +sine function, the upper side frequency a –cosine function, and the lower side frequency a +cosine function or the carrier is 90° out of phase with both the upper and lower

side frequencies, and the upper and lower side frequency are 180° out of phase with each other.


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1-40

2.

Coefficient of Modulation (m) A term used to describe the amount (modulation) present in an AM waveform.

of

amplitude

changed

Mathematically, where : Vm = peak amplitude of the

V m= m Vc

modulating signal in V Vc = peak amplitude of the carrier signal in V

In terms of AM Envelope,

m=

Vmax − Vmin Vmax + Vmin

Vmax = Vc + Vm Vmin = Vc − Vm


An AM signal has the following characteristics: carrier frequency = 150 MHz; modulating frequency = 3 MHz; peak carrier voltage = 40 volts; and peak modulating voltage is 30 volts. Calculate the peak voltage of the lower sideband frequency. 40

Solution: VSB

mV c V 30 = ⇒m= m = = 0 .75 2 Vc 40 =

0 .75 (40 ) = 15 V 2

15

fc-fm

Sample Problem:

15

fc

fc+fm

Calculate the modulation index for a standard AM transmission, if the maximum peak voltage of the modulated wave is 150 V and the modulating signal voltage is 50 V peak.

Solution: m=

Vm ⇒ Vc = Vmax − Vm Vc

50 150 − 50 = 0 .5 =

Self-Sufficient Guide to ECE by JASON AMPOLOQUIO Several Cases of m: a.

Undermodulation (m<1, Vm
b.

100% Modulation (m=1, Vm=Vc)

c.

Overmodulation (m>1, Vm>Vc)

0


1-41


1-42

Read it till it Hertz…jma OVERMODULATION creates side frequencies (harmonics) further from the carrier known as SPLATTER, BUCKSHOT, or SPURIOUS EMISSIONS which create interference to other radio services and since spectrum space is tightly controlled by law, overmodulation is illegal.

Sample Problem:

A modulating signal consists of a symmetrical triangular wave having zero dc component and peak-to-peak voltage of 11 V. Calculate the value of modulation index if it used to amplitude modulate a carrier of peak voltage 10 V.

Solution: For Emax; Vm 11 = 10 + 2 2 = 15.5 V

Emax = Vc + For Emin;

Vm 11 = 10 − 2 2 = 4.5 V

Emin = Vc − For m; m=

Emax − Emin 15.5 − 4.5 = Emax + Emin 15.5 + 4.5

= 0.55 Answer : 0.55

1-43

Self-Sufficient Guide to ECE by JASON AMPOLOQUIO B. .SIMULTANEOUS MODULATION.

mT = m12 + m22 + m32 … + mn2

where : m T = total modulation index m1 , m2 ...mn = modulation indices due to several modulating signals


An AM transmitter is modulated by two sine waves at 1.5 kHz and 2.5 kHz, with modulation of 20 percent and 80 percent respectively. Calculate the effective modulation index.

Solution: mT =

m12 + m22 =

0 .22 + 0 .82 = 0 .824

Sample Problem:

Calculate the modulating voltage of an audio signal necessary to provide 100% modulation of a 100-V carrier that is simultaneously modulated by 2 audio waves with m1 and m2 equal to 75% and 45% respectively.

Solution: For m3 at 100 % mod ulation; m3 =

(

)

m2T − m12 + m22 =

(

12 − 0.752 + 0.452

)

= 48 .5 % For Modulating Voltage; Vm3 = m3 x Vc = 0.485 x 100 V = 48 .5 V Answer : Vm = 48 .5 V



1-44

C. .SPECTRUM OF CONVENTIONAL AM SIGNAL. 1.

Frequency Spectrum

2.

Voltage Spectrum

3.

Power Spectrum

1-45

Self-Sufficient Guide to ECE by JASON AMPOLOQUIO D. .CONVENTIONAL AM SYSTEM (DSBFC). 1.

AM Bandwidth

BW = 2fm

2.

Side Frequency (Sideband)

fSB = fc ± fm

Parameter

General Solution

Transmit Power

⎛ m2 ⎞ PT = Pc ⎜⎜ 1 + ⎟ 2 ⎟⎠ ⎝ m2Pc 4

Sideband Power

PUSB = PLSB =

Transmit Voltage

VT = Vc 1 +

m2 2

Transmit Current

I T = Ic 1 +

m2 2

where: BW = AM bandwidth fm = Modulating frequency in Hz Vc = Unmodulated carrier in V VT = Modulated carrier in V Pc = Unmodulated carrier power in W PT = Modulated power in W m = modulation index



1-46

ECE Board Exam: NOV 2004

What will be the total sideband power of an AM transmitting station whose carrier power is 1200 W and a modulation of 95%?

Solution: PT = Pcarrier + PSB ⇒ where PSB = PSB =

0.95 2 (1200 ) = 541 .5 Watts 2

m2Pc 2


Calculate the power in one sideband of an AM signal whose carrier power is 50 watts. The unmodulated current is 2 A while the modulated current is 2.4 A.

Solution: I T = Ic 1 +

m2 ⇒m= 2

Pusb = Plsb =

⎡⎛ I 2 ⎢⎜⎜ T ⎢⎝ Ic ⎣

2 ⎤ ⎞ ⎟ − 1⎥ = ⎟ ⎥ ⎠ ⎦

⎡ ⎛ 2 .4 ⎞ 2 ⎤ 2 ⎢⎜⎜ ⎟⎟ − 1⎥ = 0.938 ⎢⎝ 2 ⎠ ⎥ ⎣ ⎦

m2Pc 0 .938 2 (50 ) = = 11 watts 4 4

Sample Problem:

Calculate the total power and the power in each side frequency for a standard AM transmission that is sinusoidally modulated to a depth of 80% if the unmodulated carrier power is 50 kW.

Solution: For PT ; ⎛ ⎛ m2 ⎞⎟ 0 . 82 PT = Pc ⎜1 + = 50kW ⎜1 + ⎟ ⎜ ⎜ 2 ⎠ 2 ⎝ ⎝ = 66 kW For Each Side Frequency Power ; m2Pc (0 .82 )50 kW = 4 4 = 8 kW each

Pside =

Answer : PT = 66 kW , Pside = 8 kW

⎞ ⎟ ⎟ ⎠


1-47

Sample Problem:

Calculate the amplitude and resulting side frequency if a carrier wave of frequency 10 MHz with a peak value of 10 V is amplitude modulated by a 5 kHz sine wave of amplitude 6 V.

Solution:

For m; m=

Vm 6 = Vc 10

= 0.6 For Side Frequency; fside = fc ± fm = 10 MHz ± 0.005 MHz = 10.005 MHz and 9.995 MHz For amplitude; mVc 0.6 (10 ) VSB = = = 3V 2 2 Answer : VSB = 3 V, fside = 10.005 MHz and 9.995 MHz

Sample Problem:

The output voltage of an AM transmitter is 40 V when sinusoidally modulated to a depth of 100%. Calculate the voltage at each side frequency when the modulation depth is reduced to 50%.

Solution: For Vc ; Vc =

VT 2

m 2 = 32 .65 V 1+

=

40 1+

12 2

For Each Side Frequency Voltage ; mVc (0.5)32 .65 V Vside = = 2 2 = 8 .16 V each Answer : Vc = 32 .65 V, Vside = 8 .16 V



1-48


The maximum bandwidth of an AM wave is twice of the maximum modulating frequency.

ª

The total power in an AM signal increases with modulation, reaching a value of 50% greater than that of the unmodulated carrier at 100% modulation.

ª

The extra power with modulation goes into the sidebands; the carrier power does not change with modulation.

ª

The useful power, that is, the power that carries the information, is rather small reaching a maximum of one-third of the total signal power at 100% modulation.

E. .SUPPRESSED-CARRIER AM SYSTEM. 1.

Double-Sideband Suppressed-Carrier (DSB-SC)

General Solution of Double Sideband Suppressed Carrier (DSB-SC)

ν(t) = −

mVc mVc cos 2 π(fc + fm )t + cos 2π(fc − fm )t 2 2

1-49


Parameter

General Solution m2Pc 2

Transmit Power

PT =

Sideband Power

PUSB = PLSB =

Transmit Voltage

VT =

Transmit Current

IT =

m2Pc 4

mVc 2 mIc 2


A DSB-SC system must suppress the carrier by 50 dB from its original value of 10 W. To what value must the carrier be reduced?

Solution: SdB = 10 log Pred =

2.

Poriginal Preduced

10 50 dB 10 10

⇒ Pred =

Porig 10

S dB 10

= 0.1 x10 −3 W = 100 μW

Single-Sideband Full-Carrier (SSB-FC)

General Solution of Single Sideband Full Carrier (SSB-FC)

ν(t) = Vc sin2πfc t +

mVc cos 2π(fc − fm )t 2



1-50

Parameter


Transmit Power

PT = Pc +

Sideband Power

PUSB or PLSB =

Transmit Voltage

VT = Vc 1 +

m2 4

Transmit Current

I T = Ic 1 +

m2 4

m2Pc 4


Assuming 100% modulation H3E system, what would be the transmitted power in the remaining sideband of an AM signal if the carrier power is 1000 watts?

Solution: For H3E ⇒ (SSBFC ) PSB =

3.

m2PC 12 (1000) = = 250 watts 4 4

Single-Sideband Suppressed-Carrier

General Solution of Single Sideband Suppressed Carrier (SSB-SC)

ν(t) =

mVc cos 2π(fc − fm )t 2

1-51


Parameter

4.


Transmit Power

PT =

Sideband Power

PUSB or PLSB =

Transmit Voltage

VT =

mVc 2

Transmit Current

IT =

mI c 2

m2Pc 4

Single-Sideband Reduced-Carrier (For 90% Suppression)

Transmit Power PT = 0.1Pc +

Sideband Power

m2Pc 4

PUSB = PLSB =

m2Pc 4

F. .PERCENTAGE POWER SAVING.

%PS =

PCAM − Ptx x100% PCAM

where : PCAM = Conventional or Standard AM Ptx = Transmitted AM system



1-52


Determine the power saving in percent when the carrier is suppressed in an AM signal modulated to 80%.

Solution: For Conventional AM; ⎛ m2 PT = Pc ⎜1 + ⎜ 2 ⎝ = 1 .32 Pc

2 ⎞ ⎛ ⎟ = Pc ⎜1 + 0 .8 ⎟ ⎜ 2 ⎠ ⎝

⎞ ⎟ ⎟ ⎠

For Suppressed Carrier system ; m2Pc 0 .82 Pc = 2 2 = 0 .32 Pc

Ptx =

Perce ntage Power Saved P − Ptx %℘ = CAM x 100 % PCAM =

1 .32Pc − 0 .32Pc x 100 % = 75 .75 % 1 .32Pc

Sample Problem:

Calculate the percentage power saving for J3E system at 90% modulation.

Solution: For Conventional AM; ⎛ m2 PT = Pc ⎜1 + ⎜ 2 ⎝ = 1 .405 Pc

2 ⎞ ⎛ ⎟ = Pc ⎜1 + 0 .9 ⎟ ⎜ 2 ⎠ ⎝

⎞ ⎟ ⎟ ⎠

For J3E system ; 0 .92 m2 = Pc x 4 4 = 0 .2025 Pc

PJ3E = Pc x

Perce ntage Power Saved − Ptx P x 100 % %℘ = CAM PCAM =

1 .405Pc − 0 .2025 Pc x 100 % = 85 .6 % 1 .405Pc

1-53

Self-Sufficient Guide to ECE by JASON AMPOLOQUIO Comparison between Different Systems (Applicable when m=1)

System

Carrier Power

Total Sideband Power

LSB or USB Power

Percentage Power Saving

DSB-FC (A3E)

66.67% of PT

33.33% of PT

16.67% of PT

0%

DSB-SC

0% of PT

100% of PT

50% of PT

66.67%

SSB-FC (H3E)

80% of PT

20% of PT

20% of PT

16.67%

SSB-SC (J3E)

0% of PT

100% of PT

100% of PT

83.33%


The standard AM broadcast band starts at 535 kHz and ends at 1605 kHz.

ª

Carrier assignments start at 540 kHz and continue in a succession of 10-kilohertz increments until the upper limit of the broadcast band is reached (1610 kHz). This adds up to a total of 107 carrier assignments, or CHANNELS, over the entire broadcast band.

ª

Since interference between such closely spaced stations would be nearly impossible to prevent, the FCC (U.S.) avoids assigning adjacent channels to stations in the same area.



1-54

I

H

1.

The A. B. C. D.

standard AM broadcast band ______. starts at 88 kHz and ends at 108 kHz starts at 535 kHz and ends at 1605 kHz starts at 535 MHz and ends at 1605 MHz starts at 88 MHz and ends at 108 MHz

2.

Two sinusoidal signals, V1 and V2, are fed into an ideal balanced mixer. V1 is a 20-MHz signal; V2 is a 5-MHz signal. What frequencies would you expect at the output of the mixer? A. 5 MHz and 15 MHz B. 20 MHz and 100 MHz C. 15 MHz and 25 MHz D. 5 MHz and 25 MHz

3.

An AM transmitter generates 100 watts with 0% modulation. How much power will it generate with 20% modulation? A. 50.12 watts B. 310.1 watts C. 102 watts D. 256 watts

4.

If the carrier power is 1000 watts, what is the power in the USB at 70.7% modulation? A. 333.33 watts B. 125 watts C. 666.67 watts D. 70.7 watts

5.

A carrier is modulated by three audio tones are 0.3, 0.4, and 0.5, then what A. 0.636 C. 0.707

6.

You 100 A. C.

7.

A SSB transmitter is connected to a 50-ohm antenna. If the peak output voltage of the transmitter is 20 volts, what is the PEP? A. 6 watts B. 4 watts C. 8 watts D. 2 watts

8.

The total power in an AM signal increases with modulation, reaching a value of _____ greater than that of the unmodulated carrier at _____. A. 80%, 80% B. 32%, 80% modulation D. 64%, 80% modulation C. 50%, 75%

tones. If the modulation indexes for the is the total modulation index? B. 1.2 D. 0.9

look at an AM signal with an oscilloscope and see that the maximum Vpp is volts and the minimum Vpp is 25 volts. What is the modulation index? 0.25 B. 1.25 0.6 D. 0.75


1-55

The total power in an AM signal increases with modulation, reaching a value of _____ greater than that of the modulated wave at _____. A. 13%, 50% B. 22%, 75% modulation D. 33%, 90% modulation C. 50%, 50%

10. An AM transmitter supplies a 10 kW of carrier power to a 50 ohms load. It operates at a carrier frequency of 1.2 MHz and is 85% modulated by a 3 MHz sine wave. Calculate the rms voltage of the signal. APRIL 2004 A. 547 V B. 825 V C. 327 V D. 707.1 V 11. For H3E transmitter, the useful power, that is, the power that carries the information, is rather small reaching a maximum of _____. A. one-third of the total signal power at 100% modulation B. one-fifth of the total signal power at 100% modulation C. one-fourth of the total signal power at 100% modulation D. one-tent of the total signal power at 100% modulation 12. For A3E transmitter, the useful power, that is, the power that carries the information in one of the sideband, is rather small reaching a maximum of _____. A. one-third of the total signal power at 100% modulation B. one-fifth of the total signal power at 100% modulation C. one-fourth of the total signal power at 100% modulation D. one-sixth of the total signal power at 100% modulation 13. The power amplifier of an AM transmitter draws 100 watts from the power supply with no modulation. Assuming high-level modulation, how much power does the modulation amplifier deliver for 100% modulation? A. 16.67 watts B. 50 watts C. 33.33 watts D. 66.67 watts 14. If the final RF amplifier of an AM transmitter is powered by 100 volts DC, what is the maximum collector voltage at 100% modulation? A. 400 volts B. 50 volts C. 200 volts D. 100 volts 15. Determine the power saving in percent when the carrier is suppressed in an AM signal modulated to 80%. NOV 2003 A. 75.76% B. 82.82% C. 33.33% D. 16.67% 16. The most commonly used filter in SSB generation. A. Mechanical B. RC C. LC D. Low pass 17. Which of the following is not a technique of generating an SSB signal? A. phase shift method B. filter method C. weaver method D. Armstrong method


1-56


18. An emission technique where the total current will be twice as much when the modulation index is doubled. A. R3E B. H3E C. A3E D. J3E 19. What is the maximum modulating signal frequency that can be used with an H3E system with 20-kHz bandwidth? A. 10 kHz B. 20 kHz C. 5 kHz D. 40 kHz 20. An AM broadcast transmitter is tested by feeding the RF output into a 50Ω (dummy) load. Tone modulation is applied. The carrier frequency is 850 kHz and the FCC licensed power output is 5 kW. The sinusoidal tone is set for 90% modulation. What is the average power that is being dissipated in the dummy load? A. 7.25 kW B. 7025 W C. 72.k5 W D. 725 W 21. How many percent of the total transmitted power is present in the sideband in an A3E system? A. 16.67% B. 80% C. 50% D. 33.33% 22. An intelligence signal is amplified by a 70% efficient amplifier before being combined with a 10-kW carrier to generate the AM signal. If it is desired to operate at 100% modulation, what is the dc input power to the final intelligence amplifier? A. 7.14 kW B. 5.14 kW C. 4.14 kW D. 6.14 kW 23. A transistor RF power amplifier operating class C is designed to produce 40 W output with a supply voltage of 60 V. If the efficiency is 70%, what is the average collector current? NOV 2003 A. 666.67 mA B. 952.4 mA C. 476.2 mA D. 238.1 mA 24. At 80% modulation H3E, what is the percentage power saving? A. 66.67% B. 33.33% C. 12.12% D. 16.67% 25. An AM transmitter has a 1-kW carrier and is modulated by three different sine waves having equal amplitudes. If the total modulation index is 80%, calculate the individual values of m in % and the total transmitted power. A. 56.2%, 1.32 kW B. 46.2%, 1.72 kW C. 46.2%, 1.32 kW D. 56.2%, 1.72 kW 26. The total bandwidth needed for an AM signal at 55.25 MHz with 0.5 MHz video modulation is ______. A. 0.5 MHz B. 1 MHz C. 101.5 MHz D. 10 MHz


1-57

27. A transmitter radiates 9 kW with the carrier unmodulated and 2.08 kW when the carrier is sinusoidally modulated and then suppressed. The modulation index is ______. A. 0.6 B. 0.8 C. 0.68 D. 0.58 28. The "envelope" of an AM signal is due to: A. the baseband signal B. C. the amplitude signal D. 29. The A. B. C. D.

the carrier signal none of the above

equation for full-carrier AM is: v(t) = (Ec + Em) x sin(ωct) v(t) = (Ec + Em) x sin(ωmt) + sin(ωct) v(t) = (Ec - Em) x sin(ωmt) x sin(ωct) v(t) = (Ec + Em sin(ωmt)) x sin(ωct)

30. An SSB transmitter has an average power ranging from 750-1000 W. What is the PEP? A. 9 kW B. 6 kW C. 3 kW D. 5 Kw 31. Calculate the average power of an SSB signal with 2-tone modulation if the peak voltage is 25 V, and assume that the load is 50Ω, resistive. A. 1.56 to 2.08 W B. 1.32 to 2.8 W C. 1.12 to 2.08 W D. 1.66 to 2.58 W 32. Determine the percentage power saving if the carrier and the USB is suppressed in an AM system modulated at 85%. A. 82.23% B. 66.67% C. 86.73% D. 89.71% 33. When a broadcast AM transmitter is 80% modulated, its antenna current is 15 A. What will be the new output current and the percentage increase when the modulation depth is increased by 95%? A. 15.78 A, 17% B. 13.1 A, 20.46% C. 13.1 A, 17% D. 15.78 A, 20.46% 34. In amplitude modulated wave equation, the carrier is _______ with both the upper and lower side frequencies, and the upper and lower side frequencies are _______ with each other. A. 90° out of phase, 180° out of phase B. 180° out of phase, 90° out of phase C. 90° out of phase, 270° out of phase D. 270° out of phase, 90° out of phase 35. A term used to describe the amount of amplitude changed present in an AM waveform. A. Deviation B. Coefficient of Modulation C. Shift D. Drift



1-58

36. Calculate the modulation index for a standard AM transmission, if the maximum peak voltage of the modulated wave is 150 V and the modulating signal voltage is 50 V peak. A. 25% B. 75% C. 50% D. 100% 37. What is the value of modulation index and the relation between modulating signal amplitude and carrier amplitude if the transmitted AM wave is undermodulated. B. m>1, Vm>Vc A. m=1, Vm=Vc D. m= infinite, Vm=Vc=0 C. m<1, Vm
41. Calculate the power in one sideband of an AM signal whose carrier power is 50

watts. The unmodulated current is 2 A while the modulated current is 2.4 A.

APRIL 2004 A. C.

22.1 W 50 W

B. D.

31.4 W 25 W

42. How many percent of the total transmitted power is present in the carrier in an H3E system? A. 0% B. 66.67% C. 25% D. 16.67% 43. A 200 W carrier is modulated to a depth of 75%. The power of the modulated wave is A. 56.25 W B. 228.125 W C. 256.25 W D. 200 W 44. A SSBFC broadcast radio transmitter radiates 10 kW when the modulation percentage is 65%. The carrier power is ______. A. 8.26 kW B. 12.11 kW C. 2.11 kW D. 9.04 kW 45. At 100% modulation J3E, what percentage of the total transmitted power is in sideband? A. 100% B. 0% C. 80% D. 50%


1-59

46. The antenna current of an SSBFC AM transmitter is 8 A when only the carrier is sent. It is increases to 8.5 A when the carrier is sinusoidal modulated. The percentage modulation is _____. A. 77.7% B. 71.8% C. 66.7% D. 50.7% 47. _______ is the process of impressing or imparting a low-frequency source information (voice, audio, video, or data) onto a high-frequency bandpass signal with a carrier frequency fc by the introduction of amplitude, frequency or phase perturbation. A. Mixing B. Modulation C. Heterodyning D. Multiplexing 48. _______is an analog modulation scheme in which the amplitude of a relatively high-frequency carrier signal is varied in accordance with the instantaneous amplitude of an information signal. A. Pulse Modulation B. Quadrature Amplitude Modulation C. Angle Modulation D. Amplitude Modulation 49. Fourier analysis often allows complex signals to be express as a series of _______. A. sinusoids B. pulse C. ellipse D. rectified signals 50. Mathematical process that allows complex signals to be express as a series of sinusoids. A. Bessel Function B. Taylor Series C. Fourier analysis D. Heaviside Expansion 51. An AM transmitter uses a high level modulation. The RF power amplifier runs from 12 volt source, putting out a carrier power of 85 watts, with an efficiency of 85%, what load impedance is required in order for it to deliver the rated power? APRIL 2004 B. 0.13 Ω A. 2.16 Ω D. 0.85 Ω C. 1.69 Ω 52. At 50% modulation DSBSC, what is the percentage power saving? A. 75.75% B. 88.88% C. 66.67% D. 87.87% 53. At 100% modulation H3E, what percentage of the total transmitted power is in the carrier? A. 66.67% B. 20% C. 33.33% D. 80% 54. A 500 W carrier is simultaneously modulated by two audio waves with modulation percentage of 55% and 65% respectively. Calculate the total sideband power if an additional audio wave modulates the carrier at 25%. A. 196.875 Watts B. 15.625 Watts C. 696.875 Watts D. 98.43 Watts



1-60

55. At 80% modulation J3E, what is the percentage power saving? A. 66.67% B. 87.87% C. 75.75% D. 16.67% 56. A radio transmitter, SSB-FC AM radiates 50 kW of carrier power. The radiated power at 85% modulation will be _____. A. 68.1 kW B. 9.03 kW C. 59.03 kW D. 18.1 kW 57. What is the value of modulation index and the relation between modulating signal amplitude and carrier amplitude at 100% modulation. B. m>1, Vm>Vc A. m=1, Vm=Vc C. m<1, Vm
B. D.

jitter hits

61. Calculate the modulating voltage of an audio signal necessary to provide 100% modulation of a 100-V carrier that is simultaneously modulated by 2 audio waves with m1 and m2 equal to 75% and 45% respectively. A. 24.25 V B. 32.8 V C. 50.3 V D. 48.5 V 62. A DSB-SC system must suppress the carrier by 50 dB from its original value of 10 W. To what value must the carrier be reduced? NOV 2004 A. 1 mW B. 10 mW C. 0.1 mW D. 0.01 mW 63. The total power in an AM signal increases with modulation, reaching a value of _____ greater than that of the unmodulated carrier at 100% modulation. A. 25% B. 75% C. 50% D. 100% 64. For a standard AM transmission, the maximum peak-to-peak voltage is 150 V and the minimum peak-to-peak voltage is 50 V. Calculate the modulation index. A. 75% B. 50% C. 25% D. 100%


1-61

65. An AM transmitter uses high-level modulation. The RF power amplifier draws 12 A from a 22 V supply, putting out a carrier power of 140 watts. What impedance would be seen at the modulation transformer secondary? NOV

2003

A. C.

2.16Ω 1.83Ω

B. D.

0.183Ω 1.56Ω

66. At 100% modulation, what percentage of the total transmitted power is in each sideband? A. 50% B. 66.67% C. 16.67% D. 33.33% 67. A standard AM transmission, sinusoidally modulated to a depth of 30%, produces side frequencies of 4.928 and 4.914 MHz. The amplitude of each side frequency is 75 V. Determine the amplitude and frequency of the carrier. A. 500V, 4.907 MHz B. 50V, 4.907 MHz C. 50V, 4.90 MHz D. 500V, 4.90 MHz 68. The output power of an AM transmitter is 1 a depth of 100%. Calculate the power modulation depth is reduced to 50%. A. 83.33 W B. C. 41.67 W D.

kW when sinusoidally modulated to at each side frequency when the 750 W 20.83 W

69. An audio system requires a frequency response from 50 Hz to 15 kHz for high fidelity. If this signal were transmitted using AM, what bandwidth would it require? A. 15 kHz B. 25 kHz C. 8 kHz D. 30 kHz 70. An AM transmission of 3 kW is 100% modulated. If it is transmitter as an SSB signal, what would be the power transmitted? NOV 2004 A. 2000 W B. 500 W C. 1500 W D. 250 W 71. What is the value of modulation index and the relation between modulating signal amplitude and carrier amplitude if the transmitted AM wave is overmodulated. B. m>1, Vm>Vc A. m=1, Vm=Vc D. m= infinite, Vm=Vc=0 C. m<1, Vm
must must must must

be be be be

Class A Class B linear low-power

73. The maximum bandwidth of an AM wave is _____ of the maximum modulating frequency. A. 3x B. equal to C. half D. twice


1-62


74. An AM signal has the following characteristics: carrier frequency = 150 MHz; modulating frequency = 3 MHz; peak carrier voltage = 40 volts; and peak modulating voltage is 30 volts. Calculate the peak voltage of the lower sideband frequency. APRIL 2004 A. 7.5 V B. 5 V C. 10 V D. 15 V 75. The amplitude of the upper and lower side frequencies is a function of both the _______. A. carrier frequency and the modulation frequency B. carrier amplitude and the modulation index C. modulated wave amplitude and the modulation index D. sideband amplitude and the modulation frequency 76. An AM transmitter is modulated by two sine waves at 1.5 kHz and 2.5 kHz, with modulation of 20 percent and 80 percent respectively. Calculate the effective modulation index. APRIL 2004 A. 82% B. 85% C. 80% D. 78% 77. In amplitude modulated wave equation, the carrier component is a _______ function, the upper side frequency a _______ function, and the lower side frequency a _______ function. A. -sine, +cosine, +sine B. -sine, –cosine, -sine C. +sine, –cosine, +cosine D. +sine, +cosine, -cosine 78. For conventional AM, the useful power, that is, the power that carries the information, is rather small reaching a maximum of _____. A. one-third of the total signal power at 100% modulation B. one-fifth of the total signal power at 100% modulation C. one-fourth of the total signal power at 100% modulation D. one-tent of the total signal power at 100% modulation 79. The values of Vmax and Vmin as read from an AM wave on an oscilloscope is 2.6 and 0.29 respectively. Determine the percentage modulation. NOV 2002 A. 69.2& B. 39.95& C. 79.9% D. 34.6& 80. If Va sin(ωat) amplitude modulates the carrier Vc sin(ωct), it will produce the frequencies: B. (ωc + ωa)/2 and (ωc – ωa)/2 A. ωc + ωa and ωc – ωa C. ωc + ωa and 2ωc + 2ωa D. (ωc x ωa)/2 and (ωc x ωa)/2 81. At 100% modulation, the total sideband power is: A. equal to the carrier power B. twice the carrier power C. half the carrier power D. 1.414 x carrier power 82. What will be the total sideband power of an AM transmitting station whose carrier power is 1200 W and a modulation of 95%? NOV 2004 A. 1200 W B. 541.5 W C. 270.75 W D. 483.5 W


1-63

83. The outline of the peaks of a carrier has the shape of the modulating signal and is called the A. Carrier variation B. Envelope C. Waveshape D. Trace 84. The new signals produced by modulation are called A. Spurious emissions B. Harmonics C. Sidebands D. Intermodulation products 85. What is the minimum AM signal needed to transmit information? A. Both sidebands B. Carrier plus sidebands C. Carrier only D. One sideband 86. During 100% modulation, what percentage of the average output power is in the sidebands? A. 66.67% B. 33.33% C. 22.22% D. 88.88% 87. An AM transmitter at 27 MHz develops 10 W of carrier power into a 50-Ω load. It is modulated by a 2-kHz sine wave between 20 and 90% modulation. Determine the maximum and minimum waveform voltage of the AM signal at 20% and 90% modulation. A. Vmin(20%)=25.3 V, Vmax(20%)= 37.9 V, Vmin(90%)=3.14 V, Vmax(90%)= 60.1 V B. Vmin(20%)=33.3 V, Vmax(20%)= 22.1 V, Vmin(90%)=3.14 V, Vmax(90%)= 60.1 V C. Vmin(20%)=25.3 V, Vmax(20%)= 22.1 V, Vmin(90%)=4.45 V, Vmax(90%)= 12.1 V D. Vmin(20%)=33.3 V, Vmax(20%)= 37.9 V, Vmin(90%)=4.45 V, Vmax(90%)= 12.1 V 88. The inputs to a balanced modulator are 1 MHz and a carrier of 1.5 MHz. The outputs are A. 500 kHz B. 2.5 MHz C. 1.5 MHz D. 500 kHz and 2.5 MHz 89. A widely used balanced modulator is called the A. Diode bridge circuit B. Full-wave bridge rectifier C. Lattice modulator D. Balanced bridge modulator 90. A SSB signal is generated around a 200-kHz carrier. Before filtering, the upper and lower SB are separated by 200 Hz. Calculate the filter Q required to obtain 40-dB suppression. A. Q=2.5 B. Q=25 C. Q=250 D. Q=2500 91. Amplitude modulation is the same as A. Linear mixing C. Signal summation

B. D.

Analog multiplication Multiplexing

92. The principal circuit in the popular 1496/1596 IC balanced modulator is a A. Differential amplifier B. Rectifier C. Bridge D. Constant current source 93. The most commonly used filter in SSB generators uses A. LC networks B. Mechanical resonators C. Crystals D. RC networks and op amps



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94. An unmodulated carrier is 300 Vp-p. Calculate %m when its maximum p-p value reaches 400, 500, and 600 V. A. %m1=44.4%, %m2=83.3%, %m3=100% B. %m1=22.2%, %m2=33.3%, %m3=100% C. %m1=16.67%, %m2=83.3%, %m3=100% D. %m1=33.3%, %m2=66.7%, %m3=100% 95. A balanced modulator used to demodulate a SSB signal is called a(n) A. Transponder B. Product detector C. Converter D. Modulator 96. Frequency translation is carried out by a circuit called a A. Translator B. Converter C. Balanced modulator D. Local oscillator 97. A 100-V carrier is modulated by a 1-kHz sine wave. Determine the side frequency amplitudes when m=0.75. A. Vsf=37.5 V B. Vsf=86.5 V C. Vsf=46.5 V D. Vsf=57.5 V 98. The component used to produce AM at very high frequencies is a A. Varactor B. Thermistor C. Cavity resonator D. PIN diode 99. An intelligence signal is amplified by a 70% efficient amplifier before being combined with a 10 kW carrier to generate the AM signal. If it is desired to operate at 100% modulation, what is the dc input power to the final intelligence amplifier? A. Pin/dc=34.66 kW B. Pin/dc=5.75 kW C. Pin/dc=7.14 kW D. Pin/dc=3.26 Kw 100.

The ac rms antenna current of an AM transmitter is 6.2 A when unmodulated and rises to 6.7 A when modulated. Calculate %m. A. m=11.9% B. m=33.9% C. m=57.9% D. m=78.9%

101.

The inputs to a mixer are fo and fm. In down conversion, which of the following mixer output signals is selected? B. fm A. fo D. fo+fm C. fo-fm

102.

Calculate the required Q for a 1MHz carrier, 80 dB suppression, 200 Hz frequency separation. A. Q=12 5000 B. Q=1 250 000 C. Q=1 250 D. Q=125 000

103.

A. C.

In a diode modulator, the negative half of the AM wave is supplied by a(n) Tuned circuit B. Transformer Capacitor D. Inductor


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104.

Amplitude modulators that vary the carrier amplitude with the modulating signal by passing it through an attenuator work on the principle of A. Rectification B. Resonance C. Variable resistance D. Absorption

105.

An AM transmission of 1000 W is fully modulated. Calculate the power transmitted if it is transmitted as a SSB signal. B. PSSB=826 W A. PSSB=167 W D. PSSB=478 W C. PSSB=369 W

106.

That circuit that recovers the original modulating information from an AM signal is known as a A. Modulator B. Demodulator C. Mixer D. Crystal set In the phasing method of SSB generation, one sideband is canceled out due

107.

to A. C.

B. D.

Sharp selectivity Phase inversion

A. C.

A SSB transmission drives 121 Vp into a 50-Ω antenna. Calculate the PEP. PEP=642 W B. PEP=31.7 W PEP=146 W D. PEP=35.6 W

A. C.

The desired output from a mixer is usually selected with a Phase-shift circuit B. Crystal filter Resonant circuit D. Transformer

108.

109.

110.

Phase shift Carrier suppression

A 1-MHz, 40 Vp carrier is modulated by a 5-kHz intelligence signal such that m=0.7. This AM signal is fed to a 50-Ω antenna. Calculate the power of each spectral component fed to the antenna. A. Pc=35 W, Pusb= Plsb=34.96 W B. Pc=16 W, Pusb= Plsb=1.96 W C. Pc=87 W, Pusb= Plsb=1.44 W D. Pc=11 W, Pusb= Plsb=4.22 W

111.

A. C.

The most commonly used amplitude demodulator is the Diode mixer B. Balanced modulator Envelope detector D. Crystal filter

112.

Calculate the filter’s required Q to convert DSB to SSB, given that the two sidebands are separated by 200 Hz. The suppressed carrier (40 dB) is 2.0 MHz. A. Q=12,241 B. Q=44,670 C. Q=36,250 D. Q=6,610

113.

Amplitude modulation can be produce by A. Having a carrier vary a resistance B. Having the modulating signal vary a capacitance C. Varying the carrier frequency D. Varying the gain of an amplifier



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114.

A circuit that generates the upper and lower sidebands but no carrier is called a(n) A. Amplitude modulator B. Diode detector C. Class C amplifier D. Balanced modulator A. C.

In a diode ring modulator, the diode act like Variable resistors B. Switches Rectifiers D. Variable capacitors

A. C.

The output of a balanced modulator is AM B. SSB D.

115.

116.

FM DSB

117.

An AM transmitter at 27 MHz develops 10 W of carrier power into a 50-Ω load. It is modulated by a 2-kHz sine wave between 20 and 90% modulation. Determine the sideband signal voltage and power at 20% and 90% modulation. A. Vsb(20%)=1.24 V, Psb(20%)=0.21 W, Vsb(90%)=22.06 V, Psb(90%)=64.025 W B. Vsb(20%)=4.24 V, Psb(20%)=0.14 W, Vsb(90%)=10.06 V, Psb(90%)=64.025 W C. Vsb(20%)=2.24 V, Psb(20%)=0.1 W, Vsb(90%)=10.06 V, Psb(90%)=2.025 W D. Vsb(20%)=4.24 V, Psb(20%)=0.61 W, Vsb(90%)=22.06 V, Psb(90%)=2.025 W

118.

If A. B. C. D.

119.

Distortion of the modulating signal produces harmonics which cause an increase in the signal A. Carrier power B. Bandwidth C. Sideband power D. Envelope voltage

120.

The process of translating a signal, with or without modulation, to a higher or lower frequency for processing is called A. Frequency multiplication B. Frequency division C. Frequency shift D. Frequency conversion

121.

An AM transmitter at 27 MHz develops 10 W of carrier power into a 50-Ω load. It is modulated by a 2-kHz sine wave between 20 and 90% modulation. Determine the load current at 20% and 90% modulation. A. Iload(20%)=0.924 A, Iload(90%)=0.33 A B. Iload(20%)=0.451 A, Iload(90%)=0.53 A C. Iload(20%)=0.612 A, Iload(90%)=0.33 A D. Iload(20%)=0.924 A, Iload(90%)=0.53 A

122.

A. C. 123.

m is greater than 1, what happens? Normal operation Carrier drops to zero Carrier frequency shifts Information signal is distorted

The equivalent circuit of a quartz crystal is a Series resonant circuit B. Parallel resonant circuit Neither A nor B D. Both A and B

Amplitude modulation generated at a very low voltage or power amplitude is known as A. High-level modulation B. Low-level modulation C. Collector modulation D. Minimum modulation


A. C.

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A modulator circuit performs what mathematical operation on its two inputs? Addition B. Multiplication Division D. Square root

125.

The ratio of the peak modulating signal voltage to the peak carrier voltage is referred to as A. The voltage ratio B. Decibels C. The modulation index D. The mix factor

126.

Calculate the power advantage gained by suppressing the carrier at 100% modulation. A. 7.78 dB B. 6 dB C. 4.78 dB D. 3 dB

127.

Calculate the S/N improvement by suppressing the carrier and one of the sideband at 100%. A. 7.78 dB B. 6 dB C. 4.78 dB D. 3 dB

128.

An SSB transmitter has an average power ranging from 750-1000 W. What is the PEP? A. 3 kW B. 5 kW C. 6 kW D. 9 kW

129.

Calculate the average power of an SSB signal with 2-tone modulation if the peak voltage is 25 V, and assume that the load is 50Ω, resistive. A. 1.32 to 2.8 W B. 1.66 to 2.58 W C. 1.56 to 2.08 W D. 1.12 to 2.08 W

130.

A. C.

At 80% modulation J3E, what is the percentage power saving? 16.67% B. 66.67% 75.75% D. 87.87%


ANGLE MODULATION

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Section

Angle

3

Modulation


Angle Modulation is an analog modulation scheme in which the angle of a relatively high-frequency carrier signal is varied in accordance with the instantaneous amplitude of an information signal.

DEFINITION.

DEFINITION. Frequency Deviation is the amount of frequency shifts that occurs when it is acted on by a modulating signal. DEFINITION. Phase Deviation is the amount of phase shifts that occurs when it is acted on by a modulating signal. A. .ANGLE MODULATION. 1.

Direct Frequency Modulation An angle modulation scheme in which the frequency of a constantamplitude carrier is varied in proportion to the amplitude of the modulating signal at a rate equal to the frequency of the modulating signal.

⎛ ⎞ Δf νFM ( t ) = Vc cos ⎜ ωc t + sin ωmt ⎟ fm ⎝ ⎠ 2.

Direct Phase Modulation An angle modulation scheme in which the phase of a constantamplitude carrier is varied in proportion to the amplitude of the modulating signal at a rate equal to the frequency of the modulating signal.

νPM ( t ) = Vc cos ( ωc t + Δφ sin ωmt ) where: Δf = Frequency deviation in Hz Δφ = Phase deviation in rad ωc = Carrier radian frequency in rad

s ωm = Modulating signal radian frequency in rad

s

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Self-Sufficient Guide to ECE by JASON AMPOLOQUIO B. .GENERAL SOLUTION FOR ANGLE MODULATION. Direct FM Generation

INPUT SIGNAL

Type of Modulation

OUTPUT SIGNAL

Angle-Modulated Wave Equation

νFM (t) = Vc cos ⎣⎡ ωC t + Kνm (t)⎦⎤ FM

νFM (t) = Vc cos ⎣⎡ ωC t + KVm cos(ωmt)⎦⎤ Direct PM Generation OUTPUT SIGNAL

INPUT SIGNAL


ANGLE MODULATION

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Type of Modulation

Angle-Modulated Wave Equation

∫

νPM (t) = Vc cos ⎡ ωC t + K PM νm (t)dt ⎤ ⎢⎣ ⎥⎦ PM

⎡ ⎤ K V νPM (t) = Vc cos ⎢ ωC t + PM m sin(ωmt)⎥ ωm ⎣ ⎦

C. .COMPARISON BETWEEN FM & PM.

0

0

Frequency Modulation

1.

2.

Phase Modulation

Maximum Frequency Deviation ª

FM The maximum frequency deviation occurs during the maximum positive and negative peaks of the modulating signal.

ª

PM

The maximum frequency deviation occurs during the zero crossing of the modulating signal.

Modulator Output ª

FM For FM modulator, changes would occur in the output frequency in respect to changes in the amplitude of the input voltage.

ª

PM

For PM modulator, changes would occur in the phase of the output frequency in respect to changes in the amplitude of the input voltage.


Instantaneous Frequency Deviation ª

FM The instantaneous frequency deviation is directly proportional to the amplitude of the modulation signal and inversely proportional to its frequency.

Δf ∝ ª

Vm fm

PM

The instantaneous frequency deviation is proportional to the first derivative or slope of the modulating signal. Δf ∝

4.

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d ν m (t) dt

Instantaneous Phase Deviation ª

FM The instantaneous phase deviation is proportional to the integral of the modulating signal voltage.

∫

Δφ ∝ ν m (t)dt ª

PM

The instantaneous phase deviation is directly proportional to the amplitude of the modulation signal and unaffected by its frequency. Δφ ∝ Vm

5.

Modulation Index ª

FM The modulation index is proportional to the amplitude of the modulating signal and inversely proportional to the frequency of the modulating signal.

mFM ∝

Vm fm


ANGLE MODULATION

1-72

ª

PM

The modulation index is proportional to the amplitude of the modulating signal, independent of its frequency.

mPM ∝ Vm


The amount of frequency shift is proportional to the amplitude of the modulating signal. (FM)

This simply means that if a 10-volt signal causes a frequency shift of 20 kHz, then a 20-volt signal will cause a frequency shift of 40 kHz. ª

The amount of phase shift is proportional to the amplitude of the modulating signal. (PM)

If a 10-volt signal causes a phase shift of 20°, then a 20-volt signal causes a phase shift of 40°. ª

The rate of frequency shift is proportional to the frequency of the modulating signal. (FM)

This implies that if the carrier is modulated with a 1-kHz tone, then the carrier is changing frequency 1,000 times each second. ª

The rate of phase shift is proportional to the frequency of the modulating signal. (PM)

If the carrier were modulated with a 1-kHz tone, the carrier would advance and retard in phase 1,000 times each second.

D. .ANGLE MODULATION PARAMETERS. 1.

Frequency Deviation (δ) Frequency deviation is the amount of frequency shifts that occurs when it is acted on by a modulating signal.

δ = k FM x Vm

where : δ = peak frequency deviation in Hz k FM = frequency deviation sensitivit y in Hz

V Vm = peak value of modulating signal in V

1-73


Phase Deviation (φ) Phase deviation is the amount of phase shifts that occurs when it is acted on by a modulating signal.

where: φ = peak phase deviation in rad

φ = k PM x Vm

k PM = phase deviation sensitivity in rad

V Vm = peak value of modulating signal in V


A phase modulator has kp = 2 rad/V. What RMS voltage of a sine wave would cause a peak phase deviation of 30 degrees?

Solution: φ = k PM x Vm ⇒ Vm =

3.

φ k PM

=

π 180 ° rad V

30 ° x 2

= 0.262 V ∴ Vrms =

Vm 2

= 0.185 V

Modulation Index (m)

FM

mFM =

PM δ fm

mPM = φmax

Sample Problem:

An FM modulator has a frequency deviation sensitivity of 5 kHz/V and modulating signal νm(t) = 2 cos(4000π)t . Calculate the peak frequency deviation and modulation index.

Solution: δ = k FM x Em = = 10 kHz

5 kHz x 2V V

δ 10 kHz = 4 kHz fm 2 =5

mFM =


ANGLE MODULATION

1-74

Sample Problem:

A PM modulator has a phase deviation sensitivity of 2.5 rad/V and modulating signal νm(t) = 2 cos(4000π)t . Calculate the peak phase deviation

and modulation index.

Solution: δ = k PM x Em =

2 .5 rad x 2V V

= 5 rad

4.

mPM = φmax = 5 rad =5

Percent Modulation (%M) With angle modulation, percent modulation is simply the ratio of actual frequency deviation to the maximum frequency deviation allowed by law stated in percent form.

% mod ulation =

δ actual x 100% δ max


What is the frequency swing of an FM broadcast transmitter when modulated 80%?

Solution: %M =

5.

δ actual x100 % ⇒ δ actual = %M x δmax = 0 .8 x 75 kHz = 60 kHz δmax

Carrier Swing (CS) Carrier swing is the difference between the maximum positive and negative deviation of the carrier in Hz.

CS = 2δ


1-75

Sample Problem:

Calculate the frequency deviation and % modulation under FCC standards for a given modulating signal that produces 100 kHz carrier swing.

Solution: For frequency deviation; CS 100 kHz δ= = = ±50 kHz 2 2 For % modulation; %M =

δ actual ± 50 kHz x 100 % = x 100 % δmax ± 75 kHz

= 67%

6.

Deviation Ratio (DR) The worst case (maximum) modulation index and is equal to the maximum peak frequency deviation divided by the maximum modulating signal frequency.

DR =

7.

δmax fm(max)

Bandwidth Requirements (BW)

FM System

Bandwidth Formula

Narrowband FM (m<0.25)

BW = 2fm

Wideband FM (m>100)

BW = 2δ

Carson’s Rule (Approximate formula)

BW = 2(δ + fm )

Using Bessel Table (Exact formula)

BW = 2(n x fm )

where: fm = modulating signal frequency in Hz δ = peak frequency deviation in Hz n = number of significant sidebands


ANGLE MODULATION

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A system uses a deviation of 100 kHz and a modulating frequency of 15 kHz. What is the approximate bandwidth?

Solution: BW = 2(δ + fm ) = 2(100 + 15) = 230 kHz

Sample Problem:

Calculate the deviation ratio and approximate bandwidth for the worst-case modulation index for an FM broadcast-band transmitter (FCC standard).

Solution: DR =

8.

δmax 75 kHz = =5 fm(max) 15 kHz

BW = 2(δ + fm ) = 2(75 + 15) = 180 kHz

Lock Range & Capture Range

Lock Range

Capture Range i.

ii.

f

Capture Range (CR) Capture or acquire range is the range of frequencies over which the VCO can lock onto a new signal. Lock Range (LR) Lock or track range is the span of frequencies over which the PLL can remain locked and track a signal.


1-77


A phase-locked loop has a VCO with a free running frequency of 10 MHz. As the frequency of the reference input is gradually raised from zero, the loop locks at 8 MHz and comes out of lock again at 14 MHz. Determine the lock range.

Solution: R lock = 2(fol − fvco ) = 2(14 MHz − 10 MHz ) = 8 MHz


A phase-locked loop has a VCO with a free running frequency of 14 MHz. As the frequency of the reference input is gradually raised from zero, the loop locks at 12 MHz and comes out of lock again at 18 MHz. Calculate the capture range.

Solution: R cap = 2(fvco − fl ) = 2(14 MHz − 12 MHz ) = 4 MHz

Sample Problem:

Calculate the range of frequencies will the PLL be able to capture and subsequently maintain lock if it used a VCO with a free-running frequency of 100 kHz, PLL has a 10% capture range and 20% lock range.

Solution: R cap = fc ± ΔC = 100 kHz ± 10%(100 kHz) = 90 to 110 kHz

R lock = fL ± ΔL = 100 kHz ± 2 0%(100 kHz) = 8 0 to 120 kHz

Sample Problem:

The capture range of a PLL and filter is 12%, while the lock range is 18% and the VCO has a center frequency of 20 MHz. Calculate the PLL operation with an 18.5 MHz input?

Solution: 20 − 18.5 x100% = 7.5% 20 = 20 MHz ± 7.5%(20 MHz)

f = fC ± Δ ⇒ Δ =

= 18.5 to 21.5 MHz


ANGLE MODULATION

1-78

E. .STEREOPHONIC FM. With stereophonic transmission, the information signal is spatially divided into two 50-Hz to 15-kHz audio channels (left and right channel).

For Your Information…

The word “stereo” is a Greek word for three-dimensional.

F. .PRE-EMPHASIS & DE-EMPHASIS.

1.

Pre-emphasis The process of boosting or amplifying the high-frequency components of the modulating signal prior to performing modulation for the reason of uneven signal-to-noise ratio at high frequencies.

2.

De-emphasis The reverse process that is doned in the receiver to compensate for the uneven amplification in the transmitter to restore the original level of the modulating signal.

Pre-emphasis & De-emphasis Time Constant (τ)

In terms of R and C

τ = 75μs = RC

In terms of L and R

τ = 75μs =

L R


1-79

Sample Problem:

Calculate the approximate break frequency for FM broadcast band preemphasis circuit. (FCC standards)

Solution: fb = =

1 1 L = ∴ Time Cons tan t = 75 μs = RC or 2πRC R 2π RL 1 = 2.12 kHz ⇒ FCC standards 2π(75 μs)


The original FM broadcast band in the USA in the early-1940s was on 42-50 MHz with 200 kHz channel spacing.

This band was abandoned after World War II and is now allocated to a seldom-used two-way communications service. ª

Throughout the world, 88-108 MHz (or some portion thereof) is used as a broadcast band, with one very notable exception: Japan,

which uses its own unique 76-90 MHz band with 100 kHz channel spacing.

G. .GENERATION OF FM & PM. 1.

Direct Method ª

Varactor Diode Modulator A practical direct FM generator that uses a varactor (voltage variable capacitor, a.k.a. varicap) diode to deviate the frequency of a crystal oscillator.

ª

FM Reactance Modulator The variation in the reactance of a junction FET will cause the frequency of oscillation or resonant frequency to vary in accordance with the instantaneous amplitude of the modulating signal thereby directly producing FM.

ª

Linear integrated-circuits VCO modulator LIC VCO can generate direct FM output waveform when the input modulating signal is applied directly to the input of the voltagecontrolled oscillator where it deviates the carrier frequency.


ANGLE MODULATION

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2.

Indirect Method ª

Armstrong Method A relatively low-frequency subcarrier is phase shifted 90° and fed to a balanced modulator, where it is mixed with the input modulating signal. The output from the balanced modulator is a DSBSC wave that is combined with the original carrier in a combining network to produce a low-index, phase modulated waveform.

H. .ANGLE MODULATION VERSUS AM. 1.

2.

Advantages of Angle Modulation ª

Noise Immunity Probably the most significant advantage of angle modulation over AM is noise immunity.

ª

Noise Performance With the use of limiters, FM & PM demodulators can actually reduce the noise level and improve the signal-to-noise ratio during demodulation process since most noise results in unwanted amplitude variations in the modulated wave.

ª

Capture Effect A phenomenon that allows a receiver to differentiate between two signals received with the same frequency. The receiver will capture the stronger signal and eliminate the weaker signal. With AM, if two signals are received with same frequency, both will be demodulated and produce audio signals that could be heard simultaneously.

ª

Power Utilization & Efficiency With AM most of the transmitted power (almost 67%) is contained in the carrier while with angle modulation the power remains constant regardless if modulation is present.

Disadvantages of Angle Modulation ª

Bandwidth FM and PM require much more bandwidth than AM (approximately 20 times that of AM).

200 kHz for FM/PM while only 10 kHz for AM. ª

Circuit Complexity FM and PM modulators, transmitters, demodulators, and receivers are more complex to design and build than their AM counterparts.

1-81


I

H

1.

In PM, if the carrier were modulated with a 5-kHz tone, A. the carrier would advance and retard in phase 5,000 times each second B. the carrier would advance and retard in phase 5,000 times each minute C. the carrier would advance and retard in phase 5,000 times each millisecond D. the carrier would advance and retard in phase 5,000 times each kilosecond

2.

Calculate the range of frequencies will the PLL be able to capture and subsequently maintain lock if it used a VCO with a free-running frequency of 100 kHz, PLL has a 20% capture range and 40% lock range. A. Capture Range = 80 to 110 kHz, Lock Range = 60 to 120 kHz B. Capture Range = 80 to 120 kHz, Lock Range = 60 to 140 kHz C. Capture Range = 90 to 120 kHz, Lock Range = 80 to 140 kHz D. Capture Range = 60 to 140 kHz, Lock Range = 90 to 120 kHz

3.

Determine the phase error necessary to produce a VCO frequency shift of 10 kHz for an open loop gain of 40 kHz/rad. A. 0.125 rad B. 0.25 rad C. 0.5 rad D. 25 rad

4.

Determine the worst case output signal-to-noise ratio for a broadcast FM that has a maximum intelligence frequency of 10 kHz. The input signal-to-noise ratio is 2. A. 5 B. 15 C. 10 D. 20

5.

Determine the peak frequency deviation and modulation index for an FM modulator with a deviation sensitivity of 5 kHz/V and a modulating signal of 2cos(4000πt). A. 1 kHz, 50 B. 1 kHz, 5 C. 10 kHz, 5 D. 10 kHz, 50

6.

NBFM stands for: A. National Broadcast FM C. Near Band FM

7.

B. D.

Non-Broadcast FM Narrowband FM

When FM reception deteriorates abruptly due to noise, it is called: A. the capture effect B. the threshold effect C. the noise effect D. the limit effect


ANGLE MODULATION

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8.

If an FM transmitter employs one doubler, one tripler, and one quadrupler, what is the carrier frequency swing when the oscillator frequency is 5 kHz. A. 60 kHz B. 120 kHz C. 180 kHz D. 240 kHz

9.

In PM system, A. the instantaneous phase deviation is directly proportional to the phase of the modulation signal and unaffected by its frequency B. the instantaneous phase deviation is directly proportional to the amplitude of the modulation signal and unaffected by its phase C. the instantaneous phase deviation is directly proportional to the of the frequency modulation signal and unaffected by its amplitude D. the instantaneous phase deviation is directly proportional to the amplitude of the modulation signal and unaffected by its frequency

10. The amount of frequency deviation from the carrier center frequency in an FM transmitter is proportional to what characteristic of the modulating signal? A. Shape B. Phase C. Amplitude D. Frequency 11. If the amplitude of the modulating signal decreases, the carrier deviation A. Goes to zero B. Decreases C. Remains constant D. Increases 12. Maximum frequency deviation of a PM signal occurs at A. Peak negative amplitude B. Peak positive amplitude C. Zero crossing points D. Peak positive/negative amplitudes 13. In PM, carrier frequency deviation is not proportional to: A. Carrier amplitude and frequency B. Modulating signal amplitude C. Modulating signal frequency D. Modulator phase shift 14. To compensate for increases in carrier frequency deviation with an increase in modulating signal frequency, what circuit is used between the modulating signal and the phase modulator? A. Low-pass filter B. High-pass filter C. Bandpass filter D. Phase shifter 15. De-emphasis circuit is use _____. A. prior to modulation B. after modulation C. for de-emphasing high frequency component D. for de-emphasing low frequency component 16. Direct Frequency Modulation is an angle modulation scheme in which the _______ of a constant-amplitude carrier is varied in proportion to the _______ of the modulating signal at a rate equal to the frequency of the _______. A. frequency, amplitude, modulating signal B. amplitude, frequency, carrier signal C. amplitude, frequency, modulating signal D. frequency, amplitude, carrier signal

1-83


17. 60% modulation is equivalent to _____ carrier swing of an FM broadcast transmitter. A. 110 kHz B. 80 kHz C. 90 kHz D. 100 kHz 18. In FM, ff the carrier is modulated with a 1-kHz tone, A. then the carrier is changing frequency 500 times each second B. then the carrier is changing frequency 1,000 times each minute C. then the carrier is changing frequency 60,000 times each minute D. then the carrier is changing frequency 2,000 times each second 19. Calculate the approximate bandwidth for an FM system with a 60 kHz deviation and 12 kHz modulating frequency. A. 144 kHz B. 134 kHz C. 124 kHz D. 184 kHz 20. In PM system, A. the instantaneous phase deviation is proportional to the slope of the modulating signal B. the instantaneous frequency deviation is proportional to or slope of the modulating signal C. the instantaneous frequency deviation is proportional to derivative or slope of the modulating signal D. the instantaneous phase deviation is proportional to the or slope of the modulating signal

first derivative or the first derivative the second second derivative

21. Determine the peak phase deviation for a PM modulator with a deviation sensitivity of 2.5 rad/V and a modulating signal of 2cos(4000πt). A. 25 rad B. 15 rad C. 0. 5 rad D. 5 rad 22. Lock range is also known as A. Acquire range C. Capture range

B. D.

Track range Hold-in range

23. The process of boosting or amplifying the high-frequency components of the modulating signal prior to performing modulation for the reason of uneven signal-to-noise ratio at high frequencies. A. De-emphasis B. Frequency Boosting C. Mixing D. Pre-emphasis 24. The reverse process that is doned in the receiver to compensate for the uneven amplification in the transmitter to restore the original level of the modulating signal. A. Post-emphasis B. Frequency Masking C. Heterodyning D. Squelching 25. If a 2-volt instantaneous value of modulating signal amplitude causes a 10-kHz deviation in carrier frequency, what is the deviation sensitivity of the modulator? A. 1 kHz/volt B. 10 kHz/volt C. 5 kHz/volt D. 20 kHz/volt


1-84

ANGLE MODULATION

26. If a 2-kHz audio tone causes a frequency deviation of 4 kHz, what is the modulation index? A. 1 B. 6 C. 2 D. 8 27. In FM system, A. the instantaneous phase deviation is proportional to the modulating signal voltage B. the instantaneous phase deviation is proportional to the modulating signal voltage C. the instantaneous frequency deviation is proportional to the modulating signal voltage D. the instantaneous frequency deviation is proportional to modulating signal voltage

integral of the derivative of the the derivative of the integral of the

28. In PM, a frequency shift occurs while what characteristic of the modulating signal is changing? A. Shape B. Phase C. Amplitude D. Frequency 29. What will be the deviation caused by a 3-kHz tone if the modulation index is 3? A. 4.5 kHz B. 9 kHz C. 0 kHz D. 6 kHz 30. If the deviation sensitivity of an FM modulator is 2 kHz/V, what will be the modulation index caused by a 1-volt, 1-kHz audio signal? A. 0.2 B. 1 C. 0.5 D. 2 31. At a modulation index of 2, how much power is in the carrier of a 1000-watt FM transmitter? A. 2000 watts B. 500 watts C. 48.4 watts D. 484 watts 32. At a modulation index of 2, how much power is in the first pair of sidebands of a 1000-watt FM transmitter? A. 500 watts B. 673 watts C. 57.3 watts D. 166.67 watts 33. ______ is the span of frequencies over which the PLL can remain locked and track a signal. A. Acquire range B. Track range C. Capture range D. Tune-in range 34. At a modulation index of 2, how much power is in the fifth pair of sidebands of a 1000-watt FM transmitter? A. 100 mW B. 400 mW C. 200 mW D. 800 mW 35. Using Carson's rule, what is the approximate bandwidth of an FM signal with a modulation index of 2 being modulated by a 5-kHz signal? A. 15 kHz B. 10 kHz C. 30 kHz D. 45 kHz

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Self-Sufficient Guide to ECE by JASON AMPOLOQUIO 36. Determine the deviation ratio and bandwidth for the worst case modulation index for an FM broadcast-band transmitter. A. 75, 240 kHz B. 15, 200 kHz C. 5, 240 kHz D. 5, 1500 kHz 37. For PM modulator, A. changes would occur in the phase of the output changes in the amplitude of the input voltage B. changes would occur in the phase of the input changes in the amplitude of the input voltage C. changes would occur in the phase of the output changes in the amplitude of the output voltage D. changes would occur in the phase of the input changes in the amplitude of the output voltage

frequency in respect to frequency in respect to frequency in respect to frequency in respect to

38. If a frequency modulator produces 5 kHz of frequency deviation for a 10-V modulating signal, determine how much frequency deviation is produced for a 2 V modulating signal. A. 150 kHz B. 10 kHz C. 0.5 kHz D. 1 kHz 39. Capture range is also known as A. Lock range C. Acquire range

B. D.

Tune-in range Track range

40. Determine the voltage at the output of a phase comparator with a transfer function of 0.5 V/rad and a phase error of 0.75 rad. A. 0.75V B. 3.73V C. 0.373V D. 0.73V 41. Determine the percent modulation for a TV broadcast station with a maximum deviation of 50 kHz when the modulating signal produces 40 kHz at the antenna. A. 70% B. 80% C. 60% D. 90% 42. For FM modulator, A. changes would occur in the input frequency in respect to changes in the amplitude of the output voltage B. changes would occur in the output frequency in respect to changes in the amplitude of the input voltage C. changes would occur in the input frequency in respect to changes in the amplitude of the input voltage D. changes would occur in the output frequency in respect to changes in the amplitude of the output voltage 43. The transmitter technique adopted to reduce the noise effect in FM system. A. pre-emphasis B. antinoise C. noise masking D. noise killing 44. De-emphasis the receiver in the effect attenuates components and the noise in what frequency range? A. low B. intermediate C. DC D. high

modulating


signal

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ANGLE MODULATION

45. Determine the modulation index of a standard FM broadcast having a hypothetical maximum carrier frequency deviation of + 12 kHz and a maximum modulating frequency of 4 kHz. April 2003 A. 6 B. 5 C. 3/2 D. 3 46. In FM the carrier deviation is determined by ______. A. modulating frequency B. modulating phase C. modulating voltage D. all of the above 47. A frequency doubler has an FM signal input at 13 MHz with a deviation of 5 kHz. The output frequency of the carrier will be ______. A. 260 kHz B. 260 MHz C. 26 kHz D. 26 MHz 48. In phase modulation, the modulation index is proportional to ______. A. signal strength B. carrier frequency C. carrier voltage D. modulating frequency 49. Determine the change in frequency for a VCO with a transfer function of 4 kHz/V and a dc input voltage change of 1.2Vp. A. 4.8 kHz B. 0.333 kHz C. 3.33 kHz D. 0.48 kHz 50. In PM system, A. the modulation index is proportional to the amplitude and frequency of the modulating signal, independent of its phase B. the modulation index is proportional to the amplitude of the modulating signal, independent of its frequency C. the modulation index is proportional to the phase and frequency of the modulating signal, independent of its amplitude D. the modulation index is proportional to the phase deviation of the modulating signal, independent of its phase 51. What is the modulation index of an FM transmitter whose frequency deviation is 50 kHz, while its actual frequency is 10 kHz? April 1999 A. 100 B. 5 C. 0.5 D. 3 52. A FM wave has bandwidth of 160 kHz and modulation index of 7.5, the frequency deviation will be ______. A. 115 kHz B. 75 MHz C. 70.5 kHz D. 115 MHz 53. For an FM modulator with a peak frequency deviation of 10 kHz, a modulatingsignal frequency of 110 kHz, and a 500 kHz carrier, determine the approximate bandwidth using Carson’s rule. A. 40 kHz B. 0.40 kHz C. 4 kHz D. 420 kHz


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54. In frequency modulation for a given frequency deviation, the modulation varies ______. A. inversely as the modulating frequency B. directly as the modulating frequency C. independent of modulating frequency D. all of the above 55. Determine the carrier swing for an FM modulator with a deviation sensitivity of 4 kHz/V and a modulating signal of 10sin(2π2000t). A. 40 kHz B. 75 kHz C. 80 kHz D. 20 kHz 56. ______ is the range of frequencies over which the VCO can lock onto a new signal. A. Track range B. Hold-in range C. Capture range D. Lock range 57. In a FM signal, the power______. A. increases as modulation index increases B. reduces as modulation index increases C. remains constant even when modulation index varies D. none of the above 58. In FM system, the maximum frequency deviation occurs during the _____. A. maximum positive of the modulating signal B. maximum negative peaks of the modulating signal C. maximum positive and negative peaks of the modulating signal D. zero crossing of the modulating signal 59. The carrier swing necessary to provide 80% modulation in the FM broadcasting band is ______. April 2003 & March 1996 A. 120 kHz B. 200 kHz C. 150 kHz D. 180 kHz 60. The A. B. C. D.

frequency signal in phase modulated signal is proportional to ______. only amplitude of the modulating signal only frequency of modulating signal amplitude as well as frequency of the modulating signal none of the above

61. Calculate the approximate break frequency for FM broadcast band preemphasis circuit. A. 0.212 kHz B. 2.12 kHz C. 21.2 kHz D. 212 kHz 62. In FM, the frequency deviation is ______. A. proportional to modulating frequency B. proportional to amplitude of modulating signal C. constant D. directly proportional to amplitude and inversely proportional to modulating frequency


ANGLE MODULATION

1-88

63. For an FM broadcast station, the maximum deviation produced by audio modulation is 45 KHz. The percent modulation is ______. A. 100 B. 10 C. 60 D. 45 64. The frequency of the stereo subcarrier signal in FM broadcasting is______. A. 19 kHz B. 50 kHz C. 38 kHz D. 67 kHz 65. A particular circuit that rids FM of noise. A. discriminator B. C. detector D.

limiter phase shifter

66. What approximate frequency deviation will produce a 150 kHz of available bandwidth for a 10 kHz modulating signal? A. 75 kHz B. 55 kHz C. 65 kHz D. 85 kHz 67. The A. B. C. D.

FM modulation index: increases with both deviation and modulation frequency increases with deviation and decreases with modulation frequency decreases with deviation and increases with modulation frequency is equal to twice the deviation

68. One way to derive FM from PM is: A. integrate the modulating signal before applying to the PM oscillator B. integrate the signal out of the PM oscillator C. differentiate the modulating signal before applying to the PM oscillator D. differentiate the signal out of the PM oscillator 69. The A. B. C. D.

bandwidth of an FM signal is considered to be limited because: there can only be a finite number of sidebands it is equal to the frequency deviation it is band-limited at the receiver the power in the outer sidebands is negligible

70. In FM system, A. the modulation index is proportional to the frequency deviation of the modulating signal and inversely proportional to the amplitude and frequency of the modulating signal B. the modulation index is proportional to the phase deviation of the modulating signal and inversely proportional to the amplitude and phase of the modulating signal C. the modulation index is proportional to the amplitude of the modulating signal and inversely proportional to the frequency of the modulating signal D. the modulation index is proportional to the frequency of the modulating signal and inversely proportional to the amplitude of the modulating signal 71. Mathematically, the calculation of FM bandwidth requires the use of: A. fractals B. Bessel functions C. Taylor series D. ordinary trigonometry and algebra

Self-Sufficient Guide to ECE by JASON AMPOLOQUIO 72. FM bandwidth can be approximated by: A. Armstrong's Rule B. C. Carson's Rule D.

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Bessel's Rule none of the above

73. Calculate the carrier swing necessary to provide 80% modulation using the FCC standards in the FM broadcast band. A. 140 kHz B. 110 kHz C. 120 kHz D. 130 kHz 74. An FM receiver switching suddenly between two stations on nearby frequencies is called: A. the capture effect B. the threshold effect C. the "two-station" effect D. none of the above 75. In PM system, the maximum frequency deviation occurs during the _____. A. maximum positive of the modulating signal B. maximum negative peaks of the modulating signal C. maximum positive and negative peaks of the modulating signal D. zero crossing of the modulating signal 76. Pre-emphasis is used to: A. increase the signal to noise ratio for higher audio frequencies B. increase the signal to noise ratio for lower audio frequencies C. increase the signal to noise ratio for all audio frequencies D. allow stereo audio to be carried by FM stations 77. A pre-emphasis of 75 μs refers to: A. the time it takes for the circuit to work B. the "dead time" before de-emphasis occurs C. the time delay between the L and R channels D. the time-constant of the filter circuits used 78. FM A. B. C. D.

stereo: uses DSBSC AM modulation is implemented using an SCA signal has a higher S/N than mono FM is not compatible with mono FM

79. An SCA signal: A. can use amplitude modulation C. is monaural

B. D.

can use FM modulation all of the above

80. The A. B. C. D.

modulation index of an FM signal can using measurements at points where using measurements at points where using measurements at points where only by using Bessel functions

be determined readily: J0 equals one J0 equals zero the deviation equals zero

81. The A. B. C. D.

standard FM broadcast band ______. starts at 88 kHz and ends at 108 kHz starts at 535 kHz and ends at 1605 kHz starts at 535 MHz and ends at 1605 MHz starts at 88 MHz and ends at 108 MHz


1-90

ANGLE MODULATION

82. ______ is the amount of frequency shifts that occurs when it is acted on by a modulating signal. A. frequency drift B. carrier swing C. frequency deviation D. carrier tolerance 83. Phase deviation is the amount of phase shifts that occurs when it is acted on by a _______. A. noise B. temperature variation C. carrier signal D. modulating signal 84. Direct Phase Modulation is an angle modulation scheme in which the _______ of a constant-amplitude carrier is varied in proportion to the _______ of the modulating signal at a rate equal to the frequency of the _______ signal. A. phase, amplitude, carrier B. amplitude, phase, carrier C. phase, amplitude, modulating D. amplitude, phase, modulating 85. In FM system, A. the instantaneous frequency deviation is inversely proportional to the amplitude of the modulation signal and inversely proportional to its frequency B. the instantaneous frequency deviation is directly proportional to the frequency of the modulation signal and inversely proportional to its amplitude C. the instantaneous frequency deviation is directly proportional to the amplitude of the modulation signal and inversely proportional to its frequency D. the instantaneous frequency deviation is directly proportional to the frequency of the modulation signal and directly proportional to its amplitude 86. If a 10-volt signal causes a frequency shift of 20 kHz, then a 20-volt signal will cause a frequency shift of how many kHz? A. 5 B. 20 C. 200 D. 40 87. If a 5-volt signal causes a phase shift of 40°, then a 20-volt signal causes a phase shift of how many degrees? B. 5° A. 2.5° D. 80° C. 160° 88. ______ is the difference between the maximum positive and negative deviation of the carrier in Hz. A. Carrier Range B. Carrier swing C. Frequency deviation D. Dynamic range 89. The ratio of actual frequency deviation to the maximum frequency deviation allowed by law stated in percent form. A. deviation ratio B. efficiency C. percent modulation D. percent deviation 90. The worst case modulation index and is equal to the maximum peak frequency deviation divided by the maximum modulating signal frequency. A. Deviation factor B. Carrier deviation C. Deviation Ratio D. Frequency index


1-91

91. An FM signal, 2000sin(2πx108t+2sinπx104t), is applied to a 50-Ω antenna. Determine the carrier frequency and transmitted power. A. fc=100 MHz, Pt=40 kW B. fc=110 MHz, Pt=50 kW C. fc=150 MHz, Pt=60 kW D. fc=180 MHz, Pt=70 kW 92. Decreasing the input frequency to a locked PLL will cause the VCO output to A. Increase B. Decrease C. Remain constant D. Jump to the free-running frequency 93. An FM signal, 2000sin(2πx108t+2sinπx104t), is applied to a 50-Ω antenna. Determine the modulation index, modulating frequency and BW. A. mfm=0.2, fm=2 kHz, BW=28 kHz B. mfm=2, fm=1 kHz, BW=25 kHz C. mfm=1.2, fm=1 kHz, BW=25 kHz D. mfm=2, fm=5 kHz, BW=30 kHz 94. The range of frequencies over which a PLL will track input signal variations is known as A. Circuit bandwidth B. Capture range C. Band of acceptance D. Lock range 95. An FM receiver provides 100 dB of voltage gain prior to the limiter. Calculate the receiver’s sensitivity if the limiter’s quieting voltage is 300mV. B. Sensitivity=4 μV A. Sensitivity=3 μV D. Sensitivity=6 μV C. Sensitivity=5 μV 96. Which of the following frequency demodulators requires an input limiter? A. Foster-Seeley discriminator B. Pulse-averaging discriminator C. Quadrature detector D. PLL 97. If an FM transmitter employs one doubler, one tripler, and one quadrupler, what is the carrier frequency swing when the oscillator frequency swing is 2 kHz? A. CS=22 kHz B. CS=33 kHz C. CS=48 kHz D. CS=66 kHz 98. Over a narrow range of frequencies, the PLL acts like a A. Low-pass filter B. Bandpass filter C. Tunable oscillator D. Frequency modulator 99. A certain FM receiver provides a voltage gain of 200,000 (106 dB) prior to its limiter. The limiter’s quieting voltage gain is 200 mV. Determine the receiver’s sensitivity. B. Vsensitivity=1 μV A. Vsensitivity=0.1 μV D. Vsensitivity=100 μV C. Vsensitivity=10 μV


ANGLE MODULATION

1-92

100.

A PLL is set up such that its VCO free-runs at 10-MHz. The VCO does not change frequency until the input is within 50 kHz of 10 MHz. After that condition, the VCO follows the input to +200 kHz of 10 MHz before the VCO starts to free-run again. Determine the lock range and capture range of the PLL. A. fcap=50 kHz, flock=200 kHz B. fcap=150 kHz, flock=400 kHz C. fcap=100 kHz, flock=400 kHz D. fcap=250 kHz, flock=500 kHz

101.

A. C. 102.

Both FM and PM are types of what kind of modulation? Amplitude B. Phase Angle D. Duty cycle

The local FM stereo rock station is at 96.5 MHz. Calculate the local oscillator frequency and the image frequency for a 10.7 MHz IF receiver. A. LO= 105.7 MHz, IMAGE=127.9 MHz B. LO= 107.2 MHz, IMAGE=117.9 MHz C. LO= 105.2 MHz, IMAGE=117.9 MHz D. LO= 107.2 MHz, IMAGE=127.9 MHz A. C.

Which discriminator averages pulses in a low-pass filter? Ratio detector B. PLL Quadrature detector D. Foster-Seeley discriminator

A. C.

Which frequency demodulator is considered the best overall? Radio detector B. PLL Quadrature D. Pulse-averaging discriminator

A. C.

In a pulse-averaging discriminator, the pulses are produced by a(n) Astable multivibrator B. One shot multivibrator Zero-crossing detector D. Low-pass filter

103.

104.

105.

106.

Determine the worst-case output S/N for a narrowband FM receiver with δmax=10 kHz and a maximum intelligence frequency of 3 kHz. The input S/N is 3:1. A. S/N=5 B. S/N=10 C. S/N=20 D. S/N=30

107.

What is the frequency swing of an FM broadcast transmitter when modulated 60%? A. +15 kHz B. +25 kHz C. +35 kHz D. +45 kHz

108.

An FM transmitter delivers, to a 75-Ω antenna, a signal of υ = 1000 sin 109 t + 4 sin104 t . Calculate the modulation index, deviation, and BW.

(

A. B. C. D. 109.

m=2, m=4, m=6, m=8,

δ=5.37 δ=6.37 δ=6.37 δ=4.37

)

kHz, kHz, kHz, kHz,

BW=8 kHz BW=16 kHz BW=24 kHz BW=32 kHz

An FM demodulator that uses a differential amplifier and tuned circuits to convert frequency variations into voltage variations is the A. Quadrature detector B. Foster-Seeley discriminator C. Differential peak detector D. Pulse-locked loop


1-93

110.

Determine the permissible range in maximum modulation index for commercial FM that has 30 Hz to 15 kHz modulating frequencies. A. m=5 to 2500 B. m=5 to 500 C. m=2.5 to 250 D. m=25 to 2500

111.

Determine the worst case output S/N for a broadcast FM program that has a maximum intelligence frequency of 5 kHz. The input S/N is 2. A. S/N=10 B. S/N=20 C. S/N=30 D. S/N=40

112.

The band of frequencies over which a PLL will acquire or recognize an input signal is called the A. Capture range B. Circuit bandwidth C. Band of acceptance D. Lock range

113.

Which oscillators are preferred for carrier generators because of their good frequency stability? A. LC B. RC C. LR D. Crystal

114.

Calculate the amount of frequency deviation caused by a limited noise spike that still causes undesired phase shift of 35° when the intelligence frequency (fi) is 5 kHz. A. 1.05 kHz B. 2.05 kHz C. 3.05 kHz D. 4.05 kHz

115.

In a broadcast FM system, the input S/N=4. Calculate the worst-case S/N at the output of the receiver’s internal noise effect is negligible. A. S/N=16.8:1 B. S/N=17.8:1 C. S/N=18.8:1 D. S/N=19.8:1


NOISE ANALYSIS and dB calculations

1-94

Section

Noise Analysis

4

and dB Calculations

Noise: Any passband of wanted signal.

DEFINITION.

undesirable

energy

that

falls

Read it till it Hertz! within

the

DEFINITION. Interference: Is a form of external noise and happens when information signal from source produce frequencies that fall outside their allocated bandwidth and interfere with information signal from another source. DEFINITION. Distortion: The alteration of information in which the original proportions are changed, resulting from a defect in communication system.

NOISE ANALYSIS

A. .2 GENERAL CLASSIFICATIONS. 1.

Correlated Noise Correlated noise is noise that is correlated to the signal and cannot be present in a circuit unless there is an input signal present.

2.

Uncorrelated Noise Noise that is present regardless of whether there is a signal present or not.

B. .TYPES OF CORRELATED NOISE. 1.

Harmonic distortion Results when unwanted harmonics of a signal are produced through non-linear amplification (mixing).

1-95


%THD =

νhigher ν fun

X100%

where: %THD = %Total Harmonic Distortion νhigher = Quadratic sum of the r.m.s. harmonics = ν 22 + ν 23 + ... νn2 ν fun = rms voltage of the fundamental frequency

Sample Problem:

Calculate the total harmonic distortion if the %2nd order and %3rd order are 2.5% and 1.25 % respectively and fundamental amplitude of 8 V.

Solution:

%2 nd Order

=

ν2 X100 νf

ν2 x 100 8 = 0 .2 V

2 .5 % = ν2

%3 rd Order

=

ν3 X100 νf

ν3 x 100 8 = 0 .1 V

1 . 25 % = ν3

0.2 2 + 0.1 2 X100 8 = 2.795% Answer : 2.795% %THD

2.

=

Intermodulation distortion Results when unwanted sum and difference frequencies are generated when two or more signals are amplified in a non-linear device.



1-96

C. .TYPES OF UNCORRELATED NOISE. 1.

2.

External Noise that is generated outside the device or circuit. i.

Atmospheric noise - A naturally occurring electrical disturbance that originate within the Earth’s atmosphere.

ii.

Extra-terrestrial noise - Noise that consists of electrical signals that originate from outside Earth’s atmosphere. a.

Solar noise - Noise generated directly from the sun’s activity.

b.

Cosmic noise - Noise that originate from nearby stars, and galaxies.

iii.

Man-made noise - Noise generated by mankind.

iv.

Impulse noise - Consist of sudden burst of irregularly shaped pulses that generally last between a few microseconds and a fraction of a millisecond.

Internal Electrical interference generated within a device or circuit. i.

Shot noise - Shot noise is a random fluctuation that accompanies any direct current crossing a potential barrier caused by the random arrival of carrier at the output element of electronic devices.

ii.

Partition noise - Partition noise occurs wherever current has to divide between two or more electrodes and results from the random fluctuation in the division.

iii. Flicker noise - Flicker Noise is associated with crystal surface defects in semiconductor and also found in vacuum tubes. Flicker noise is almost exactly 1/f for low frequency. It is often referred to as pink noise because most of the power is concentrated at the lower end of the frequency spectrum. iv. Burst noise - Burst noise is another low frequency noise that seems to be associated with heavy-metal ion contamination. Burst noise produce popping sound if amplified in an audio system, hence the name popcorn noise or 1/f2. v.

Transit-time noise - Transit time noise occurs when the time taken by charge carrier to cross a junction is comparable to the period of the signal.


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vi. Thermal noise - Thermal noise is associated with the rapid and random movement of electrons within a conductor due to thermal agitation.

Read it till it Hertz…jma Thermal noise is known in several names such as ª

Brownian noise after its discoverer Robert Brown.

ª

Johnson noise after the man who related Brownian particle movement to electron movement.

ª

White noise because thermal noise is equally distributed throughout the frequency spectrum, which is analogous to white light, which contains all visible frequencies.

ª

Gaussian noise because thermal noise exhibit a Gaussian

distribution.

NOISE CALCULATIONS A. .NOISE POWER & SPECTRUM DENSITY. PN = kTB

SN = kT

where: PN = noise power in W SN = noise spectrum density in W/Hz k = Boltzmann's constant (1.38 x 10-23 J/K) B = bandwidth in Hertz T = ambient temperature in K



1-98

Sample Problem:

Calculate the spectrum density and thermal noise power for a certain communication system with an IF bandwidth of 10.7 MHz.

Solution: For Spectrum Density

For Noise Power

SN = kT = (1.38 x 10 − 23 ) x 290 ° = 4 x 10

− 21

W

PN = kTB = SN x B = (4 x 10− 21) x 10.7 MHz

Hz

Answer: SN = 4 x 10−21 W

= 4.28 x 10−14 W

Hz

, PN = 4.28 x 10 −14 W

B. .NOISE VOLTAGE ANALYSIS.

VN = 4kTBnR L

where: VN = noise voltage in V RL = load resistor in Ω

C. .ADDITION OF NOISE DUE TO SEVERAL SOURCES.

2 2 2 2 Vn(total) = Vn(1) + Vn(2) + Vn(3) + ...Vn(N)

where: Vn(total) = total noise voltage in V


1-99


An amplifier operating over a 5 MHz bandwidth has a 100 ohms input resistance. It is operating at 27 degrees Celsius, has a voltage gain of 200 and an input signal of 6 μVrms. Calculate the output rms noise.

Solution: Vn(input) = 4KTBNR t = 4(1.38 x10 −23 )(27 + 273)(5 x10 6 )(100) = 2.88 μVrms Vn(output) = A x Vn(input) = 200 x 2.88 μV = 5.76 μVrms Vn(output)rms = 5.76 μVrms


The resistor R1 and R2 are connected in series at 300 oK and 400 oK temperatures respectively. If R1 is 200Ω and R2 is 300Ω, find the power produced at the load (RL = 500Ω) over a bandwidth of 100 kHz.

Solution: Vn(total) = =

Pn(load) =

Vn(1)2 + Vn(2)2 =

4kB{T1R 1 + T2R 2 }

4(1.38 x 10 − 23 )(100 x 103 ){(300 x 200) + (400 x 300)} = 996.8 nV Vn(rms)2 2R L

⎛ 996.8 nV ⎞ ⎜ ⎟ 2 ⎠ = ⎝ 2(500)

2

= 0.496 fW

D. .CASCADED AMPLIFIER.



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1.

Equivalent Noise Resistance (Req)

R eq = R1 +

R2

A12

+

R3

A12 A22

…+

RN

2 A12 A22 … AN −1

Sample Problem:

The first stage of a two-stage amplifier has a voltage gain of 10, a 600-Ω input resistor, a 1600-Ω equivalent noise resistance and a 27kΩ output resistor. For the 2nd stage, these values are 25, 81kΩ, 10kΩ and 1MΩ, respectively. Calculate the equivalent input-noise resistance.

Solution:

For R 1 R1 = R IN1 + R eq1 = 600 + 1600 = 2.2 kΩ For R 2 R2 =

R OUT1 x R IN2 R OUT1 + R IN2

+ R eq2 =

27 x 81 + 10 27 + 81

= 30.2 kΩ For R 3 R3 = 1 MΩ For R eq R eq = R1 +

R2

A12

= 2.2 kΩ + = 2518Ω Answer : 2.518 kΩ

+

R3

A12 A22

30.2 kΩ 102

+

1 MΩ 102 x 252


Equivalent Noise Temperature (Teq)

Teq = Te1 +

Te2 A12

+

Te3 A12 A 22

…+

TeN A12 A22 … A 2N−1

E. .NOISE IN REACTIVE CIRCUITS. 1.

Noise Voltage (VN)

VN = 4KTBNR D

RD =

Q = Qω0L = Q2r ω0 C

where: R D = dynamic resistance in Ω Q = quality factor ω 0 = angular frequency in rad C = capacitance in Farad L = coil inductance in Henry r = coil resistance in Ω


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1-102

Sample Problem:

A parallel tuned circuit at the input of a radio receiver is tuned to resonate at 125 MHz by a capacitance 23.5 pF. The Q-factor of the circuit is 40 and with a channel bandwidth of the receiver limited to 10 kHz by the audio sections. Determine the effective noise voltage of this radio receiver tuned circuit.

Solution: For the Dynamic Impedance Q 40 = ω0C 2π (125 x 106 ) (23.5 x 10−12 ) = 2.17 kΩ

RD =

For Noise Voltage VN = =

4kTBRD 4 (1.38 x 10−23 ) (17 + 273) (10 x 103 ) (2.17 x 103 )

= 0.589 μV Answer : 0.589 μV

F. .SHOT NOISE CURRENT.

IN = 2qI 0B

w h e re : I N = s h o t n o is e c u rre n t in A q = c h a rg e o f s in g le e le c tro n = 1 .6 x 1 0 - 1 9 C I 0 = d c b ia s c u rre n t in A B n = n o is e b a n d w id th in H z


A diode noise generator is required to produce 10 μV of noise in a receiver with an input impedance of 75 Ω, resistive, and a noise power bandwidth of 200 kHz. What must the current through the diode be?

Solution:

For Noise Current V 10 μV IN = N = R 75 Ω = 0.133 μA

For Diode Current IN = 2qI0B IN2 (0.133 μA)2 = 2qB 2 (1.6 x 10−19 C) (200 x 103 ) = 276 mA

I0 =

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Self-Sufficient Guide to ECE by JASON AMPOLOQUIO G. .SIGNAL-to-NOISE POWER RATIO. 1.

Ideal case

⎛S⎞ ⎛S⎞ =⎜ ⎟ ⎜ ⎟ ⎝ N ⎠ out ⎝ N ⎠in

2.

Practical case

⎛ ApSin ⎛S⎞ ⎜ N ⎟ = ⎜⎜ A N + N ⎝ ⎠out ⎝ p in int ernal

⎞ ⎟ ⎟ ⎠

Small Signal-to-Noise Ratio SIGNAL

NOISE

SIGNAL + NOISE

Large Signal-to-Noise Ratio SIGNAL

NOISE

SIGNAL + NOISE



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H. .NOISE FACTOR & NOISE FIGURE. 1.

Noise Factor (F) Figure of merit used to indicate how much the signal-to-noise ratio deteriorates as a signal passes trough a circuit or series of circuit.

( NS )IN F= S ( N ) OUT 2.

Noise Figure (NF) Noise factor expressed in dB

NF = 10 logF


A transistor has a measured S/N power of 100 at its input and 20 at its output. Determine the noise figure of the transistor.

Solution: NFdB = 10 log

() ()

S N in S N ou

= 10 log

100 ≈ 7 dB 20

For Cascaded Network

FRIISS Formula

FT = F1 +

F2 − 1 F3 − 1 Fn − 1 + + ... A1 A1 A 2 A1 A 2 ...A n −1

Total S/N Ratio

⎛S⎞ ⎛S⎞ ⎜ N ⎟ = ⎜ N ⎟ − 10 logn ⎝ ⎠ T ⎝ ⎠1


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Equivalent Noise Temperature (Teq) The absolute temperature of a resistor that, connected to the input of a noiseless amplifier of the same gain, would produce the same noise at the output of a real amplifier. F = Noise Factor

Teq = Ta (F − 1)

Ta = Ambient Temperature in K


A 3-stage amplifier is to have an overall noise temperature no greater than 70 degrees K. The overall gain of the amplifier is to be at least 45 dB. The amplifier is to be built by adding a low-noise first stage to an existing amplifier with existing characteristics as follows: Stage 2 has 20 dB power gain; 3 dB noise figure. Stage 3 has 15 dB power gain and 6 dB noise figure. Calculate the maximum noise figure (in dB) that the first stage can have.

Solution:

A1 = A T − (A2 + A3 ) = 45 − (20 + 15) = 10 dB FT = 1 +

Teq Ta

=1+

70 = 1.24 290

⎡ F − 1 F3 − 1 ⎤ F1 = FT − ⎢ 2 + ⎥ A1 A2 ⎦ ⎣ A1 4 −1 ⎤ ⎡2 − 1 = 1.24 − ⎢ + 10 10 x100 ⎥⎦ ⎣ NF1 = 10 log F1 = 0.56 dB Answer :0.56 dB


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NOISE ANALYSIS and dB calculations 4.

Equivalent Noise Resistance (Req) F = Noise Factor

R eq = R a (F − 1)

5.

R a = Antenna Resistance in Ω

Relation between Equivalent Noise Temperature & Equivalent Noise Resistance

Teq Ta

=

R eq Ra

NOISE LEVEL CALCULATIONS A. .DECIBEL & NEPER. 1.

Decibel (dB) A unit of measure (abbreviated dB) originally used to compare sound intensities and subsequently electrical or electronic power outputs; now also used to compare voltages. An increase of 10 dB is equivalent to a 10-fold increase in intensity or power, and a 20-fold increase in voltage.

2.

Neper (Np) A transmission unit used in Northern European countries originally used to express the attenuation of current along a transmission line, using natural logarithm.

⎛P ⎞ dB = 10 log ⎜ 1 ⎟ ⎝ P2 ⎠

⎛I ⎞ Np = ln ⎜ 1 ⎟ ⎝ I2 ⎠

Relation between decibel & Neper

1Neper = 8.686 dB

1 dB = 0.115 Neper


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B. .NOISE MEASUREMENT LEVEL & UNITS. 1.

Relative Level Point (RLP) i. For two-wire switching systems, the sending end terminals of a long distance have been long considered to be at a point of zero relative level.

ii.

For four-wire switching, these are theoretical points; the CCITT adopted a relative level of -3.5 dBr (0dBr) for the sending end of a four-wire circuit. In American system, -2 dBr (0 dBr) is widely used).

2.

Transmission Level Point (TLP) The American term for relative level point.

3.

0 dBr Any point in a circuit with the same relative level as the sending terminal is a point of zero relative level 0 dBr.

4.

0 TLP Zero transmission level point (0 TLP) is the point at which the test tone level should be 0 dBm. Relation between 0 TLP and 0 dBr

0 TLP = 0 dBr

5.

dBm dBm is the dB in reference to 1 mW.

6.

dBm0 An absolute unit of power in dBm measured at or referred to a point of 0 TLP. Relation between dBm0 and dBm

dBm0 = dBm − 0 TLP ( 0 dBr )



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Sample Problem:

Calculate the strength of a signal in dBmO if it has an absolute power level of -27 dBm at -24 dBm TLP.

Solution:

Note that 0 dBr = 0 TLP dBm 0 = dBm − dBr = dBm − TLP

= −27 dBm − (− 24 dBm TLP ) = −3 dBm 0

Answer : − 3 dBm 0

7.

dBrn dBrn is the unit of measurement of noise power used in the Western Electric 144-type handset with a sensitivity of -90 dBm at 1000 Hz.

8.

dBrn0 dBrn0 relates noise power reading in dBrn to 0 TLP to establish a common reference point throughout the system. Relation between dBrn0 and dBrn

dBrn0 = dBrn − 0 TLP ( 0 dBr ) 9.

dBa Subsequent to the 144 handset, the Western Electric developed the F1A handset that is 5 dB less sensitive (-85 dBm at 1000 Hz) to the older handset (144-type). The noise measurement unit was the dBa.

10. dBa0 dBa0 relates noise power reading in dBa to 0 TLP to establish a common reference point throughout the system. Relation between dBa0 and dBa

dBa0 = dBa − 0 TLP ( 0 dBr )


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11. dBrnC A third more sensitive handset unit was developed by Western Electric (500-type) giving rise to the C-message line weighting curve and its companion unit, the dBrnC. 12. dBrnC0 dBrnC0 relates noise power reading in dBrnC to 0 TLP to establish a common reference point throughout the system. Relation between dBrnC0 and dBrnC

dBrnC0 = dBrnC − 0 TLP ( 0 dBr )


When measuring a voice channel at a -4 dB test point level, the meter reads -76 dBm. Calculate the reading in dBrnCO.

Solution: dBrnC 0 = dBrnC − 0 TLP

∴ dBrnC = dBm + 90

= dBm + 90 − 0 TLP = −76 + 90 − (−4) = 18 dBrnC 0

Sample Problem:

A 1 kHz tone has a level of 70 dBrnC at a point that is –9 dB TLP. What would be the maximum C-message weighted noise level at the 0 TLP for a signal-to-noise ratio of 30 dB?

Solution: dBrnC 0 = dBrnC − 0 TLP = 70 − (− 9 ) = 79 ⇒ signal level in dBrnC0 ⎛S⎞ = SdBrnC 0 − NdBrnC 0 ⎜ ⎟ ⎝ N ⎠dB 30 dB = 79 − NdBrnC 0 N = 49 dBrnC 0 Answer : 49 dBrnC 0



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13. dBmp The CCITT weighting unit for commercial circuit which is nominally identical to the American F1A weighting unit. 14. dBm0p dBmp0p relates noise power reading in dBmp to 0 TLP to establish a common reference point throughout the system. Relation between dBmp and dBmp0p

dBm0p = dBmp − 0 TLP ( 0 dBr )

Sample Problem:

A -42 dBmp of noise at a -5 dBr point would be reported as ____ dBm0p.

Solution:

dBm 0p = dBmp − 0 TLP (0 dBr ) = − 42 − ( − 5) = − 35 Answer : − 35 dBm 0p

15. pWp The unit of noise power measured with the CCITT recommended psophometer. The reference tone is -90 dBm (1 picowatt) at 800 Hz. 16. pWp0 pWp0 relates noise power reading in pWp to 0 TLP to establish a common reference point throughout the system. Relation between pWp and pWp0

pWp0 = 10 logpWp − 0 TLP ( 0 dBr )

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Self-Sufficient Guide to ECE by JASON AMPOLOQUIO C. .NORTH AMERICAN STANDARD. Handset

Weighting Curve

Reference Frequency

Reference Level

Noise Unit

144 type Handset

144-Line Weighting

1000 Hz

-90 dBm

dBrn

F1A type Handset

F1A-Line Weighting

1000 Hz

-85 dBm

dBa

500 type Handset

C-message Weighting

1000 Hz

-90 dBm (retained)

dBrnC

D. .EUROPEAN STANDARD (CCITT). Weighting Curve

Reference Frequency

Reference Level

Noise Unit

Psophometric Weighting

800 Hz

-90 dBm

dBmp & pWp


dBrn correspond to dB above reference noise.

ª

dBa correspond to dB adjusted.

ª

dBrnC correspond to dB above reference noise using C-message line weighting.

ª

dBmp correspond to dB psophometrically weighted.

ª

pWp correspond to picowatts psophometrically weighted.

ª

A 3000 Hz of white noise (not weighted) is attenuated by 8 dB when measured by 144 weighting network, 3 dB using F1A weighting, 2 dB using C-message weighting, and 2.5 dB for psophometric weighting.



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E. .NOISE LEVEL COMPUTATIONS. 1.

Pure Test Tone

Handset Type

General Solution

144 Handset

dBrn = dBm + 90

F1A Handset

dBa = dBm + 85 dBrnC = dBm + 90

500 type Handset Relation between dBrnC & dBa

dBrnC = dBa + 5

Sample Problem:

A 1 kHz test tone is inserted at a local loop with an amplitude of +4 dBm and is transmitted towards the central office. In this direction the loop has a level of +10 dB TLP, because the signal will be attenuated as it moves towards the central office (about 5 dB). Express the level of the tone in dBrnCO.

Solution: dBrnC = dBm + 90 = 4 + 90 = 94

dBrnC0 = dBrnC − 0 TLP = 94 − 10 dB = 84

And since the signal will be attenuated on its way to the C.O. dBrnC 0 = 84 − 5 dB = 79 Answer : 79 dBrnC 0


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3 kHz White Noise (Random Noise)

Handset Type

General Solution

144 Handset

dBrn = dBm + 90

F1A Handset

dBa = dBm + 82

500 type Handset

dBrnC = dBm + 88

Relation between dBrnC & dBa

dBrnC = dBa + 6

F. .VOLUME UNIT COMPUTATIONS. The VU or volume unit is a unit used to measure the power level (volume) of program channels (broadcast) and certain types of speech or music. Approximate Talker Power in dBm

Pt(dBm) = VU − 1.4

ECE Board Exam: APRIL 2004 Calculate the approximate talker power in dBm for a complex signal with VU meter readout of 5 VU. Solution: Pt(dBm) = 5 − 1.4 = 3.6 dBm



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I

H

1.

A deep space communications system dish antenna receives a signal from a space satellite with -100 dBm of power. The special pre-amplifer boosts this by 50 dB, and another pre-amplifier boosts it an additional 40 dB. Further amplifier stages add 20, 25, and 10 dB to the signal. What is the signal level, in mW, after the last stage? A. 28.525 W B. 8.12 W C. 31.6 W D. 60.8 W

2.

Calculate the input S/N for an amplifier with an equivalent noise resistance of 300Ω, equivalent shot noise current of 5 μA if the amplifier is fed from a 150Ω, 10 μV rms sinusoidal source over a bandwidth of 10 MHz. A. 5.71 dB B. 3.568 dB C. 9.68 dB D. 1.4 dB

3.

0 dBrn is equal to how many dBm at 1000 Hz? A. 0dBm B. -90 dBm C. -88 dBm D. -85 dBm

4.

The resistor R1 and R2 are connected in series at 300oK and 400oK temperatures respectively. If R1 is 200Ω and R2 is 300Ω, find the power produced at the load (RL = 500Ω) over a bandwidth of 100 kHz. APRIL 2003 A. 0.496 fW B. 5.78 pW C. 0.15 fW D. 52.48 pW

5.

A series of noise measurement was conducted using a voltmeter. If the noise registered 20 mV six times; -17 mV two times; -35 mV five times; 25 mV three times and 10 mV four times; determine the rms noise value. A. 27.82 mVrms B. 48.69 mVrms C. 16.12 mVrms D. 23.85 mVrms

6.

Relates noise power reading (dBrnc) to 0 TLP to establish a common reference point throughout the system. A. 0 dBr B. dBrnC0 C. 0 TLP D. 0 dBm

7.

An amplifier has an impedance of 50Ω. Using a matched-impedance diode noise generator, it is found that the DUT (device under test) has doubled noise output power when the diode has a dc current of 14 mA. Determine the noise figure for the DUT. A. 8.25 dB B. 11.46 dB C. 6.82 dB D. 10.12 dB


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8.

Determine the noise figure for a non-ideal amplifier with the following parameters: Input S/N=80 dB; Power gain=60 dB; Input Noise=2x10-18 W Internal Noise=6x10-12 W A. 1.85 dB B. 6 dB C. 4.65 dB D. 2.5 dB

9.

Calculate the total noise voltage at the input and the input S/N ratio in dB for an amplifier with an equivalent noise resistance of 300Ω, and the equivalent shot noise current is 5 μA. The amplifier is fed from a 150Ω, 10 μV rms sinusoidal source and the noise BW is 10 MHz. A. VN=12.34μV, S/N=1.4 dB B. VN=12.34μV, S/N=26.4 dB C. VN=8.51μV, S/N=26.4 dB D. VN=8.51μV, S/N=1.4 dB

10. What is the noise unit in C-message weighting curve? A. dBrn B. dBmp C. dBrnC D. dBa 11. What is the effect on the signal-to-noise ratio of a system (in dB) if the bandwidth is doubled, considering all other parameters to remain unchanged except the normal thermal noise only. The S/N will be____. APRIL 2003 A. S/N ratio is decreased by 1/16 B. S/N ratio is decreased by ¼ C. S/N ratio is decreased by 1/8 D. S/N ratio is decreased by ½ 12. Calculate the approximate talker power in dBm for a complex signal with VU meter readout of 5 VU. A. 6.4 dBm B. 3.6 dBm C. 8.6 dBm D. 1.4 dBm 13. The absolute temperature of a resistor that, connected to the input of a noiseless amplifier of the same gain, would produce the same noise at the output of a real amplifier. A. Ambient noise temperature B. Equivalent noise temperature C. Room temperature D. Curie temperature 14. Calculate the output signal-to-noise ratio in dB for three identical links, given that the S/N ratio for any one links is 60 dB. A. 21.83 dB B. 43.21 dB C. 55.23 dB D. 18.79 dB 15. Calculate the diode’s equivalent voltage if the forward bias current is 1mA over a 100 kHz bandwidth. (Assume the dynamic resistance of the diode equal to 26mV/Idc) A. 10.851 μVrms B. 4.25 μVrms C. 23.56 μVrms D. 0.147 μVrms 16. Calculate the strength of a signal in dBmO if it has an absolute power level of 27 dBm at -24 dBm TLP. A. 3dBm0 B. -3dBm0 C. -51 dBm0 D. -24 dBm0


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17. Calculate the overall noise figure referred to the input for a mixer stage that has a noise figure of 20 dB, and this is preceded by an amplifier that has a noise figure of 9 dB and an available power gain of 15 dB. A. 9.45 dB B. 10.44 dB C. 11.78 dB D. 12.56 dB 18. Calculate the noise factor of an attenuator pad that has an insertion loss of 6 dB. A. 6 B. 8 C. 4 D. 2 19. A diode generator is required to produce 12 microV of noise in a receiver with an input impedance of 75 ohms and a noise power bandwidth of 200 kHz. Determine the current through the diode in milliamperes. APRIL 2004 A. 39.8 mA B. 0.398 mA C. 3.98 mA D. 398 mA 20. Noise that consists of electrical signals that originate from outside Earth’s atmosphere. A. static B. Extra terrestrial noise C. Pink noise D. Shot noise 21. A 20,000 ohms resistor is at room temperature (290 deg K). threshold noise voltage for a bandwidth of 100 kHz. NOV 2002 A. 56.58 μV B. 5658 μV C. 565.8 μV D. 5.658 μV

Calculate the

22. Calculate the overall noise temperature of the receiving system for a receiver that has a noise figure of 12 dB, and it is fed by an LNA that has a gain of 50 dB and a noise temperature of 90 K. A. 90 K B. 76.12 K C. 122 K D. 43.89 K 23. Consist of sudden burst of irregularly shaped pulses that generally last between a few microseconds and a fraction of a millisecond. A. static B. Popcorn noise C. Impulse noise D. Shot noise 24. A tone measures 80 dBrnC at the output of a four wire central office. What will it be at a local loop that is –8 dB TLP? A. 70 dBrnC B. 72 dBrnC C. 74 dBnrnC D. 18 dBrnC 25. In measurement of noise temperature, an avalanche diode source is used, the ENR being 14 dB. The measured Y factor is 9 dB. Calculate the equivalent noise temperature of the amplifier under test. A. 670 K B. 706 K C. 760 K D. 607 K 26. Zero transmission level point is the point at which the test tone level should be 0 dBm. A. 0 dBm B. 0 TLP C. pWp D. dBmO


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27. A single tuned circuit has a Q-factor of 70 and is resonant at 3 MHz with a 470 pF tuning capacitor. Calculate the equivalent noise BW for the circuit. A. 657.69 Hz B. 67.32 kHz C. 145.12 kHz D. 538.19 kHz 28. An amplifier with an overall gain of 20 dB is impressed with a signal whose power level is 1 watt. Calculate the power output in dBm. APRIL 2003 A. 50 dBm B. 62.61 dBm C. 31.45 dBm D. 45.67 dBm 29. What is the reference level and frequency in F1A-line weighting curve? A. -85 dBm and 1000 Hz B. -88 dBm and 3000 Hz C. -85 dBm and 800 Hz D. -90 dBm and 1000 Hz

30. Two resistors rated 5 ohms and 10 ohms are connected in series and are at 27 degrees Celsius. Calculate their combined thermal noise voltage for a 10 kHz bandwidth. NOV 2003 A. 50 μV B. 5 μV C. 0.005 μV D. 0.05 μV 31. A unit of measure originally used to compare sound intensities and subsequently electrical or electronic power outputs; now also used to compare voltages. A. SPL B. dB C. Log D. Neper 32. Three telephone circuits, each having an S/N ratio of 44 dB, are connected in tandem if a fourth circuit is added that has an S/N ratio of 34 dB. Determine the overall S/N ratio. A. 32.86 dB B. 41 dB C. 46.84 dB D. 50 dB 33. The ENR for an avalanche diode is 13 equal to room temperature, determine A. 612.3 K C. 6,076.3 K

dB. Given that the cold temperature is the hot temperature of the source. B. 63,769.3 K D. 16,348.3 K

34. A random fluctuation that accompanies any direct current crossing a potential barrier caused by the random arrival of carrier at the output element of electronic devices. A. Shot noise B. Impulse noise C. Transit-time noise D. Popcorn noise 35. A tone introduced at a -3.5 dBr point with an absolute power level of -18.5 dBm may be referred to as _____ dBm0 signal. A. -21 B. 75 C. -15 D. 81.5 36. A 3000 Hz white noise is attenuated how many dB when measured by a noise measurement set using an F1A weighting. A. 1.5 dB B. 3 dB C. 5 dB D. 8 dB


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37. A three stage amplifier system has a 3-dB bandwidth of 200 kHz, determined by an LC tuned circuit at its input, and operates at 22°C. The first stage has a power gain if 14 dB and a noise figure of 3 dB. The second and third stages are identical, with power gains of 20 dB and noise figure of 8 dB. The output load is 300Ω. The input noise is generated by a 10-kΩ resistor. Calculate the overall noise figure of the system. A. 7.29 dB B. 3.45 dB C. 6.36 dB D. 7.41 dB 38. In measuring noise in a voice channel at -46 dB transmission point level, the meter reads -70 dBm (F1A weighted), convert the reading into pWp. A. 100 pWp B. 57 pWP C. 85 pWP D. 79 pWp 39. An attenuator has a loss of 26 dB. If a power of 3 W is applied to the attenuator, find the output power. April 2003 A. 12.82 mW B. 17.61 mW C. 7.54 mW D. 6.28 mW 40. A transmission unit used in Northern European countries originally used to express the attenuation of current along a transmission line, using natural logarithm. A. SPL B. dB C. Log D. Neper 41. You are measuring a voice channel at a -4 dB transmission point level, the meter reads -73 dBm (pure test tone) convert the reading into dBrnCO. A. 16 dBrnCO B. 13 dBrnCO C. 8 dBrnCO D. 21 dbrnCO 42. Noise, which is characterized by a uniform distribution of energy over the frequency spectrum and a normal distribution of levels. A. Pink noise B. Shot noise C. Gaussian noise D. Mixer noise 43. In a microwave communications system, determine the noise power in dBm for an equivalent noise bandwidth of 10 MHz. APRIL 2003 A. -121.4 dBm B. -169.28 dBm C. 117.89 dBm D. -103.98 dBm 44. What is the equivalent signal in dBm for 4 different signals +10 dBm each? A. 40 dBm B. 4 dBm C. 16 dBm D. 14 dBm 45. The alteration of information in which the original proportions are changed, resulting from a defect in communication system. A. Interference B. Distortion C. Attenuation D. Noise 46. Determine the shot noise for a diode with a forward bias of 1.40 mA over an 80 kHz bandwidth. (q=1.6x10 raised to minus 19 C). APRIL 2003 A. 0.06 μA B. 0.006 μA C. 0.6 μA D. 6 μA


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47. Calculate the strength of a signal in dBa when it is equal to -74 dBrnC. A. -69 dBa B. 11 dBa C. -79 dBa D. 16 dBa 48. Results when unwanted harmonics of a signal are produced through non-linear amplification (mixing). A. Amplitude distortion B. Pink Noise C. White noise D. Intermodulation distortion 49. Two resistors, 20kΩ and 50kΩ are at ambient temperature. Calculate for a bandwidth equal to 100 kHz, the thermal voltage for the two resistors connected in parallel. Nov 1997 A. 55.12 μV B. 15.29 μV C. 4.78 μV D. 13.12 μV 50. A –10dBmO signal introduce at the 0dBr point (0 TLP) has an absolute value of signal level of ____? A. -80 dBm B. -10 dBm C. -70 dBm D. 0 dBm 51. An absolute unit of power in dBm referred to the 0 TLP. A. dBm B. dBW0 C. dBm0 D. dBmC 52. A satellite receiving system include a dish antenna with a Teq equal to 35 K is connected via a coupling network with a Teq of 40 K to a microwave receiver with a Teq of 52 K referred to its input. What is the noise power to the receiver’s input over a 1-MHz frequency range and also determine the receiver’s noise figure. A. 1.75 fW, 0.716 dB B. 17.5 fW, 0.76 dB C. 0.75 fW, 0.16 dB D. 1.75 fW, 0.16 dB 53. Figure of merit used to indicate how much the signal-to-noise ratio deteriorates as a signal passes trough a circuit or series of circuit. A. Noise voltage B. Signal-to-noise ratio C. Noise factor D. Carrier-to-noise ratio 54. Express the ratio in decibels of power ratio 50 is to 10 Watts. NOV 2002 A. 5.66 dB B. 6.99 dB C. 3.87 dB D. 8.13 dB 55. Given a noise factor of 10, what is the noise figure in dB? APRIL 2003 A. 10 dB B. 15 dB C. 16 dB D. 21 dB 56. The equivalent noise resistance for an amplifier is 300Ω, and the equivalent shot noise current is 5 μA. the amplifier is fed from a 150Ω, 10μVrms sinusoidal signal source. If the noise bandwidth is 10 MHz, calculate the input S/N ratio in dB. A. 1.4 dB B. 3.78 dB C. 4.2 dB D. 10.5 dB


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57. 25 dBrnC in C-message weighting A. 25 C. 20

NOISE ANALYSIS and dB calculations is equal to ____ dBa in F1A-line weighting. B. 60 D. 65

58. A naturally occurring electrical disturbance that originate within the Earth’s atmosphere. A. Static B. Gaussian Noise C. Transit-time noise D. Amplitude distortion 59. Calculate the output S/N ratio in dB for three links, the first two of which have S/N ratios of 45 dB and the third has 40 dB. A. 31.4 dB B. 37.87 dB C. 40.2 dB D. 50.5 dB 60. Is another type of low-frequency noise observed in bipolar transistor that appears as a series of burst at two or more levels. When present in an audio system, the noise produces popping sounds, and for this reason is also known as “popcorn noise”. A. Impulse noise B. Pink noise C. Burst noise D. Mixer noise 61. An RF stage uses a single amplifier with 1000 gain that adds 50 μV noise, on a received input signal of 20μV+2μV noise. Calculate the output signal-to-noise ratio. A. 21.4 dB B. 19.8 dB C. 14.3 dB D. 15.7 dB 62. The signal power of the input to an amplifier is 100 microW and the noise power is 1 microW. At the output, the signal power is 1W and the noise power is 40 mW. What is the amplifier noise figure? APRIL 2003 A. 87.5 dB B. 31.67 dB C. 6.02 dB D. 16.13 dB 63. What is the signal strength of a -18 dBm 1000 Hz tone in dBa? A. 67 B. 64 C. 72 D. 70 64. An amplifier with 20 dB gain is connected to another with 10 dB gain by means of a transmission line with a loss of 4 dB; if a signal with a power level of -14 dBm were applied to the system, calculate the power output. APRIL 2003 A. 15 dBm B. 14 dBm C. 10 dBm D. 12 dBm 65. Results when unwanted sum and difference frequencies are generated when two or more signals are amplified in a non-linear device. A. Transit-time noise B. Amplitude distortion C. Gaussion Noise D. Intermodulation distortion 66. A -50 dBmp of noise at a -7 dBr point would be reported as ____ dBm0p. A. -57 B. -43 C. -47 D. 47


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67. Three amplifiers are cascaded together. Determine the total noise figure in dB if the noise figure of each amplifier is equal to 3 dB and the amplifier’s gain is 10 dB each. NOV 2002 A. 3.24 dB B. 30.24 dB C. 0.324 dB D. 13.24 dB 68. An amplifier operating over a 4 MHz bandwidth has 100-Ω input resistance. It has a voltage gain of 46 dB, and has an input signal of 5μVrms Determine the rms output signal-to-noise ratio in dB. A. 5.78 dB B. 3.55 dB C. 2.89 dB D. 4.12 dB 69. Is a form of external noise and happens when information signal from source produce frequencies that fall outside their allocated bandwidth and interfere with information signal from another source. A. Interference B. Distortion C. Attenuation D. Noise 70. Noise that is present regardless of whether there is a signal present or not. A. Amplitude distortion B. Correlated Noise C. Harmonic distortion D. Uncorrelated Noise 71. Three amplifiers are cascaded together. Determine the total noise figure in dB if the noise figure of each amplifier is equal to 3dB and the amplifier’s gain is 10 dB each. NOV 2002 A. 0.24 dB B. 2.24 dB C. 3.24 dB D. 1.24 dB 72. When measuring a voice channel at a -4 dB test point level, the meter reads 76 dBm. Calculate the reading in dBrnCO. APRIL 2004 A. 14 dBrnCO B. 18 dBrnCO C. 16 dBrnCO D. 20 dBrnCO 73. Noise that occurs wherever current has to divide between two or more electrodes and results from the random fluctuation in the division. A. Shot noise B. Impulse noise C. Transit-time noise D. Partition noise 74. A transistor has a measured S/N power of 100 at its input and 20 at its output. Determine the noise figure of the transistor. NOV 2004 A. 9.98 dB B. 3.18 dB C. 6.98 dB D. 4.49 dB 75. Any point in a circuit with the same relative level as the sending terminal is a point of zero relative level abbreviated as. A. 0 dBr B. 0 dBm C. 0 pWp D. dBrnCO 76. A parallel LC tank circuit is made up of an inductor of 3 mH and a winding resistance of 2Ω. The capacitor is 0.47μF. Determine the maximum impedance and BW. A. Zmax=5.12 kΩ, BW=579 Hz B. Zmax=6.27 kΩ, BW=618 Hz C. Zmax=3.19 kΩ, BW=106 Hz D. Zmax=7.73 kΩ, BW=316 Hz


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77. The noise produce by a resistor is to be amplified by a noiseless amplifier having a voltage gain of 75 and a BW of 100 kHz. A sensitive meter at the output reads 240 μVrms. Assuming operation at 37° C, calculate the resistor’s resistance. If the BW were cut to 25 kHz, determine the expected output meter reading. A. R=5.257 kΩ, readout= 381 μV B. R=5.984 kΩ, readout= 120 μV C. R=5.914 kΩ, readout= 815 μV D. R=5.679 kΩ, readout= 187 μV 78. Determine the noise current and equivalent noise voltage for a diode with forward bias of 1 mA over a 1-MHz bandwidth. A. In=56.3 nA, Vn=0.718 μV B. In=89.2 nA, Vn=0.648 μV C. In=17.9 nA, Vn=0.465 μV D. In=47.9 nA, Vn=0.482 μV 79. Calculate the S/N ratio for a receiver output of 4V signal and 0.48V noise both as a ratio and in decibel form. A. 24.58, 64.52 dB B. 33.72, 97.15 dB C. 89.34, 56.23 dB D. 69.44, 18.42 dB 80. An amplifier with NF=6 dB has Si/Ni of 25 dB. Calculate the So/No in dB and as a ratio. A. 13 dB, 66.82 B. 16 dB, 93.24 C. 19 dB, 79.4 D. 11 dB, 61.57 81. A single stage amplifier has a 200-kHz BW and a voltage gain of 100 at room temperature. Assume that the external noise is negligible and that a 1-mV signal is applied to the amplifier’s input. Calculate the output noise voltage if the amplifier has a 5-dB NF and the input noise is generated by a 2-kΩ resistor. A. 813 μV B. 458 μV C. 945 μV D. 678 μV 82. Calculate the noise voltage of a 1-kΩ resistor at 17° C over a 1-MHz BW. Repeat for the series and parallel combination of three 1-kΩ resistors. A. 4 μV, 6.93 μV, 2.31 μV B. 5 μV, 1.55 μV, 5.53 μV C. 2 μV, 5.78 μV, 5.52 μV D. 8 μV, 3.31 μV, 1.16 μV 83. Calculate the noise voltage for a 1-kΩ resistor at 17° C “tuned” by an LC circuit with a BW of 1 MHz. A. 50.1 μV B. 15.01 μV C. 535.01 μV D. 5.01 μV 84. A three-stage amplifier has an input stage with noise ratio NR=5 and power gain equal to 50. Stages 2 and 3 have NR=10, and Ap=1000. Calculate the NF for the overall system. A. 4.71 dB B. 8.33 dB C. 14.79 dB D. 32.84 dB 85. Calculate the noise power at the input of a microwave receiver with an equivalent noise temperature of 45 K. It is fed from an antenna with a 35 K equivalent noise temperature and operates over a 5-MHz BW. A. Pn=97.33x10-15 W B. Pn=5.52x10-15 W -15 C. Pn=6.38x10 W D. Pn=25.3x10-15 W


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86. Generally referred to as the primary cause of atmospheric noise A. Lightning B. Thunderstorm C. El Niño phenomenon D. Rain effect 87. Absence of one of the following components in a parametric amplifier renders this circuit a low-noise amplifier A. Capacitance B. More circuit stages C. Modulation D. Resistance 88. What is the major cause of atmospheric or static noise? A. Meteor showers B. Airplanes C. Thunderstorm D. Sunspots 89. Where is the noise generated that primarily determines the signal-to-noise ratio in a VHF (150 MHz) marine-band receiver? A. In the atmosphere B. In the receiver front end C. In the receiver rear end D. In the ionosphere 90. Extra-terrestrial noise is observable at frequencies from A. 8 to 1.43 GHz B. Above 2 GHz C. 5 to 8 GHz D. 0 to 20 kHz 91. Noise from random acoustic or electric noise that has equal per cycle over a specified total frequency band. A. White noise B. Gaussian noise C. Thermal noise D. All of these 92. Referred to as the stage in the radio contributes most of the noise A. IF amplifier B. Mixer C. Supply stage D. Speaker 93. Industrial noise frequency is between A. 200 to 3000 MHz C. 15 to 160 MHz

B. D.

0 to 10 MHz 20 GHz

94. At what power level does a 1 kHz tone cause zero interference (144 weighted)? A. -90 dBm B. 90 dBm C. 90 dB D. -90 dB 95. Background noise is the same as the following except A. White noise B. Thermal noise C. Impulse noise D. Gaussian noise 96. Noise reduction system used for film sound in movie A. dBa B. dBm C. dolby D. dBx 97. A receiver connected to an antenna whose resistance is 50 ohms has an equivalent noise resistance of 30 ohms. What is the receiver’s noise temperature? A. 464 K B. 174 K C. 293 K D. 754 K



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98. What is the reference frequency of CCITT psophometric noise measurement? A. 1500 Hz B. 1000 Hz C. 3400 Hz D. 800 Hz 99. Atmospheric noise is less severe at frequencies above A. 1 GHz B. 10 GHz C. 30 MHz D. 20 KHz

A. C.

NIF stands for noise interference figure Non-intrinsic figure

A. C.

Which does not affect noise in a channel? Bandwidth B. Temperature Quantizing level D. None of these

100.

101.

102.

B. D.

Noise improvement factor narrow intermediate frequency

What does the noise weighting curve show? Power levels of noise founding carrier systems The interfering effect of other frequencies in a voice channel compared with a reference frequency of one kilohertz C. Noise signals measured with a 144 handset D. Interfering effects of signals compared with a 3-kHz tone A. B.

103.

A. C.

The most common of noise measurement in white noise voltage testing dBm B. dBW dBrn D. NPR

104.

The random and unpredictable electric signals from natural causes, both internal and external to the system is known as A. attenuation B. noise C. distortion D. interference

105.

The ratio of the level of the modulated output with no modulation standard test modulation to the level demodulated output with no modulation applied both measured with the same bandwidth A. Residual noise level B. Reference audio output C. Audio frequency response D. None of these A. C.

A figure of merit to measure the performance of a radiation detector Safety factor B. Quality factor Ripple factor D. Noise equivalent power

A. C.

Considered as the main source of an internal noise Temperature change B. Thermal agitation Flicker D. Device imperfection

A. C.

Unit of noise power of psophometer dBm pWp

106.

107.

108.

B. D.

dBmO dBa


Section

5

Transmitters and Receivers

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DEFINITION. Selection is the ability of the receiver to select a particular frequency of a station from all other station frequencies appearing at the antenna of the receiver. DEFINITION. Reception occurs when a transmitted electromagnetic wave passes through the receiver antenna and induces a voltage in the antenna. DEFINITION. Detection is the action of separating the low (audio) frequency intelligence from the high (radio) frequency carrier. A detector circuit is used to accomplish this action. DEFINITION. Reproduction is the action of converting the electrical signals to sound waves, which can then be interpreted by your ear as speech, music, and the like. A. .AM MODULATOR CIRCUITS. 1.

2.

High-Level Modulation The FCC defines high-level modulation in the Code of Federal Regulations as "modulation produced in the plate circuit of the last radio stage of the system." ª

Plate Modulator The plate modulator is a high-level modulator. The modulator tube must be capable of varying the plate-supply voltage of the final power amplifier.

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Collector-Injection Modulator A collector-injection modulator is a transistorized version of the plate modulator.

Low-Level Modulation The FCC defines low-level modulation in the Code of Federal Regulations as "modulation produced in an earlier stage than the final." ª

Control-Grid Modulator A control-grid modulator is a low-level modulator that is used where a minimum of a.f. modulator power is desired. It is less efficient than a plate modulator and produces more distortion.


BASIC Transmitters AND Receivers

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Base-Injection Modulator A base-injection modulator is used to produce low-level modulation in equipment operating at very low power levels. It is often used in small portable equipment and test equipment.

ª

Cathode Modulator The cathode modulator is a low-level modulator employed where the audio power is limited and the inherent distortion of the grid modulator cannot be tolerated.

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Emitter-Injection Modulator The emitter-injection modulator is an extremely modulator that is useful in portable equipment.

low-level

B. .BLOCK DIAGRAM OF LOW-LEVEL AM DSBFC TRANSMITTER.

AM Transmitter Functional Blocks ª

Information Source Generally an acoustical transducer, such as microphone, a magnetic tape, a CD, or a phonograph record.

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Pre-amp Stage Typically a sensitive, class A linear voltage amplifier with a high input impedance whose function is to raise the amplitude of the source signal to a usable level while producing non-linear distortion.

ª

Driver Stage A linear amplifier that simply amplifies the information signal to an adequate level to sufficiently drive the modulator.

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Power Amplifier/Modulator Stage A transistor operating in class C is likely to be used in the power amplifier, with collector modulation accomplished by means of a push-pull Class AB audio amplifier.

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Self-Sufficient Guide to ECE by JASON AMPOLOQUIO Transmitter Efficiency (η)

When no modulation is applied

When modulation is applied General Solution

P η= c Ps

η=

Psb Pmod

When m=1 (50% increased)

η=

Psb 0.5Pc = Pmod Pmod

where: η = conversion efficieny Pc = unmodulated carrier power in W Ps = dc input power to the collector in W Psb = sideband power due to modulation in W Pmod = average(audio) power supplied by the modulator in W


A transistor RF power amplifier operating class C is designed to produce 40 W output with a supply voltage of 60 V. If the efficiency is 70%, what is the average collector current?

Solution: η=

Pc P 40 ⇒ Ps = c = = 57.14 W Ps η 0.7

∴ Ic =

Ps 57.14 = = 0.952 A Vs 60


The power amplifier of an AM transmitter has an output carrier power of 25 W and an efficiency of 70% and is collector-modulated. How much audio power will have to be supplied to this stage for 100% modulation?

Solution: η =

Psb ; when m = 1; PT = 1.5Pc and Psb = 0.5Pc = 0 .5 x 25 = 12 .5 W Pmod

Pmod = Paudio =

Psb 12.5 = = 17.85 W η 0.7



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RF Carrier Oscillator Stage Generates a high frequency signal that will be modulated by a low-frequency information in order for it to carry the information. Stability of Crystal Oscillator

fout = fc + kfc (T − To ) where: fc = center frequency in MHz k = temperature coefficient in Hz per MHz per degree C = Hz/MHz/° C ⇒ ppm


A portable radio transmitter has to operate at temperature from -5°C to 35°C. If its signal is derived from a crystal oscillator with a temperature coefficient of +1ppm/°C and it transmits at exactly 146 MHz at 20°C, find the transmitting frequency at the lower extreme of the operating temperature range.

Solution: fout = fc + kfc (T − To ) = 146 MHz + 10−6 x 146 MHz (−5 − 20) = 145.99635 MHz

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Buffer Amplifier Stage A low gain, high input impedance linear amplifier used to isolate the oscillator from the high power amplifier and provides frequency stability.

ª

Intermediate and Power Amplifier Stage Either class A or class B push-pull amplifier required for low-level transmitters to maintain symmetry in the AM envelope.


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Sample Problem:

A quartz watch is guaranteed accurate to 125 seconds per year. Assuming a year has 8760 hrs, calculate the accuracy of the crystal oscillator in the watch in parts per million.

Solution: a=

125 sec x (1x106 ) = 4 ppm 3600 sec 8760 hrsx 1 hr

C. .RADIO RECEIVER.

Collection of electronic devices that transform the transmitted message back into a form understandable by human. 1.

Two Basic Types Of Radio Receiver i.

Tuned Radio Frequency (TRF) Receiver Probably the simplest designed radio receiver available.

Disadvantage of TRF Receiver a. Inconsistent bandwidth that varies with center frequency when tuned over a wide range of input frequencies. b.

Instability due to large number of RF amplifiers all tuned the same center frequency.

c.

Non-uniform gain over a very wide frequency range.

TRF Selectivity where:

f Q= r BW

fr = resonant frequency in Hz =

1 2 π LC



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Sample Problem:

A TRF receiver is to be designed with a single tuned circuit using a 10μH inductor. Calculate the capacitance range of the variable capacitor required to cover the entire AM band (535-1605 kHz) and also calculate the BW of this receiver at 540 kHz and 1600 kHz (assume Q=110).

Solution:

At 535 kHz 1 1 fr = ⇒C = 2 π LC L (2 πfr )2 1 C = 10 μH(2 π x 535 kHz )2 = 8 .85 nF

At 535 kHz fr 535 kHz = Q 110 = 4 . 86 kHz

BW =

ii.

2.

At 1605 kHz fr =

1 2 π LC

C =

⇒ C =

1

L (2 π fr )2

1

10 μ H(2 π x1605 kHz )2 = 983 pF

At 1605 kHz fr 1605 kHz = Q 110 = 14 .6 kHz

BW =

Super-heterodyne Receiver Invented by Major Edward Armstrong that downconverts the incoming RF signal to IF before processing and before the extraction of the information signal.

Receiver Components i.

Antenna Collect very weak signal radiated by a broadcast station.

ii.

RF Section The RF amplifier amplifies the modulated RF signal received by the antenna. This section determines the sensitivity of the receiver.

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Image Frequency Any frequency other than the selected radio frequency carrier that, if allowed to enter a receiver will produce a cross-product frequency that is equal to the I.F.

High-Side Injection

Low-Side Injection

IM = LO + IF IM = RF + 2IF

IM = LO − IF IM = RF − 2IF

Image Frequency Rejection Ratio (IMRR)

IFRR = (1 + Q2ρ2 )

ρ=

fimage fRF

−

fRF fimage


An AM receiver is tuned to broadcast station at 600 kHz. Calculate the image rejection in dB, assuming that the input filter consists of one tuned circuit with a Q of 60.

Solution: fimage = fRF + 2IF = 600 kHz + 2(455 kHz)

ρ=

= 1,510 kHz

fimage fRF

−

fRF fimage

1510 600 − 600 1510 = 2.12

=

IFRR = 20 log 1 + Q2ρ2 = 20 log 1 + (602 x2.122 ) = 42.1 dB



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A receiver has two uncoupled tune circuit before the mixer, each with a Q of 75. The signal frequency is 100.1 MHz. The IF is 10.7 MHz. The local oscillator uses high-side injection. Calculate the image rejection ratio in dB.

Solution: fimage = fRF + 2IF = 100.1 MHz + 2(10.7 MHz) = 121.5 MHz

ρ=

fimage fRF

−

fRF fimage

121.5 100.1 − 100.1 121.5 = 0.39

=

IFRR T = IFRR1 x IFRR2 ∴ Qp = Qs IFRR T(dB) = IFRR1(dB) + IFRR2(dB) = 20 log 1 + Qp2ρp2 + 20 log 1 + Qs2ρs2 ⎡ ⎤ = 2⎢20 log 1 + (752 x0.392 ) ⎥ = 58.65 dB ⎣ ⎦

Double Spotting Double Spotting occurs when a receiver picks up the same station at two nearby points on the receiver dial caused by poor front-end selectivity or inadequate image-frequency rejection. iii. Mixer/Converter Section The mixer/converter section down-converts the incoming RF signal to a fixed IF signal. iv. IF Section The IF section provides most of the receiver gain and selectivity.

Gain k=kc k=1.5kc k=3kc f

kc =

1 Qp Q s

k o = 1.5k c

BW = k fo


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where: Qp = Q of primary circuit Q s = Q of secondary circuit k c = critical coupling factor k o = optimum coupling factor fo = resonant frequency in Hz BW = bandwidth in Hz


An IF transformer of a radio receiver operates at 455 kHz. The primary circuit has a Q of 50 and the secondary has a Q of 40. Find the bandwidth using the optimum coupling factor.

Solution: kc =

1 QpQd

=

1 50 x 40

= 0.0224

∴ k o = 1.5 k c = 1.5 x 0.0224 = 0.0336

BW = k o x fo = 0.0336 x 455 kHz = 15.288 kHz

Some Popular IF

System AM Broadcast AM Broadcast (automobiles) FM Broadcast FM Two-way radios Picture IF Sound IF Radar RX/TX Satellite RX v.

IF 455 kHz 262.5 kHz 10.7 MHz 21.4 MHz 41.25 MHz 45.75 MHz 30 or 60 MHz 70 MHz

Detector Section The purpose of the detector section is to extract or demodulate the original source information from the IF signal.

viii. Audio Section Compose of several cascaded audio amplifier that amplifies the demodulated information to a level that will drive the speaker.



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D. .DEMODULATOR CIRCUITS. 1.

2.

AM Demodulators i.

Diode or Envelope Detectors The simplest and most widely used amplitude demodulator is the diode detector and since it recovers the envelope of the AM signal which is the modulating signal, the circuit is sometimes referred to as an envelope detector.

ii.

Product Detectors (Balance Diode Mixer) The AM signal and local oscillator are coupled into the diode, which is a nonlinear device and the nonlinear mixing produce sum and difference frequencies and the output tank circuit is tuned to the difference (IF) frequency.

FM Demodulator i.

Slope Detector The simplest form of tuned-circuit frequency discriminator which is basically composed of a tuned-circuit and a peak detector that converts the amplitude variations to an output voltage that varies at a rate equal to that of the input frequency changes and whose amplitude is proportional to the magnitude of the frequency changes.

ii.

Foster-Seeley Discriminator Foster-Seeley Discriminator (a.k.a. phase-shift discriminator) is a tuned-circuit frequency discriminator whose operation is very similar to that of the balanced slope detector.

iii. Ratio Detector A ratio detector is relatively immune to amplitude variations in its input signal which is the basic disadvantage of slope detector and Foster-Seeley discriminator. iv. Phase Lock Loop Essentially, a PLL is a closed-loop feedback control system in which the input frequency of the feedback signal voltage is the parameter of interest rather than simply a voltage.

PLL IC is consists of a phase comparator (mixer), a LPF, a low-gain amplifier and a VCO.

Self-Sufficient Guide to ECE by JASON AMPOLOQUIO v.

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Quadrature FM Demodulator Sometimes called as coincidence detector which extracts the original information signal from the composite IF waveform by multiplying two quadrature (90° out-of-phase) signals

A quadrature detector uses a 90° phase shifter, a single tuned circuit, and a product detector to demodulate FM signals. vi. Pulse-Averaging Discriminator The FM signal is applied to a zero-crossing detector or clipper/limiter which generates a binary voltage level change each time the FM signal varies from minus to plus, or plus to minus. The result is a rectangular wave containing all the frequency variations of the original signal but without amplitude variations which is then applied to a one-shot multivibrator whose output is then fed to a simple RC low-pass filter.

E. .BASIC RECEIVER CHARACTERISTICS. 1.

Sensitivity The minimum RF signal level that can be detected at the input to the receiver and still produced usable demodulated information.

The ability of a receiver to reproduce weak signals is a function of the sensitivity of a receiver and expressed in μV. 2.

Selectivity The measure of the ability of a receiver to accept a given band of frequencies and reject all others.

Selectivity is the degree of distinction made by the receiver between the desired signal and unwanted signals.


When measuring the selectivity of a receiver, you discover that a signal level of 450 microV on an adjacent channel is required to give the same output as 1 microV signal on the channel to which the receiver is tuned. Calculate the adjacent channel selectivity in dB.

Solution: SdB = 20 log

Vadj Vr

= 20 log

450 μV = 53.06 dB 1 μV



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3.

Shape Factor (SF) The ratio of the -60dB bandwidth to the -3dB(-6dB) bandwidth of a tuned circuit or filter.

SF =

4.

BW−60dB BW − 3dB

Fidelity The fidelity of a receiver is its ability to accurately reproduce, in its output, the signal that appears at its input.

Ability of a communication system to produce an exact replica of the original source information. 5.

Dynamic Range (DR) The input power range over which the receiver is useful.

DR dB = 20 log

DR dB = 10 log

V(max_ input ) sensitivity (volts)

P(max_ input ) threshold( watts )


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A receiver has a sensitivity of 0.6 microV and a blocking dynamic range of 60 dB. What is the strongest signal that can be present along with a 0.6 microV signal without blocking taking place?

Solution: DR dB = 20 log

Vinput Vsensitivity

⎡ 60 ⎤ ⎡ DR ⎤ ⇒ Vinput = Vsensitivity ⎢10 20 ⎥ = 0.6 μV ⎢10 20 ⎥ = 600 μV ⎣ ⎦ ⎣ ⎦

Sample Problem:

A receiver has a dynamic range of 85 dB. It has 1.5 nW sensitivity (threshold). Determine the maximum allowable input signal.

Solution: DR dB = 10 log 85 dB = 10 log

P(max_ input) threshold(watts) P(max_ input) 1.5 x 10− 9 W

P(max_ input) = 0.474 W

For Your Information… st The 1 radio transmitter was the Spark-Gap oscillators coupled to long-wire antennas. MOPA or Master Oscillator Power Amplifier consists of Hartley oscillator capacitively coupled to a neutralized triode amplifier inductively coupled to a tunable “Marconi” antenna.

F. .RECEIVER CONTROL CIRCUITS. 1.

Automatic Gain/Volume Control (AGC/AVC) The function of an automatic gain control, also referred to as an automatic volume control, is to limit unwanted variations in the output of the receiver caused by variations in strength of the received signal input. ª

Reverse AGC When we use an AGC voltage to cause degeneration by driving the amplifiers toward cutoff, it is referred to as reverse AGC.



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Forward AGC For forward AGC the amplifier is driven toward the saturation region of its characteristic curve.

2.

Automatic Frequency Control (AFC) A type of circuit that sense the difference between the actual oscillator frequency and the frequency that is desired and produces a control voltage proportional to the difference.

3.

Squelch Circuit The sensitivity of a receiver is maximum when no signal is being received. This condition occurs, for example, when a receiver is being tuned between stations. The purpose of a squelch circuit is to quiet a receiver in the absence of a received signal.

4.

Blanking Circuit In essence, a blanking circuit detects the occurrence of a highamplitude, short duration noise spike, then mutes the receiver by shutting off a portion of the receiver for the duration of the pulse.

5.

S-meter The S-meter is designed to indicate signal strength for comparison and to aid in tuning the receiver and sometimes in adjusting the antenna.

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I

H

1.

The oscillator of a CB transmitter is guaranteed accurate to +0.005%. What is the maximum frequency at which it could actually transmit if it is set to transmit on channel 20, with a nominal carrier frequency of 27.205 MHz. A. 27.2064 MHz B. 38.6128 MHz C. 24.1289 MHz D. 14.8123 MHz

2.

For A. B. C. D.

3.

The power amplifier of an AM transmitter has an output carrier power of 25 W and an efficiency of 70% and is collector-modulated. How much audio power will have to be supplied to this stage for 100% modulation? APRILL 2003 A. 54.12 W B. 17.9 W C. 33.81 W D. 23.6 W

4.

Determine the overall bandwidth for three single-tuned amplifiers each with a BW of 10 kHz. A. 10 kHz B. 5.098 kHz C. 30 kHz D. 2.08 kHz

5.

If a superhetordyne receiver is tuned to a conversion or local oscillator is operating frequency of the incoming signal that would A. 1673 kHz B. C. 1812 kHz D.

desired signal at 1000 kHz and its at 1300 kHz, what would be the possibly cause image reception? 1600 kHz 1340 kHz

6.

The Colpitts VFO uses: A. a tapped inductor C. an RC time constant

a two-capacitor divider a piezoelectric crystal

7.

a "frequency multiplier" to work, it requires: a nonlinear circuit a linear amplifier a signal containing harmonics an input signal that is an integer multiple of the desired frequency

B. D.

A receiver has a sensitivity of 0.5μV and a blocking dynamic range of 80 dB. What is the strongest signal that can be present along with a 0.5μV signal without blocking taking place? A. 15 mV B. 75 mV C. 5 mV D. 35 mV



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8.

A "frequency synthesizer" is: A. a VCO phase-locked to a reference frequency B. a VFO with selectable crystals to change frequency C. a fixed-frequency RF generator D. same as a mixer

9.

A common AM detector is the: A. PLL C. ratio detector

B. D.

envelope detector Foster Seeley discriminator

10. A tuned circuit tunes the AM radio broadcast band or from 540 to 1700 kHz. If its bandwidth is 10 kHz at 535, what is it at 1605 kHz? A. 17.32 kHz B. 51.46 kHz C. 31.56 kHz D. 917.32 kHz 11. The A. B. C. D.

two basic specifications for a receiver are: the sensitivity and the selectivity the number of converters and the number of IFs the spurious response and the tracking the signal and the noise

12. An FM detector produces a peak-to-peak voltage of 1.5 V from an FM signal that is modulated to 15 kHz deviation by a sinewave. What is the detector sensitivity? A. 61.2 μV/Hz B. 34.1 μV/Hz C. 136.12 μV/Hz D. 50 μV/Hz 13. An oscillator has a frequency of 100 MHz at 20°C, and a temperature coefficient of +10 ppm per degree Celsius. What will be the shift in frequency at 70°C? What percentage is that? A. 50 kHz, 0.5% B. 25 kHz, 0.05% C. 25 kHz, 0.5 % D. 50 kHz, 0.05% 14. In a diode modulator, the negative half of the AM wave is supplied by a(n) A. Capacitor B. Inductor C. Transformer D. Tuned circuit 15. A receiver has a sensitivity of 0.6 microV and a blocking dynamic range of 60 dB. What is the strongest signal that can be present along with a 0.6 microV signal without blocking taking place? NOV 2004 A. 0.06 μV B. 0.6 μV C. 600 μV D. 60 μV 16. If a varactor has a capacitance of 90 pF at zero volts, what will be the capacitance at 4 volts? A. 53.12 pF B. 30 pF C. 12.6 pF D. 581 pF 17. The two inputs to a mixer are the signal to be translated and a signal from a(n) A. Filter B. Local Oscillator C. Modulator D. Antenna


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18. A PLL FM detector uses a VCO with a deviation sensivity of 85 kHz/V. If it receives an FM signal with a deviation of 65 kHz and sinewave modulation, what is the rms output voltage from the detector? A. 7.18 V B. 0.54 V C. 1.82 V D. 0.192 V 19. Amplitude modulation can be produce by A. Varying the carrier frequency B. Varying the gain of an amplifier C. Having the modulating signal vary a capacitance D. Having a carrier vary a resistance 20. Determine the quality factor necessary for a single sideband filter with a 1-MHz carrier frequency, 80 dB unwanted suppression and 200-Hz frequency separation. A. 15,000 B. 125,000 C. 12,000 D. 25,000 21. The Hartley oscillator uses: A. a tapped inductor C. an RC time constant

B. D.

a two-capacitor divider a piezoelectric crystal

22. What value of transformer coupling would a double-tuned 10-MHz IF amplifier with optimal coupling need to get a bandwidth of 100 kHz? A. 0.001 B. 0.01 C. 1 D. 0.1 23. The function of a limiter is: A. to remove amplitude variations C. to limit dynamic range

B. D.

to limit spurious responses to limit noise response

24. Suppose a SSB receiver requires an injected frequency of 1.5 MHz. What would be the acceptable frequency range of the BFO if the maximum acceptable baseband shift is 100 hertz? A. 1.499 MHz + 100 Hz B. 1.5 MHz + 1000 Hz C. 1.5 MHz + 100 Hz D. 1.4999 MHz + 100 Hz 25. A receiver has a sensitivity of 0.3 μV. The same receiver can handle a signal level of 75 mV without overloading. What is its AGC range in dB? A. 108 dB B. 125 dB C. 51 dB D. 83 dB 26. If two signals, Va = sin(ωat) and Vb = sin(ωbt), are fed to a mixer, the output: A. will contain ω1 = ωa + ωb and ω2 = ωa – ωb B. will contain ω1 = ωa / ωb and ω2 = ωb / ωa C. will contain ω = (ωa + ωb ) / 2 D. will contain ω1 = ωa x ωb and ω2 = ωb x ωa


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27. When measuring the selectivity of a receiver, you discover that a signal level of 450 microV on an adjacent channel is required to give the same output as 1 microV signal on the channel to which the receiver is tuned. Calculate the adjacent channel selectivity in dB. NOV 2003 A. 16.87 dB B. 74.95 dB C. 26.5 dB D. 53 dB 28. A receiver’s IF filter has a shape factor of 2.5 and a bandwidth of 6 kHz at the 3 dB point. What is its bandwidth at -60 dB point? A. 23 kHz B. 48 kHz C. 15 kHz D. 62 kHz 29. In a balanced mixer, the output: A. contains equal (balanced) amounts of all input frequencies B. contains the input frequencies C. does not contain the input frequencies D. is a linear mixture of the input signals 30. When measuring the selectivity of a receiver, you discover that a signal level of 500 μV on an adjacent channel is required to give the same output as a 1 μV signal on the adjacent channel to which the receiver is tuned. What is the adjacent channel selectivity in dB? A. 43 dB B. 54 dB C. 38 dB D. 23 dB 31. Basically, selectivity measures: A. the range of frequencies that the receiver can select B. with two signals close in frequency, the ability to receive one and reject the other C. how well adjacent frequencies are separated by the demodulator D. how well the adjacent frequencies are separated in the mixer 32. The conditions for sinusoidal oscillation from an amplifier are called: A. the loop-gain criteria B. the Hartley criteria C. the Bode criteria D. the Barkhausen criteria 33. A radio transmitter has to operate at a temperature of 34 degrees Celsius. If its signal is derived from a crystal oscillator with a temperature coefficient of plus 1ppm per degree Centigrade, and it transmits at exactly 150 MHz at 20 degrees Centigrade, find the transmitting frequency. APRIL 2004 A. 149.9967 MHz B. 150.0021 MHz C. 150.9971 MHz D. 149.0589 MHz 34. In an AM transmitter with 100% modulation, the voltage of the final RF stage will be: A. approximately half the DC supply voltage B. approximately twice the DC supply voltage C. approximately four times the DC supply voltage D. approximately 1.414 times the DC supply voltage


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35. An FM broadcast-band receiver tunes to 88-108 MHz. Calculate the range of local oscillator frequencies assuming the receiver uses high-side injection. A. 96.7 to 119.7 MHz B. 95.7 to 115.7 MHz C. 98.7 to 118.7 MHz D. 99.7 to 119.7 MHz 36. The A. B. C. D.

Clapp oscillator is: a modified Hartley oscillator a modified Colpitts oscillator a type of crystal-controlled oscillator only built with FETs

37. The and A. C.

transformer of a double-tuned IF amplifier has a Q of 25 for both primary secondary. What value of kc do you need to achieve optimal coupling? 0.06 B. 6 0.6 D. 0.006

38. A crystal oscillator is accurate within 0.0005%. How far off frequency could its output be at 30MHz? A. 250 Hz B. 120 Hz C. 200 Hz D. 150 Hz 39. Suppose you have an FM modulator that puts out 1 MHz carrier with a 100hertz deviation. If frequency multiplication is used to increase the deviation to 400 hertz, what will be the new carrier frequency? A. 2 MHz B. 6 MHz C. 4 MHz D. 8 MHz 40. When comparing values for shape factor: A. a value of 1.414 dB is ideal B. C. a value of 1.0 is ideal D. 41. The A. B. C. D.

a value of 0.707 is ideal there is no ideal value

main function of the AGC is to: keep the gain of the receiver constant keep the gain of the IF amplifiers constant keep the input to the detector at a constant amplitude to automatically track the frequency of the RF signal

42. An IF filter has a –60 dB bandwidth of 25 kHz and a –3 dB bandwidth of 20 kHz. What is the shape factor value? A. 0.8 B. 1.25 C. 0.5 D. 0.125 43. In high-level AM, the power in the sidebands comes from: A. the modulating amplifier B. the RF amplifier C. the driver stage D. the carrier 44. Basically, sensitivity measures: A. the weakest signal that can be usefully received B. the highest-frequency signal that can be usefully received C. the dynamic range of the audio amplifier D. the ability of the receiver to attenuate noise signals



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45. A collector-modulated class C amplifier has a carrier output power of 120 W and an efficiency of 85%. Calculate the transistor power dissipation with 100% modulation. APRIL 2003 A. 11.76 W B. 31.76 W C. 41.76 W D. 21.76 W 46. Frequency translation is done with a circuit called a A. Mixer B. Multiplier C. Filter D. Summer 47. For a non-ideal amplifier with the following parameters: Input signal = 2 x 10-10W, Input noise = 2 attoWatts, Power gain = 60 dB, Internal noise = 6 pW Determine the noise figure. A. 4 dB B. 3 dB C. 6 dB D. 2 dB 48. For a 10-MHz crystal with a temperature coefficient of +10 Hz/MHz/°C, determine the frequency of operation if the temperature decreases by 5°C. A. 9.95 MHz B. 9.9995 MHz C. 9.995 MHz D. 9.5 MHz 49. An AM transmitter uses high-level modulation. The RF power amplifier draws 12 A from a 22 V supply, putting out a carrier power of 140 watts. What impedance would be seen at the modulation transformer secondary? NOV

2003 A. C.

2.16Ω 1.83Ω

B. D.

0.183Ω 1.56Ω

50. Using high-side injection for a 1-MHz IF, what is the frequency of the local oscillator when the receiver is tuned to 5 MHz? A. 4 MHz B. 5 MHz C. 6 MHz D. 3 MHz 51. Suppressing the audio when no signal is present is called: A. AGC B. squelch C. AFC D. limiting 52. Determine the voltage at the output of a phase comparator with a transfer function of 0.5 V/rad and a phase error of 0.75 rad. A. 0.75 V B. 3.73 V C. 0.373 V D. 0.73 V 53. The power amplifier of an AM transmitter draws 100 watts from the power supply with no modulation. Assuming high-level modulation, how much power does the modulation amplifier deliver for 100% modulation? A. 16.67 watts B. 50 watts C. 33.33 watts D. 66.67 watts


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54. Determine the phase error necessary to produce a VCO frequency shift of 10 kHz for an open loop gain of 40 kHz/rad. A. 25 rad B. 0.125 rad C. 0.25 rad D. 0.5 rad 55. Suppose the bandwidth of a tuned circuit is 10 kHz at 1 MHz. Approximately what bandwidth would you expect it to have at 4 MHz? A. 5 kHz B. 10 kHz C. 40 kHz D. 20 kHz 56. Determine the change in frequency for a VCO with a transfer function of 4 kHz/V and a dc input voltage change of 1.2Vp. A. 4.8 kHz B. 0.48 kHz C. 3.33 kHz D. 0.333 kHz 57. Suppose a receiver uses a 5-MHz IF frequency. Assuming high-side injection, what would be the image frequency if the receiver was tuned to 50 MHz? A. 60 MHz B. 40 MHz C. 55 MHz D. 45 MHz 58. Determine the hold-in range for a PLL with an open loop gain of 25. kHz/rad A. 9.271 kHz B. 39.271 kHz C. 9.1 kHz D. 3.271 kHz 59. A receiver has a sensitivity of 0.6 microV and a blocking dynamic range of 60 dB. What is the strongest signal that can be present along with a 0.6 microV signal without blocking taking place? APRIL 2003 A. 0.06 μV B. 0.6 μV C. 600 μV D. 60 μV 60. Mixing for frequency conversion is the same as A. Linear summing B. Filtering C. AM D. Rectification 61. One input to a conventional AM modulator is a 500 kHz carrier with an amplitude of 20 Vp. The second input is a 10 kHz modulating signal that is sufficient to cause an change in the output wave of +7.5 Vp. Determine the peak amplitude of the USF and LSF voltages. A. 3.75Vp B. 2.75Vp C. 1.75Vp D. 4.75Vp 62. Suppose you have generated a USB SSB signal with a nominal carrier frequency of 10 MHz. What is the minimum frequency the SSB signal can be mixed with so that the output signal has a nominal carrier frequency of 50 MHz? A. 20 MHz B. 10 MHz C. 40 MHz D. 60 MHz 63. Determine the improvement in the noise figure for a receiver with an RF bandwidth equal to 200 kHz and an IF bandwidth equal to 10 kHz. A. 0.13 dB B. 1.3 dB C. 3 dB D. 13 dB



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64. If the final RF amplifier of an AM transmitter is powered by 100 volts DC, what is the maximum collector voltage at 100% modulation? A. 400 volts B. 50 volts C. 200 volts D. 100 volts 65. Suppose the output of a balanced modulator has a center frequency of 10 MHz. The audio modulation frequency range is 1 kHz to 10 kHz. To pass the USB, what should be the center frequency of an ideal crystal filter? A. 9.95 MHz B. 10.005 MHz C. 9.995 MHz D. 10.05 MHz 66. For 54, A. C.

an AM commercial broadcast band receiver with an input filter Q-factor of determine the bandwidth at the low and high end of the spectrum. 1 kHz, 29.630 kHz B. 10 kHz, 29.630 kHz 1 kHz, 2.630 kHz D. 10 kHz, 2.630 kHz

67. For an AM broadcast-band superheterodyne receiver with an RF of 600 kHz and high-side injection, calculate the IFRR for a preselector Q of 100. A. 46.52 dB B. 2.5 dB C. 23.25 dB D. 3.2 dB 68. For the A. C.

a 20 MHz crystal with a temperature coefficient of -8 Hz/MHz/°C, determine frequency of operation if the temperature increases by 20°C. 19.8 MHz B. 19.68 MHz 19.78 MHz D. 19.9968 MHz

69. Image frequencies occur when two signals: A. are transmitted on the same frequency B. enter the mixer, with one being a reflected signal equal to the IF frequency C. enter the mixer, one below and one above the local oscillator by a difference equal to the IF D. enter the mixer, and the difference between the two signals is equal to twice the IF 70. An FM detector that is not sensitive to amplitude variations is: A. Foster-Seeley detector B. a quadrature detector C. a PLL detector D. coherent detector 71. The A. B. C. D.

local oscillator and mixer are combined in one device because: it gives a greater reduction of spurious responses it increases sensitivity it increases selectivity it is cheaper

72. An image must be rejected: A. prior to mixing C. prior to detection

B. D.

prior to IF amplification images cannot be rejected

73. For an AM receiver with a -80 dBm RF input signal level and the following gains and losses, determine the audio signal level Gains: RF = 33 dB, IF = 47 dB, Audio amp = 25 dB Losses: Preselector = 3 dB, Mixer loss = 6 dB, detector loss = 8 dB A. 8 dBm B. 0.8 dBm C. 80 dBm D. 18 dBm


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74. For a citizen band receiver using high-side injection with an RF carrier of 27 MHz, determine the preselector Q required to achieve an IFRR of 40.32 dB. A. 540 B. 120 C. 353 D. 84 75. The A. B. C. D.

typical squelch circuit ______. cuts off an audio amplifier when the carrier is absent cuts off an IF amplifier when the AGC is minimum cuts off an IF amplifier when the AGC is maximum eliminates the RF interference when the signal is weak

76. When a super heterodyne receive with an IF of 450 kHz is tuned to a signal at 1200 kHz, the image frequency will be ______. A. 1650 kHz B. 900 kHz C. 2100 kHz D. 750 kHz 77. Which of the following circuit cannot be used to demodulate SSB? A. BFO B. Phase discriminator C. Product detector D. Balanced modulator 78. A phase-locked loop has a VCO with a free-running frequency of 12 MHz. As the frequency of the reference input is gradually raised from zero, the loop locks at 10 MHz and comes out of lock again at 16 MHz. Find the capture range and lock range. NOV 2003 & NOV 2004 A. CR=12 MHz, LR=16 MHz B. CR=8 MHz, LR=12 MHz C. CR=4 MHz, LR=8 MHz D. CR=6 MHz, LR=10 MHz 79. A quartz watch is guaranteed accurate to 15 seconds per month. Assuming a month has 30 days, calculate the accuracy of the crystal oscillator in the watch in parts per million. A. 3.56 ppm B. 12.71 ppm C. 5.79 ppm D. 61.40 ppm 80. Image frequency rejection mainly depends upon ______. A. detector B. IF C. mixer D. RF 81. A crystal oscillator is accurate within 0.0005%. How far off frequency its output be at 27MHz? A. 125 Hz B. 115 Hz C. 155 Hz D. 135 Hz 82. Which would be best for DSBSC: A. carrier detection C. envelope detection

B. D.

coherent detection ratio detection

83. With mixing: A. the carrier frequency can be raised B. the carrier frequency can be lowered C. the carrier frequency can be changed to any required value D. the deviation is altered


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84. A phase-locked loop has a VCO with a free running frequency of 16 MHz. As the frequency of the reference input is gradually raised from zero, the loop locks at 12 MHz and comes out of lock again at 20 MHz. Find the capture range and lock range. A. 8 MHz, 8 MHz B. 4 MHz, 8 MHz C. 2 MHz, 4 MHz D. 8 MHz, 4 MHz 85. A quartz watch is guaranteed accurate to 125 seconds per year. Assuming a month has 30 days, calculate the accuracy of the crystal oscillator in the watch in ppm. A. 8.12 ppm B. 3.23 ppm C. 1.24 ppm D. 4.02 ppm 86. A tuned circuit tunes the AM radio broadcast band (540 to 1700 kHz). If its bandwidth is 10 kHz at 540 kHz, what is it at 1700 kHz? A. 17.7 kHz B. 1700 kHz C. 10.5 kHz D. 21.7 kHz 87. A receiver has a sensitivity of 0.5 μV and a blocking dynamic range of 70 dB. What is the strongest signal that can be present along with a 0.5 μV signal without blocking taking place? A. 1.58 mV B. 15.8 mV C. 158 mV D. 0.158 mV 88. Image frequency problems would be reduced by: A. having an IF amplifier with the proper shape factor B. having a wideband RF amplifier after the mixer C. having a narrowband RF amplifier before the mixer D. inserting a low-pass filter before the mixer stage 89. A SSB transmitter is connected to a 50-ohm antenna. If the peak output voltage of the transmitter is 20 volts, what is the PEP? A. 6 watts B. 4 watts C. 8 watts D. 2 watts 90. The 10 MHz crystal oscillator in a PLL frequency synthesizer has been calibrated to a reference standard known to be accurate to 5 ppm. The oscillator is guaranteed to drift no more 10 ppm per month. Suppose this oscillator is used in a simple single-loop PLL to generate a frequency of 45 MHz. What are the maximum and minimum values that the output frequency could have after one year? A. 10.0025 MHz, 9.99875 MHz B. 9.0025 MHz, 10.99875 MHz C. 10.25 MHz, 9.75 MHz D. 9.25 MHz, 10.75 MHz

Section 6 Acoustics Fundamentals w/ Basic Illumination Section 7 Television Fundamentals Section 8 AM, FM and TV Broadcasting Standards Section 9 Microphones and Loudspeakers

Acoustics & Broadcasting


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Section

Acoustics

6

Fundamentals


DEFINITION. Acoustics is defined as the generation, transmission, and reception of energy in the form of vibrational waves in matter. Sound is a physical wave, or a mechanical vibration, or simply a series of pressure variation, in an elastic medium.

DEFINITION.

A. .MAIN SUB-DISCIPLINE OF ACOUSTICS. ª

Aeroacoustics -- is the study of aerodynamic sound, generated when a fluid flow interacts with a solid surface or with another flow. It has particular application to aeronautics, examples being the study of sound made by jets and the physics of shock waves (sonic booms).

ª

Architectural acoustics -- is the study of how sound and buildings interact including the behavior of sound in concert halls and auditoriums but also in office buildings, factories and homes.

ª

Bioacoustics -- is the study of the use of sound by animals such as whales, dolphins and bats.

ª

Biomedical acoustics -- is the study of the use of sound in medicine, for example the use of ultrasound for diagnostic and therapeutic purposes.

ª

Loudspeaker acoustics -- is an engineering discipline behind the design of the loudspeaker.

ª

Psychoacoustics -- is the study of how people react to sound, hearing, perception, and localization.

ª

Psychological Acoustics -- is the study of the mechanical, electrical and biochemical function of hearing in living organisms.

ª

Physical acoustics -- is the study of the detailed interaction of sound with materials and fluids and includes, for example, sonoluminescence (the emission of light by bubbles in a liquid excited by sound) and thermoacoustics (the interaction of sound and heat).

ª

Vibration acoustics -- Structural Acoustics and Vibration is the study

of how sound and mechanical structures interact; for example, the transmission of sound through walls and the radiation of sound from vehicle panels.


ACOUSTICS FUNDAMENTALS

2-2

ª

Wolffian Acoustics -- is the study of salient features of pediatric ultrasound insofar as it reviews technologic factors, technique, and the normal anatomy used to evaluate the pediatric tract for abnormality.

ª

Musical acoustics -- is the study of the physics of musical instruments.

ª

Underwater acoustics -- is the study of the propagation of sound in the oceans. Closely associated with sonar research and development.

ª

Acoustic engineering -- is the study of how sound is generated and measured by loudspeakers, microphones, sonar projectors, hydrophones, ultrasonic transducers, sensors.

B. .THEORY OF SOUND WAVES. Sound is an alteration in pressure, stress, particle displacement, or particle velocity that is propagated in an elastic medium. General Subdivision i. Infrasonic -- below 20 Hz ii. Sonic -- 20 Hz to 20,000 Hz iii. Ultrasonic -- above 20,000 Hz 1.

Velocity of Propagation Sound travels at different velocities depending upon the medium. Since sound travels not only in air but also through parts of the structure it is of interest to know the velocities in other media. i.

In a gas medium

s=

γPO ρO

where : PO = is the steady pressure of the gas in N/m 2 ρ O = is the steady or average density of the gas in kg/m 3 γ = ratio of the specific heat of constant pressure to that of the constant volume

ii.

In normal dry air

s = 331.45 + 0.607TC

for TC ≤ 20°C


TK 273

s = 331.45

2-3

for TC > 20°C

Sample Problem:

Determine the speed of sound in air at STP. (Used γ = 1.4 for air, ρ = 1.29 kg/m3, Ρ = 1 atm = 1.013 x 105 Pa)

Solution: s=

γP0 = ρ0

N ⎤ ⎡ 1.4⎢1.013 x 105 2 ⎥ m ⎦ ⎣ = 332 m kg s 1.29 3 m

Sample Problem:

A camera focuses using ultrasound. If it focuses precisely at 20° C, how far off (%) will it be at 0° C?

Solution: s = 331.45 + 0.607(0o ) = 331.45 m ⇒ for 0 degrees s = 331.45 + 0.607(20o ) = 343.59 m ⇒ for 20 degrees 343.59 − 331.45 %100 343.59 = 3.53%

%miss =

Answer : 3.53%

B. .CLASSES OF SPEECH. 1.

Voiced Sounds Voiced Sounds are produced when our vocal chords vibrate as a result of our lungs generating sufficient pressure to open our vocal folds. For most person, the vibration frequency of their folds is within 50 to 400 Hz range and is referred to as the pitch frequency component of voice.



2-4

2.

Unvoiced Sounds Unvoiced Sounds refer to the period of time when our vocal folds are normally open, allowing air to pass from our lungs freely into the rest of our vocal tract. Examples of unvoiced sounds include s, f, and sh, generated by constricting the vocal tract by slightly closing our lips.

3.

Plosive Sound Plosive Sound results from the complete closure of our vocal tract, resulting in air pressure becoming extremely high behind the closure.

C. .ATTRIBUTES OF SOUND. 1.

Pitch Pitch (music), highness or lowness of a musical tone as determined by the rapidity of the vibrations producing it.

The qualitative perception of pitch is closely related to the measurable frequency of a periodic pressure oscillation associated with sound waves. 2.

Timbre Timbre is a more elusive qualitative attribute of sound.

Timber is a quality of a sound that distinguishes it from others of the same pitch or volume. Timbre can be considered the texture or color of a sound. Quantitatively, it is related to waves with different shapes. 3.

Loudness Loudness is the human impression of the strength of a sound. The loudness of a noise does not necessarily correlate with its sound level.

Loudness level of any sound, in phons, is the decibel level of an equally loud 1 kHz tone, heard binaurally by an ontologically normal listener.

Phon = 40 + 10 log2 Sone A 10dB sound level increase is perceived to be about "twice as loud" in many cases. The sone is a unit of comparative loudness with 0.5 sone 1 sone 2 sones 4 sones

= = = =

30 40 50 60

phons, phons phons phons

The sone "10dB rule" is inappropriate at very low and high sound levels where human subjective perception does not follow it.


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Duration This is probably the quality that is simplest to relate to a measurable quantity: the duration of a sound is the time interval between its beginning and end points. Summary

Sound Quality

Measurable Quantity

Pitch Timbre loudness duration

frequency of sound waves pressure oscillation waveform amplitude of pressure oscillation time interval

Sample Problem: Two public address speaker systems are being compared, and one is perceived to be 64 times louder than the other. What will be the difference in sound intensity levels between the two when measured by a dB-meter? Solution: Remember: A 10dB sound level increase is perceived to be about "twice as loud" ln 64 64 = 2 α ⇒ α = = 6 ln 2 The intensities differs by 6 factors of 10, ΔSIL = α x (10 dB rule) = 6 x 10 = 60 dB

Sample Problem: Given the following data:

Compute for

SIL (dB) 60 60 60 60 a. b.

Loudness (sone) 3.2 5.4 5.9 4.7

Loudness Level (phon) 55 62 63 60

Total Loudness (L) Total Loudness Level (L

)

Solution: L = 3.2 + 5.4 + 5.9 + 4.7 = 19.2 sones

L = 40 + 10 log2 19.2 = 82.63 phons



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D. .FREQUENCY PARAMETERS. 1.

Harmonics Harmonics is defined as the integral multiples of the fundamental.

Nth Harmonics = nxfF

2.

fF = fundamenta l frequency in Hz

Overtone Musical term for harmonics.

N th Overtone = (n + 1)xfF

3.

Octave Frequency interval of 2.

N th Octave = 2 n xfF

4.

Decade Frequency interval of 10.

Nth Decade = 10n xfF

Sample Problem: A flute with all of the holes closed can be considered as a tube with both ends open. It has a fundamental frequency of 261.6 Hz (which is middle C). Calculate the 3rd harmonics, 3rd overtones, 3rd octaves, and 3rd decades of this fundamental generated. Solution:

3rd Harmonics = 3 x (261.6 Hz) ⇒ 784.8 Hz

3rd Overtones = (3 + 1) x (261.6 Hz) ⇒ 1.0464 kHz 3rd Octaves = 23 x (261.6 Hz) ⇒ 2.0928 kHz

3rd Decades = 103 x (261.6 Hz) ⇒ 261.6 kHz


2-7

E. .SOUND FIELDS. 1.

Near Field The area where the direct sound dominates and the sound pressure may vary significantly with just small changes in position.

2.

Far Field The area beyond the near field. This is made up of two sections; i.

Free field Free field where the direct sound still dominates and the sound pressure level decreases 6 dB for each doubling of distance.

ii.

Reverberant Field Reverberant field where the reflected sound adds to the direct sound and the decrease per doubling of distance will be less than 6 dB.

F. .LEVELS OF SOUND.

1.

Sound Power (W) and Sound Power Level (PWL) Sound power is the amount of energy emitted by a sound source per unit time.

Sound Power Level (PWL) is often quoted on machinery to indicate the total sound energy radiated per second. It is quoted in decibels with respect to the reference power level. i.

General Solution

PWL dB = 10 log

W W0

W = Acoustic power in W W0 = Reference acoustic power = 10 − 12 Watts



2-8

In dB

PWL dB = 10 log W + 120 dB

ii.

For Multiple Sources but different PWL

PWL T = 10 log

iii.

WT W0

WT = W1 + W2 + W3 … + WN

For Multiple Sources having the Same PWL

PWL T = PWL + 10 logn

Since : WT = nWN

Sample Problem: Determine the total PWL of 3 motors with a radiated acoustic power of 25 mW, 40 mW, and 85 mW respectively. 85 mW

Solution: 25 mW

40 mW

WT = W1 + W2 + W3 … + Wn = 25 mW + 40 mW + 85 mW = 150 mW

PWL T = 10 log = 10 log

WT W0 150 mW

10−12 W = 111.76 dB

2-9


Sound Intensity (I) and Sound Intensity Level (SIL) Sound intensity in a specified direction at a point in a sound field is defined as the average rate of flow of sound energy through a unit area normal to this direction at the point considered.

Sound Intensity Level is the ratio between a given intensity and reference intensity. i.

General Solution

SIL dB = 10 log

I I0

I0 = 10− 12

W m2

In dB

SIL dB = 10 logI + 120 dB

ii.

For Multiple Sources but different SIL

SIL T = 10 log

ii.

IT I0

I T = I1 + I2 + I3 … + IN

For Multiple Sources having the Same SIL

SIL T = SIL + 10 logn

Since : IT = nIN



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Sample Problem: You are at a party and talking with a group of people. They all produce sound levels of the same magnitude at your position. The combined level when all four talks at once is 70 dB. a. What is the sound level from one person? b. What is the sound intensity from one person? c. If two additional people joined the conversation with an individual level of 68 dB each, compute for the new combined intensity level. Solution: a.) SIL T = SIL + 10 log n

b.) SIL = 10 log

70 = SIL + 10 log 4 SIL = 64 dB

64 = 10 log

I I0 I 10

I = 2.51

−12

W m2

μW m2

c.) IT = I(4 people) + I(2 people) = 10

μW m2

= 22.59

3.

+ 12.59 μW m2

μW m2

⇒ SIL T = 73.54 dB

Sound Pressure (P) and Sound Pressure Level (SPL) Sound Pressure is the root mean square of the instantaneous sound pressures in a stated frequency band during a specified time intervals, unless another time-averaging process is indicated. i.

General Solution P0 = reference pressure

SPL dB = 20 log

P P0

= 20 μPa = 0 .0002 μbar = 2 .089

In dB

SPL dB = 20 logP + k

lb ft 2

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Self-Sufficient Guide to ECE by JASON AMPOLOQUIO ii.

For Multiple Sources But Different SPL

SIL T = 20 log

iii.

PT P0

PT = P12 + P22 + P32 … + PN2

For Multiple Sources Having The Same SPL

SPL T = SPL + 10 logn

Since : PT = P n

Sample Problem:

Calculate the effective SPL for a telephone ring, an operating vacuum cleaner, and an air conditioner with an individual SPL of 82 dB, 76 dB, and 80 dB respectively relative to 20 μPa.

Solution: Telephone Ring

Vacuum Cleaner

Air conditioner

P SPL1 = 20 log 1 P0

P SPL2 = 20 log 2 P0

SPL 3 = 20 log

P1 20 μPa P1 = 251 .78 mPa

82 = 20 log

SPL T = 20 log = 20 log

P2 20 μPa P2 = 126 .2 mPa

76 = 20 log

P3 P0

P3 20 μPa P3 = 200 mPa

80 = 20 log

P12 + P22 + P32 PT = 20 log P0 P0 (1 x 10−3 ) 251.782 + 126.22 + 2002 = 84.75 dBSPL 20 μPa

Sample Problem: APRIL 2004

If the RMS sound pressure is 5 lb/ft2, what is the sound pressure level?

Solution: SPL = 20 log

P 5 = 20 log = 7.58 dB Po 2.089



2-12

Read it till it Hertz…jma To calculate Sound Pressure Level (SPL) in dB for several units of sound pressure just add the following constant to their corresponding dB value.

Given Unit

To dB

Atmosphere (Standard) Atmosphere (Technical)

4.

20log P + 194.1 20log P + 193.8

lb/in2

20log P + 170.8

Torr

20log P + 136.5

millimeter of Mercury

20log P + 136.5

Pascal

20log P + 94

Microbar

20log P + 74

Relationship between Sound Intensity (I) and Sound Pressure (P) As most measurements of sound are in terms of sound pressure (P), it is useful to know the relationship between sound intensity and sound pressure:

i.

General Solution where :

I=

ii.

2

P ρc

I = is the sound intensity in W/m 2 P = is the sound pressure in Pa ρ = is the density of the medium in kg/m 3 c = is the speed of sound in m/s

For dry air at 17°C (ρ=1.2kg/m3 , c=341.7m/s)

I=

P2 410 rayls


For Your Information… The dB scale for SPL and SIL are equal thus the dB value of the two can be used interchangeably but always keep in mind that the actual intensity and actual pressure corresponding to a particular dB level are completely different in magnitude and units.

Sample Problem:

What is the intensity of sound whose RMS pressure is 81.2 x10-3N/m2?

Solution: I= =

P2 ⇒ c = 331,45 + 0.607(25°C) ρc (81.2 mPa)2 W = 15.85 x 10−6 2 1.2 x 346.625 m

SIL = 10 log

5.

15.85 x 10−6 10−12

= 72 dBSIL

Relationship between Sound Intensity (I) and Sound Power (W)

i.

General Solution

WQ I= 4 πr 2

where: W = sound power in W Q = Directivity factor r = distance from the source in m

In dB

PWL dB = SIL dB + 20 logrm − QdB + 11


2-13


2-14

ii.

For Airborne Sound

I=

W 4 πr 2

where : Q = 1 PWL = 10 log W SIL = 10 log I

In dB

PWL dB = SIL dB + 20 log rm + 11


The sound power level of a certain jet plane flying at a height of 1 km is 160 dB. Find the maximum sound pressure level on the ground directly below the flight path assuming that the aircraft radiates sound equally in all directions.

Solution: 1 km

PWL dB = SIL dB + 20 log rm + 11 ⇒ since SIL dB = SPL dB SPL dB = PWL dB − 20 log rm − 11 = 160 − 20 log 1000 − 11 = 89 dB


2-15

iii. For Ground-originating Sound

where : Q =2

W I= 2 πr 2

PWL = 10 log W SIL = 10 log I

In dB

PWL dB = SIL dB + 20 logrm + 8

Sample Problem:

A car horn outdoors produces a sound intensity level of 90 dB at 10 m away. At this distance, what is the sound power in watt?

Solution: SIL = 90 dB

10 m

PWL dB = SIL dB + 20 log rm + 8 ⇒ since SILdB = SIL dB PWL dB = SPL dB + 20 log rm + 8 = 90 − 20 log 10 + 8 = 118 dBPWL ⇒ W = 0.63 W



2-16

iv. For enclose space with reverberant field a.

General Solution

I T = ID + IR

b.

For sound source with directional characteristics (Small enclosure)

4⎞ ⎛ Q SPL dB = PWL dB + 10 log ⎜ + ⎟+K 2 R⎠ ⎝ 4 πr c.

For sound source with no directivity (Large space)

SPL dB = PWL dB − 20 logr − 10 logR − 5 + K

where : R = S = α = K = =

Sα ⇒ room constant 1− α Room surface area absorption coefficien t 0 . 2 dB ⇒ metric 10.5 dB ⇒ English

I T = total sound intensity in ID = IR =

WQ

W m2

⇒ direct sound intensity

4 πr 2 4W ⇒ R

reverberan t sound intensity

G. .FUNDAMENTALS OF ROOM ACOUSTICS. Room acoustics is concerned with the behavior of sound within an enclosed space with a view to obtaining the optimum effects on the occupants. Reverberation Time (RT60) Reverberation time is the time required for the mean square sound pressure of a given frequency in an enclosure, initially in a steady state, to decay after the source is stopped, to 60 dB or one-millionth of its initial value. 1.

Stephen and Bate Equation The Stephen and Bate equation is used for computing the ideal reverberation time in second.

2-17


Sabine Equation The equation governing the decay of uniformly diffuse sound in a live room for an average absorption less than or equal to 0.2 Metric System

RT60 = 0.161

V A

where: V = Room volume A = room total absorption ⇒ A = Sα

English system

RT60 = 0.049

V A

S = room total surface area α = ave. room absorption coefficient

Past Board Problem:

A church has an interval volume of 90.05 ft3 (2,550 m3). When it contains 2,000 customary sabins of absorption (186 metric sabins), what will be its reverberation time in seconds.

Solution: 0.16V A 0.16(2,550) = = 2.19 s 186

RT60 =

3.

RT60 = =

0.049V A 0.049(90.05) 2000 x10−3

= 2.19 s

Norris-Eyring Equation The equation governing the decay of uniformly diffuse sound in a live room for an average absorption greater than 0.2 Norris-Eyring Equation is based on the mean free path between reflections. Metric System

RT60 = 0.161

V −S ln(1 − α)

English system

RT60

V = 0.049 − S ln(1 − α)

where: V = Room volume A = room total absorption ⇒ A = Sα S = room total surface area α = ave. room absorption coefficient



2-18

Relation between Sabine & Norris-Eyring Equation

RT60 (Norris − Eyring) =

RT60 (Sabine) − ln(1 − α)

Sample Problem:

Calculate the Norris-Eyring reverberation time of uniformly diffuse sound in a live room if the average absorption is 0.8 and the Sabine reverberation time is 1.75 s.

Solution: RT60(Norris − Eyring) =

RT60(Sabine) 1.75 = = 1.087 s − ln(1 − α) − ln(1 − 0.8)

H. .NOISE ANNOYANCE. 1.

Noisiness Defined as the subjective impression of the unwantedness of a sound or annoyance.

2.

Noy A subjective unit of noisiness. A sound of 2 noys is twice as noisy as a sound of 1 noy and half as noisy as a sound of 4 noys.


Human listeners can detect the difference between two sound sources that are placed as little as three degrees apart, about the width of a person at 10 meters.

ª

The primary advantage of having two ears is the ability to identify the direction of the sound.

Self-Sufficient Guide to ECE by JASON AMPOLOQUIO I.

.NOISE-INDUCED HEARING LOSS. 1.

2 Main Types of Deafness i.

ii.

J.

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Nerve deafness Caused by damage or degeneration of hair cells in organ of Corti in the cochlea (inner ear). Such hearing loss is typically uneven, and there is greater loss of higher frequencies than lower frequencies.

This is the type of hearing loss normally found with aging and once degeneration has occurred it usually cannot be remedied.

Conduction deafness Caused by some problem in the outer or middle ear that affects transmission of sound waves to the inner ear. Conduction deafness is more even across frequencies and doesn't result in complete hearing loss. Conduction deafness is only partial hearing loss because sound waves can be transmitted to inner ear by conduction through bones of the skull.

.TYPES OF HEARING LOSS. 1.

Presbycusis Hearing loss due to normal aging process. Typically this is greater for men than women.

2.

Sociocusis Hearing loss due to non-occupational noise sources e.g. household noises, TV, road traffic, etc. (criteria for occupational/ non-occupation fuzzy).

3.

Nosocusis Hearing loss due to a pathological condition.

K. .OCCUPATIONAL HEARING LOSS. 1.

Temporary Hearing Loss After exposure to continuous noise of sufficient intensity, some temporary hearing loss which is usually recovered a few hours or days after exposure. This hearing loss is usually measured 2 minutes after the end of exposure and is termed the temporary threshold shift at 2 minutes (TTS2)

2.

Permanent Hearing Loss Repeated exposure to noise of sufficient intensity produces a permanent threshold shift (PTS). This is the same thing as noise induced permanent threshold shift (NIPTS). PTS is usually noticed at



2-20

around 4-KHz and with further exposure, gradually hearing loss spreads over wider frequency range.

L.

.ILLUMINATION BASIC.

Important Terminology 1.

Luminous Intensity Luminous intensity is the amount of luminous flux per unit solid angle in a given direction.

2.

The luminous efficacy The luminous efficacy of a light source is the measure of lightproducing efficiency of the source. It is the ratio of the total luminous flux output to the total input power of the source expressed in lumens/watt.

3.

Incandescence Emission of radiation due to the temperature of the source.

4.

Luminescence Emission of radiation due to causes other than temperature of the source.

5.

The lumen The amount of luminous energy (flux) emanating from one square foot (meter) of surface on the sphere is one lumen. Unit of light quantity. 1 Lumen = 1.5 mW

6.

Fluorescence Luminescence stimulated by radiation, not continuing more than about 10-8 s after the stimulating radiation is cut-off.

2-21

Self-Sufficient Guide to ECE by JASON AMPOLOQUIO Unit Conversion Parameter Illumination Luminance Intensity

Multiply

By

To obtain

ft-candle Lux ft-lambert candela/in2 Candlepower

10.764 0.0929 0.00221 452 1

Lux ft-candle candlela/in2 ft-lambert Candela


A 500 candle power that is located 12 feet from a surface provides how many lux?

Solution: Illu min ation =

Luminous flux in Lumen 500 candle − power = Area (12 ft)2

= 3.47

candle − power ft2

x

1 Candela 10.764 lux x candle − power foot − candle

= 37.375 lux



2-22

I

H

1.

You are at a party and talking with a group of people. They all produce sound levels of the same magnitude at your position. The combined level when all four talks at once is 76 dB. If another group of people join the conversation with an individual level of 68 dB each and produce a total level of 82.12 dB together with the 1st group, determine how many people join the conversation. A. 15 B. 5 C. 30 D. 20

2.

Assume the speed of sound is 1,130 ft/s. What frequency has a wavelength of 1 foot, 1.5 inches? A. 500 Hz B. 1000 Hz C. 1500 Hz D. 2000 Hz

3.

During a fireworks show a rocket explodes. To a listener who is 640 m away, the sound has an intensity of 0.10 mW/m2. What is the intensity for a listener who is 160 m away? A. 5.6 mW/ m2 B. 6.6 mW/ m2 2 C. 1.6 mW/ m D. 3.6 mW/ m2

4.

At room temperature, what is the velocity of sound in meters/seconds? A. 348.03 cm/s B. 341.8 m/s C. 980 cm/s D. 980 m/s

5.

The sound power level of a jet plane flying at a height of 1 km is 160 dB (ref 10 exponent -12 W). What is the maximum sound pressure level on the ground directly below the plane assuming that the aircraft radiates sound equally in all directions? APRIL 1997 A. 59.1 dB B. 69.1 dB C. 79.1 dB D. 89.1 dB

6.

Calculate the velocity of sound in ft/sec. If the temperature is 149 degrees Celsius? A. 1357.03 ft/sec B. 1920.435 ft/sec C. 1530.03 ft/sec D. 1320 ft/sec

7.

A periodic wave passes an observer, who records that there is a time of 0.5 s between crests. A. the frequency is 0.5 Hz B. the speed is 0.5 m/s C. the wavelength is 0.5 m D. the period is 0.5 s


2-23

8.

A method of expressing the amplitude of a complex non-periodic signals such as speech. NOV 1997 A. Frequency B. Pitch C. Wavelength D. Volume

9.

For a harmonic wave of a certain type in a given medium, doubling the frequency has the effect of A. halving the speed B. halving the wavelength C. doubling the amplitude D. doubling the period

10. Quite generally, doubling the amplitude of the wave A. doubles the frequency B. halves the period C. quadruples the energy D. doubles the speed 11. A dB-meter is placed in front of an audio system reads 62 dB. The volume is turned up and the meter now reads 72 dB. Evidently, the sound got A. twice as loud B. half as loud C. 20x louder D. 10x louder 12. A sound that has an intensity level of 100 dB is how many times more intense than a sound of 20 dB? A. 5 B. 8 C. 1000 D. 108 13. ______ is the transmission of sound from one room to an adjacent room through common wall, floors or ceilings. APRIL 1997 A. Reflection B. Flanking transmission C. Refraction D. Reverberation 14. A sound-level meter is placed in front of a loudspeaker of a 60-W audio system reads 70 dB. All else being equal, when placed in front of a 120-W system, the meter will read A. 120 dB B. 140 dB C. 63 dB D. 73 dB 15. ______ is the average rate of transmission of sound energy in a given direction through a cross sectional area of 1square meter at right angle to the direction.

NOV 1996 A. C.

Pressure variation Sound pressure

B. D.

Loudness Sound intensity

16. If one stereo system sound 16x louder than another, the difference in their sound levels is about A. 40 dB B. 30 dB C. 16 dB D. 160 dB 17. A speaker cabinet has an internal volume of 84,950 square cm. It has a port area on the baffle of 3,230 square cm and a baffle thickness of 19 mm. What is the Helmholtz resonance in hertz of this speaker enclosure? MARCH 1996 A. 245 Hz B. 250 Hz C. 260 Hz D. 265 Hz


2-24


18. A person standing at one side of a playing field on a cold winter night emits a brief yell. The short acoustical wavetrains returns 1 s later as an echo having “bounced off” a distant dormitory. Approximately how far away is the building. A. 100 meters B. 165 meters C. 500 meters D. 332 meters 19. Two audio systems at a demonstration are blasting away, with one putting out 10x the acoustic power of the other. What is the difference in their sound levels? A. 10 dB B. 5 dB C. 13 dB D. 16 dB 20. The sound energy per unit area at right angles to the propagation direction per unit time. APRIL 1997 A. Sound stress B. Loudness C. Coherence D. Sound intensity 21. A small but noisy printer produces an acoustic intensity of 560 μW/m2 at a point 5 m away. Approximately what value will a dB-meter read 20 m from the printer? A. 85 dB B. 65 dB C. 75 dB D. 95 dB 22. Calculate the effective SPL for a washing machine, a radio receiver, and a blender with an SPL of 76 dB, 84 dB, and 92 dB respectively relative to 2x10-5 N/m2. A. 63 dBSPL B. 93 dBSPL C. 83 dBSPL D. 73 dBSPL 23. The midrange frequency range of sound is from MARCH 1996 A. 16 to 64 Hz B. 256 to 2048 Hz C. 2048 to 4096 Hz D. 512 to 2048 Hz 24. What is the combined SPL produce by 2 guitars and 4 violins with an individual SIL of 82 dB and 80 dB each? A. 77.35 dBSPL B. 88.56 dBSPL C. 84.64 dBSPL D. 37.71 dBSPL 25. One violin playing produces a sound level of 68 dB at your ears. If a second violin plays, producing the same sound level at your ear, what is the sound level from both violins playing? A. 75 dB B. 71 dB C. 77 dB D. 79 dB 26. Two sound sources produce intensity levels of 50 dB and 60 dB, respectively, at a point. When functioning simultaneously, what is the total SIL? A. 60.4 dB B. 110 dB C. 50.4 dB D. 70 dB 27. An instrument for recording waveform of audio frequency. APRIL 1997 A. Audioscope B. Radioscope C. Phonoscope D. Oscilloscope


2-25

28. The speed of a wave on a stretched string ______ when the tension in it is doubled A. doubles B. increases by a factor of 2 C. decreases by a factor of 2 D. increases by a factor of 1.414 29. What is the increase in sound pressure level in dB, if the pressure is double? A. 2 dB B. 3 dB C. 6 dB D. 4 dB 30. If the RMS sound pressure is 355 lb/ft2, what is the sound pressure level? A. 10 dB B. 45 dB C. 108 dB D. 88 dB 31. One hundred twenty μbars of pressure variation is equal to _____. A. 120 dBSPL B. 41.58 dBSPL C. 57.58 dBSPL D. 115.56 dBSPL 32. Calculate the brightness of a light source that causes 100 lumens on a 1 sq-ft surface with a diffuse reflectance of 60%. A. 6 footlamberts B. 60 footlamberts C. 12 footlamberts D. 120 footlamberts 33. The reverberation time of a 184.2 cubic meters broadcast studio is 0.84 s. Find the absorption effect of the materials used in metric sabines. A. 35.3 B. 379.8 C. 109.6 D. 10.96 34. Lowest frequency produced by a musical instrument. APRIL 1997 A. Midrange B. Fundamental C. Harmonic D. Period 35. A 40-W, 430 mA fluorescent tube produces 3200 lumens. What is the illumination on the floor of a 10-ft2 room assuming 40% overall efficiency and uniform illumination? A. 17.7 lux B. 37.7 lux C. 137.7 lux D. 13.7 lux 36. Tendency of sound energy to spread. APRIL 1997 A. Refraction B. Rarefaction C. Diffraction D. Reflection 37. What is the increase in sound pressure level in dB, if the intensity is doubled? A. 2 dB B. 3 dB C. 6 dB D. 4 dB 38. If the distance between the listener and the source of the sound is decreased to ½ the original amount, the intensity is reduced to _____. A. 2 times as great B. 4 times as great C. 3 times as great D. 5 times as great



2-26

39. If the distance between the listener and the source of the sound is doubled, the intensity is reduced to _____. A. ½ B. 1/8 C. 1/16 D. ¼ 40. A car horn outdoors produces a sound intensity level of 90 dB at 10 ft away. At this distance, what is the sound power in watt? MARCH 1996 A. 0.012 W B. 1.2 W C. 0.12 W D. 12 W 41. What is the sound pressure level of a sound having an RMS pressure of 200 N/m2? A. 140 dB B. 170 dB C. 150 dB D. 160 dB 42. ______ used to measure speech volume. APRIL 1997 A. Speech meter B. Volume meter C. Volume unit meter D. Audio frequency meter 43. What is the intensity of sound whose RMS pressure is 200 N/m2? A. 96.8 W/m2 B. 95.8 W/m2 C. 97.8 W/m2 D. 94.8 W/m2 44. Sound intensity level is _______. APRIL 1997 A. 10 log I/ref/I B. 30 log I/ref C. 10 log I/Iref D. 20 log I/ref 45. Determine the total PWL of 3 motors with a radiated acoustic power of 22 mW, 43 mW, and 79 mW respectively and calculate the SPL 15 m from an observer. A. 116.7 dBPWL, 83.87 dBSPL B. 83.87 dBPWL, 116.7 dBSPL C. 56.53 dBPWL, 111.6 dBSPL D. 111.6 dBPWL, 56.53 dBSPL 46. A sound intensity that could cause painful sensation in the human ear. APRIL

1999 A. C.

Sensation intensity Threshold of pain

B. D.

Hearing threshold Threshold of sense

47. What is the individual PWL of 7 air-conditioned with a total PWL of 116 dB? A. 127. 55 dB B. 107. 55 dB C. 97. 55 dB D. 117. 55 dB 48. The sound power level of a certain jet plane flying at a height of 1 km is 160 dB. Find the maximum intensity of sound on the ground directly below the flight path assuming that the aircraft radiates sound equally in all directions. A. 8128.3 μW/m2 B. 8.1283 μW/m2 C. 794.32 μW/m2 D. 81.283 μW/m2 49. Sound intensity is given as. NOV 1997 A. df/dP B. C. dE/dP D.

dP/dA dA/dP


2-27

50. What is the sound power from a motor car whose SPL at a distance of 7.5 m is 87 dB assuming that it radiates sound uniformly? A. 0.15 W B. 0.18 W C. 0.21 W D. 0.24 W 51. Designates the sensation of low or high in the sense of the base and treble.

NOV 1997 A. C.

Pitch Frequency

B. D.

SPL Intensity

52. A car horn outdoors produces a sound intensity level of 90 dB at 10 m away. At this distance, what is the sound power in watt? A. 0.56 x 10-6 B. 0.315 C. 1.26 D. 0.63 53. A church has an interval volume of 90.05 ft3 (2,550 m3). When it contains 2,000 customary sabins of absorption (186 metric sabins), what will be its reverberation time in seconds. A. 3.0 B. 2.5 C. 2.0 D. 2.2 54. Calculate the total loudness level of sound sources with a combined loudness of 22.5 sones. A. 53.52 phons B. 84.92 phons C. 48.39 phons D. 99.5 phons 55. Noise reduction system used for film sound in movie. NOV 1996 A. dolby B. dBm C. dBa D. dBx


Television fundamentals

2-28

Section

7

Television Fundamentals


Television is a system for transmitting images and sound by converting them into electrical or radio waves which are converted back into images and sound by a receiver.

DEFINITION.

DEFINITION.

Broadcasting means to send out in all directions.

Motion Picture is a series of real or fictional events recorded by a camera and projected onto a screen as a sequence of moving pictures, usually with an accompanying soundtrack.

DEFINITION.


William Crookes in England invented the Crookes tube, which produce cathode rays.

1884

Paul Nipkow in Germany built a mechanical scanning device, the Nipkow disc, a rotating disc with a spiral pattern of holes in it.

1897

Karl Ferdinand Braun, also in Germany, modified the Crookes tube to produce the ancestor of the modern TV receiver picture tube.

1900

The Russian scientist Constantin Perskyi is credited with coining the word "television".

1906

Boris Rosing in Russia began experimenting with the Nipkow disc and cathode-ray tube, eventually succeeding in transmitting some crude TV pictures.

1923

Vladimir Zworykin in the USA invented the first electronic camera tube, the ICONOSCOPE.

1926

John Logie Baird demonstrated a workable TV system (30 lines/frame, 5 frames/s) using mechanical scanning by Nipkow disc.

1928

Baird demonstrated color TV.

1929

The BBC began regular broadcasting using Baird’s system (405 lines/frame, 25 frames/s).

1975

Sony introduced their videocassette tape recorder system, BETAMAX, for domestic viewers.


2-29

1979

Matshusita in Japan developed a pocket-sized flat-screen TV set, using a liquid crystal display (LCD).

1989

Japanese begin broadcasting high-definition television (HDTV).

A. .PICTURE REPRODUCTION. The picture is divided into the elementary areas of black and white. When each picture element is transmitted to the right side of the figure and reproduced in the original position with its shade of black or white, the image is duplicated.

B. .PICTURE QUALITIES. 1.

Brightness The overall or average intensity, which determines the background level in the reproduced picture.

2.

Contrast The difference in intensity between black-and-white parts of the reproduced picture.

3.

Detail The quality of detail, which is also called resolution or definition, depends on the number of picture elements that can be reproduced.

4.

Color Level or Saturation The color information superimposed on a monochrome picture that depends on the amplitude of the 3.58 MHz chrominance signal.

5.

Hue The color of an object that depends on the phase angle of the 3.58 MHz chrominance signal.

6.

Aspect Ratio The ratio of width to height of the picture frame.

a=

WIDTH HEIGHT



2-30

Sample Problem:

Calculate the number of active pixel in a line in an NTSC TV system if 42 lines out of every 525 are blanked out to leave time for the vertical retraces.

Solution: # of pixel = a x no. of active lines 4 = x (525 − 42) 3 = 644 pixels

C. .TABLE OF PICTURE QUALITIES. Quality

Picture

Signal

Range between black and white

Amplitude of ac video signal

Brightness

Background illumination

DC bias on picture tube

Resolution

Sharpness of detail

Frequency response of video signal

Intensity or level of color

Amplitude of 3.58MHz chroma signal

Tint of color

Phase angle of 3.58MHz chroma signal

Contrast

Color saturation Hue


The standard television image has an aspect ratio of 4:3.

ª

High-Definition TV (HDTV) has an aspect ratio of 16:9.

ª

A motion picture has an aspect ratio of 25:9.

ª

Charge Coupled Device (CCDs) used for scientific applications often have an aspect ratio of 1:1.


2-31

D. .PHOTO-ELECTRIC EFFECT IN CAMERA TUBES. The liberation of an electric charge by electromagnetic radiation incident on a substance; includes photoemission, photoconduction and photovoltaic effect. 1.

Photoconduction In photoconduction, the conductance or resistance is change; more light decreases the resistance.

2.

Photoemission In photoemission, electron are emitted when light strikes the surface; more light produce more electrons

For Your Information… The image orthicon operates by photoemission, while the vidicon and plumbicon depend on photoconduction to produce the required camera signal.

E. .TYPES OF CAMERA TUBES. 1.

Image Orthicon -- Consist of three main sections: the image section, scanning section, and the electron multiplier.

2.

Vidicon -- The vidicon consists of a glass envelope with an optically

flat faceplate at the end to receive the light input. In this basic camera tube, the photosensitive target, or image plate, is made of antimony trisulfide.

3.

Plumbicon -- This name is a trademark of N.V. Philips. The camera tube is similar to the basic vidicon, but the image plate of plumbicon is made of lead oxide (Pbo).

4.

Saticon -- This name is a trademark of Hitachi Ltd. The image plate is made of selenium, arsenic and tellurium.

5.

Silicon Dioxide Vidicon -- A silicon semiconductor junction is used for the target material in the silicon vidicon.

6.

Chalnicon -- This name is a trademark of Toshiba Electric Co. Ltd. The target is a complex multi-layer arrangement consisting of tin oxide, cadmium selenide, and arsenic trisulfide

7.

Newvicon -- This name is a trademark of Matsushita Electric. The target is made of an amorphous zinc-selenium layer backed by antimony trisulfide.



2-32

F. .SUMMARY OF CAMERA TUBES. Size

Image Plate

Notes

Life (Hrs)

Image Orthicon

Length: 15-20 in Diameter: 3-4 in

Photocathode

High quality, high sensitivity

1,500 to 6,000

Vidicon

Length: 5-8 in Diameter: 0.6-1.6 in

Selenium Photoconductor

Simple construction, used for film pickup

5,000 to 20,000

Plumbicon

Length: 8 in Diameter: 1.2 in

PbO Photoconductor

Simple construction, low lag, sensitivity low for red light

2,000 to 3,000

Silicon Diode Array Vidicon

Length: 6 in Diameter: 1 in

Silicon diode array

Low lag, sensitive to red and infrared light

2,000 to 3,000

Type

G. .STANDARD SCANNING PATTERN. 1.

Horizontal Scanning - The linear rise of current in the horizontal deflection coils deflects the beam across the screen with continuous, uniform motion for the trace from left to right.

2.

Vertical Scanning - The sawtooth current in the vertical deflection coils moves the electron beam from top to bottom of the raster.

H. .DIRECTION FOR TRACE & RETRACE. s

HORIZONTAL

VERTICAL


2-33

INTERLACE SCANNING… When an image is scanned line by line from top to bottom, the top of the image on the screen will begin to fade by the time the electron beam reaches the bottom of the screen. With interlaced scanning, odd-numbered lines are scanned first, and the remaining even-numbered lines are scanned next. A full image is still produced 30 times a second, but the electron beam travels from the top of the screen to the bottom of the screen twice for every time a full image is produced.

Concept of Progressive Scanning The horizontal scanning lines are interlace in the television system in order to provide two views of the images for each picture frame. All the odd lines are scanned, omitting the even lines. Then the even lines are scanned to complete the whole frame without losing any picture information.

Concept of Interlace Scanning The horizontal scanning lines are interlace in the are scanned, omitting the even lines. Then the television system in order to provide two views of even lines are scanned to complete the whole frame the images for each picture frame. All the odd lines without losing any picture information. *

*Read first the odd lines and then the even lines.



2-34

I.

.PICTURE DEFINITION.

Parameter Width of Line

General Solution

ω=

Hp

Hp N

a=

Number of Horizontal Visible Lines

NV = N − NS

(# of Pixels in the Vertical Direction)

Vp

=

NH NV

Aspect Ratio

NV = 0.7(N − NS ) If Kell factor is considered

Number of Vertical Visible Lines

NH = a x NV

(# of Pixels in the Horizontal Direction)

NH = BWvideo x 2TH If Kell factor is considered NH NL = 0.835

Total Number of Pixel per Horizontal Line

Total Number of Pixel in a Frame

NL = NH If Kell factor is considered NP = NL xNV NP = NH xNV If Kell factor is considered

2-35


BWvideo = Video Bandwidth

NL xfH 2

NV 80 If Kell factor is considered BWvideo =

BWvideo = 0.35NH xfH If Kell factor is considered where: N = Total number of scanlines period per frame period (525 for NTSC ) NS = Number of scanlines suppressed during retraced ( 40 to 42 ) ω = width of each scan line Vp = Vertical dimension of the viewing area of the CRT Hp = Horizontal dimension of the viewing area of the CRT a = Aspect ratio ( 4/3 ) N V = Vertical resolution in lines/details or pixels NH = Horizontal resolution in lines/details or pixels TH = Horizontal Tra ce period (53 to 53.5 μ sec ) fH = 15, 750 Hz

Sample Problem:

Find the picture height and width in terms of number of pixels also find the total number of pixels and the highest video bandwidth (assuming NTSC system).

Solution: For number of active lines NV = N − NS = 525 − 40 = 485 lines For number of active pixels NH = a x NV 4 x 485 3 = 647 lines =

For total number of pixels NH 647 NL = = 0.835 0.835 = 775 pixels For the video bandwidth BWV = 0.35 x fH x NL = 0.35 x 15,750 Hz x 775 pixels = 4.27 MHz



2-36

Parameter Time to Scan a Pixel

General Solution

t=

Time to Scan N Pixels in a Line

53.5 μs NH

t=

1 BWv

tN = N x t

ECE Board Exam: April 2003

Calculate the amount of time to scan 20 pixels.

Solution: NH = BW x 2T = 4MHz x 2(53.5 μs) = 428 pixels tN = (# of pixels) x (time to scan a pixel) = (20 pixels) x

53 .5μs 428 pixels

= 2.5μs

J.

.DETAILS OF HORIZONTAL BLANKING.

Period

Time(μs)

Total Line (H) H blanking H sync pulse Front Porch Back Porch Visible Line time

63.5 9.5 to 11.5 (10.5 typical) 4.75 + 0.5 1.27 3.81 52 to 54

Self-Sufficient Guide to ECE by JASON AMPOLOQUIO K. .DETAILS OF VERTICAL BLANKING.

PERIOD Total Field (V) V blanking V sync pulse Total of six V sync pulses

16.7 ms 0.05 to 0.08 V 0.5H= 31.76 μs 3H= 190.5 μs

Each Equalizing pulse

0.04H= 2.54 μs

Each Serration pulse

0.04H= 2.54 μs

Visible Field time

L.

TIME

0.92 to 0.95 V

.COMPOSITE VIDEO SIGNAL.


2-37


2-38

Signal

IRE Level

Carrier Level

-40

100 %

0

75% + 2.5%

Black Level

7.5

67.5%

White Level

100

12.5% + 2.5%

Tip Of Sync Blanking Level

Luminance Level to IRE Unit Conversion

IRE = 7.5 + A x (A max − A min )

where : A = Luminance of a video signal

Sample Problem:

Find the equivalent IRE units of a video signal which has 80% of the maximum luminance level.

Solution: IRE = 7.5 + 80% (100 − 7.5) = 81.5 IRE units

M. .TELEVISION CHANNEL ALLOCATION. Channel Number 1 2-4 5-6 7-13 14-83

Low Band VHF

Frequency Band (MHz) 48-54 54-72 76-88

High Band VHF

174-210

UHF

470-884


2-39

N. .SPECTRUM OF STANDARD TV BROADCAST CHANNEL. 1.

Monochrome TV

2.

Color TV

Sample Problem:

Calculate the picture carrier, exact color subcarrier, and sound carrier of the ff channels: a. Ch 2 b. Ch 6

Solution:

Channel 2 (54 to 60 MHz) Channel 6 (82 to 88 MHz) P = 54 + 1.25 = 55.25 MHz P = 82 + 1.25 = 83.25 MHz C = P + 3.579545 = 58.829545 MHz C = P + 3.579545 = 86.829545 MHz S = 60 − 0.25 = 57.75 MHz S = 88 − 0.25 = 87.75 MHz



2-40


The tolerance for picture carrier frequency is +1000 Hz.

ª

The sound carrier frequency must be 4.5 MHz + 1000 Hz above the picture carrier frequency.

ª

The chroma subcarrier frequency is 3.579545 MHz + 0 Hz.

ª

The exact carrier frequencies for different stations on the same channel are offset by +10 kHz or -10 kHz.

O. .FREQUENCY INVERSION OF I.F. OUTPUT.

IF Signal Frequencies

Parameter

Frequency (MHz)

Picture IF

45.75

1st Chrominance IF

42.17

st

1 Sound IF nd

2

Chrominance IF

2nd Sound IF

41.25 3.58 4.50


2-41

P. .RECEIVER TROUBLES. Troubles

Caused

No raster, with normal sound No picture and no sound, with normal raster No picture, with normal raster and normal sound No sound, with normal raster and normal picture Only a horizontal line on the screen No raster and no sound

Since the sound is normal, the trouble of no brightness is usually the result of no anode voltage from the high voltage supply. This trouble is in the signal circuits, before the sound take-off point, because both the picture and sounds are affected. The most probable section with trouble is the video amplifier section. The trouble must be in the sound circuits, after the sound take-off points, because only the sound is affected. The trouble must be in the vertical deflection oscillator and the vertical output stage. This trouble means the raster and the signal circuits are not operating.

Ghosting (Double-image distortion) Noise in TV receiver that is cause by two signals arriving at the receiver at two different times.

Sample Problem:

Determine the difference in path lengths between the direct and multipath signal which produces a TV ghost displaced by 2 centimeters from the direct image on a 19-inch screen.

Solution: For the picture width a=

W 4 3 = ; H = W; Diagonal = H2 + W2 H 3 4 2

9 25 2 ⎛3 ⎞ 192 = ⎜ W ⎟ + W2 = W2 + W2 = W 16 16 ⎝4 ⎠ 4 4 W = D = (19) = 15.2 inches (38.6 cm) 5 5 For the 2cm delay Ghost displacement 2 cm x visible trace = x 53 μ sec Width 38.6 cm = 2.7 μ sec

td =

For the path length difference d = Vp x t d = (3 x 108 m / s) x 2.75 μ sec = 825 meters



2-42

Q. .Y, I, Q MATRIX WEIGHTINGS.

Color Video Signals ª

Luminance The luminance signal contains all information required to construct a black and white picture from the signal.

ª

In-Phase/ Quadrature Chrominance Chrominance is a combination of both hue and saturation.

Y = 0.30R + 0.59G + 0.11B I = 0.60R − 0.28G − 0.32B Q = 0.21R − 0.52G + 0.31B Primary Color Signals

R=

0.62Q + 0.95I + Y

G = −0.64Q − 0.28I + Y B = 1.73Q − 1.11I + Y

where: Y = Luminance signal I = In-phase signal Q = Quadrature signal

R = Red signal G = Green signal B = Blue signal


Sample Problem:

2-43

An RGB video signal has normalized value of R=0.2, G=0.4, B=0.8. Find the value of Y, I, Q, and C.

Solution:

Luminance signal Y = 0.30R + 0.59G + 0.11B = (0.30 x 0.2) + (0.59 x 0.4) + (0.11 x 0.8) = 0.384 In-phase signal I = 0.60R − 0.28G − 0.32B = (0.60 x 0.2) − (0.28 x 0.4) − (0.32 x 0.8) = −0.248 Quadrature signal Q = 0.21R − 0.52G + 0.31B = (0.21 x 0.2) − (0.52 x 0.4) + (0.31 x 0.8)

= 0.082


Note that the components (negative) of -0.28G and -0.32B total -0.60, which equals the positive value of 0.60R.

These values are chosen to make the amplitude of the I-video signal become zero (0) for white. ª

Note that the components (posititive) of 0.21G and 0.31B total 0.52, to equal the negative component of -0.52G.

These values are chosen to make the amplitude of the Q-video signal become zero (0) for white. ª

Both the I and Q signals are zero for white, since there is no chrominance information in white.



2-44

R. .LUMINANCE & IRE UNIT OF COLOR VIDEO SIGNAL.

Green Red Blue Wh Yel Cyn Grn Mg Rd

Bl Blk

Sample Problem:

Calculate the luminance signal and the IRE equivalent unit for saturated green and yellow.

Solution:

For saturated green; Y = 0.3R + 0.59G + 0.11B = (0.3 x 0) + (0.59 x 1) + (0.11 x 0) = 0.59 ⇒ 59% of max lu min ance IRE = 0% lu min ance + 59% of max luminance = 7.5 + 0.59(100 − 7.5) = 62.075

For saturated yellow; Y = 0.3R + 0.59G + 0.11B = (0.3 x 1) + (0.59 x 1) + (0.11 x 0) = 0.89 ⇒ 89% of max lu min ance IRE = 0% lu min ance + 89% of max luminance = 7.5 + 0.89(100 − 7.5) = 89.825

Self-Sufficient Guide to ECE by JASON AMPOLOQUIO Summary of Different IRE units

Color

Luminance

White Yellow Cyan Green Magenta Red Blue Black

100% 89% 70% 59% 41% 30% 11% 0%

IRE Equivalent Unit 100 89.825 72.25 62.075 45.425 35.25 17.675 7.5

S. .CHROMINANCE OF COLOR VIDEO SIGNAL.

Chrominance Magnitude & Phase

CMAGNITUDE =

I2 + Q2

Q⎞ ⎛ CPHASE = 33° − ⎜ tan−1 ⎟ I⎠ ⎝


2-45


2-46

Sample Problem:

Calculate the chrominance signal for saturated red vector.

Solution: In-phase signal Quadrature signal I = 0.60R − 0.28G − 0.32B Q = 0.21R − 0.52G + 0.31B = (0.60 x 1) − (0.28 x 0) − (0.32 x 0) = (0.21 x1) − (0.52 x 0) + (0.31 x 0) = 0.6 = 0.21 Saturated Red Magnitude CMAGNITUDE = =

2

2

I +Q

0.62 + 0.212

Saturated Red Phase ⎛ CPHASE = 33° − ⎜ tan−1 ⎝ ⎛ ° = 33 − ⎜ tan−1 ⎝

= 0.635

Q⎞ ⎟ I⎠ 0.21 ⎞ ⎟ 0.6 ⎠

= 13.7°

Summary of Chrominance Information

Color

Q Signal

In-Phase Signal

Chrominance Vector Notation

White Black Yellow Blue Green Magenta Cyan Red

-0 -0 -0.31 -0.31 -0.52 -0.52 -0.21 -0.21

-0 -0 -0.32 -0.32 -0.28 -0.28 -0.6 -0.6

0 0 0.45∠77° 0.45∠257° 0.59∠151° 0.59∠331° 0.64∠193.7° 0.64∠13.7°

2-47

Self-Sufficient Guide to ECE by JASON AMPOLOQUIO T. .NTSC COLOR DECODER FOR TELEVISION BROADCASTING.

U. .TELEVISION BROADCAST STANDARDS. Parameters

FCC Standard

European Standard

Channel Width

6 MHz

7 or 8 MHz

4.2 MHz

5, 5.5 and 6 MHz

+ 25 kHz

+ 50 kHz

525/frame, 2:1 Interlaced 15,750 Hz for Monochrome 15,734.264 + 0.044 Hz for Color 60 Hz for Monochrome

625/frame, 2:1 Interlaced

Video BW Aural Deviation Scanning Lines Horizontal Scan Frequency Vertical Scan Frequency Reference Blanking Level Reference Black Level Reference White Level Transmission Polarity

15,625 Hz + 0.1%

50 Hz

59.94 Hz for Color 75 + 25% of peak carrier level 7.5 + from blanking to ref white level 12.5 + of peak carrier

75 + 2.5% of peak carrier level

Negative

Negative

3 to 6.5% 10-12.5%



2-48

V. .COMPARISON OF SOME TV BROADCAST STANDARDS. North & South America, Japan

Spain, Italy, England, Germany, CCIR System

France

USSR

Lines/Frame

525

625

625

625

Lines/second

15,750 Hz

15,625 Hz

15,625 Hz

15,625 Hz

30

25

25

25

6

7

8

8

4.2

5

6

6

Polarity of AM video modulation

Negative

Negative

Positive

Negative

Type of Aural carrier

FM

FM

AM

FM

Color system

NTSC

PAL

SECAM

SECAM

Color subcarrier frequency (MHz)

3.58

4.43

4.43

4.43

Parameter

Frames/second Channel BW (MHz) Video BW (MHz)

1.

National Television Standards Committee (NTSC) ª

ª

Advantages a. Higher Frame Rate - Use of 30 frames per second (really 29.97) reduces visible flicker.

b.

Atomic Color Edits - With NTSC it is possible to edit at any 4 field boundary point without disturbing the color signal.

c.

Less inherent picture noise - Almost all pieces of video equipment achieve better signal to noise characteristics in their NTSC/525 form than in their PAL/625.

Disadvantages a. Lower Number of Scan Lines - Reduced clarity on large screen TVs, line structure more visible.

b.

Smaller Luminance Signal Bandwidth - Due to the placing of the color sub-carrier at 3.58MHz, picture defects such as cross-color, and dot interference become more pronounced.


2.

2-49

c.

Susceptibility to Hue Fluctuation - Variations in the color subcarrier phase cause shifts in the displayed color, requiring that the TV receivers be equipped with a Hue adjustment to compensate.

d.

Lower Gamma Ratio - The gamma value for NTSC/525 is set at 2.2 as opposed to the slightly higher 2.8 defined for PAL/625. This means that PAL/625 can produce pictures of greater contrast.

Phase Alternate/Alternation by Line (PAL) ª

ª

Advantages a. Greater Number of Scan Lines - more picture detail.

b.

Wider Luminance Signal Bandwidth - The placing of the color Sub-Carrier at 4.43MHz allows a larger bandwidth of monochrome information to be reproduced than with NTSC/525.

c.

Stable Hues - Due to reversal of sub-carrier phase on alternate lines, any phase error will be corrected by an equal and opposite error on the next line, correcting the original error.

d.

Higher Gamma Ratio - The gamma value for PAL/625 is set at 2.8 as opposed to the lower 2.2 figure of NTSC/525. This permits a higher level of contrast than on NTSC/525 signals.

Disadvantages a. More Flicker - Due to the lower frame rate, flicker is more noticeable on PAL/625 transmissions; particularly so for people used to viewing NTSC/525 signals.

b.

Lower Signal to Noise Ratio - The higher bandwidth requirements cause PAL/625 equipment to have slightly worse signal to noise performance than it's equivalent NTSC/525 version.

c.

Loss of Color Editing Accuracy - Due to the alternation of the phase of the color signal, the phase and the color signal only reach a common point once every 8 fields/4 frames. This means that edits can only be performed to an accuracy of +/4 frames (8 fields).



2-50

3.

Sequential Couleur Avec Memoire (SECAM) Sequential Color with Memory ª

Advantages a. Stable Hues and Constant Saturation - SECAM shares with PAL the ability to render images with the correct hue, and goes a step further in ensuring consistent saturation of color as well.

b.

ª

Higher Number of Scan Lines - SECAM shares with PAL/625, the higher number of scan lines than NTSC/525.

Disadvantages a. Greater Flicker – Same with PAL/625

b.

Patterning Effects - The FM subcarrier causes patterning effects even on non-colored objects.

c.

Lower monochrome Bandwidth - Due to one of the two color sub-carriers being at 4.25MHz (in the French Version), a lower bandwidth of monochrome signal can be carried.

Read it till it Hertz…jma NTSC ª ª

The oldest existing standard, developed in the USA. First used in 1954. Consists of 525 horizontal lines of display and 60 vertical lines. Sometimes irreverently referred to as "Never Twice the Same Color." Only one type exists, known as NTSC-M.

PAL ª

ª

PAL was developed by German engineer Walter Bruch and the German electronic corporation Telefunken. Walter Bruch patented his invention 1963 and the first commercial application of the PAL system was in August 1967. Also a 625/50-line display and variant of NTSC. Proponents call it "Perfection At Last." Due to the cost of the enormous circuit complexity, critics often refer to it as "Pay A Lot".

SECAM ª ª

SECAM was developed in France. First used in 1967. A 625-line vertical, 50-line horizontal display. Sometimes referred to by wags as "Something Essentially Contrary to the American Method" or SEcond Colour Always Magenta!"

2-51

Self-Sufficient Guide to ECE by JASON AMPOLOQUIO W. .DIGITAL TELEVISION (DTV).

Since 1987, the FCC has been encouraging the development of Highdefinition television (HDTV) system to replace the NTSC system. In May of 1993, the proponents of these four competing system joined forces to develop a single digital television (DTV) system that used the best ideas of the four competing system. 1.

Comparison of Different Definition Standard

i.

Video Specification

Format

HD

HD

SD

SD

SD

Aspect Ratio

16:9

16:9

4:3

16:9

4:3

Active Scan lines/frame

1080

720

480

480

480

Pixel/line

1920

1280

704

704

640

24p, 24p, 30p, 30p, 30i, 30i, 60p 60p HD = High definition, SD = Standard definition p = progressive scan, i = interlace scan

24p, 30p, 30i, 60p

Frame rates (Hz)

24p, 30p, 30i

24p, 30p, 60p



2-52

ii.

Audio Specification

DOLBY AC-3 20 kHz 48 kHz 100 dB 5.1 384 kbps

Method Audio BW Sampling Frequency Dynamic Range Number of Surround channel Compressed data rate iii. Data Transport System

Packet MPEG-2 188 bytes 4 bytes

Type TDM technique Packet size Packet header size X. .HIGH-DEFINITION PARAMETERS.

θ

Diagonal Dimension of Viewing Area 1 ⎞ ⎛ D = H2 ⎜ 1 + 2 ⎟ a ⎠ ⎝

Minimum Viewing Distance

X=

H ⎛θ⎞ 2 tan ⎜ ⎟ ⎝2⎠

where: D = diagonal dimension of the viewing area a = aspect ratio X = distance from screen to observer/viewer θ = angle subtended by the horizontal dimension H


2-53

Sample Problem:

Find the minimum viewing distance a viewer should sit from the tube to eliminate seeing any scan lines and the viewing angle subtended by the screen for HDTV system, assuming the projector screen with diagonal dimension of 1.55 m with 1125 vertical lines (90 suppressed).

Solution: 2

H=

2 2

aD

1 + a2

=

⎛ 16 ⎞ ⎜ ⎟ + 1.552 ⎝ 9 ⎠ = 1.35 m 2 ⎛ 16 ⎞ 1+⎜ ⎟ ⎝ 9 ⎠

16 x(1125 − 90) = 1840 pixels 9 1 min ute 1 θ = NH x = 1840 x = 30.7° 60 60 H 1.35 m X= = = 2.45 m ⎛θ⎞ ⎛ 30.7° ⎞ 2 tan⎜ ⎟ 2 tan⎜ ⎟ ⎝2⎠ ⎝ 2 ⎠

NH = aNV = a(N − Ns ) =

Y. .VIDEO STANDARDS. 1.

JPEG JPEG (Joint Photographic Experts Group) was developed jointly by the ISO and ITU-T. JPEG is a compression standard that is used for editing of still images, as well as color facsimile, desktop publishing, graphic arts, and medical imaging. JPEG is not appropriate for video transmission, as the compression rate is in the range of only 20:130:1. JPEG transmission in support of videoconferencing requires bandwidth in the range of 10 to 240 Mbps.

2.

MPEG MPEG (Motion Picture Experts Group) standards are several and still in final development stages. MPEG standards provide very high compression levels and excellent presentation quality. MPEG is a joint technical committee of the International Standards Organization (ISO) and the IEC (International Electrotechnical Commission).



2-54

ª

MPEG 1 Standardized in 1992, provides VHS (videotape) quality at 1.544 Mbps and is compatible with single-speed CD-ROM technology. In fact, it was designed to put movies on compact disc. MPEG 1 integrates synchronous and isochronous audio with video, and allows the random access required by interactive multimedia applications.

ª

MPEG 2 (1994) is the proposed standard for digital video at 4 to 100 Mbps over transmission facilities capable of such support (fiber optics, hybrid fiber/coax and satellite). MPEG 2 already has found application in Direct Broadcast Satellite (DBS) services, also known as Direct Satellite Systems (DSS).

ª

MPEG 3 Designed for HDTV application, was folded into MPEG-2 in 1992.

ª

MPEG-4 A low bit-rate version intended for application in videophones and other small-screen devices. It is still under development.

2-55


I

H

1.

Calculate the equivalent IRE unit of a video signal with a luminance of 72%. A. 4.1 IRE units B. 53.8 IRE units C. 74.1 IRE units D. 66.4 IRE units

2.

Calculate the increase in horizontal resolution possible if the video modulating signal bandwidth was increased to 4.5 MHz. A. 582 lines B. 525 lines C. 338 lines D. 482 lines

3.

Determine the possible increase in vertical details if the video frequency bandwidth were allowed up to 5 MHz. (assume 70% Kell factor) A. 408 pixels B. 420 pixels C. 442 pixels D. 480 pixels

4.

Calculate the decrease in horizontal details if the video signal bandwidth were reduced from 4 to 3.5 MHz. A. 458 to 322 pixels B. 458 to 375 pixels C. 428 to 375 pixels D. 428 to 322 pixels

5.

Calculate the vertical resolution if the video signal bandwidth were reduced from 4 to 3.5 MHz assuming that the horizontal resolution was not to change. A. 307 lines B. 337 lines C. 327 lines D. 317 lines

6.

Given a 635-μs vertical retrace time, the number of complete horizontal lines scanned during vertical flyback is A. 10 B. 20 C. 30 D. 63

7.

One-half line spacing between the start positions for scanning even and odd fields produces A. linear scanning B. exact interlacing C. fishtailing D. line pairing

8.

Calculate the minimum recorder wavelength for an audio cassette recorder with a tape speed of 1,875 ips, 12 kHz top recording frequency. A. 0.364 mil B. 0.567 mil C. 0.156 mil D. 0.87 mil

9.

What is the exact picture carrier frequency for channel 2 offset by -10 kHz? A. 55.24 MHz B. 55.34 MHz C. 51.25 MHz D. 55.75 MHz



2-56

10. The number of scanning lines is ______ per second. A. 60 B. 525 C. 15,750 D. 15,650 11. Calculate the sound and picture carrier frequency for channel 10 after frequency translation to the IF. A. P=45.75 MHz, S=41.25 MHz B. P=45.25 MHz, S=41.75 MHz C. P=41.75 MHz, S=45.25 MHz D. P=41.25 MHz, S=45.75 MHz 12. 10.5 μs of retrace is equal to how many % of H? A. 6.5% B. 13.5% C. 16.5% D. 12.5%

13. Color mixture is close to blue. A. Y C. R-Y

B. G-Y D. B-Y

14. The I color video signal has a bandwidth of ____ MHz. A. 1.3 B. 4.5 C. 4.2 D. 3.58 15. Calculate the percentage of time occupied by vertical blanking. A. 0.8% B. 8% C. 28% D. 18% 16. What is the horizontal resolution in lines of a monochrome receiver which has a 2.8 MHz video bandwidth? April 2003 & Nov 2004 A. 324 lines B. 254 lines C. 524 lines D. 224 lines 17. Calculate the total number of picture elements (pixels) in an NTSC frame if the video BW is 4.2 MHz. (Assume 0.7 Kell factor, 4/3 aspect ratio, and 10 μs retrace time) A. 151,761 pixels B. 211,161 pixels C. 181,661 pixels D. 192,416 pixels 18. A ghost is displaced 1 inch on a 13-inch diagonal TV screen. Determine the time between signal receptions, and the difference in path lengths. A. 3.53 km B. 1.53 km C. 2.53 km D. 0.53 km

19. Weak emission from one cathode in a tricolor picture tube causes _____, _____, and _____ in the raster and picture. A. a stronger picture, a long warm up time, color balance B. a weak picture, a long warm up time, and color imbalance C. a stronger picture, a short warm up time, and color imbalance D. a weak picture, a short warm up time, and color balance 20. Degaussing means _____ the picture tube with 60-Hz alternating current from the power line. A. de-energizing B. magnetizing C. energizing D. demagnetizing


2-57

21. _____ correction signals can be inserted in series with the V and H coils of the deflection yoke, to straighten the edges of the raster. A. Gamma B. Pincushion C. Comb D. Contrast 22. The PAL system in Europe uses a 625 lines/frame and 25frames/sec. Assuming a video bandwidth of 5 MHz, determine the number of distinguishable picture element, assuming 84% of each horizontal line and 90% percent of the 625 lines are visible. A. 143,400 B. 122,400 C. 330,400 D. 302,400 23. Calculate the minimum viewing distance for an HDTV system assume that a projector screen with a diagonal dimension of 1.4 m and an aspect ratio of 16:9 is used with 1125 vertical lines (90 suppressed). A. 2.7 m B. 2.23 m C. 3.61 m D. 1.8 m

24. The hue 180° out of phase with red is A. yellow B. green C. blue D. cyan 25. Greater p-p amplitude of the 3.58-MHz chrominance signal indicates more A. saturation B. yellow C. white D. hue 26. Which of the following statements is true? A. Negative transmission means that the carrier amplitude decreases for white B. Vestigial sideband transmission means that both upper and lower sidebands are transmitted for all modulating frequencies C. Vestigial sideband transmission means that the modulated picture carrier signal has only the upper envelope D. Negative transmission means that the carrier amplitude decreases for black 27. With a 2-MHz video signal modulating the picture carrier signal for channel 4, which of the following frequencies are transmitted? A. 67.25-MHz carrier frequency with 65.25-MHz upper side frequencies B. 66-MHz carrier frequency and 68-MHz upper side frequency C. 67.25-MHz carrier frequency and 69.25-MHz upper side frequency D. 71.75-MHZ carrier frequency with 69- and 73-MHz side frequency 28. With a 0.5-MHz video signal modulating the picture carrier, A. no side frequencies are transmitted B. both upper and lower side frequencies are transmitted C. only the upper side frequency is transmitted D. only the lower side frequency is transmitted


2-58


29. Compared to a monochrome CRT, the accelerating voltage on a color CRT is: A. about the same B. much higher C. much lower D. color CRTs use magnetic acceleration 30. The horizontal lines scanned during vertical retraces are the ______ lines. A. horizontal flyback B. vertical sync C. vertical flyback D. horizontal blanking 31. Interfaced scanning eliminates _____ because of the ____ blanking rate, while maintaining the _____ rate for complete picture frames. A. flicker, 60-Hz vertical, 30-Hz B. flicker, 30-Hz vertical, 15,750-Hz C. ghost, 60-Hz vertical, 15,750-Hz D. ghost, 15,750-Hz vertical, 30-Hz 32. The equalizing pulses and the serrations in the vertical pulse occur at halfline intervals with the frequency of _____ Hz. A. 30 B. 60 C. 31,500 D. 15,750 33. Determine the ghost width on a TV screen 15 in wide when a reflected wave results from an object ½ mi “behind” a receiver. A. 151 in B. 15.1 in C. 1.51 in D. 0.151 in 34. Calculate the ghost width for a 17-in wide TV screen when a reflected wave results from an object 3/8 mi “behind” a receiver. A. 12.8 in B. 128 in C. 0.128 in D. 1.28 in

35. A horizontal resolution of 300 lines corresponds to ____ pixels of picture information. A. 210 B. 150 C. 400 D. 200 36. Horizontal resolution of 240 lines corresponds to a video-frequency response of ___. A. 3 MHz B. 4.2 MHz C. 16 MHz D. 24 MHz 37. In the CED system, the disk capacitance varies with the A. disk size B. wavelength of the scanning light C. speed of rotation D. pit depth 38. The part that rotates to meter out the tape at constant speed is the A. control head B. entrance guide C. capstan D. erase head


2-59

39. The modulated picture carrier wave includes the composite video signal as the A. upper envelope without the lower envelope B. lower sideband without the upper sideband C. average carrier level D. symmetric envelope of amplitude variations 40. Calculate the horizontal resolution possible if the video modulating signal bandwidth is 5 MHz. A. 515 lines B. 433 lines C. 535 lines D. 483 lines 41. Calculate the magnitude chrominance for color yellow. A. 0.45 B. 0.69 C. 0.65 D. 0.21 42. Calculate the exact horizontal frequency if the sound carrier of 4.5 MHz is made to be the 286th harmonics of the horizontal line frequency. A. 15,735.27 Hz B. 15,610.27 Hz C. 15,750 Hz D. 15,625 Hz

43. The average voltage value of the 3.58-MHz modulated chrominance signal is A. the saturation of the color B. zero for most colors C. the brightness of the color D. close to black for yellow 44. The second IF value for color in receivers, for any station, is A. 0.5 MHz B. 1.3 MHz C. 4.5 MHz D. 3.58 MHz 45. If the 3.58-MHz C amplifier in the receiver does not operate, the result will be A. no color B. too much blue C. too much yellow D. no red 46. _____ is done in the camera to compensate for the nonlinear drive characteristics of the picture tube. A. Error correction B. Beam correction C. Gamma correction D. Contrast correction 47. Light flux for cameras usually is measured in _____. A. Tesla B. footcandles C. luminous intensity D. lux 48. Typical horizontal resolution in television receivers is _____ lines, corresponding to ____ video-frequency response. A. 331, 3.58-MHz B. 428, 4.25-MHz C. 312, 2.50-MHz D. 250, 3.125-MHz


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49. A _____ is a special camera tube with a test pattern printed on the image plate. A. plumbicon B. iconoscope C. monoscope D. vidicon 50. What is the exact vertical frequency if there are 262.5 lines per field? A. 50 Hz B. 49.94 Hz C. 60 Hz D. 59.94 Hz 51. Calculate the exact color subcarrier frequency if this value is made to be the 455th harmonic of half of the horizontal frequency. A. 3.58 MHz B. 4.2 MHz C. 3.579545 MHz D. 4.5 MHz

52. The ______ is the current that flows in the target circuit when the camera lens is capped for no light input. A. flash current B. leakage current C. dark current D. reverse current 53. _____ is an exponential value that specifies how white light values are expanded or compressed, and this affects the contrast ratio. A. Alpha B. Gamma C. Beta D. Sigma 54. Which of the following camera tubes uses lead oxide (PbO) for the photoconductive target plate? A. plumbicon B. vidicon C. image orthicon D. saticon 55. What is the exact frequency separation between sound carrier and color subcarrier? A. 0.92 MHz B. 0.920455 MHz C. 0.920945 MHz D. 0.920495 MHz

56. Ability of monochrome receiver to use Y signal for picture in black and white. A. monotonacity B. adaptability C. compatibility D. selectivity 57. The complementary color for blue, red and green are ____, ____, and ____ respectively. A. cyan, yellow, magenta B. magenta, yellow, cyan C. cyan, magenta, yellow D. yellow, cyan, magenta 58. Determine the total number of details for an NTSC frame if 42 lines are blanked out because of vertical flyback. Used 0.7 Kell factor, 4.2 MHz video bandwidth. A. 161, 751 B. 121, 221 C. 151, 761 D. 131, 231


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59. Converting hue and saturation in the C signal to R, G, and B primary color video signals for the tricolor picture tube. A. compandoring B. decoding C. multiplexing D. encoding 60. Converting the R, G, and B primary color video signals to hue and saturation in the C signal. A. pre-emphasis B. decoding C. encoding D. degaussing 61. How many flyback lines are produced during vertical retrace for each field and each frame when the retrace time is 0.03V? A. 8 and 16 B. 16 and 18 C. 18 and 16 D. 16 and 8

62. The hue of color sync phase is A. red C. yellow-green

B. blue D. cyan

63. Which signal has color information for 1.3-MHz bandwidth? A. R-Y B. Y C. I D. B-Y 64. Calculate the maximum number of vertical details that can be reproduced with 483 visible scanning lines with a 70% utilization factor. A. 303 B. 438 C. 348 D. 338 65. Determine the number of horizontal details for a 4 MHz video bandwidth. A. 416 picture elements B. 446 picture elements C. 426 picture elements D. 486 picture elements

66. The color with the most luminance is A. yellow B. blue C. green D. red 67. Pictures frames are repeated at the rate of ____ per second. A. 60 B. 50 C. 30 D. 25 68. The number if scanning lines is ______ per frame. A. 262.5 B. 625 C. 15,750 D. 525 69. In all standard television broadcast channels, the difference between the picture and sound carrier frequencies is A. 0.25 MHz B. 1.25 MHz C. 4.5 MHz D. 6 MHz


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70. The beat frequency between the 3.58-MHz color subcarrier and the 4.5MHz sound signal is A. 3.58 MHz B. 4.8 MHz C. 4.5 MHz D. 0.92 MHz 71. Which control varies the phase angle of the demodulated color video signal? A. picture B. drive C. tint D. color level 72. Which of the following stages must be on during horizontal flyback time? A. R – Y vidéo amplifier B. Chroma BPA C. Burst separator D. Y video amplifier 73. Picture tubes have two black conductive coating made of ______. A. Tellurium B. Phospor C. Aquadag D. Arsenic 74. The _____ of the picture tube is the total angle that the beam can be deflected without touching the sides of the envelope. A. deflection angle B. critical angle C. optical angle D. envelope angle 75. _____ adjustments cause the three electron beams to excite the correct colors on the screen. A. Comb B. Color C. Convergence D. Gamma 76. Picture tubes with an anode voltage of _____ can emit x-rays. A. 20 to 30 GV B. 200 to 300 MV C. 20 to 30 kV D. 200 to 300 kV 77. What are the lower and upper side frequencies for 200-kHz color video modulation in the 3.58-MHz bandpass amplifier? A. 3.28 and 6.78 MHz B. 3.38 and 3.78 MHz C. 3.28 and 5.48 MHz D. 3.38 and 6.28 MHz 78. The number of lines scanned per frame in the raster on the picture tube screen is A. 10 B. 262 ½ C. 20 D. 525 79. In the frame for which interlaced scanning is used, alternate lines are skipped during vertical scanning frequency because A. the frame has a 4:3 aspect ratio B. the vertical scanning frequency is doubled from 30 to 60 Hz C. the horizontal scanning is slower than vertical scanning D. the trace is slower than the retrace


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80. If the horizontal flyback is 10 percent, this time equals A. 10 μs B. 56 μs C. 6.4 μs D. 83 μs 81. Which of the following frequencies is wrong? A. 15,750 Hz for horizontal sync and scanning B. 31,500 for the vertical scanning frequency C. 31,500 for equalizing pulses and serrations in the vertical sync pulse D. 60 Hz for vertical sync and scanning 82. How much time elapses between the start of one horizontal sync pulse and the next? A. 8% of V B. V C. H D. 16.5% of H 83. For a 4-MHz video signal, the number of horizontal details in a line is ____. A. 525 B. 331 C. 426 D. 645 84. Assume a facsimile reproduction with specifications of 200 lines per frame, progressive scanning, and 5 frames per second. Calculate the video frequency corresponding to 100 total black-and-white elements in a line. A. 52 kHz B. 5.2 kHz C. 0.52 kHz D. 520 kHz 85. The circuit that combines C signal and Y luminance signal. A. color stacker B. Colorplexer C. Combiner D. comb filter 86. Retraces are not visible because of _____ pulses. A. sync B. blanking C. black D. serration 87. Black on the picture tube screen results from _____ beam current. A. negative B. maximum C. zero D. positive 88. Determine the total number of picture elements possible for the entire image if the video bandwidth is 4 MHz and 70% utilization factor is used. A. 104,000 B. 144,000 C. 184,000 D. 124,000 89. Assume a facsimile reproduction with specifications of 200 lines/frame, progressive scanning, and 5 frame/sec. Calculate the visible trace time for one line with 4% blanking. A. 9.6 μs B. 960 μs C. 0.96 μs D. 90 μs



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90. The modulation used for the chroma signal in a standard NTSC color TV receiver is: A. SSB C. suppressed-carrier AM

B. vestigial sideband AM D. FM

91. Determine the maximum luminance level for magenta. A. 87% B. 48% C. 41% D. 34% 92. A 3.125 MHz of video signal bandwidth is equal to how many horizontal resolutions? A. 250 lines B. 320 lines C. 205 lines D. 225 lines

93. Picture tubes use ______ focusing. A. electromagnetic C. electroacoustic

B. electrostatic D. magnetostatic

94. The color with the least luminance A. red C. cyan

B. blue D. magenta

95. If the phase chrominance for color yellow is 167°, what is the phase chrominance for color blue? A. 3.47° B. 34.7° C. 37° D. 347° 96. What is the horizontal scanning time for 125 pixels? A. 10.5 μs B. 15.625 μs C. 63.5 μs D. 0.125 μs 97. Calculate the video frequency response corresponding to the horizontal resolution of 320 lines. A. 4.75 MHz B. 4 MHz C. 4.5 MHz D. 4.2 MHz 98. Consider a video signal that has a resolution of 640x480 pixels, with a frame rate 0f 30 Hz using progressive scanning. The luma signal is sampled using 8 bits/sample. The two chroma channels also use 8 bits/sample, but the color resolution is ¼ that is used for luma. Find the approximate bit rate for this signal, neglecting synchronization, error correction, and compression. A. 110.6 Mbps B. 1.6 Mbps C. 10.6 Mbps D. 11.6 Mbps

99. Luminance is measured in: A. foot-candles C. IRE units

B. lumens D. NTSC units

100. The maximum luminance level is called: A. max white B. peak white C. blanking D. whiter than white


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101.What proportion of the maximum transmitter power is used to transmit a black setup level? A. 73.4% B. 22.2% C. 61.1% D. 45.6%

102. The _____ is deflected horizontally and vertically to fill the screen area. A. permanent magnet B. yoke C. electron beam D. raster 103. What is the hue of a color 90° leading sync burst phase? A. blue B. cyan C. orange D. yellow 104. How much time passes between one vertical pulse in an odd field and the next in an even field? A. V B. H C. 8% of V D. 8% of H 105. The back porch is ____ longer than the front porch A. 3x B. 4x C. 2x D. 8x 106. Which of the following camera tubes has minimum lag? A. plumbicon B. saticon C. vidicon D. iconoscope 107.Calculate the total percentage of the signal time occupied by vertical, horizontal, and active video respectively. A. 21%, 42%, 88% B. 8%, 15.7%, 77.6% C. 12.75%, 100%, 77.6% D. 16%, 21%, 88% 108.A certain receiver has a vertical retrace time of 0.025V, where V is the time for one field. How many horizontal lines does this represent? A. 2.1 B. 3.3 C. 6.6 D. 4.2

109. _____ deflection is used for picture tubes. A. Static B. Magnetic C. Electrostatic D. Magnetostatic 110.Suppose that the aspect ratio of a TV system were change from 4:3 to 2:1, with the horizontal resolution keeping the same ratio to vertical resolution as at present. Assume the number of scan lines remains at present. How many details would have to be shown on a horizontal line? A. 428 B. 338 C. 674 D. 525



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111.Under ideal conditions, the eye can resolve two details subtending 0.5 minutes of arc. Suppose a monitor screen 30 cm wide is located 1.5 m from a viewer. What video bandwidth would be required to separate horizontal details in the center of the screen by 0.5 minutes of arc assuming NTSC video? A. 4.5 MHz B. 1.5 MHz C. 5.1 MHz D. 12.9 MHz

112. The blanking level corresponds to a luminance of: A. white B. black C. whiter than white D. blacker than black 113. The sync pulse level corresponds to a luminance of: B. black A. white C. whiter than white D. blacker than black 114. In a color TV receiver, Y I Q refers to: A. luminance signal, in-phase color component, quadrature phase color

B. C. D.

component composite color signal, in-phase color component, quadrature phase color component composite video signal, in-phase video component, quadrature video color component a method of demodulating stereo sound

115. The modulation used for the video signal in a standard NTSC color TV receiver is:

A. C.

SSB suppressed-carrier AM


116. The horizontal output transformer is also called: A. the isolation transformer B. the video transformer C. the flyback transformer D. the yoke 117. ____ convergence adjustments are made by permanent magnets. A. Static B. Active C. Dynamic D. Moving 118. Deflection in CRTs used in TV receivers is done: A. magnetically for both vertical and horizontal B. electrostatically for both vertical and horizontal C. electrostatically for vertical and magnetically for horizontal D. magnetically for vertical and electrostatically for horizontal 119. The number of fields is _____ per frame. A. 2 B. 30 C. 60 D. 525 120. The vertical field-scanning frequency is ______ A. 525 Hz B. 15,750 Hz C. 60 Hz D. 30 Hz


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121. Video signal amplitude determines the picture quality called _____ A. contrast B. resolution C. color saturation D. brightness 122. Scanning in the receiver is timed correctly by _____ pulses. A. retrace B. serration C. sync D. blanking 123. The modulation used for the audio signal in a standard NTSC color TV receiver is:

A. C.

SSB suppressed-carrier AM


124. The amount of color saturation in the picture depends on the amount of _____ signal. A. blanking B. chrominance C. luminance D. contrast 125. _____ convergence adjustments vary the correction current in the coils of the convergence yoke. A. Passive B. Static C. Dynamic D. Fixed 126. Camera signal output without sync is called A. black burst B. generator lock video C. noncomposite video D. composite video 127. A low-contrast picture in which white seems flat and lacking in detail suggests. A. excessive gamma B. high gain in the preamplifier C. insufficient scanning width D. low beam current 128. The part of the visible spectrum where camera pickup tubes have the greatest output is A. infrared B. yellow-green C. blue D. red 129. The horizontal line-scanning frequency is _____ A. 30 Hz B. 60 Hz C. 525 Hz D. 15,750 Hz 130. The number of scanning lines is _____ per field. A. 30 B. 262.5 C. 525 D. 15,625 131. The gamma of the picture tube is A. 1.0 C. 0.4545

B. 1.4 D. 2.2



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132. A typical value of vidicon dark current is A. 8 mA B. About 200 μA C. 800 mA D. 0.2 μA 133. A permanent-magnet ring on the neck of the tube is used for _____. A. scanning B. centering C. deflecting D. focusing 134. Picture tubes with an anode voltage of 20 to 30 kV can emit _____. A. cosmic-rays B. UV-rays C. X-rays D. gamma rays 135. The gamma of 2.2 for the picture tube that means that the ____ values are stretched. A. green B. black C. yellow D. white 136. A picture tube with deflection angle of 110° needs a yoke with a ____ angle. A. 10° B. 220° C. 110° D. 55° 137. Color mixture close to green. A. R-Y C. B-Y

B. G-Y D. Y-R

138. The Q video signal has a bandwidth of _____ MHz. A. 4.2 B. 0.5 C. 1.3 D. 4.5 139. The linear rise on the sawtooth waveform is the_____; the sharp drop in amplitude is for the ____. A. Trace part, retrace B. Flyback part, retrace C. Retrace part, blanking D. Trace part, blanking 140. The frequency of the sawtooth waveform for horizontal deflection is ____. A. 30 Hz B. 15,750 Hz C. 31,500 Hz D. 60 Hz

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Broadcasting

Section

8


Standards

Medium Frequency Broadcast Station: An AM Broadcast station licensed for aural or sound transmission intended for direct reception by the general public and operated on a channel in the MF band.

DEFINITION.

DEFINITION.

Operating Frequency is the carrier frequency at any particular

time.

DEFINITION.

Authorized Frequency is the carrier frequency authorized by the

authority.

DEFINITION.

Authority: The National Telecommunications Commission (NTC).

A. .AMPLITUDE MODULATION STANDARDS. Parameter

Philippine Technical Standards

FCC Technical Standards

Occupied Spectrum

535-1,605 kHz

540-1,700 kHz

1070 kHz

1160 kHz

9 kHz

10 kHz

118 station

116 station

36 kHz

30 kHz

+ 20 Hz of the assigned frequency


+ 500 Hz

+ 500 Hz

455 kHz

455 kHz

AM

AM

A3E

A3E

Superheterodyne

Superheterodyne

50-15,000 Hz

50-15,000 Hz

Allocated BW BW per Station Number of Station Spacing between Station Carrier Frequency Tolerance Guardband Intermediate Frequency Modulation Scheme Type of Emission Receiver Characteristic Audio Frequency Response


BROADCASTING standards

2-70

B. .FREQUENCY ALLOCATION FOR AM BROADCAST STATIONS. Channel Number 1 2 3 4 . . . 131

Frequency (kHz) 531 540 549 558 . . . 1701

C. .SERVICE AREAS. 1.

Primary Service Area The term “primary service area” of a broadcast station means the area in w/c the groundwave field of 1mV/m (60 dBu) is not subject to objectionable interference or objectionable fading.

2.

Secondary Service Area The term “secondary service area” of a broadcast station means the area served by the skywave and not subject to objectionable interference. The signal is subject to intermittent in intensity.

3.

Intermittent Service Area The term “intermittent service area” of a broadcast station means the area receiving service from the groundwave but beyond the primary service area and subject to some interference and fading.

D. .AM BROADCASTING RATIO. 1.

Audio-Frequency Signal-to-Interference Ratio The ratio (in dB) between the values of the voltage of the wanted signal and the voltage of the interference, measured under specified conditions, at the audio-frequency output of the receiver.

2.

Radio-Frequency Wanted-to-Interference Signal Ratio The ratio (in dB) between the values of the radio-frequency voltage of the wanted signal and the interfering signal, measured at the input of the receiver under specified conditions.

3.

Audio-Frequency Protection Ratio The agreed minimum value of the audio frequency signal-tointerference ratio considered necessary to achieve a subjectively defined reception quality.

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Radio-Frequency Protection Ratio The value of the radio frequency wanted-to-interference signal ratio that enables, under specified conditions, the audio-frequency protection ratio to be obtained at the output of the receiver.


AM broadcast station shall not operate more than 5% and not lower than 10% of its authorized operating power.

ª

The carrier shift at any percentage of modulation shall not exceed 5%.

ª

AM transmitter must be capable of maintaining the operating frequency within the limits of +10 Hz of the assigned frequency.

E. .BROADCAST AUXILIARY SERVICES. 1.

2.

Studio-to-Transmitter Link (STL)

Band

Operating Frequency

Maximum Power

A

300 to 315 MHz

15 W

B

734 to 752 MHz

15 W

C

942 to 952 MHz

15 W

Remote Pick-up Broadcast Station

Band

Operating Frequency

Maximum Power

A

315 to 325 MHz

35 W

B

450 to 451 MHz

35 W

C

455 to 456 MHz

35 W



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3.

4.

Communications, Coordination and Control Link

Band

Operating Frequency

Maximum Power

A

4 to 12 MHz

100 W (SBS)

B

25.67 to 26.1 MHz

160 W (ERP)

C

162.235 to 162.615 MHz 166.25 MHz,170.15 MHz

160 W (ERP)

D

432.5 to 433 MHz 437.5 to 438 MHz

200 W (for repeater)

Maximum Power Allocation

Area

Maximum Power

Metro Manila

50 kW

All other areas

10 kW

C

---

D

10 W authorized

F. .IMPORTANT TERMINOLOGY. 1.

Operating Frequency The carrier frequency at any particular time.

2.

Authorized Frequency The carrier frequency authorized by the Authority.

4.

Operating Power “Operating Power” is the transmitter output power.

5.

Maximum Rated Carrier Power “Maximum Rated Carrier Power” is the maximum power at which the transmitter can be operated satisfactorily and is determined by the design of the transmitter.

6.

Authorized Operating Power “Authorized Operating Power” is the power authorized by the Authority.

7.

Modulator Stage “Modulator Stage” means the last audio amplifier stage of the modulating wave which modulates a radio-frequency stage.


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8.

Modulated Stage “Modulated Stage” means the radio frequency stage to which the modulator is coupled and in which the continuous carrier wave is modulated in accordance with the system of modulation and the characteristics of the modulating wave.

9.

Daytime The term “Daytime” refers to the period of time between 2200 Universal Time Coordinates (UTC) to 1000 Universal Time Coordinates (UTC) or 6:00 AM to 6:00 PM local standard time.

10. Nighttime The term “Nighttime” refers to the period of time between 1000 Universal Time Coordinates (UTC) to 2200 Universal Time Coordinates (UTC) or 6:00 PM to 6:00 AM local standard time. 11. Experimental Period The term “Experimental Period” means that time between 12 midnight to 5:00 AM local standard time or 1600 to 2100 Universal Time Coordinates (UTC) 12. Spurious Emission The emission on any frequency outside of the assigned channel or authorized band of frequencies and tolerances allowed by these regulations. Emission outside of the assigned channel, as a result of the modulating process, is not considered spurious, unless it is due to overmodulation. 13. Authority The National Telecommunications Commission.



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I

H

A. .FM BROADCAST STANDARDS. Parameter



Occupied Spectrum

88-108 MHz

88-108 MHz

20 MHz

20 MHz

In 200 kHz increment from 88.1 MHz (FM channel 201) to 107.9 MHz (FM channel 300)

In 200 kHz increment from 88.3 MHz (FM channel 2) to 107.5 MHz (FM channel 98)

800 kHz

800 kHz



+ 75 kHz

+ 75 kHz

+ 25 kHz 75 μsec with break frequency of 2122 Hz

+ 25 kHz 75 μsec with break frequency of 2122 Hz

Allocated BW

BW per Station

Spacing between Station Carrier Frequency Tolerance Maximum Frequency Deviation Guardband Pre-emphasis Intermediate Frequency Receiver Characteristic Audio Frequency Response

10.7 MHz

10.7 MHz

Superheterodyne

Superheterodyne

50-15,000 Hz

50-15,000 Hz

Parameter

FM 2-way (Mobile Radio)

Analog TV (Aural Portion)

+ 5 kHz

+ 25 kHz

21.4 MHz

45.75 MHz

1 for Δf= 5 kHz and fm=5 kHz

1.67 for Δf= 25 kHz and fm=15 kHz

Maximum Frequency Deviation Intermediate Frequency Modulation Index

2-75


Center Frequency of nth Broadcast Station

FMn(MHz ) = FM1(MHz ) + 0.2 ( n − 1 )

2.

3.

Frequency Allocation for FM Broadcast Stations

Channel Number

Frequency (MHz)

201 202 203 204 . . . 300

88.1 88.3 88.5 88.7 . . . 107.9

Table of Assignments The frequency assignments for the cities of Manila, Laoag, Legaspi, Cebu, Davao, and Zamboanga shall be

Channel Number

Frequency (MHz)

Channel Number

Frequency (MHz)

202 206 210 214 218 222 226 230 234 238 242 246 250

88.3 89.1 89.9 90.7 91.5 92.3 93.1 93.9 94.7 95.5 96.3 97.1 97.9

254 258 262 266 270 274 278 282 286 290 294 298

98.7 99.5 100.3 101.1 101.9 102.7 103.5 104.3 105.1 105.9 106.7 107.5



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4.

Class of FM Broadcast Stations

Class

Service

Authorized Power

ERP

AHAAT

A

Commercial

25 kW authorized 10 kW minimum

125 kW

2000 ft

B

Commercial

10 kW authorized 1 kW minimum

30 kW

500 ft

C

Non-commercial community station

---

1 kW

---

D

Educational Station

10 W authorized

---

---

AHAAT=Antenna Height above Average Terrain


FM broadcast station are not authorized to operate in the same city or in nearby cities with a frequency separation of 800 kHz.

ª

A commercial broadcast entity may establish only one primary FM radio station within the geographical boundary of any province.

ª

FM transmitter shall operate satisfactorily in the operating power range with frequency swing of +75 kHz, which is defined as 100% modulation. The carrier shift at any percentage of modulation shall not exceed 5%.

B. .BROADCAST AUXILIARY SERVICES. 1.

Studio-to-Transmitter Link (STL)

Band

Operating Frequency

Maximum Power

A

300 to 315 MHz

15 W

B

734 to 752 MHz

15 W

C

942 to 952 MHz

15 W

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3.

Remote Pick-up Broadcast Station

Band

Operating Frequency

Maximum Power

A

315 to 325 MHz

35 W

B

450 to 451 MHz

35 W

C

455 to 456 MHz

35 W

Communications, Coordination and Control Link

Band

Operating Frequency

Maximum Power

A

4 to 12 MHz

100 W (SBS)

B

25.67 to 26.1 MHz

160 W (ERP)

C D

162.235 to 162.615 MHz 166.25 MHz 170.15 MHz 432.5 to 433 MHz 437.5 to 438 MHz

160 W (ERP) 200 W (for repeater)

C. .IMPORTANT TERMINOLOGY. 1.

Frequency Swing The instantaneous departure of the frequency of the emitted wave from the center frequency resulting from modulation.

2.

Antenna Height Above Average Terrain (AHAAT) The height of the radiation center of the antenna above the terrain 3 to 16 kilometers from the antenna.

3.

Antenna Field Gain The ratio of the effective free space field intensity produced at 1.6-km in the horizontal plane expressed in mV/m for 1 KW antenna input power to 137.6 mV/m.

4.

Antenna Power Gain The square of the ratio of rms free space field strength produce at 1.6km in the horizontal plane expressed in mV/m for 1 KW antenna input power to 137.6 mV/m.



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5.

Effective Radiated Power (ERP) The product of the transmitter power multiplied by (a) the antenna power gain

ERP = PT x GT

(b) the antenna field gain squared

ERP = PT x GF2


Determine the effective radiated power of 20 kW TV broadcast transmitter whose antenna has a field gain of 2.

Solution: ERP = PT x G F 2 = (20 x 10 3 W) x 2 2 = 80 kW

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I

H

A. .TV BROADCAST STANDARDS. Parameter Occupied Spectrum BW per Channel Visual Carrier Frequency Aural Carrier Frequency Chrominance Subcarrier Frequency Visual Modulation Aural Modulation Baseband Video BW Baseband Aural BW



54-890 MHz

54-890 MHz

6 MHz

6 MHz

1.25 MHz + 1 kHz above the lower boundary of the channel 4.5 MHz + 1 kHz above the visual carrier frequency

1.25 MHz + 1 kHz above the lower boundary of the channel 4.5 MHz + 1 kHz above the visual carrier frequency

3.579545 MHz + 10 Hz

3.579545 MHz + 10 Hz

AM with negative transmission FM using a 75μsec pre-emphasis

AM with negative transmission FM using a 75μsec pre-emphasis

4 MHz

4.2 MHz

50-15,000 Hz

50-15,000 Hz


2-80

BROADCASTING standards 1.

Characteristics of the NTSC Video Frequency


Parameter Number of lines/frame

525

Field frequency (field/sec) Interlace

2:1

Frame frequency (picture/sec) Horizontal frequency (lines/sec)

29.97 Hz 15,734.264 Hz

Tolerance (lines/sec)

+0.044

Scanning sequence

left-to-right

Scanning sequence

top-to-bottom

Approximate gamma of picture signal

0.45

Nominal Video BW

4.2 MHz

Chrominance sub-carrier Tolerance 2.

59.94 Hz

3.579545 MHz +10 Hz

Details of Line-Synchronizing Signals

Parameter

%H

μsec

Line Period (H)

100

63.556

16.5 to 18

10.5 to 11.4

2

1.27

6.6 to 8

4.2 to 5.1

Line Blanking interval Front Porch Synchronizing pulse


Details of Synchronizing Signals

Parameter

Duration

Field Period (V)

16.683 msec

Field Blanking Period

0.07 to 0.08V

st

Duration of 1 equalizing pulse Duration of synchronizing pulse Synchronizing pulse 4.

3H 3H 4.2 to 5.1

Radio Frequency Characteristics

Parameter Nominal RF Bandwidth Sound carrier-to-Visual carrier separation Nearest edge of channel relative to visual-carrier Nominal width of main sideband Nominal width of vestigial sideband Type of polarity of visual modulation Synchronizing levels as a % of peak carrier Blanking level as a % of peak carrier Difference between black level and blanking level as a % of peak carrier Peak-white level as a % of peak carrier Type of sound modulation

Philippine Technical Standards 6 MHz +4.5 MHz 1.25 MHz 4.2 MHz 0.75 MHz A5C, negative 100% 72.5 to 77.5% 2.875 to 6.75% 10 to 15% F3 + 25 kHz, 75 μsec pre-emphasis


2-81

2-82

BROADCASTING standards 5.

Maximum Power Allocation

Channel

Maximum Visual ERP in dB above 1-kW

2 to 6

20 dBk (100 kW)

7 to 13

25 dBk (316 kW)

21 to 32 62 to 69

37 dBk (5000 kW)

In Metro-Manila and Metro-Cebu, the maximum effective radiated powers of 350 KW for channels 2 to 6 and 1,000 KW for channel 7 to 13 are allowed.


Antenna towers represent hazard to aircraft. As a result they must be painted with equal width stripes of aviation orange (TT-P-59) and white (TT-P-102), each stripe approximately one-seventh the height of the tower.

ª

To mark the tower at night, towers up to 150 ft in height must have two steady burning 116 or 125 W (32.5-candela) lamps in an aviation red light globe at the top of the tower.

ª

For towers higher than 150 ft the top beacon light consists of two 620 or 700 W PS-40 Flashing Code beacon lamps (2000 candelas) with aviation red filters. Midway between flashing and top beacons, steady burning red beacon light are installed.

6.

Protection from Interference

Channel

Duration

2 to 6

48 dBu

7 to 13

55 dBu

21 to 32

65 dBu

62 to 69

70 dBu


2-83

B. .IMPORTANT TERMINOLOGY. 1.

Aspect Ratio The ratio of picture width to picture height as transmitted.

2.

Aural Transmitter The radio equipment for the transmission of the aural signal only.

3.

Aural Center Frequency The frequency of the emitted wave without modulation.

4.

Blanking Level The level of the signal during the blanking interval, except the interval between scanning synchronizing pulse and the chrominance subcarrier synchronizing burst.

5.

Chrominance The colorimetric difference between any color having a specific chromacity.

6.

Chrominance Subcarrier The carrier which is modulated by the chrominance information.

7.

Field Scanning through the picture area once in the chosen scanning pattern.

8.

Frame Scanning all of the picture area once, in the line interlaced scanning pattern of two to one, a frame consists of two fields.

9.

Luminance Luminous flux emitted, reflected, or transmitted per unit solid angle per unit projected area of the source.

10. Monochrome Transmission The transmission of TV signal which can be reproduced in gradations of single color only. 11. Color Transmission The transmission of color TV signal which can be reproduced with different hue, saturation, and luminance. 12. Negative Transmission Where a decrease in initial light intensity causes an increase in the transmitted power. 13. Reference Black Level The level corresponding to specified maximum excursion of the luminance signal in the black direction.


2-84

BROADCASTING standards 14. Reference White Level The level corresponding to specified maximum excursion of the luminance signal in the white direction. 15. Scanning The process of analyzing successively, according to a predetermined method, the light values of picture elements constituting the total picture area. 16. Visual Carrier Frequency The frequency of the carrier which is modulated by the picture information. 17. Visual Transmitter The radio equipment for the transmission of the visual signal only.

2-85


I

H

1.

One of the following is a possible cause of an abrupt frequency variation in a self exited transmitter oscillator circuits resulting to poor frequency stability to hold a constant frequency oscillation. NOV 1998 A. Heating of capacitor in the oscillator B. Heating and expansion of oscillator coil C. DC and RF heating of resistors which change in values D. Loose connections in the oscillator, amplifier, or antenna circuits

2.

It is an average power of a radio transmitter supplied to the antenna transmission line taken during a long sufficient interval of time and compared with the lowest frequency encountered in the modulation, taken under the normal operating conditions. NOV 1998 A. Rated power B. Mean power C. Carrier power D. Peak envelope power

3.

Determine the effective radiated power of 20 kW TV broadcast transmitter whose antenna has a field gain of 2. NOV 1998 A. 40,000 watts B. 80 kW C. 40 kW D. 8,000 watts

4.

Refers to one of the front end circuit of a VHF TV superheterodyne receiver which is usually a separate circuit coupled to the mixer. NOV 1998 A. Local oscillator B. AGC C. RF amplifier D. Antenna feed

5.

The actual vertical resolution of broadcast TV with reference to the number of horizontal lines. NOV 1999 A. 428 lines B. 485 lines C. 525 lines D. 339 lines

6.

How many visible horizontal lines are left due to vertical retrace interval? NOV

1999 A. C.

7.

525 lines 40 lines

B. D.

485 lines 500 lines

One of the following is the referred to as the number of scanning lines per frame for NTSC standard. NOV 1999 A. 625 lines B. 325 lines C. 425 lines D. 525 lines


2-86


8.

When does broadcast station conduct an equipment test? NOV 1999 A. At any time B. During night time C. During day time D. During experimental period

9.

One of the following is a high powered RF ac basic transmitter that has two or more stages, the oscillator stage which determines the frequency of operation and RF amplifier stage or stages that develop the high power output. The purpose of which is to develop a good frequency stability. NOV 1998 A. Hartley B. Alexanderson C. MOPA D. Goldsmith

10. One of the following is a possible cause for a slow drift of frequency in a selfexcited transmitter oscillator circuits resulting to poor frequency stability. NOV

1998

A. B. C. D.

heating of capacitor in the oscillator loose shielding faulty capacitor, resistor, tubes or resistors poor soldered connections

11. The final amplifier of either FM or AM transmitter operates as

NOV 1998 A. C.

class C class B

B. D.

class A class D

12. TV channels 2, 4 and 5 are known as ________. APRIL 1997 A. mid band UHF B. high band VHF C. low band UHF D. low band VHF 13. The allowable deviation ratio of commercial FM broadcast. APRIL 1997 A. 15 B. 5 C. 25 D. 75 14. ______ is the time duration for one horizontal trace. APRIL 1997 A. 62 μs B. 48 μs C. 52 μs D. 50 μs 15. An antenna that can only receive a television signal. APRIL 1997 A. TVRO B. Reference antenna C. Isotropic antenna D. Yagi antenna 16. Referred to as an average power of a radio transmitter supplied to the antenna transmission line taken during one radio frequency cycle under no modulation.

APRIL 1998

A. C.

Rated power Peak envelope power

B. D.

Carrier power Mean power

17. With which emission type is the capture-effect more pronounced? APRIL 1998 A. CW B. FM C. SSB D. AM


2-87

18. How can intermodulation interference between two transmitters in close proximity reduced or eliminated? APRIL 1998 A. Through installing a band-pass filter in the antenna feed line B. Through installing a terminated circular or ferrite isolator in the feed line of the transmitter and duplexer C. By using a class C final amplifier with high driving power D. By installing a low-pass filter in the antenna feed line 19. What is emission F3F? A. Facsimile C. Modulated CW

APRIL 1998 B. D.

Television RTTY

20. Referred to an oscillator signal “leak through” from a properly neutralized amplifier such as a master oscillator power amplifier. APRIL 1998 A. stray signal B. carrier C. back wave D. loss wave 21. Which frequency band is the standard AM radio broadcast? APRIL 1998 A. UHF B. MF C. HF D. VHF 22. The maximum power suggested by KBP on 919-1312 AM broadcast station in Metro Manila is. NOV 1996 A. 20 kW B. 5 kW C. 10 kW D. 15 kW 23. What is the maximum color TV bandwidth? NOV 1996 A. 1.6 MHz B. 1.3 MHz C. 0.5 MHz D. 1.0 MHz 24. Which of the following filters block FM radio band for TV channels (2 to 13)?

NOV 1996

A. C.

Low-pass filter Band reject filter

B. D.

High-pass filter Band-pass filter

25. The lowest resistance grounding on earth. NOV 1996 A. Sand B. Limestone C. Surface loam soil D. Clay 26. The method of generating FM used by broadcast station is.

NOV 1996 A. C.

direct indirect

B. D.

insertion all of these

27. The frequency spectrum of the stereophonic FM signal. NOV 1996 A. 67 kHz B. 19 to 38 kHz C. 59.5 to 74.5 kHz D. 30 to 53 kHz 28. After the IF stages have been aligned, the next stage to align in FM receiver is.

NOV 1996 A. C.

local oscillator limiter stage

B. D.

Mixer stage RF amplifier


2-88


29. Type of modulation used in TV broadcast visual transmitter. NOV 1996 A. Vestigial sideband B. DSBFC C. SSBFC D. Pulse modulation 30. The father of color television A. J.L. Baird C. Vladimir Zworykin

B. D.

Paul Nipkow Alexander Bain

2-89


Section

Microphones and

9

Loudspeakers


DEFINITION. Microphone is an energy converter that changes sound energy into electrical energy. A diaphragm in the microphone moves in and out in accordance with the compression and rarefaction of the atmosphere caused by sound waves. The diaphragm is connected to a device that causes current flow in proportion to the instantaneous pressure delivered to it. DEFINITION. Loudspeaker is device that converts an electrical signal into sound. A. .ELECTRO-ACOUSTIC TRANSDUCERS. 1.

Electromagnetic transducer The electromagnetic transducer consists of a permanent magnet, soft magnetic pole-shoe, an anchor and a spring.

2.

Electrodynamic transducer The electrodynamic transducer has a fixed air gap. Inside the air gap, there is a magnetic field in which a conductor is moving.

3.

Electrostatic transducer The electrostatic transducer is essentially a capacitor consisting of one fixed and one moving electrode.

4.

Piezoelectric transducer The piezoelectric transducer makes use of the ability of certain materials to produce unbalanced charge distribution on their surface when they are deformed. In the inverse effect, deformation appears when electrical field is applied.

B. .MICROPHONE BASIC. Microphone is an energy converter that changes sound energy into electrical energy. A diaphragm in the microphone moves in and out in accordance with the compression and rarefaction of the atmosphere caused by sound waves. The diaphragm is connected to a device that causes current flow in proportion to the instantaneous pressure delivered to it. Many devices can perform this function.


2-90

Microphones and loudspeakers 1.

2.

Characteristics ª

Frequency Response - The range of frequencies over which the microphone will normally pick up the sound.

ª

Sensitivity - Ability to pick-up high and low level signals.

ª

Directivity - Range pick-up i. Omni-directional ii. Bi-directional iii. Unidirectional

General Classification ª

According to Power: i. Generator - one which does not require external power source. ii.

ª

Modifier - requires an external power source.

According to manner of Coupling: i. Pressure type - conversion depends on the sound pressure acting on the microphone. ii.

Contact type - conversion depends on the transmitted vibration.

iii. Velocity type - conversion depends on the velocity of sound wave on the metallic ribbon. ª

According to Element used: i.

Carbon Microphone - Sound waves striking the diaphragm vary the pressure on the button which varies the pressure on the pile of carbon granules. The changing pressure deforms the granules, causing the contact area between each pair of adjacent granules to change, and this causes the electrical resistance of the mass of granules to change (lose contact). Since the voltage across a conductor is proportional to its resistance, the voltage across the capsule varies according to the sound pressure.

ii.

Crystal Microphone - The crystal microphone uses the piezoelectric effect of Rochelle salt, quartz, or other crystalline materials. This means that when mechanical stress is placed upon the material, a voltage electromagnetic force (EMF) is generated.


2-91

iii. Dynamic Microphone - A coil of fine wire is mounted on the back of the diaphragm and located in the magnetic field of a permanent magnet. When sound waves strike the diaphragm, the coil moves back and forth cutting the magnetic lines of force. This induces a voltage into the coil that is an electrical reproduction of the sound waves iv. Magnetic Microphone - the magnetic or moving armature microphone consists of a coil wound on an armature that is mechanically connected to the diaphragm with a driver rod. The coil is located between the pole pieces of the permanent magnet. Any vibration of the diaphragm vibrates the armature at the same rate. v.

Electret Microphone - An electret microphone is a relatively new type of condenser microphone invented at Bell laboratories in 1962 by Gerhard Sessler and Jim West, and often simply called an electret microphone.

vi. Ribbon Microphone - In ribbon microphones a thin, usually corrugated metal ribbon is suspended in a magnetic field: vibration of the ribbon in the magnetic field generates a changing current. Basic ribbon microphones detect sound in a bidirectional (also called a figure-of-eight) pattern because the ribbon, which is open to sound both front and back, responds to the pressure gradient rather than the sound pressure. C.

.LOUDSPEAKERS. Loudspeaker is a device that converts electrical signal into sound.

Loudspeaker types ª

Woofers A woofer is a loudspeaker capable of reproducing the bass frequencies. The frequency range varies widely according to design and hence while some woofers can cover the audio band from 50 Hz to 3 kHz, yet others may only work up to 1 kHz.

ª

Mid-ranges A mid-range loudspeaker, also known as a squawker is designed to cover the middle of the audio spectrum, typically from about 200 Hz to about 4-5 kHz.

ª

Tweeters A tweeter is a loudspeaker capable of reproducing the higher end of the audio spectrum, usually from about 1 kHz to 20 or 35 kHz.


2-92

Microphones and loudspeakers

ª

Full-ranges A full-range speaker is designed to have as wide a frequency response as possible. There exist full-range drivers with pure titanium diaphragm which are capable of reproducing a frequency range from 50 Hz to 20 kHz. Typically 2" to 5" in diameter.

ª

Subwoofers A subwoofer driver is a woofer optimised for the lowest range of the audio spectrum. A typical subwoofer only reproduces sounds below 120 Hz (although some subwoofers allow a choice of the cross-over frequency).

1.

Loudspeakers are basically comprised of two things: i. One or more drivers to push and pull the air in the room. ii. An enclosure (fancy word for "box") to make the driver sound and look good.

2.

Broad Categories i.

Dynamic a. Moving Coil - A moving coil loudspeaker uses a coil of wire suspended in a stationary magnetic field (complements of a permanent magnet). b.

ii. 3.

Ribbon - Ribbon loudspeakers use the same basic principle where you put current into a wire (or sheet of metal in this case...) and it creates a magnetic field.

Electrostatic Operates on the principles of electrostatic attraction.

Enclosure Design

i.

Dipole Radiator (no enclosure) a.k.a. Double radiator As the diaphragm moves "outwards" (i.e. towards the listener), the resulting air pressure at the "front" of the diaphragm is positive while the pressure at the back of the diaphragm is of equal magnitude but opposite polarity.


2-93

Infinite Baffle This is called an "infinite baffle because the diaphragm is essentially mounted on a wall of infinite dimensions.

iii. Finite Baffle A finite baffle causes the energy from the rear of the driver to reach the listener with a given delay time determined by the dimensions of the baffle.

iv. Folded Baffle A folded baffle is essentially a large flat baffle that has been "folded" into a tube which is open on one end (the back) and sealed by the driver at the other end as is shown below. v.

Sealed Cabinet (a.k.a. Acoustic Suspension) The acoustic suspension eliminates the resonance of an openback cabinet by sealing it up, thus turning the tube into a box. Now the air sealed inside the enclosure acts as a spring which pushes back against the rear of the diaphragm.

vi. Ported Cabinet (a.k.a. Bass Reflex) A bass reflex can be achieved by simply cutting a hole in sealed cabinet enclosure.


The word woofer in loudspeaker comes from ‘woofs’ which are infrasonic frequencies below 20 Hz.

ª

The word tweeter in loudspeaker comes from ‘tweets’ which are frequencies from 15, 000 Hz to 20, 000 Hz and extend beyond ultrasonic range.


2-94

Microphones and loudspeakers

Sample Problem:

A loudspeaker is driven by a voltage of 0.7 V (RMS). The resulting pressure (measured under standard conditions at a distance of 1 meter) is 0.3 Pa. Calculate the voltage sensitivity of the loudspeaker.

Solution: First, we calculate the sound pressure level for 0.7 Volts: ⎛ ⎞ 0 .3 ⎟ = 83.5 20 log⎜ ⎜ 20 x 10 −6 ⎟ ⎝ ⎠ The SPL rise for 2.83 volts is ⎡ 2 .83 ⎤ 20 log ⎢ ⎥ = 12 .1 ⎣ 0 .7 ⎦ Hence Voltage Sensitivity is 83.5 + 12.1 = 95.6 dB/2.83V @1m

Sample Problem:

A loud speaker with a sensitivity of 85dB/2.83V@1m is driven by 1.5V (RMS). What is the resulting SPL at a. 1 meter b. 5 meters

Solution: a.

⎛ 1.5 ⎞ ΔSPL = 20 log⎜ ⎟ = −5.5dB ⎝ 2.83 ⎠

hence SPL=85-5.5dB=79.5dB at 1meter b.

⎛1⎞ ΔSPL = 20 log⎜ ⎟ = −14dB ⎝5⎠

hence SPL=79.5-14dB=65.5dB at 5 meters

Section 10 Transmission Lines and Waveguides Section 11 Fiber Optics Communications Section 12 Telephone Networks and System Section 13 Facsimile Transmission

Wire Communications



Section

10

Transmission Lines and Waveguides

3-1


Transmission Lines: Metallic conductor system that is used to guide or transfer electrical energy from one point to another.

DEFINITION.

DEFINITION. Characteristic Impedance: Impedance seen looking at an infinitely long transmission lines or the impedance seen into a finite length of line that is terminated in a purely resistive load equal to the characteristic impedance of the line. Velocity Factor is defined as the ratio of the actual velocity of propagation through a given medium to the velocity of propagation through free space.

DEFINITION.

A. .2 BROAD CATEGORIES. 1.

Balance Impedance to ground from each of the two wires is equal and this ensures that the current in both wires are equal in magnitudes but opposite in sign or travel in opposite directions.

2.

Unbalance One wire is at ground potential, while the other is at signal potential.

B. .PARALLEL-CONDUCTOR TRANSMISSION LINES.

1.

Two-Wire Open Line The line consists of two wires that are generally spaced from 2 to 6 inches apart by insulating spacers. This type of line is most often used for power lines, rural telephone lines, and telegraph lines.


3-2

TRANSMISSION LINES and waveguides 2.

Two-Wire Ribbon (Twin Lead) This type of transmission line is commonly used to connect a television receiving antenna to a home television set. This line is essentially the same as the two-wire open line except that uniform spacing is assured by embedding the two wires in a low-loss dielectric, usually polyethylene.

3.

Twisted Pair As the name implies, the line consists of two insulated wires twisted together to form a flexible line without the use of spacers. It is not used for transmitting high frequency because of the high dielectric losses that occur in the rubber insulation.

4.

Shielded Pair Consists of parallel conductors separated from each other and surrounded by a solid dielectric. The conductors are contained within a braided copper tubing that acts as an electrical shield. The assembly is covered with a rubber or flexible composition coating that protects the line from moisture and mechanical damage.

Read it till it Hertz…jma Electrocution of the Human Body…

Contrary to popular belief, it is the current - not the voltage - level which is responsible for effects. According to Ohm's Law, of course, a certain voltage is required to cause the necessary currents to flow. Values show varies depending on the body.

Onset Current Level (mA) 1 8 10 13 21 20 38 42 70 90 100

Effect Threshold of sensation Mild sensation Painful Cannot let go Muscular paralysis Severe shock Breathing labored Breathing upset Extreme breathing difficulties Ventricular fibrillation Death


3-3

C. .CONCENTRIC or COAXIAL TRANSMISSION LINES.

1.

Rigid (Air) Coaxial Line The rigid coaxial line consists of a central, insulated wire (inner conductor) mounted inside a tubular outer conductor. The spacers are made of Pyrex, polystyrene, or some other material that has good insulating characteristics and low dielectric losses at high frequencies.

2.

Solid Flexible Coaxial Line Solid flexible coaxial lines are made with an inner conductor that consists of flexible wire insulated from the outer conductor by a solid, continuous insulating material. The outer conductor is made of metal braid, which gives the line flexibility.

D. .TRANSMISSION LINE LOSSES. 1.

Copper Losses One type of copper loss is I2R loss. In RF lines the resistance of the conductors is never equal to zero. Whenever current flows through one of these conductors, some energy is dissipated in the form of heat. This heat loss is a POWER LOSS.

Another type of copper loss is due to SKIN EFFECT where currents tend to flow near the surface of the conductor. A special type of wire called Litzendraht wire (Litz wire) is often used to reduce skin effect. 2.

Dielectric Losses Result from the heating effect on the dielectric material between the conductors.

3.

Radiation and Induction Losses Radiation and induction losses are similar in that both are caused by the fields surrounding the conductors. Induction losses occur when the electromagnetic field about a conductor cuts through any nearby metallic object and a current is induced in that object. As a result, power is dissipated in the object and is lost.


3-4


Coupling Loss Occurs whenever a connection is made to or from a transmission line or when two separate pieces of transmission lines are connected together.

E. .TRANSMISSION LINE EQUIVALENT CIRCUIT.

Distributed Parameters ª

Series Resistance (R)

ª

Series Inductance (L)

ª

Shunt Conductance (G)

ª

Shunt Capacitance (C)

The transmission line has electrical resistance along its length. This resistance is usually expressed in ohms per unit length and is shown as existing continuously from one end of the line to the other. When current flows through a wire, magnetic lines of force are set up around the wire. The energy produced by the magnetic lines of force collapsing back into the wire tends to keep the current flowing in the same direction. This represents a certain amount of inductance, which is expressed in nanohenrys per unit length. Since any dielectric, even air, is not a perfect insulator, a small current known as leakage current flows between the two wires. In effect, the insulator acts as a resistor, permitting current to pass between the two wires. This property is called conductance (G) and is the opposite of resistance which is expressed in Siemens per unit length. Capacitance also exists between the transmission line wires. Notice that the two parallel wires act as plates of a capacitor and that the air between them acts as a dielectric. The capacitance between the wires is usually expressed in picofarads per unit length.


3-5

F. .PRIMARY LINE CONSTANTS. They are constant in the sense that they do not vary with voltage and current; however, they are frequency dependent to some extent.

2-wire line

Coaxial line

S d

Types

Series Inductance (Henry/meter)

D

d

Series Capacitance (Farad/meter)

2-Wire

L≅

μ 2S ln π d

C≅

Coaxial

L≅

μ D ln 2π d

C≅

πε 2S ln d 2 πε D ln d

where : μ = Permeability of medium in H/m ε = Permittivity of medium in F /m S = center to center spacingbetween conductors in m d = diameter of conductor in m For Coaxial line D = Diameter of outer conductor in m d = diameter of inner conductor in m


3-6

TRANSMISSION LINES and waveguides

G. .SECONDARY LINE CONSTANTS. 1.

Characteristic Impedance (Zo) Impedance seen looking at an infinitely long transmission lines or the impedance seen into a finite length of line that is terminated in a purely resistive load equal to the characteristic impedance of the line. ª

In terms of Primary Line Constant

ZO =

where: Z = series impedance

Z Y

= R + jω L Y = shunt admittance = G + jω C

At Low Frequency

ZO =

ª

At High Frequency

R G

L C

ZO =

In terms of Physical Dimensions

2-Wire Line Z o = 276 log10

Coaxial S r

Zo =

138 εr

Sample Problem:

log10

D d

The primary line constant for a coaxial cable at a frequency of 10 MHz were determine approximately as follows; L = 234 nH/m, C = 93.5 pF/m, R = 0.568 Ω/m, G = 0. Determine the characteristic impedance.

Solution: Since f = 10 MHz ZO =

L = C

= 50 Ω

234 x 10− 9 93.5 x 10− 12


3-7

Propagation Constant (γ) Used to determine the reduction in voltage or current with distance as a TEM propagates down a transmission lines.

In terms of

ª

Series Impedance (Z) & Shunt Admittance (Y)

γ = ZY

Attenuation Coefficient (α) & Phase Shift Constant (β)

γ = α + jβ

Attenuation Coefficient (α)

α=

1.

Propagation Constant

R 2Z 0

α = attenuation coefficient in Neper/m R = series resistance in Ω /m

Phase Shift Coefficient (β) 1.

General Solution

β=

2.

2π λ

β = Phase shift coefficient in rad/m λ = Signal wavelength in m

In Free space (Homogenous)

In terms of Velocity of light (c) Permeability (μ) & Permittivity (ε)

Phase Shift Coefficient β=

ω c

β = ω με


3-8


In other medium (Non-Homogenous)

In terms of

Phase Shift Coefficient

Velocity of Propagation (Vp) Inductance (L) & Capacitance (C)

β=

ω Vp

β = ω LC

Sample Problem:

A signal will undergo a phase shift of how many rad/m when propagating on a 25-m coaxial cable with a velocity of 0.66c and operating at 5 MHz. Also compute for the total phase delay in degrees.

Solution:

Phase shift coefficient; β=

2π(5 x 106 ) rad ω = = 0.16 Vp m 0.66(3 x 108 )

Total phase delay in degrees; rad 180° βA° = 0.16 x x 25 m = 229.18° delay m π

H. .TRANSMISSION LINE PARAMETERS. 1.

Velocity Factor (F) Defined as the ratio of the actual velocity of propagation through a given medium to the velocity of propagation through free space.

In terms of

Velocity Factor

Light velocity and velocity of propagation

F=

Relative permittivity (dielectric constant)

F=

Refractive Index

F=

Vp c 1 εr 1 n


3-9


Calculate the velocity factor of a coaxial cable used as a transmission line, with the characteristic impedance of 50 ohms, capacitance is 40 pF/m, and inductance equal to 50 μH/m.

Solution: F= =

Vp c

=

1

1

∴ Vp =

c LC 1

LC

;

c (50 μH)(40 pF)

= 0.0745

2.

Electrical Length The length of a transmission line relative to the length of the wave propagating down the line.

β A° =

360°L λ

β A rad =

ωL λ

where : β A° = electrical length in deg β A rad = electrical length in rad L = physical length in m λ = operating wavelength in m

Sample Problem:

What length of standard RG-8/U coaxial cable would required to obtain a 45° phase shift at 300 MHz?

Solution: β A° =

A=

ω 360° 360° 360° xA = xA = = x A = 45° 0.66c Vp λ 0.66(3 x 108 ) f 300 x 106 45° 0.66(3 x 108 ) x = 0.0825 m 360° 300 x 106


3-10


Reflection A transmission line phenomena wherein a portion of the incident energy is delivered back to the source due to impedance mismatched.

4.

Reflection Coefficient (Γ) A vector quantity representing the ratio of reflected voltage to incident voltage or reflected current to incident current.

In terms of Reflected & Incident Signal Reflected Power & Incident Power Load Impedance & Characteristic Impedance

Reflection Coefficient

Γ=

Vref I = − ref Vinc Iinc Γ=

Γ=

Pref Pinc

ZL − Z 0 ZL + Z 0

Sample Problem:

Calculate the magnitude of reflection because of the mismatch between a 75Ω line and (50-j25)Ω load.

Solution: Γ =

ZL − ZO (50 − j25) − 75 −25 − j25 = = ZL + ZO (50 − j25) + 75 125 − j25

= −0.154 − j0.231 = 0.277∠ − 123° ⇒ Γ = 0.277

5.

Standing Waves An interference pattern that exists (because of impedance mismatched) when two sets of traveling waves going on opposite direction meets.


3-11

Standing Wave Ratio (SWR) A scalar quantity that represent the degree of impedance mismatch or defines as the ratio of the maximum voltage to the minimum voltage or the maximum current to the minimum current on a TL.

In terms of

Standing Wave Ratio

Reflected & Incident Signal

SWR =

Transmission line voltage measurement Load Impedance & Characteristic Impedance

Vinc + Vref Iinc + Iref = Vinc − Vref Iinc − Iref

SWR =

SWR =

ZL Z0

Incident and Reflected Power

Vmax Imax = Vmin Imin SWR =

ZL > Z 0

1+

Pref Pinc

1−

Pref Pinc

SWR =

Z0 ZL

Z 0 > ZL

Relationship between SWR and Γ

SWR =

1+Γ 1−Γ

Γ=

SWR − 1 SWR + 1

Sample Problem:

Calculate the SWR & reflection coefficient of the line if the forward power is 250 W, and the reverse power is 45 W.

Solution: Γ2 = SWR =

Pref ⇒Γ = Pinc

Pref = Pinc

45 = 0.424 250

1 + Γ 1 + 0.424 = = 2.47 1 − Γ 1 − 0.424


3-12


Return Loss (RL) The ratio of the power in the reflected wave to that in the incident wave.

RL =

7.

1 Γ2

RL dB = −20 log Γ

Transmission Loss (TL)

In terms of

Transmission Loss

Standing Wave Ratio

2 ⎧⎪ ⎡ SWR − 1 ⎤ ⎫⎪ TL dB = −10 log ⎨1 − ⎢ ⎥ ⎬ ⎣ SWR + 1 ⎦ ⎭⎪ ⎪⎩

Reflection Coefficient

TL dB = −10 log(1 − Γ2 )

Sample Problem:

A coaxial TL with a Z0 of 50-Ω is connected to the 50-Ω output of a signal generator, and also to a 20-Ω load impedance. Calculate the mismatch loss.

Solution: ⎧ Z − ZO ⎪ Mismatch(dB) = −10 log(1 − Γ2 ) = −10 log⎨1 − L ZL + ZO ⎪⎩

2⎫

⎪ ⎬ ⎪⎭

2 ⎧⎪ 20 − 50 ⎫⎪ = −10 log⎨1 − ⎬ = 0.88 dB 20 + 50 ⎪ ⎪⎩ ⎭

8.

Voltage and Current Transmission Coefficient (τV and τC) Voltage Transmission Coefficient (τV) The ratio of the transmitted wave voltage to the incident wave voltage.

τV = 1 +

Vref Vinc

τV = 1 + Γ


S O U R C E

3-13

L O A D

Incident Reflected

Transmitted

Current Transmission Coefficient (τC) The ratio of the transmitted wave current to the incident wave current.

τC = 1 +

Iref Iinc

τC = 1 − Γ

Sample Problem:

A 10 V positive going pulse is sent down a 50 m of lossless 50Ω cable with a velocity factor of 0.8. The cable is terminated with a 150Ω resistor. Calculate the voltage and current transmission coefficient and the amount of transmitted voltage to the load.

Solution: Γ =

ZL − ZO 150 − 50 1 = = ZL + ZO 150 + 50 2

τV = 1 + Γ = 1 + Vtransmitted = τ V x Vinc =

1 2

=

3 2

∴ τC = 1 − Γ = 1 −

1 1 = 2 2

3 (10) = 15 V 2

One may be puzzled to note that the transmitted voltage can be greater than the incident voltage; however, this is not of concern since the transmitted current will be less than the incident current. In fact, what is important is that the transmitted power is always less than (or equal to) the incident power irrespective of whether Γ is positive or negative. 9.

Transmitted Power (PT) The portion of the incident power consumed by the load or radiated by an antenna.

PT = Pinc − Pref

PT = Pinc (1 − Γ2 )


3-14


Sample Problem:

A generator sends a 75 mW down a 50Ω line. The generator is matched to the line, but the load is not. If the coefficient of reflection is 0.6, how much power is reflected and how much is dissipate in the load?

Solution: Pabs = Pinc − Pref = Pinc − Γ2Pinc = Pinc (1 − Γ2 ) = 75 x 10−3(1 − 0.62 ) = 48 mW Pref = Pinc − Pabs = 75 mW − 48 mW = 27 mW

10. Input Impedance (Zin) The impedance seen at the input of a lossless transmission line.

Zin = Z 0

ZL + jZ 0 tan β A Z 0 + jZL tan β A

Sample Problem:

Calculate the effective inductance seen at the input of an open circuit TL of length 0.12 m at 3 GHz. Assume Z0=75Ω, velocity factor of 0.65

Solution: Zin = ZO

ZL + jZO tan βA 2πf ⇒ ZL = ∞(open ckt); β = ZO + jZL tan βA Vp

⎫⎪ ⎧⎪ 2π(3 x 109 ) x 0.12⎬ = j51.77Ω = − jZO cot β A = − j(75) cot ⎨ 8 ⎪⎭ ⎪⎩ 0.65(3 x 10 ) from Z = R ± jX ⇒ XL = 51.77Ω = 2πfL L =

XL 51.77 = = 2.75 nH 2πf 2π(3 x 109 )


3-15

.TRANSMISSION LINE WAVE PROPAGATION. 1.

Infinite Transmission Line Condition

If the transmission is uniform and infinite, the wave in the +z (forward) direction will continue indefinitely and never return in the –z (reverse) direction. 2.

Matched Impedance Condition (ZO=RLOAD)

If the uniform transmission line is truncated and connected instead to a lumped resistive load RL = ZO, the entire +z wave(forward traveling wave) is dissipated in the load, which has the same effect as if an infinite line of characteristic impedance ZO were attached at the same point. This matched impedance condition is a unique situation in which all the power of the +z wave is delivered to the load just as if it were an infinite transmission line, with no reflected waves generated in the -z direction. 3.

Short-Circuit Load Condition (ZLOAD=0) For a shorted-load condition once the surge energy reaches the shorted end, the total voltage was compelled to be zero while the current can have any value, as determined by other constraints on the system.


3-16


4.

Open-Circuit Load Condition (ZLOAD=∞) For an open load condition, the incoming surge of energy cannot simply disappear, because there is nothing capable of dissipating the energy at this point. What happen is that the energy reflects from the open end of the line back to the load.

Read it till it Hertz…jma For open-load TL ª

The reflected voltage is the same (in phase) as the incident voltage and the total voltage at the destination end of the line is twice the incident voltage.

ª

But the incident current and reflected current is equal in magnitude but opposite in directions (out of phase) and the total current is zero.

For shorted-load TL ª

The reflected current is the same (in phase) as the incident current and the total current at the destination end of the line is twice the incident current.

ª

But the incident voltage and reflected voltage is equal in magnitude but opposite in directions and the total voltage is zero.

Self-Sufficient Guide to ECE by JASON AMPOLOQUIO J.

3-17

.COMPARISON BETWEEN MATCHED, OPEN, & SHORTED TL. Parameter

Matched Condition

Short-Circuit Condition

Open-Circuit Condition

Load Impedance

Zload = Z 0

Zload = 0

Zload = ∞

Vref = 0

Vref = − Vinc

Reflected Signal

Iref = 0 Vload = IL Z 0

Load Signal

Iload =

Voltage Reflection Coefficient Current Reflection Coefficient Standing Wave Ratio Return Loss

Input Impedance 5.

Iref = Iinc =

VL Z0

Iload 2

Vload = 0 (min) Iload = Iinc + Iref = 2Iinc (max)

Vref = Vinc =

Vload 2

Iref = − Iinc Vload = Vinc + Vref = 2Vinc (max)

Iload = 0 (min)

Γ=0

Γ = 1∠180° = −1.0

Γ = 1∠ 0° = +1.0

Γ=0

Γ = 1∠0° = 1.0

Γ = 1∠180° = −1.0

SWR = 1

SWR = ∞

SWR = ∞

RL = ∞

RL = 1

RL = 1

Zin = Z 0

Z in = jZ o tan(βA )

Zin = − jZ o cot(βA)

Mismatched Impedance Condition (ZO≠ZLOAD) For mismatched impedance conditions waves can exist traveling independently in either direction on a linear transmission line. If a wave in the –z (from the load to the source) direction is formed by a complete or partial reflection of the +z wave by some discontinuity such as a lumped load of RLOAD, the two waves are by definition coherent and an interference pattern will exist. i.

Voltage And Current Phasor At Any Point On The Line

V(z) = V + e − jβ z + Vo− e + jβ z o

I(z) = Io+ e − jβ z + I o− e + jβ z


3-18

TRANSMISSION LINES and waveguides ii.

Voltage & Current Relation

I0+ =

V0+ Z0

I0− = −

V0− Z0

iii. Distance of the 1st Voltage Minimum from the Load

dmin =

θ+π λ = (θ + π) 2β 4π

When θ=0, which occurs when ZL is purely real and greater than Z0, dmin=λ/4 and a voltage maximum exists right at the load. When θ=-π, which occurs when ZL is purely real and less than Z0, dmin=0 and a voltage minimum exists right at the load.

Sample Problem:

Calculate the voltage reflection coefficient, SWR, and determine the position of the first voltage minimum from the load which has an impedance of (15j20)Ω. (Use Z0=50Ω)

Solution: Γ=

ZL − Z0 (15 − j20) − 50 −35 − j20 = = ZL − Z0 (15 − j20) + 50 65 − j20

=

8.06∠ − 150.26° = 0.593∠ − 133.16° 13.6∠ − 17.1°

= 0.593e− j0.74π ∴ θ = −0.74π 1 + Γ 1 + 0.593 SWR = = = 3.914 1 − Γ 1 − 0.593 λ dmin = (θ + π) = λ (−0.74π + π) = 0.065λ 4π 4π


3-19

K. .TRANSMISSION LINE MATCHING.

1.

Quarter-Wave Transformer Matching A short piece (quarter wavelength) of wire inserted between the load and the transmission line. Disadvantages of using QWTs ª A TL must be placed between the load and the feedline. ª A special characteristic impedance for the QWT is required, which depends both on the load resistance and the characteristic impedance of the feedline. ª QWTs work perfectly only for one load at one frequency.

One Section

Z 0 ' = Z 0 ZL

Z 0 ' = impedance of the transformer in Ω Z 0 = impedance of the line in Ω ZL = impedance of the load in Ω

Two Sections

Z1 = 4 Z 03 xZL

2.

Z2 = 4 Z 0 xZL 3

Stub Matching i.

Single Stub Tuning Transmission line stubs is simply a piece of additional transmission line that is placed across the primary line to remove the reactive component of the load.


3-20

TRANSMISSION LINES and waveguides The transform load impedance at the stub position z=-d is

Z(z = − d) = Z 0

ZL + jZ 0 tan β d Z 0 + jZL tan β d

The distance d is chosen so that G=Y0, this condition leads to the solutions

d=

ii.

[

R XL ± ⎛⎜ L ⎞⎟ (Z0 − RL )2 + XL2 ⎝ Z0 ⎠ t= RL − Z0

λ tan−1 t 2π

]

Length Of Stub To Be Used a.

For Short-Circuit Stub

LS =

b.

λ ⎛Y ⎞ tan−1 ⎜ 0 ⎟ 2π ⎝ B ⎠

For Open-Circuit Stub

LS =

⎛ B ⎞ λ tan−1 ⎜ ⎟ 2π ⎝ Y0 ⎠

Sample Problem:

Determine the relative position of the stub needed to match the load with an impedance of 35-j47.5Ω to a TL with a characteristic impedance of 50Ω using a shunt, short-circuited single-stub tuner.

Solution: XL ± t =

{

RL (Z0 − RL )2 + XL2 Z0 RL − Z0 d=

}

− 47.5 ± =

{

}

35 (50 − 35)2 + 47.52 ⎧0.388 50 =⎨ 35 − 50 ⎩5.945

⎧1 tan−1(0.388) = 0.0589λ 1 ⎪⎪ tan−1 t = ⎨ 2π 2π ⎪ 1 tan−1(5.945) = 0.2235λ ⎪⎩ 2π

Self-Sufficient Guide to ECE by JASON AMPOLOQUIO L.

3-21

.MISCELLANEOUS TRANSMISSION LINE. 1.

Microstrip A miniaturized version of stripline best suited to circuit integration of semiconductor devices.

Parameter

Equation

Inductance

L=μ

T L

Capacitance

C=ε

L T

Z o = 120 π

Characteristic Impedance

Zo =

T T when << 1 L L

200 εr + 1.41

6T ⎡ ⎤ log ⎢ ⎥ ⎣ 0.8L + w ⎦

Z0 = characteristic impedance,Ω εr = dielectric constant T = dielectric thickness L = width of the conducting copper trace w = thickness of conducting copper trace

Sample Problem:

What is the value of Zo for a single 0.1-in wide, 0.005-in thick track plus groundplane microstrip line? Assume that the PC board is 0.075-in thick and that the dielectric constant of the board is 2.

Solution: Zo = =

200 εr + 1.41 200

2 + 1.41 = 78.4Ω

⎤ ⎡ 6T log⎢ ⎥ ⎣ 0.8L + w ⎦

⎡ ⎤ 6(0.075) log⎢ ⎥ ⎣ 0.8(0.1) + 0.005 ⎦


3-22

TRANSMISSION LINES and waveguides For Multi-track Microstrip

Z0 =

276 εr

⎡ πx ⎤ log ⎢ ⎥ ⎣L + w ⎦

Sample Problem:

A microstrip line is formed using a 0.095-in thick PC board (εr=1.8) with a bottom groundplane and a double 0.15-in wide, 0.008-in thick track on the top separated 0.2-in. What is the characteristic impedance?

Solution: Zo =

2.

276 εr

276 0.2π ⎧ πx ⎫ ⎧ ⎫ log⎨ log⎨ ⎬= ⎬ = 123.3Ω 1.8 ⎩L + w ⎭ ⎩ 0.15 + 0.008 ⎭

Stripline Consists of a printed conductor between two ground planes, typically formed from copper-clad polyethylene sheets.


Zo =

3-23

⎧ ⎫ ⎪⎪ ⎪⎪ 2T log ⎨ ⎬ w εr ⎪ L(0.8 + ) ⎪ ⎪⎩ t ⎭⎪

138

Z0 = characteristic impedance,Ω εr = dielectric constant T = dielectric thickness L = width of the conducting copper trace w = thickness of conducting copper trace t = distance between copper trace and the ground plane

Sample Problem:

A stripline is formed using multilayer board (εr=2). The center track is 0.15in wide and 0.005-in thick, and the PC board first layer thickness is 0.05-in thick, with an overall board thickness of twice the single layer. What is the characteristic impedance?

Solution: ⎧ ⎫ ⎧ ⎫ ⎪ ⎪ ⎪ ⎪ 2(2 x 0.05) 138 2T ⎪ ⎪ 138 ⎪ ⎪ log⎨ Zo = log⎨ ⎬= ⎬ ≈ 17Ω εr 2 ⎪ L ⎛⎜ 0.8 + w ⎞⎟ ⎪ ⎪ 0.15⎛⎜ 0.8 + 0.005 ⎞⎟ ⎪ ⎪⎩ ⎝ ⎪⎩ 0.05 ⎠ ⎪⎭ t ⎠ ⎪⎭ ⎝

3.

Balanced 4-wire

Z0 =

⎧ ⎪ 2D log10 ⎨ 2 εr ⎪ d ⎩

138

2⎤ ⎡ ⎢ 1 + ⎛⎜ D2 ⎞⎟ ⎥ ⎢ ⎝ D1 ⎠ ⎥⎥ ⎣⎢ ⎦


−1 ⎫

⎪ ⎬ ⎪ ⎭

3-24


Balanced, near ground

Z0 =

5.

Wires in parallel, near ground

Z0 =

6.

−1 ⎫ ⎧ 2⎤ ⎡ ⎛ D ⎞ ⎥ ⎪ ⎪ 2D ⎢ log10 ⎨ 1+⎜ ⎬ ⎟ εr ⎝ 2h ⎠ ⎥ ⎪ ⎪ d ⎢⎣ ⎦ ⎩ ⎭

276

−1 ⎧ 2⎤ ⎫ ⎡ ⎛ 2h ⎞ ⎥ ⎪ ⎪ 4h ⎢ log10 ⎨ 1+⎜ ⎬ ⎟ εr ⎝ D ⎠ ⎥ ⎪ ⎪ d ⎢⎣ ⎦ ⎩ ⎭

69

Balance 2-wire near ground

Z0 =

−1 ⎫ ⎧ ⎛ D2 ⎞ ⎤ ⎪ ⎪ 2D ⎡⎢ ⎥ log10 ⎨ 1 + ⎜⎜ ⎟⎟ ⎬ εr ⎪ d ⎢⎣ ⎝ 4h1h2 ⎠ ⎥⎦ ⎪ ⎩ ⎭

276


Single wire, near ground

Z0 =

8.

138 εr

⎛ 4h ⎞ log10 ⎜ ⎟ ⎝ d ⎠

Single wire between grounded parallel planes (ground return)

Z0 =

9.

3-25

138 εr

⎛ 4h ⎞ log10 ⎜ ⎟ ⎝ πd ⎠

Balanced line between grounded parallel planes

Z0 =

⎧ ⎡ πD ⎤ ⎫ ⎪ 4h tanh ⎢ 2h ⎥ ⎪ ⎪ ⎣ ⎦⎪ log10 ⎨ ⎬ π d εr ⎪ ⎪ ⎩⎪ ⎭⎪

276


3-26

TRANSMISSION LINES and waveguides 10. Balanced line between grounded parallel planes

Z0 =

M. .TRANSMISSION LINE SUMMARY.

276 εr

⎛ 2h ⎞ log10 ⎜ ⎟ ⎝ πd ⎠


3-27

N. .WAVEGUIDE THEORY. Waveguides are hollow metal “pipes” that guide electromagnetic waves.

1.

2.

Waveguide Dimension ª

Widest dimension of a waveguide is called the "a" dimension and determines the range of operating frequencies.

ª

The narrowest dimension determines the powerhandling capability of the waveguide and is called the "b" dimension.

Types of Electromagnetic (EM) Wave Propagation ª

Transverse Electromagnetic (TEM) The electric and magnetic field components are transverse to the direction of propagation.

ª

Transverse Electric (TE) All electric field components are transverse to the direction of propagation.


3-28


ª

3.

Transverse Magnetic (TM) All magnetic field components are transverse to the direction of propagation.

Boundary Conditions ª

For an electric field to exist at the surface of a conductor it must be perpendicular to the conductor.

An electric field CANNOT exist parallel to a perfect conductor. ª

For a varying magnetic field to exist, it must form closed loops in parallel with the conductors and be perpendicular to the electric field.

The magnetic field CANNOT exist normal to a perfectly conducting plate. 4.

Dominant Modes & Cut-off frequency

For TE10 mode (dominant) in rectangular waveguide

General Solution

fc =

2

⎛ mπ ⎞ ⎛ nπ ⎞ ⎜ a ⎟ +⎜ b ⎟ 2 π με ⎝ ⎠ ⎝ ⎠ 1

λc =

2

fc =

c 2a

2 2

2

⎛m⎞ ⎛n⎞ ⎜ a ⎟ + ⎜b⎟ ⎝ ⎠ ⎝ ⎠

λ c = 2a

where: m = # of

λ 2

variations of field in the "a" dimension

n = # of

λ 2

variations of field in the "b" dimension



A rectangular waveguide has a width of 1.2 in and a height of 0.7 in. waveguide will pass all signals above ____ GHz.

3-29

The

Solution: fc =

c 3 x 108 = = 4.92 GHz 2a 2(1.2 in x 0.0254 m ) 1 in


A rectangular waveguide has a width of 0.6 inch. Calculate the waveguide cut-off frequency.

Solution: fc =

c 3 x 108 = = 9.84 GHz 2a 2(0.6 x 0.0254 m ) 1 in

Sample Problem:

Determine the high pass cut-off of a rectangular waveguide which has a broad dimension wall of 1.5 inches.

Solution: ⎧ 0.0254 m ⎫ λ c = 2a = 2⎨1.5 in x ⎬ 1 in ⎩ ⎭ = 0.0762 m

c 3 x 108 = 2a 2(0.0762) = 3.94 GHz

fc =


3-30


Field patterns for the TE10 mode in rectangular wave guide

6.

Waveguide Velocities ª

Relation between Group and Phase velocity

νG = sin2 τ νP

ν G ν P = c2

where : c = speed of light τ = angle of incidence

ª

Group velocity (νG) The velocity of propagation of a wave along a waveguide is less than its velocity through free space (speed of light). This lower velocity is caused by the zigzag path taken by the wavefront.

The forward-progress velocity of the wavefront in a waveguide is called group velocity and is somewhat slower than the speed of light. In terms of

Equation 2

Waveguide Dimension

⎛ λ ⎞ νG = c 1 − ⎜ ⎟ ⎝ 2a ⎠

Critical Frequency

⎛f ⎞ νG = c 1 − ⎜ c ⎟ ⎝ f ⎠

2

Self-Sufficient Guide to ECE by JASON AMPOLOQUIO ª

3-31

Phase Velocity (νP) Phase velocity is the rate at which the wave appears to move along the wall of the guide, based on the way the phase angle varies along the walls.

In terms of

Equation

νP =

Waveguide Dimension

c ⎛ λ ⎞ 1−⎜ ⎟ ⎝ 2a ⎠ c

νP =

Cut-off Frequency

2

⎛f ⎞ 1−⎜ c ⎟ ⎝ f ⎠

2

Sample Problem:

Determine the group and phase velocities within a rectangular waveguide with an internal dimension of 1.52 x 0.9 in and is fed by a 12 GHz carrier using a coaxial probe.

Solution: 2

⎛ λ ⎞ νG = c 1 − ⎜ ⎟ ⎝ 2a ⎠

νGνP = c2 ⇒ νP =

7.

2

⎛ ⎞ 0.025 ⎟⎟ = c 1 − ⎜⎜ 2 x 0 . 03861 ⎝ ⎠ 2

c

283.84 x 106

= 283.84 x 106 m

= 317.08 x 106 m

s

s

Impedance and Guide Wavelength i.

Waveguide Impedance (Zo)

In terms of Waveguide Dimension

Cut-off Frequency

Equation

Zo =

Zo =

120 π ⎛ λ ⎞ 1−⎜ ⎟ ⎝ 2a ⎠

2

120 π ⎛f ⎞ 1−⎜ c ⎟ ⎝ f ⎠

2


3-32



Calculate the characteristic impedance of a waveguide with a cut-off frequency of 3.75 GHz, at a frequency of 5 GHz?

Solution: Zo =

ii.

120π 2

⎛f ⎞ 1 − ⎜⎜ c ⎟⎟ ⎝ f ⎠

=

120π 2

= 570Ω

⎛ 3.75 ⎞ 1−⎜ ⎟ ⎝ 5 ⎠

Guide Wavelength (λg)

In terms of

Equation λg =

Phase Velocity

λg =

Waveguide Dimension

Cut-off Frequency

λg =

νP f

λo ⎛ λ ⎞ 1−⎜ ⎟ ⎝ 2a ⎠

2

λo ⎛f ⎞ 1−⎜ c ⎟ ⎝ f ⎠

2

Sample Problem:

A 5.6 GHz microwave signal is propagated in a waveguide. Assuming that the internal angle of incidence to the waveguide surfaces is 42°. Calculate the wavelength of the signal in the waveguide.

Solution:

Compute first the phase velocity; Vp =

c 3 x 108 = = 4.5 x 108 m s sin τ sin 42°

then for the guide wavelength λg =

Vp f

=

4.5 x 108 5.6 x 109

= 0.08 m


3-33

O. .CIRCULAR WAVEGUIDES. 1.

TE (Transverse Electric) Mode The lower cutoff frequency (or wavelength) for a particular TE mode in circular waveguide is determined by the following equation:

λ c, mn =

2.

where :

2πr k mn

r = waveguide radius

m

km1

km2

km3

0

3.832

7.016

10.174

1

1.841

5.331

8.536

2

3.054

6.706

9.970

TM (Transverse Magnetic) Mode The lower cutoff frequency (or wavelength) for a particular TM mode in circular waveguide is determined by the following equation:

λ c,mn =

2 πr k 'mn

m

k’m1

k’m2

k’pm3

0

2.405

5.520

8.654

1

3.832

7.016

10.174

2

5.135

8.417

11.620


3-34


P. .MICROWAVE TUBES & SEMICONDUCTOR DEVICES. 1.

Microwave Tubes i.

Magnetron A combination of diode vacuum tube with built in cavity resonators and an extremely powerful permanent magnet.

Invented by Randall and Boot ii.

Klystron A microwave vacuum tube using two cavity resonators to produce velocity modulation.

Invented by the Varian brother before WWII Velocity modulation is the speeding and slowing down of the beam.

electron

Reflex Klystron - single cavity Klystron iii. Traveling Wave Tube It was invented by Rudolf Kompfner and developed into a viable device by J.R. Pierce and L.M. Field at Bell Lab. iv. Cross Field Tubes A cross-field microwave tube is a device that converts dc electric power into microwave power using an electronic energyconversion process similar to that of magnetron oscillator. 2.

Microwave Semiconductor Devices i.

Point Contact diode The oldest microwave semiconductor device, also called as CAT whisker.

Application: microwave mixer & detectors ii.

Schottky barrier (Hot-carrier) diode Application: microwave mixer & detectors

iii. Varactor (Voltage Variable Capacitor) The application of reverse bias controlled the thickness of the depletion region which in turn makes the diode a variable capacitor.

Application: harmonic generators (frequency multipliers), parametric amplification & electronic tuning.


3-35

iv. Step Recovery diode Design to store charged in the depletion region when it is conducting in the forward bias direction and produce a sharp step or snap pulse when reverse biased.

Application: harmonic generators (frequency multipliers), mixer v.

PIN (Positive Intrinsic Negative) diode Proposed by R.N. Hall in 1952 and was first recognized by Uhlir in 1958.

Application: microwave switches, modulators & protectors vi. IMPATT (Impact Avalanche and Transit Time) diode The electron and hole velocity increases so high that these carriers form additional holes and electrons by knocking them out of the crystal structure, so called impact ionization.

Application: microwave power generation or amplification vii. Backward diode (Tunnel rectifier)

Application: video detection circuit & low level mixing viii. MASER (Microwave Amplification by Stimulated Emission of Radiation) Developed by Charles Townes in 1954 and provided extremely low noise amplification of microwave signals by a quantum mechanical process.

Application: LNA(low-noise amplifier) ix. RIMPATT (Read Impact Avalanche and Transit Time) diode Used at very high frequency microwave oscillator and now commonly known as double-drift IMPATT. x.

TRAPATT (Trapped Plasma Avalanche Triggered Transit) diode Same operation with IMPATT but the drift velocity is much less than in an IMPATT diode.

Application: narrowband amplification

microwave

high

power

generation


or

3-36

TRANSMISSION LINES and waveguides xi. Gunn diode Exhibit negative resistance characteristic and has a threshold of 3.3kV/cm for oscillation to take place. Gun Effect is a bulk property of semiconductor that occurs in N-type materials only, so it is associated rather than holes.

The term diode is a misnomer for Gunn device since there is no junction, nor rectification involved. Application: microwave power generation or amplification, Police Radar, CW Doppler Radar xii. SAW (Surface Acoustic Wave) device SAW resonator employs thin lines etched on a metallic surface electrodeposited on a piezoelectric substrate, which is commonly quartz.

Application: narrowband bandpass filter


I

3-37

H

1.

What is the impedance of two sections of quarter-wave transformer (connected in series) needed in order to match a line 54 Ω to a load of 300 Ω A. 28.9 Ω, 958.4 Ω B. 82.9 Ω, 195.4 Ω C. 8.9 Ω, 15.4 Ω D. 182.9 Ω, 295.4 Ω

2.

A pattern of voltage and current variations along a transmission line not terminated in its characteristic impedance is called A. A magnetic field B. Radio waves C. An electric field D. Standing waves

3.

A transmitter is required to deliver 100 W to an antenna through 45 m of coaxial cable with a loss of 4 dB/100m. What must be the output power of the transmitter, assuming the line is matched? A. 34.51 W B. 345 W C. 1.51 W D. 151 W

4. A shorted quarter-wave line at the operating frequency acts like a(n) A. Capacitor B. Inductor C. Series resonant circuit D. Parallel resonant circuit 5.

A TDR display shows a discontinuity 1.4 μs from the start. If the line has a velocity factor of 0.8 how far is the fault from the reflectometer? A. 16.8 m B. 168 m C. 32.5 m D. 325 m

6.

Two adjacent minima on a slotted line are 23 cm apart. Find the frequency, assuming a velocity factor of 95%. A. 620 MHz B. 230 MHz C. 452 MHz D. 134 MHz

7.

The forward power in a transmission line is 150 W, and the reverse power is 20 W. Calculate the SWR on the line. A. 0.55 B. -1.6 C. 2.15 D. 0.15

8. A ratio expressing the percentage of incident voltage reflected transmission line is known as the A. Line efficiency B. Standing-wave ratio C. Velocity factor D. Reflection coefficient


on

a

3-38

9.

TRANSMISSION LINES and waveguides A transmission line of unknown impedance is terminated with two different resistances, and the SWR is measured each time. With a 75 Ω termination, the SWR measures 1.5. With a 300 Ω termination, it measures 2.67. What is the impedance of the line A. 1.12 Ω B. 32 Ω C. 50 Ω D. 112 Ω

10. A positive voltage pulse sent down a transmission line terminated in a shortcircuit: would would would would

A. B. C. D.

reflect as a positive pulse reflect as a negative pulse reflect as a positive pulse followed by a negative pulse not reflect at all

11. A 10 V positive going pulse is sent down a 50 m of lossless 50 Ω cable with a velocity factor of 0.8. The cable is terminated with a 150Ω resistor. Calculate the length of time it will take the reflected pulse to return to the start and the amplitude of the reflected pulse. A. 381 ns, 7.5 V B. 587 ns, 6 V C. 417 ns, 5 V D. 256 ns, 12 V 12. A shorted half-wave line at the operating frequency acts like a(n) A. Capacitor B. Inductor C. Series resonant circuit D. Parallel resonant circuit 13. A transmitter delivers 50 W into a 600 Ω lossless line that is terminated with an antenna that has an impedance of 275 Ω, resistive. How much power actually reaches the antenna? A. 43.1 W B. 10.71 W C. 22.42 W D. 38.43 W

14. A balanced load can be connected to an unbalanced cable: A. directly B. by using a filter C. by using a "balun" D. cannot be connected 15. A properly matched transmission line has a loss of 1.5 dB/m. If 10 W is supplied to one end of the line, how many watts reach the load, 27 m away? A. 9.8 W B. 6.1 W C. 7.8 W D. 9.1 W 16. A receiver requires 0.5 μV of signal for satisfactory reception. How strong must the signal be at the antenna if the receiver is connected to the antenna by 25 m of matched line having an attenuation of 6 dB/100m? A. 0.814 μV B. 0.671 μV C. 0.594 μV D. 0.36 μV 17. Calculate the impedance looking into a 50 Ω line 1 m long, terminated in a load impedance of 100 Ω, if the line has velocity factor of 0.8 and operates at a frequency of 30 MHz. A. 10 – j50 Ω B. 40 - j30 Ω C. 30 – j60 Ω D. 20 – j40 Ω


3-39

18. A positive voltage-pulse sent down a transmission line terminated in an opencircuit: would would would would

A. B. C. D.

reflect as a positive pulse reflect as a negative pulse reflect as a positive pulse followed by a negative pulse not reflect at all

19. A coaxial cable has a capacitance of 90 pF/m and a characteristic impedance of 50 Ω. Find the inductance of a 1 m length. A. 225 nH/m B. 22.5 nH/m C. 2.25 nH/m D. 225 pH/m 20. Find the characteristic impedance of a coaxial cable using a polyethylene dielectric with an inner conductor 2 mm in diameter and on outer conductor 8 mm in diameter. A. 548 Ω B. 5.48 Ω C. 5480 Ω D. 54.8 Ω

21. As frequency increases, the resistance of a wire: A. increases B. decreases C. stays the same D. changes periodically 22. What is the minimum impedance of a 2-wire parallel line? A. 128 Ω B. 83 Ω C. 130 Ω D. 89 Ω 23. The attenuation coefficient of a line is 0.0006 N/m. Determine the attenuation coefficient in dB/m A. 0.0521 dB/m B. 0.521 dB/m C. 0.00521 dB/m D. 5.21 dB/m

24. When analyzing a transmission line, its inductance and capacitance are considered to be: lumped equal reactances

A. C.

B. distributed D. ideal elements

25. The primary line constant for a coaxial cable at a frequency of 10 MHz were determine approximately as; L = 234 nH/m, C = 93.5 pF/m, R = 0.568 Ω/m, G = 0 Calculate the attenuation coefficient in dB/m. A. 0.00568 dB/m B. 0.493 dB/m C. 0.568 dB/m D. 0.0493 dB/m 26. When the load impedance matches the line impedance, the value of standing wave ratio and reflection coefficient is _____, _____ respectively. A. 1, 0 B. ∝, 1 C. -1, ∝ D. 0, -1

27. On a Smith Chart, you "normalize" the impedance by: A. assuming it to be zero B. dividing it by 2π C. multiplying it by 2π D. dividing it by Z0


3-40


28. What is the effective capacitance of a shorted TL 3λ/7 long at 25 MHz (assume Zo=54Ω)? A. 52.1 pF B. 245 pF C. 0.79 pF D. 4.5 pF

29. The velocity factor of a cable depends mostly on: A. the wire resistance B. the dielectric constant C. the inductance per foot D. dimension of the line 30. What is the characteristic impedance of a 3.6-cm parallel-strip line, which has a spacing 0.8 cm? A. 346.2 Ω B. 145 Ω C. 84 Ω D. 103 Ω 31. In ferrite, what does YIG mean? A. Yttrium-Immense-Garnet C. Yttrium-Iron-Gold

B. D.

Yttrium-Iron-Garnet Yttrium-Immense-Gold

32. Calculate the characteristic impedance of a uniform transmission line which has the following constants R=12 Ω/m, G=1.4 μS/m, L=1.5 μH/m, and C=1.4 nF/m at 7 kHz. A. 33-j2.5 Ω B. 18.5-j34 Ω C. 0.63-j16.7 Ω D. 27-j48.3 Ω 33. For a shorted load TL, the value of reflection coefficient and standing wave ratio is _____, _____ respectively. A. -1, ∝ B. 0, -1 C. ∝, 1 D. -1, 0 34. A transmitter supplies 50 W to a load through a line with an SWR of 2:1. Find the power absorbed by the load. A. 34.51 W B. 25 W C. 44.4 W D. 38.5 W 35. Determine the magnitude of the reflected voltage if a 10 V signal is applied to a 50 Ω coaxial transmission line terminated in a 200 Ω load. A. 8V B. 6V C. 7.5V D. 12V 36. If a finite length of TL is terminated in a load impedance equal to the characteristic impedance of the line this will appear as A. Open line B. Quarter wavelength line C. Shorted line D. Infinite line 37. RG-59U, a common type of transmission line for microwave applications, has an open circuit impedance of 150∠25° Ω and a short circuit impedance of 37.5∠25° Ω. What is the characteristic impedance of the line? A. 55∠-35° Ω B. 35∠-15° Ω C. 95∠-55° Ω D. 75∠-5° Ω

38. The impedance "looking into" a matched line: A. is infinite B. is zero C. is the characteristic impedance D. 50 ohms


3-41

39. Find the impedance on a 50 Ω transmission line at a distance of λ/8 from a 400 Ω load. A. 130∠-51° Ω B. 350∠-81° Ω C. 950∠-83° Ω D. 50∠-76° Ω 40. What is the effective inductance of an open-circuited stub 4λ/9 long at 42 MHz (assume Zo=50 Ω)? A. 76.34 nH B. 130 nH C. 520 nH D. 89.3 nH

41. The effect of frequency on the resistance of a wire is called: A. I2R loss B. the Ohmic effect C. the skin effect D. the flywheel effect 42. The A. B. C. D.

characteristic impedance of a cable depends on: the resistance per foot of the wire used the resistance per foot and the inductance per foot the resistance per foot and the capacitance per foot the inductance per foot and the capacitance per foot

43. The optimum value for SWR is: A. zero C. as large as possible

B. one D. there is no optimum value

44. Calculate the line wavelength in feet for a 1.8 GHz signal propagating in a medium with an index of refraction equal to 1.68 A. 0.325 ft B. 0.919 ft C. 0.546 ft D. 0.422 ft 45. Calculate the velocity factor for a transmission line with a series inductance of 280 nH/m and a shunt capacitance of 82.5 pF/m A. 1.24 B. 1.42 C. 0.693 D. 0.936 46. The ideal return loss value is ___ A. Infinity C. 0

B. D.

+1 -1

47. A non-optimum value for SWR will cause: A. standing waves B. loss of power to load C. higher voltage peaks on cable D. all of the above 48. Calculate the capacitance of a line with a dielectric constant of 2.21 and a series inductance of 321 nH/m A. 7.5 pF B. 76.5 pF C. 57.6 pF D. 67.5 pF 49. Calculate the characteristic impedance of a line with a shunt capacitance of 95 pF/m and having a dielectric medium with a 1.55 index of refraction A. 72.38 Ω B. 54.38 Ω C. 64.8 Ω D. 74 Ω


3-42


50. ____ determines the variation of current or voltage with distance along a TL A. Distributed parameters B. Reflection coefficient C. Propagation coefficient D. Characteristic impedance 51. Determine the amount of distributed resistance of a 15-m coaxial cable with a total capacitance of 1.4025 nF, total inductance of 3.51 μH and total attenuation of 0.7395 dB A. 0.569 Ω/m B. 5.69 Ω/m C. 569 Ω/m D. 56.9 Ω/m

52. For best matching, the load on a cable should be: A. lower than Z0 B. higher than Z0 C. equal to Z0 D. 50 ohms 53. Calculate the amount of phase shift velocity factor of 0.85 and operating at A. 0.296 rad/m C. 296 rad/m

coefficient in rad/m for a line with a 12 MHz. B. 29.6 rad/m D. 2.96 rad/m

54. Calculate the operating frequency that will produce a phase shift of 0.05 rad/m in a coaxial cable with a Zo of 55 Ω and a C=92.5 pF/m A. 43.7 MHz B. 5.43 MHz C. 81.2 MHz D. 1.56 MHz

55. A positive voltage pulse sent down a transmission line terminated with its characteristic impedance: would reflect as a positive pulse would reflect as a negative pulse would reflect as a positive pulse followed by a negative pulse would not reflect at all

A. B. C. D.

56. The ideal reflection coefficient value is ___ A. +1 B. Infinity C. 0 D. -1 57. What is the equivalent capacitance for a 50 Ω shot-circuited line 3λ/8 in length at 500 MHz? A. 56 pF B. 90.2 pF C. 6.4 pF D. 23.5 pF 58. Calculate the impedance of a quarter-wave transformer necessary to match a TL with a Zo of 75 Ω to a load of 125 Ω. A. 100 Ω B. 48.4 Ω C. 193.36 Ω D. 96.82 Ω 59. Find the characteristic impedance of the waveguide if the cut-off frequency is 3.75 GHz and will operate at 5GHz. A. 50 Ω B. 570 Ω C. 5.7 Ω D. 57 Ω

60. As frequency increases, the loss in a cable's dielectric: A. increases B. decreases C. stays the same D. there is no loss in a dielectric


3-43

61. Calculate the guide wavelength for the waveguide whose cut-off is 3.75 GHz and operates at 5 GHz. A. 98 cm B. 908 cm C. 9.08 cm D. 90.8 cm 62. A signal with a level of 20 dBm enters the main waveguide of a directional coupler. The coupler has an insertion loss of 1 dB, coupling of 20 dB, and directivity of 40 dB. Find the strength of the signal emerging from each guide. A. 19 dBm (main guide), 0 dBm (secondary guide) B. 19 dB (main guide), 0 dB (secondary guide) C. 0 dBm (main guide), 19 dBm (secondary guide) D. 0 dB (main guide), 19 dB (secondary guide) 63. For an open load TL, the value of return loss and reflection coefficient is _____, _____ respectively. A. -1, ∝ B. 1, 1 C. -1, 1 D. ∝, -1

64. A Smith Chart is used to calculate: A. transmission line impedances B. propagation velocity C. optimum length of a transmission line D. transmission line losses 65. A GUNN device has a thickness of 7 μm. At what frequency will it oscillate in the transit-time mode? A. 1.3 GHz B. 1.43 GHz C. 14.3 GHz D. 143 GHz 66. What is the RL in dB for a 50 Ω transmission line connected to a 100 Ω load? A. 5.65 dB B. 4.6 dB C. 8.13 dB D. 9.5 dB 67. A microstrip line is formed using a 0.095-in thick PC board (εr=1.8) with a bottom groundplane and a single 0.15-in wide, 0.008-in thick track on the top. What is the characteristic impedance? A. 112.14 Ω B. 72.4 Ω C. 85.34 Ω D. 66.8 Ω 68. Two series-connected λ/4 sections are line. The first λ/4 section has an impedance of the second λ/4 section. A. 438 Ω C. 283 Ω

used to match a 400-Ω load to a 100-Ω impedance of 141.4-Ω , calculate the B. D.

8333 Ω 338 Ω

69. Calculate the group and phase velocities for an angle of incidence of 33° A. 1.6x108 m/s, 5.51x108 m/s B. 2.8x108 m/s, 3.2x108 m/s C. 5.51x108 m/s , 1.6x108 m/s D. 3.2x108 m/s , 2.8x108 m/s 70. A rectangular waveguide has dimensions of 3cm x 5cm. Calculate the TE10 mode cut-off frequency. A. 0.3 GHz B. 3 GHz C. 13 GHz D. 33 GHz


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71. A waveguide with a 4.5 GHz cut-off frequency is excited with a 6.7 GHz signal. Find the wavelength in free space, and the wavelength in the waveguide. A. 0.716 m, 0.224 m B. 0.136 m, 0.448 m C. 0.448 m, 0.136 m D. 0.224 m, 0.716 m 72. Calculate the spacing of a 300-Ω folded dipole when ¾ inch tubing is used in its construction. A. 2.5 inches B. 4.5 inches C. 7.5 inches D. 8.5 inches 73. A 50-Ω short circuited line is 0.1λ in length, at a frequency of 500 MHz. Calculate the equivalent inductance. A. 24.5 nH B. 0.245 nH C. 245 nH D. 2.45 nH 74. A rectangular waveguide has a broad dimension wall of 0.9-in and is fed by a 10-GHz carrier from a coaxial cable. Determine the guide wavelength, phase, and group velocities. A. 2.264 cm, 2.264x108 m/s, 3.975x108 m/s B. 2.264 cm, 3.975x108 m/s, 2.264x108 m/s C. 3.975 cm, 3.975x108 m/s, 2.264x108 m/s D. 3.975 cm, 2.264x108 m/s, 3.975x108 m/s 75. At very high frequencies, transmission lines are used as A. Antennas B. Resistors C. Insulators D. Tuned circuits 76. An ideal directional coupler has a directivity of 25 dB and an isolation of 40 dB. What is its coupling value? A. 5 dB B. 15 dB C. 40 dB D. 65 dB 77. While we're on the subject of Smith charts, what is the impedance of the point at the dead center line? A. 0 - j50 Ω B. 50 ± j0 Ω C. 0 ± j0 Ω D. 0 ± j∞ Ω 78. What is the impedance seen at the input when the transmission line is open? jZ o tan(β A) B. − jZ o cot(β A) A.

C.

0

D.

∞

79. On a Smith chart, what does a point in the bottom half of the chart represent? A. A capacitive impedance B. Power saturation C. Resistive impedance D. An inductive impedance 80. What is the power of a 2 Vpk-pk sine wave across a 50 ohm load? A. +19.0 dBm B. +10.0 dBm C. -10.0 dBm D. -20.0 dBm 81. What are the minimum and maximum combined VSWR limits at an interface characterized by a 1.25:1 VSWR and a 2.00:1 VSWR? A. 3.75:1 (min), 1.25:1 (max) B. 1.60:1 (min), 2.50:1 (max) C. 0.75:1 (min), 3.25:1 (max) D. 1.75:1 (min), 2.25:1 (max)


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82. While we're on the subject of Smith charts, what is the impedance of the point at the far left edge of the center horizontal line? A. 50 ohm match B. Infinite ohms (open circuit) C. Zero ohms (short circuit) D. pure inductive reactance 83. While we're on the subject of Smith charts, what is the impedance of the point at the far right edge of the center horizontal line? A. 0 - j50 Ω B. 50 ± j0 Ω C. 0 ± j0 Ω D. 0 ± j∞ Ω 84. What is the impedance of free space? A. 75 Ω C. 377π Ω

B. D.

120π Ω 50 Ω

85. What happens to the noise figure of a receiver when a 10 dB attenuator is added at the input? A. Noise figure increases by 10 dB B. Noise figure doesn't change C. Noise figure decreases by 10 dB D. All of the above 86. While we're on the subject of Smith charts, what is the impedance of the point at the top edge of the center vertical line? A. pure inductive reactance B. open circuit C. short circuit D. pure capacitive reactance 87. What is the impedance seen at the input when the transmission line is shorted? jZ o tan(β A) B. − jZ o cot(β A) A.

C.

0

D.

∞

88. The rate at which the wave appears to move along the wall of the guide, based on the way the phase angle varies along the walls. A. Velocity of propagation B. Phase velocity C. Wave velocity D. Group velocity 89. The forward-progress velocity of the wavefront in a waveguide is called as ______. A. Velocity of propagation B. Phase velocity C. Wave velocity D. Group velocity 90. While we're on the subject of Smith charts, what is the impedance of the point at the bottom edge of the center vertical line? A. 0 - j50 Ω B. 50 ± j0 Ω C. 0 ± j0 Ω D. 0 ± j∞ Ω 91. A standing wave ratio is a measure of A. power out compared to noise out B. the amount of power received C. power radiated from the surface D. power out compared to reflected power back


3-46


92. What do you call the phenomenon in digital circuits that describe the duration of time digital signal passes a circuit? A. Elapsed time B. Travel delay C. Propagation delay D. Transmission time 93. An SWR reading which has a short circuit termination. A. One B. Infinity C. Zero D. Unstable 94. In an open wire transmission line, what is the normal separation between its two (2) conductors? A. 2.5 to 5 feet B. 2 to 3 meters C. 2 to 6 cm D. 0.001 to 0.01 cm 95. Which of the following term is used to describe the attenuation and phase shift per unit length of a transmission line? A. Propagation constant B. Degree of shift C. Phase shift D. Line constant 96. If voltage change is equal to twice its original what is the corresponding change in dB? A. 10 dB B. 3 dB C. 6 dB D. 9 dB 97. Energy that has neither been radiated into space nor completely transmitted A. Modulated waves B. Captured waves C. Standing waves D. Incident waves 98. To a couple a coaxial line, it is better to use a A. balun B. slotted line C. directional coupler D. quarter-wave transformer 99. What is the main reason why coaxial cable is not used in microwave signal transmission? A. Number of repeaters B. High attenuation C. Wide bandwidth D. Low impedance 100.

A. C. 101.

In wire communications, non-resonant transmission lines are referred to as Lose lines B. Flat lines Non-reactive lines D. Reactive lines

An electronic equipment used in radio communications to measure standing wave ratio A. Oscilloscope B. Reflectometer C. Wave meter D. Radio meter

102.

A. C.

How does a shorted half-wave line act a certain operating frequency? Series resonant circuit B. Capacitor Inductor D. Parallel resonant circuit


A. B. C. D.

Referred as the dielectric constant of a transmission line material Inductance and capacitance Propagation velocity Characteristic impedance Velocity factor

A. C.

The characteristic impedance of a transmission line does not depend upon its Conductor diameter B. Length Conductor radius D. Conductor spacing

103.

104.

105.

What is the input impedance equivalent of an open ended transmission line which is longer than a quarter wavelength? A. Equivalent to reactive circuit B. As pure inductor C. As resistive equivalent D. Open equivalent A. C.

In the study of transmission cable, twin lead is also referred to as a Double cable B. Open pair Ribbon cable D. Twisted pair

A. C.

The greater the diameter of a wire, the _____ is the resistance. unstable B. lesser stable D. higher

A. C.

Which determines the velocity factor in transmission line? The termination impedance B. Dielectrics in the line The line length D. The center conductor resistivity

106.

107.

108.

109.

Technical study which deals with production, transport and delivery of a quality signal from source to destination A. Communication System Engineering B. Telephony Engineering C. Telegraphic Engineering D. Transmission System Engineering

110.

A. C. 111.

3-47

The outer conductor of a coaxial transmission line is always grounded at the input and output B. output only point of high SWR D. input only

What is the meaning of the term velocity factor of a transmission line? The velocity of the wave on the transmission line multiplied by the velocity of light in vacuum B. The index of shielding for coaxial cable C. The ratio of the characteristic impedance of the line to the terminating impedance D. The velocity of the wave on the transmission line divided by the velocity of the light in a vacuum A.

112.

If the terminating impedance is exactly equal to the characteristic impedance of the transmission line the return loss is A. Infinite B. One C. Zero D. None of these


3-48


A. B. C. D.

Best To To To To

A. B. C. D.

Referred to as a cause of crosstalk Nonlinear envelope delay Improper level setting Electric coupling between transmission media Mechanical coupling between transmission media

113.

114.

115.

What is the term for the ratio of actual velocity at which a signal travels through a line to the speed of light in a vacuum? A. Velocity factor B. Characteristic impedance C. Standing wave ratio D. Surge impedance

116.

A. B. C. D. 117.

What is the main purpose of a communications system? To have a frequency assignment For modulation To provide an acceptable replica of the information at the destination None of these

In a transmission line, if the maximum current to minimum current ratio is 2:1 what is the ratio of the maximum voltage? A. 1:2 B. 2:1 C. 4:1 D. 1:4 A. C.

A single conductor running from the transmitter to the antenna Twin-lead B. Microstrip Single line wire D. RG-8/U

A. B. C. D.

Transmission line must be matched to the load to transfer maximum power to the load reduce the load current transfer maximum current to the load transfer maximum voltage to the load

A. B. C. D.

When electromagnetic waves are propagated through a waveguide, they travel through the dielectric without touching the walls travel along four walls of the waveguide travel along the broader walls of the guide are reflected from the walls but do not travel along them

A. C.

Which of the following determine the characteristic of a transmission line? Physical dimensions B. Capacitance Length D. Inductance

118.

119.

120.

121.

122.

reason for pressurizing waveguides with dry air reduce the possibility of internal arcing maintain temperature of the waveguide increase the speed of propagation maintain propagation

The standing wave ratio is equal to ____ if the load is properly matched with the transmission line. A. 50 B. 1 C. 0 D. 2


Characteristic impedance of a transmission line is the impedance measured at the ______when its length is infinite. A. output B. shorted end of the line C. midsection D. input

124.

A. C. 125.

3-49

When a transmission line uses ground return, it is called a/an _______line. balanced B. unbalanced ungrounded D. grounded

Transmission lines when connected to antenna have resistive load whose resistance is greater than the characteristic impedance of the line B. resistive load whose resistance is less than characteristic impedance C. resistive load at the resonant frequency D. capacitive load

A.

A. B. C. D.

Termination means a result of cutting both ends of a conductor looking back impedance of a line with no load load connected to the output end of a transmission line a result of disconnecting a line from a transmitter

A. C.

Transmission lines are either balanced or unbalanced will respect to positive terminal B. input ground D. negative terminal

126.

127.

128.

What is the velocity factor for non-foam dielectric 50 or 75 ohms flexible coaxial cable such as RG 8, 11, 58, and 59? A. 0.66 B. 0.10 C. 2.70 D. 0.30

129.

A. C. 130.

Propagation mode of microwave in a waveguide is known as TM B. SW TEM D. TE

A transmission line consisting of two conductors that have equal resistance per unit length A. Balanced line B. Open-wire line C. Unbalanced line D. Coaxial line

.


FIBER OPTICS COMMUNICATIONS

3-50

Section

Fiber Optics

11

Communications

DEFINITION.


Optics: Branch of physical science dealing with the propagation and

behavior of light.

Fiber Optics: The technology of transferring information, for example, in communications or computer technology, through a number of thin flexible glass or plastic tubes (optical fibers) using modulated light waves.

DEFINITION.

Optical Fibers: In the simplest form, they are cylindrical dielectric waveguides made up of central cylinder of glass (core) with one index of refraction, surrounded by an annulus (clad) with a slightly different index of refraction.

DEFINITION.


John Tyndall, a British physicist, demonstrated that light can be guided along a curved stream of water using Total Internal Reflection.

1880

Alexander Graham Bell experimented with an apparatus he called photophone.

1930

John L. Baird and C. W. Hansell were granted patent for scanning and transmitting television images through uncoated fiber cables.

1951

A.C.S. van Heel, H.H. Hopkins and N.S. Kapany experimented with light transmission through bundles of fiber that led to the development of the flexible fiberscope.

1956

N.S. Kapany coined the term “fiber optics”.

1958

Charles Townes and Arthur Schawlow wrote a technical paper about LASER and MASER.

1960

Theodore Maiman, built the first optical maser.

1967

K.C. Kao and G.A. Bockham proposed a cladded fiber cables.

1988

ANSI published the SONET standards.


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A. .BENEFITS OF FIBER-BASED SYSTEM. 1.

Tremendous Bandwidth - An optical fiber can easily support 100 Mbps while advanced systems are carrying beyond 1 Gbps.

2.

No Interference - The light pulses travel entirely within the fiber causing no harmful interference, known as electromagnetic interference (EMI) and radio-frequency interference (RFI) in nearby wire cables or adjacent optical fiber.

3.

Noise Immunity - The optical fiber system is also immune to nearby signals and EMI/RFI, regardless of interference magnitude

4.

No Electrical Hazard - There is complete electrical isolation between ends of the link. This eliminates ground loops which affects performance, as well as the danger of shock at one end if there is a misconnection or failure at the other end.

5.

Secure Communication - Since the light energy stays entirely within the fiber, the only way to intercept the signal is to tap physically into the line since there is no radiated energy field to intercept. Taps are difficult to accomplish physically, and a tap in the line causes a loss in signal power that can easily be detected.

6.

Safe to use in Dangerous Environment - Since there is no electrical energy present; fiber optics can be used wherever, even there is a danger of explosion from sparks.

7.

Lightweight - The weight and bulk of fiber optical cable is much less than the equivalent wire cables for the same effective bandwidth and number of users. Disadvantages: 1. Cost - The cost of the fiber is a little greater than that of basic copper wire in some configurations. 2.

Complex deployment & repairs - It is difficult to splice optical fibers to make them longer or to repair breaks.

3.

Complex connectors - Connectors for fibers are more complex to attach to the cable and require precise physical alignment.

4.

Complex network - Switching, routing and distribution of fiber optic signals are difficult.

5.

Complicated test equipments - Fiber-based system needs special test equipment.



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B. .NATURE OF LIGHT. 1.

Wave Nature of Light Light is an electromagnetic wave having a very high oscillation frequency and a very short wavelength.

In fiber optics and any other field of expertise concerning light signals, it is more pronounced to express it in wavelength rather than frequency. Relation between micron, nanometer and Angstrom

Unit

Value

micron

10-6 m

nanometer

10-9 m

Angstrom

10-10 m

Multiply by

To Obtain

103 m

nanometer

104 m

Angstrom

-3

10 m

micron

10 m

Angstrom

10-4 m

micron

10-1 m

nanometer

ECE Board Exam: APRIL 2003 a. b.

20 Angstrom is equal to how many microns? 100 Angstrom is equal to how many microns?

Solution: a. 20 Angstrom x

10 −4 micron = 0.002 micron 1 Angstrom

b. 100 Angstrom x

10 −4 micron = 0.01 micron 1 Angstrom

Self-Sufficient Guide to ECE by JASON AMPOLOQUIO i.

ii.

3-53

General Subdivision a.

Infrared Band of light wavelengths that are too long to be seen by the human eye. (770 nm to 100,000 nm)

b.

Visible Light Band of light wavelengths to which the human eye respond. (390 nm to 770 nm)

c.

Ultraviolet Band of light wavelengths that are too short to be seen by the human eye. (10 nm to 390 nm)

Wavelength, Frequency, & Amplitude a.

Wavelength The wavelength of a monochromatic wave is the distance between two consecutive wave peaks.

b.

Frequency Corresponds to the number of wavelengths that pass by a certain point in space in a given amount of time.

c.

Amplitude The amplitude of an electromagnetic wave corresponds to the maximum strength of the electric and magnetic fields and to the number of photons in the light.

iii. Wave Properties of Light a.

Reflection Phenomenon of wave motion, in which a wave is returned after impinging on a surface. When energy, such as light, traveling from one medium encounters a different medium, part of the energy usually passes on while part is reflected.

Rayleigh Criterion - States that if the cosine of the angle of incidence is greater than the ratio of the signal wavelength with respect to eight times of the irregularities height will result to specular reflection. Loading ECE SUPERBook


3-54

b.

Refraction The change in direction that occurs when a wave of energy such as light passes from one medium to another of a different density, for example, from air to water.

Less dense

Less dense More dense

Less dense More dense

More dense

If n1=n2, then θ1=θ2 and Vp1=Vp2 ------------------------------------If n1θ2 and Vp1>Vp2 ------------------------------------If n1>n2, then θ1<θ2 and Vp1
Read it till it Hertz…jma Lightning speed information… ª

Olaf Roemer, a Danish astronomer, made the first rough estimate of the speed of light in 1676 by measuring the length of time an eclipse occurred on one of Jupiter’s moon

ª

In 1849, the French physicist A.H. Fizeau developed the first nonastronomical method of measuring the speed of light with the use of an apparatus.

ª

Jean Bernard Leon Foucault, French physicist, improved Fizeau’s method of measuring the speed of light by substituting a rotating mirror for the toothed wheel

ª

The precise measurement of the speed of light was done by A.A. Michelson, an American physicist using the same method as Foucault’s. His calculated value of c is 2.997025 x 108 m/s.

ª

The designation c for the speed of light originates from the Latin word “celeritas”, which means velocity.

ª

The Scottish physicist, James Clerk Maxwell formulated equations combining the theories of electricity and magnetism and later proved that electromagnetic waves such as light travel at a speed of 3x108 m/s.


3-55

c.

Diffraction The bending or spreading out of waves as they pass around the edge of an obstacle or through a narrow aperture.

d.

Absorption The reduction in the intensity of radiated energy within a medium caused by converting some or all of the energy into another form.

e.

Dispersion The separation of visible light or other electromagnetic waves into different wavelengths.



3-56

2.

Particle Nature of Light Light behaves as though it were made up of very small particles called photons. Energy of a Single Photon in Joules (J) & electron-Volt (eV)

E(J) = hf

E(eV) =

1.241 λ

where: h = Planck's constant = 6.625 x 10-34 Js f = Frequency in Hz λ = wavelength in μm

Sample Problem:

Find the number of photons incident on a detector in 1 s if the optic power is 1μW and the wavelength is 0.8μm.

Solution: Energy of a single photon EP = hf = =

hc λ

(6.625 x 10 − 34 Js)(3 x 10 8 m/s ) 0.8 μm

= 2.48 x 10 − 19 J Compu ting for Total Energy E = Power x time = 1 μW x 1 s = 10 − 6 J The number of photon required is E 10 − 6 J = EP 2.48 x 10 − 19 J = 4.03 x 1012 photons Answer : 4.03 x 1012 photons


3-57

Ray Theory of Light A number of phenomena are adequately explained by considering light as narrow rays and this area of optical science concerns the application of laws of reflection and refraction of light in the design of lenses. ª ª ª ª

In a vacuum, ray travel at a velocity of 3x108 m/s. Rays travel in a straight path unless deflected by some change in the medium. When ray is reflected, the angle of incidence is equal to the angle of refraction (Specular Reflections) If any power crosses the boundary (refraction occurs), the transmitted ray directions is given by Snell’s Law or Fresnel Law.

C. .LENSES. Lens, in optical systems, glass or other transparent substance so shaped that it will refract the light from any object and form a real or virtual image of the object. 1.

Ray Paths through a Thin Lens.

ª ª ª ª 2.

Rays traveling through the center of the lens are undeviated. Ray1 Incident rays traveling parallel to the lens axis pass trough the focal point after emerging from the lens. Ray2 An incident ray traveling parallel to a central ray in the focal plane after transmission through the lens. Ray3 An incident ray passing trough the focal point travels parallel to the lens axis after it merges from the lens. Ray4

Object Position The position of the objects and focused image are related by the thin lens equation.



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αo

1 1 1 + = do di f

αim

M=

di do

where: do = Object distance di = Image distance M = Magnification factor

Sample Problem:

Find the object and image distance for a lens with a magnification factor equal to 4 and the focal length is 20 mm.

Solution: If M is equal to 4

Comp uting for di

d M= i = 4 do

1 1 1 + = di 4di 20

di = 4do

5 1 = 4di 20

Answer : di = 25 mm , do = 6.25 mm


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Beam Focusing i. Spot Size (wo) The distance at which the beam intensity has dropped to 1/e2 of its peak value.

w0 =

ii.

λf wπ

where : wo = spot size in mm w = collimated spot size

Beam Diffraction (θ) For longer distances, diffraction theory shows that the beam diverges at a constant full angle.

θ=

2λ wπ

where : θ = divergence angle in rad

Sample Problem: Consider a Gaussian beam whose spot size is 1 mm when collimated. The wavelength is 0.82 μm. Compute the divergence angle. Also compute for the spot size at 10 km. Solution:

The divergence angle is 2λ 2(0.82 x 10−6 ) θ= = wπ π(10−3 ) = 0.522 x 10−3 rad ≈ 0.03D

At 10 km, the spot size is λfL (0.82 x 10−6 )(10 km) = wπ π(10−3 ) = 2.6 meters

wo =



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D. .OPTICAL CONFIMENT IN A FIBER. 1.

Critical Angle (θc) Defined as the minimum angle of incidence at which a light ray may strike the interface of two media and results in an angle of refraction of 90° or greater.

θC = sin−1

2.

n2 n1

Snell’s Law This important law, named after Dutch mathematician Willebrord Snell, states that the product of the refractive index and the sine of the angle of incidence of a ray in one medium is equal to the product of the refractive index and the sine of the angle of refraction in a successive medium. where:

n1 sin θ1 = n2 sin θ2

n1 = index of the 1st medium n2 = index of the 2nd medium θ1 = angle of incidence in degrees θ2 = angle of refraction in degrees

Other Relation

General Solution

In terms of propagation velocity

v2 sin θ2 = v1 sin θ1

In terms of velocity factor

F2 sin θ2 = F1 sin θ1

In terms of dielectric constant

k 1 sin2 θ2 = k 2 sin2 θ1


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Index of Refraction (n) The refractive index of a substance measures how the substance affects light traveling through it. It is equal to the speed of light in a vacuum divided by the speed of light in that substance. Substance

Refractive Index

Vacuum Air Ice Water Ethyl Alcohol Magnesium Fluoride Glass (fused quarts) Glass (crown) Sodium Chloride (salt) Diamond

1.0000 1.0003 1.309 1.33 1.36 1.38 1.46 1.52 1.54 2.42

Sample Problem: What striking angle in the fiber surface is needed to produce a minimum angle of incidence (critical angle) between the core-clad boundary that will effectively confine the light signals within the fiber if n1=1.55 and n2=1.45? Solution:

θ Computing for the Critical angle θC = sin−1

n2 n2

1.45 1.55 = 69.3° = sin−1

Computing for the striking angle n sinθ0 = 1 sinθ1 ⇒ θ1 = 90° − θc n0 ⎧1.55 ⎫ θ0 = sin−1 ⎨ sin(90° − 69.3°)⎬ ≈ 1 ⎩ ⎭ = 33.21°



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Sample Problem:

Calculate the resulting deviation from the normal line for a light signal that travels from air-to-glass-to-diamond if the striking angle is 38°.

Solution: From air - to - glass : n1 = air, n2 = glass ⎧n ⎫ ⎧ 1 ⎫ θ2 = sin−1 ⎨ 1 sin θ1 ⎬ = sin−1 ⎨ (sin 38°)⎬ = 24.94° ⎩1.46 ⎭ ⎩ n2 ⎭ From glass - to - diamond : n1 = glass, n2 = diamond ⎧n ⎫ ⎧1.46 ⎫ θ'2 = sin−1 ⎨ 1 sin θ1 ⎬ = sin−1 ⎨ sin(24.94°)⎬ = 14.74° ⎩ 2.42 ⎭ ⎩ n2 ⎭

4.

Index Profile A graphical representation of the value of the refractive index across the fiber.

5.

Numerical Aperture (NA) The figure of merit used to describe the light gathering or lightcollecting ability of an optical fiber.

NA = n12 − n22

where : n1 = index of the 1st medium n2 = index of the 2nd medium

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Acceptance Angle or Acceptance Cone Half Angle (θmax) The maximum angle in which the external light rays may strike the air/fiber interface and still propagate down the fiber.

θMAX = sin−1 ( NA ) = sin−1 n12 − n22

7.

Fractional Index Change (Δ) The normalized difference between the index of the core and cladding.

General Solution

Δ=

Approximate Solution

n12 − n22

Δ≈

2n12

n1 − n2 n1

E. .OPTICAL FIBER WAVEGUIDES.

FIBER TYPES

SINGLE MODE

MULTI MODE

STEP INDEX

GRADED INDEX



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1.

Single-Mode Step-Index Fiber

i.

Maximum Radius for Single Mode propagation (V=2.405)

General Solution rmax =

Alternate Solution

0.383λ NA

rmax =

0.383λ n1 2 Δ

Sample Problem:

Calculate the maximum core radius to support single mode operation for a fiber with a NA of 0.15 and λ=0.82μm.

Solution: rmax =

2.

0 .383 λ 0 .383 (0 .82 μm) = = 2 .1μm NA 0 .15

Multi-Mode Step-Index Fiber

i.

Typical Step-Index Fiber Characteristics

Construction

n1

n2

NA

α

Δ

All-Plastic PCS SCS

1.49 1.46 1.48

1.41 1.40 1.46

0.48 0.41 0.24

29°.0 24.2° 13.9°

0.0540 0.0410 0.0135


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Normalized Frequency (V-parameter)

General Solution V=

πd NA λ

Alternate Solution V=

πd n1 2 Δ λ

iii. Number of Modes (M)

#M ≈

1 2 V 2

Sample Problem:

Compute the number of modes for a fiber whose core diameter is 50 μm. Assume that n1=1.48, n2=1.46, and λ=0.82μm.

Solution: πd π(50 ) NA = x 1 .48 2 − 1 .46 2 = 46 .45 λ 0 .82 Number of modes 1 1 # M ≈ V 2 ≈ (46 .45)2 ≈ 1,079 modes 2 2 V =

3.

Multi-Mode Graded-Index Fiber (GRIN) The graded-index core fiber has a core material whose refractive index varies with distance from the fiber.

F. .FIBER LOSSES. 1.

Absorption Absorption is caused by three different mechanisms: a. Absorption by atomic defects in the glass composition. b. Extrinsic absorption by impurity atoms in the glass material.



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2.

c.

Intrinsic absorption by the basic constituent atoms of the fiber material.

ª

Material Absorption Are those due to the molecules of the basic fiber material either glass or plastic that can be overcome only by changing the fiber material.

ª

Ultraviolet absorption Caused by valence electrons in the silica material from which fibers are manufactured.

ª

Infrared absorption Result of photons of light that are absorbed by the atoms of the glass core molecule.

ª

Ion resonance absorption Caused by OH- ions in the material that has been trapped in the glass during manufacturing process which can be minimized by drying the glass in chlorine gas to leach out the water vapor

ª

Hydrogen Effects The hydrogen either can interact with the glass to produce hydroxyl ions and their losses or it can infiltrate the fiber and produce its own loss. The solution is to eliminate the hydrogenproducing source or to add coating to the fiber that is impermeable to hydrogen.

Scattering Losses Occurs when a wave interacts with a particle in a way that removes energy in the directional propagating wave and transfer it to other directions. i.

Linear Scattering Primarily characterized by having no change in frequency in the scattered wave ª

Rayleigh Scattering Results from light interacting with the inhomogeneities (submicroscopic irregularities) in the medium that are much smaller than the wavelength of the light.

L=

0.887 λ4

where : λ = signal wavelength in μm


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Sample Problem:

Calculate the Rayleigh scattering loss in dB for a 50/125 step-index fiber operating at 1200 nm. Also compute for the attenuation in neper.

Solution: The Rayle igh Loss is ⎛ 0 .887 L dB = −10 log ⎜⎜ ⎝ λ4 ⎛ 0 .887 = −10 log ⎜⎜ ⎝ 1 .2 4 = 3 .68 dB

ª

ii.

The attenuation is 1 Neper 8.686 dB 1 Neper = 3.68 dB x 8.686 dB

⎞ ⎟⎟ ⎠ ⎞ ⎟⎟ ⎠

α = L dB x

= 0.423 Neper

Mie Scattering Occurs at inhomogeneities that are comparable in size to a wavelength and can be reduce by carefully controlling the quality and cleanliness of the manufacturing process.

Non-Linear Scattering High values of electric field within the fiber lead to the presence of non-linear scattering interactions that causes significant power to be scattered in the forward, backward, or sideways directions. ª

Brillouin Scattering Modeled as a modulation of the light by the thermal energy in the material mainly in the backward directions.

PB = (17.6 x 10 −3 )(a2 )(λ 2 )(α)(Δυ)

where : λ = signal wavelength in μm a = core radius in μm α = signal attenuation Δυ = line frequency in THz



3-68

Sample Problem:

Consider an 8/125 single-mode fiber operating at 1300 nm with a loss of 0.8 dB/km. The line width of the source is 0.013 nm. Calculate the Brillouin scattering threshold.

Solution: Δυ =

c 2

λ

(Δ λ ) =

3 x 10 8 (1300 x 10 − 9 )2

(0 .013 x 10 − 9 )

= 2 .31 x 10 9 Hz Brillouin Scattering PB = (17 .6 x 10 − 3 )( a2 )( λ2 )( α )( Δ υ) = (17 .6 x 10 − 3 )( 42 )(1 .32 )(0 .8)(2 .31) = 0 .879 Watts

ª

Raman Scattering The non-linear interaction produces a high-frequency phonon and a scattered photon predominately in the forward directions.

PR = (23.6 x 10−2 )(a2 )(λ2 )(α)

Sample Problem:

Consider an 8/125 single-mode fiber operating at 1300 nm with a loss of 0.8 dB/km. The line width of the source is 0.013 nm. Calculate the ratio of the Brillouin scattering threshold to the Raman scattering threshold.

Solution: PR = (23 .6 x 10 −2 )( a2 )( λ2 )( α ) = (23 .6 x 10

−2

2

2

)( 4 )(1 .3 )(0 .8)

= 5 .105 Watts

Ratio of threshold power PB 0.879 = PR 5.105 = 0.172 = 17.2%

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Macrobending Refers to a large-scale bending, such as that which occurs intentionally when wrapping the fiber on a spool or pulling it around a corner.

4.

Microbending Occurs when a fiber is sheathed within a protective cable. The stresses set up in the cabling process cause small axial distortions to appear randomly along the fiber.

Developed during deployment of the fiber, or can be due to local mechanical stresses placed on the fiber often referred to as cabling or packaging losses.

Critical Radius of Curvature

rCRITICAL =

3n2 λ

4 π ( NA )

3

=

0.24n2 λ NA 3

Sample Problem:

Calculate the critical radius of curvature for a multimode 50/125 fiber with an NA of 0.2, n2 of 1.48 and operating at 850 nm.

Solution: rCRITICAL = =

0.24n2λ NA3

(

0.24(1.48) 850 x 10−9

0.23 = 37.74 μm

)



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5.

Connector Losses

ª

Lateral misalignment Lateral misalignment loss is simply due to the non-overlap of the transmitting and receiving fiber cores

ª

Angular misalignment

θ

Coupling Loss (LdB)

n θ ⎞ ⎛ L = −10 log ⎜ 1 − o ⎟ πNA ⎠ ⎝ where: θ = misalignment angle in radians n0 = refractive index of the material filling the groove

Sample Problem:

Calculate the coupling loss for a fiber facility with a misalignment angle of 2.4° and 0.24 NA.

Solution: n θ ⎞ ⎛ L = −10 log ⎜⎜1 − o ⎟⎟ π NA ⎠ ⎝ ⎛ ⎜ (1)⎛⎜⎜ 2.4D x π D ⎜ 180 = −10 log ⎜1 − ⎝ x 0 . 24 π ⎜ ⎜ ⎝ = 0.248 dB

⎞⎞ ⎟⎟ ⎟ ⎠⎟ ⎟ ⎟ ⎟ ⎠


Gap between ends

ª

Non flat ends (Imperfect finish)

3-71

G. .PULSE SPREADING IN FIBER. Dispersion The spreading (in time-domain) of light pulses as it propagates down the fiber end. 1.

Material Dispersion (DM) Pulse at different wavelengths has different velocities.

Δt MAT = DM x Δλ A km DM = Dispersive coefficient in ps /nm − km Δλ = −3dB wavelength (line or spectral width) in nm



3-72

Sample Problem:

For the step-index fiber 12.5 km long is to be used with a 0.8μm light source with a spectral width of 1.5 nm. What value of material dispersion might be expected assuming DM = 0.15ns/nm-km.

Solution: ΔtMAT = DM xΔλ A 0.15 ns = x 1.5nm nm − km 0.225 ns = km

2.

For 12.5 km length 0.225 ns x 12.5 km km = 2.81 ns

ΔtMAT =

Waveguide Dispersion (Chromatic dispersion) Pulses at different wavelengths (but propagating in the same mode) must travel at slightly different angles.

Δt WAVE = DW x Δλ A km DW = Peak Waveguide Dispersive coefficient in ps /nm − km 6.6ps nm − km Δλ = −3dB wavelength (line or spectral width) in nm =

3-73


Sample Problem:

A 12.5-km single-mode fiber is used with a 1.3μm light source which has a spectrum width of 6 nm. Find the total expected waveguide dispersion.

Solution: Δt WAVE = DM xλ A 6.6 ps = x 6nm nm − km 39.6 ps = km

3.

For 12.5 km length 39.6 ps x 12.5 km km = 495 ps

Δt WAVE =

Modal Dispersion (Modal Delay Spreading) A pulse at a single wavelength splits power into modes that travel at different axial velocities because of the path differences.

Δt MODAL =

Ln1 Δ Ln Δ ≈ 1 c 1−Δ c

Sample Problem:

Consider a 50/125 step-index fiber with n1=1.47 and Δ=1.5%. Calculate the group delay (modal dispersion) for this fiber at an operating wavelength of 850 nm.

Solution: ΔtMODAL ≈

Ln1Δ (12.5 km)(1.47)(0.015) ≈ 918.75 ns ≈ c 3 x 105 km / s



3-74

4.

Total Dispersion At any wavelength the total dispersion is the root mean square combination of material, modal and waveguide dispersion.

Δt TOTAL = Δt MAT 2 + ΔtMODAL2 + Δt WAVE2

:FACTS TO REMEMBER:. Modal dispersion is only present for multi-mode fiber.

Sample Problem:

A single-mode fiber operating at 1.3 μm is found to have a total material dispersion of 2.81 ns and a total waveguide dispersion of 0.495 ns. Determine the receive pulse width approximate bit rate for the fiber if the transmitted pulse has a width of 1.5 ns.

Solution: Computing Δt TOTAL ΔtMAT 2 + ΔtMODAL 2 + Δt WAVE2

Δt TOTAL =

= 2.812 + 02 + 0.4952 = 2.85 ns Max Bit Rate fB ≅

1 1 ≅ ≅ 175.44 Mbps 2Δt TOTAL 2(2.85 ns )

H. .RECEIVER RISE TIME & BANDWIDTH. 1.

Rise Time (t) The rise time is the time for the detector output (e.g. current) to change from 10 to 90% of its final value when the optic input power variation is a step.

For Your Information… 2

2

t s = t tx + t f + t rx

2

The fiber rise time is equal to the total dispersion within the fiber.

3-75

Self-Sufficient Guide to ECE by JASON AMPOLOQUIO where: ts = system rise time in ns t tx = source rise time in ns trx = receiver rise time in ns t f = fiber rise time in ns = rise time owing to material, modal, and waveguide dispersion

2.

3.

Maximum Data Rate (fb)

UPRZ

fb =

1 2t s

UPNRZ

fb =

1 ts

fb ≅

1 2 Δt

fb ≅

1 Δt

Bandwidth (BW)

Electrical

BWe =

0.35 Δt

BWo = 2 x BWe

Optical BWo = 4.

1 2 Δt

Bandwidth-Distance product

BW xA =

1 xA km 2 Δt



3-76

Sample Problem:

A fiber optic system uses a detector with a rise time of 1.5 ns and a source with a rise time of 4ns. If an RZ code is used with a data rate of 100 Mbps over a distance of 20 km, calculate the maximum acceptable dispersion for the fiber and the equivalent BW-Distance product.

Solution: fb =

1 1 1 1 = ⇒ ts = = = 5 ns 2 Δt 2t s 2fb 2(100 x 106 bps)

Fiber rise time; ts =

t tx + t f + trx ⇒ t f =

t s − (t tx + trx ) =

52 − (1.52 + 42 ) = 2.6 ns

Dispersion per unit length; Δt ns t 2.6 ns = s = = 0.13 A A km 20 km BW - Distance product BW x A =

I.

1 1 x 20 km = 3.846 GHz − km xA = 2(2.6 ns) 2 Δt

.REFLECTION AT A PLANE BOUNDARY. 1.

Reflection Coefficient (ρ) The ratio of the reflected electric field to the incident electric field.

ρ=

2.

n1 − n2 n1 + n2

n1 = core index n 2 = cladding index

Reflectance (R) The ratio of the reflected beam intensity to the incident beam intensity. ª

Uncoated fiber 2

⎛ n − n2 ⎞ R=⎜ 1 ⎟ ⎝ n1 + n2 ⎠


3-77

Sample Problem:

Calculate the reflectance for an air to glass interface assuming the refractive index of glass is 1.5

Solution: 2

2

⎡ n − n2 ⎤ ⎡1.5 − 1 ⎤ R =⎢ 1 ⎥ =⎢ ⎥ = 0.04 + n n ⎣1.5 + 1 ⎦ 2⎦ ⎣ 1

ª

Coated fiber 2

⎡n1n3 − n22 ⎤ ⎦ R= ⎣ 2 ⎡n1n3 + n22 ⎤ ⎣ ⎦

n1 = 1st medium index n 2 = coating material index n 3 = 2nd medium index

Sample Problem:

Determine the refractive index of a coating layer place between fibers whose index is 1.5 and 1.57 respectively, to produce a zero reflectance.

Solution: R =

3.

[n n [n n

1 3

− n22

1 3

+ n22

] ]

2 2

= 0 ⇒ n2 =

n1n3 = 1.5 x 1.57 = 1.534

Brewster Angle (θB) Named after British physicist David Brewster, the reflectance of the component vibrating parallel to the plane of incidence is zero. At this angle of incidence, the reflected ray would be perpendicular to the refracted ray, and the tangent of this angle of incidence is equal to the refractive index of the second medium if the first medium is air.

⎛n ⎞ θB = tan−1 ⎜ 2 ⎟ ⎝ n1 ⎠



3-78

Sample Problem:

Find the Brewster angle for the air-to-glass and glass-to-air interface.

Solution: air - to - glass(n1 = air, n2 = glass) ⎛n ⎞ ⎛ 1.5 ⎞ θB = tan−1 ⎜⎜ 2 ⎟⎟ = tan−1 ⎜ ⎟ = 56.3° n ⎝ 1 ⎠ ⎝ 1⎠ glass - to - air (n1 = glass, n2 = air) ⎛n ⎞ ⎛ 1 ⎞ θB = tan−1 ⎜⎜ 2 ⎟⎟ = tan−1 ⎜ ⎟ = 33.7° ⎝ 1.5 ⎠ ⎝ n1 ⎠

J.

.DIODE LIGHT SOURCES. 1.

Light-emitting semiconductors

Material GaAs AlGaAs InGaAs InGaAsP 2.

Wavelength Range (μm)

Bandgap Energy (eV)

0.9 0.8-0.9 1.0-1.3 0.9-1.7

1.4 1.4-1.55 0.95-1.24 0.73-1.35

Typical characteristics of diode light sources

Property

LED

Laser Diode

Single-Mode Laser Diode

Spectral Width (nm)

20-100

1-5

0.2

2-250

0.1-1

0.05-1

300

2000

6000

Very low

Moderate

High

105

104-105

104-105

Rise time (ns) BW Coupling Efficiency Lifetime (hours)

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Self-Sufficient Guide to ECE by JASON AMPOLOQUIO K. .DIODE LIGHT DETECTORS. 1.

Typical Characteristics of diode light Detectors

Material

Structure

Rise Time (ns)

Wavelength (nm)

Responsitivity (A/W)

Silicon Germanium InGaAS Silicon Germanium InGaAs

PIN PIN PIN APD APD APD

0.5 0.1 0.3 0.5 1 0.25

300-1100 500-1800 900-1700 400-1000 1000-1600 1000-1700

0.5 0.7 0.6 75 35 12

2.

Quantum Efficiency (η)

η=

3.

where :

e p

e = # of emitted electrons p = # of incident photons

Responsivity (R) The ratio of the output current of the detector to its optic input current.

R=

I P

R=η

q E

where: q = charge of an electron = 1.6 x 10-19 C E = energy of incident photon = hx f

Sample Problem:

Compute the responsivity of a detector having a quantum efficiency of 1% at 0.8 μm.

Solution: R = η = η

q c ⇒ E = h x f = hx E f qλ (1.6 x 10 −19 )(0.8 x 10 − 6 ) = 0.01 = 0.0064 A W hc (6.625 x 10 − 34 )(3 x 10 8 )



3-80

4.

Radiance (R) where:

P r= ΩA

r = radiance in mW /st − cm2 Ω = solid angle in steradians A = aperture area of light source in cm2

5.

Irradiance (ir)

ir =

L.

P A

where: ir = irradiance in W /cm2 A = aperture area of light source in cm2

.CONSTRUCTIONS OF OPTICAL FIBERS. 1.

Double Crucible Molten core-glass is placed in the inner vessel and molten claddingglass occupies the outer vessel and later forms a glass-cladded core.

2.

Rod in Tube In the rod-in-tube procedure a rod of core-glass is placed inside of a tube of cladding-glass where the end of this combination is heated, softening the glass so that a thin fiber can be pulled from it.

3.

Doped Deposited Silica (DDS) The most extensively used fiber fabrication process involve building up a fiber perform by vapor deposition of the glass constituents.

4.

External Deposition External deposition by hydrolysis is referred to as external chemical vapor deposition (external VCD).

5.

Axial Deposition The deposition occurs on the end of the rotating bait, which is withdrawn as the perform builds up.

6.

Internal Deposition In this process the chemical vapors are deposited on the inside of a glass tube that is rotating in a glass lathe where a traveling oxyhydrogen torch moves along the tube, fusing the deposited material to form a transparent glassy film.


3-81

M. .LASER FUNDAMENTALS. LASER is an acronym for Light Amplification by Stimulated Emission of Radiation and was first predicted by Albert Einstein near the beginning of the 20th century, the first working laser was not demonstrated until 1960 when Theodore Maiman did so using a ruby. 1.

2.

Laser Characteristics ª

Coherent The property of laser light wherein corresponding points on the wavefront are in phase.

ª

Collimated Property of laser light wherein light rays travel parallel with each other.

ª

Monochromatic Laser emits light signal with single color, frequency or wavelength.

Laser Types ª

Gas Gas lasers use a mixture of helium and neon enclosed in a glass tube.

ª

Solid-state optically pumped Liquid lasers use organic dyes enclosed in a glass tube for an active medium. A powerful pulse of light excites the organic dye.

ª

Liquid dye Solid lasers use a solid cylindrical crystal, such as ruby, for the active medium. The ruby is excited by a tungsten lamp tied to an ac power supply.

ª

Semiconductor Semiconductor lasers are made from semiconductor p-n junctions and are commonly called injection laser diodes or ILDs. The excitation mechanism is a dc power supply that controls the amount of current to the active medium.



3-82

Read it till it Hertz…jma LASER SAFETY CLASS

DESCRIPTION

Class 1

Low power / non-hazardous

Class 2 Class 2a

Class 3a Class 3b

Class 4

Low power / minor controls necessary Emit less than 1 mW visible CW radiation. Not considered hazardous for momentary (<0.25 sec) unintentional exposure. Class 2a lasers are those class 2 lasers not intended to be viewed, i.e. supermarket scanners. Medium power / direct viewing hazard / little diffuse reflection hazard. Class 3a is visible lasers with 1-5 mW power output, invisible lasers, and those having 1-5 times the Accessible Emission Limit (AEL) of class 1 lasers. Class 3b is all other class 3 lasers at all wavelengths which have a power output less than 500 mW. High power / eye and skin hazard / potential diffuse reflection hazard or fire hazard

N. .FIBER-BASED TECHNOLOGY.

1.

SONET/SDH Transmission Hierarchy SONET/SDH is a set of international standards for broadband communications over single-mode fiber optic transmission systems, allowing manufacturers to build equipment to support full interconnectivity and interoperability.

3-83

Self-Sufficient Guide to ECE by JASON AMPOLOQUIO Comparison between SONET & SDH

Optical Carrier (OC-n)

SONET STS-n

OC-1

STS-1

51.84

672

OC-2

STS-2

103.68

1,344

OC-3

STS-3

STM-1

155.52

2,016

OC-4

STS-4

STM-3

207.36

2,688

OC-9

STS-9

STM-3

466.56

6,048

OC-12

STS-12

STM-4

622.08

8,064

OC-18

STS-18

STM-6

933.12

12,096

OC-24

STS-24

STM-8

1,244.16

16,128

OC-36

STS-36

STM-12

1,866.24

24,192

OC-48

STS-48

STM-16

2,488.32

32,256

OC-96

STS-96

STM-32

4,976

64,512

OC-256

STS-256

STM-64

13,219.20

171,360

2.

SDH STM-n

Signal Level (Mbps)

Equivalent DS-0 (64 kbps)

SONET Terms ª

Optical Carrier (OC) is the definition of the SONET optical signal. The defined OC levels begin at OC-1 (51.84 Mbps) and culminate in OC-255 (13.2192 Gbps).

ª

Synchronous Transport Signal (STS) is the electrical equivalent of the SONET optical signal; it is known as Synchronous Transport Module (STM) in SDH. The signal begins as electrical and is converted to optical for transmission over the SONET fiber facilities. Each STS-1 frame is transmitted each 125 μs, yielding raw bandwidth of 51.84 Mbps. The STS frame includes five elements, Synchronous Payload Envelope, Section Overhead, Line Overhead Path Overhead, and Payload.



3-84

3.

ª

Synchronous Payload Envelope (SPE) is the envelope that carries the user payload data. It is analogous to the payload envelope of a X.25 packet. The SPE consists of 783 octets (87 columns and 9 rows of data octets).

ª

Payload is the actual data content of the SONET frame and rides within the SPE. Total usable payload at the OC-1 level consists of up to 49.54 Mbps, into which a T3 frame fits quite nicely. The balance of the 51.84 Mbps is consumed by Transport Overhead and Path Overhead.

ª

Multiplexing is on the basis of direct time division multiplexing. Either full SONET speeds or lesser asynchronous and synchronous data streams can be multiplexed into the STS-N payload, which is then converted into an OC-N payload.

SONET Network Elements and Overheads

i.

Network Elements ª

Terminal Multiplexer The PTE, an entry level path terminating terminal multiplexer, acts as a concentrator of DS1 signals and tributaries.

ª

Regenerator These are optical amplifier that boosts the signal level in the fiber due to significant distance between multiplexers.

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Add/Drop Multiplexer Provides interfaces between the different network signals and SONET signals.

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4.

Overheads ª

Path Overhead (POH) contained within the SPE, comprises 9 octets for the relay of OAM&P information in support of endto-end network management.

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Transport Overhead (TOH) consists of Section Overhead and Line Overhead.

Section Overhead (SOH) of 9 octets is dedicated to the transport of status, messages, and alarm indications for the maintenance of SONET links.

Line Overhead (LOH) of eighteen (18) bytes controls the reliable transport of payload data between network elements.

Virtual Containers and Tributaries

ª

Virtual Containers (Virtual Paths) Virtual Containers are simply end-to-end communications paths, routes or circuits, which carry traffic from one end point to another. The path is not fixed or dedicated, neither is it dedicated to a particular conversation or user. A virtual path consists of many virtual tributaries.

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Virtual Tributaries Virtual Tributaries carry one form of signal, such as a DS-1, DS-2 or DS-3 signal within a byte-interleaved frame. Virtual tributaries can be mapped into a single virtual path. A virtual tributary may be channelized (e.g., a 48-channel T-1 for voice) or unchannelized (e.g, a clear channel DS-1 for full motion video).

Summary for various VTs

VT Type VT 1.5 VT 2 VT 3 VT 6

Bit Rate 1.728 2.304 3.456 6.912

Mbps Mbps Mbps Mbps

VT Size 9 rows, 3 columns 9 rows, 4 columns 9 rows, 6 columns 9 rows, 12 columns



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Virtual Channels (Tributary Units) Virtual Channels exist within virtual tributaries. For example, a virtual tributary might carry a T1 frame. Within that tributary, there might exist 24 channels with each channel carrying a single voice or data communication in multiple time slots.

O. .MEASUREMENT OF LIGHT. ª

SI Photometry Units

Quantity

Symbol

SI unit

Abbr.

Luminous flux

F

lumen (cd·sr)

lm

Luminous energy

Qv

lumen-second

lm·s

Luminous intensity

Iv

candela (lm/sr)

cd

Luminance

Lv

candela/square metre

cd/m2

Illuminance

Ev

lux (lm/m2)

lx

Luminous emittance

Mv

lux (lm/m2)

lx

Luminous efficacy

Κ

lumens / watt

lm/W

Notes also called luminous power units are sometimes called Talbots --also called luminosity Used for light incident on a surface Used for light emitted from a surface ratio of luminous flux to radiant flux

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SI Radiometry Units

Quantity

Symbol

SI unit

Abbr.

Radiant energy

Q

joule

J

Radiant intensity

I

watt per steradian

W·sr−1

power per unit solid angle

Irradiance

E

watt per square metre

W·m−2

power incident on a surface

Radiant flux

Φe

watt

W

Radiance

L

watt per steradian per square metre

W·sr−1·m−2

Radiant emittance / Radiant exitance

M

watt per square metre

W·m−2

Spectral radiance

Lλ or Lν

Spectral irradiance

Eλ or Eν

watt per steradian per metre3 or watt per steradian per square metre per Hertz watt per metre3 or watt per square metre per hertz

W·sr−1·m−3

or W·sr−1·m−2·Hz−1

W·m−3

or W·m−2·Hz−1

Notes energy

radiant energy per unit time, also called radiant power power per unit solid angle per unit projected source area. power emitted from a surface

commonly measured in W·sr−1·m−2·nm−1

commonly measured in W·m−2·nm−1



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I

H

1.

A fiber has a NA = 0.2588. A light source us coupled to it which emits 75% of its light into a 60o full-cone angle , 50% into a 30o cone, and 25% into a 15o cone. What is the coupling efficiency when this source and fiber are connected? A. 25% B. 75% C. 50% D. 85%

2.

Find the number of photons incident on a detector in 1 s if the optic power is 1 μW and the wavelength is 0.8 μm. B. 2.12 x 10 12 photons A. 42.0 x 10 12 photons D. 54.10 x 10 12 photons C. 4.03 x 10 12 photons

3.

Medium 1 is made of silicon and medium 2 is made of glass. Their refractive indexes are 3.4 and 1.4, respectively. For an angle of refraction of 30 degrees, determine the angle of incidence. April 2003 A. 11.88 degrees B. 10.68 degrees C. 10.17 degrees D. 11.53 degrees

4.

In multimode graded-index fibers, what is the relation between the index of refraction of the glass at the center of the fiber core with respect to the index of refraction of the cladding glass? A. Lower B. Higher C. Approximately equal D. Either A or B

5.

How many photons are arriving per second at a receiver if the power is 1 nW at wavelength of 1.3 μm? B. 6.54 x 109 photons/sec A. 0.654 x 109 photons/sec D. 654 x 109 photons/sec C. 65.4 x 109 photons/sec

6.

For a multimode graded-index fiber, the numerical aperture is at its maximum value at which of the following locations? A. At the fiber axis B. At the core-cladding interface C. Half way between the fiber center and the core-cladding interface D. One-fourth of the way between the fiber center and the core- cladding interface


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7.

A SI fiber has n1 =1.5, n2 =1.49, and core diameter 50 μm. Consider the guided ray traveling at the steepest angle with respect to the fiber axis. How many reflections are there per meter for this ray? A. 321 reflections/meter B. 232 reflections/meter C. 2321 reflections/meter D. 12321 reflections/meter

8.

A step-index multimode fiber and a graded-index multimode fiber have the same core and cladding sizes and the same refractive index difference. Which fiber type, if either, will accept light more easily and have more propagating modes? A. Step-index fiber B. Graded-index fiber C. Neither; they will behave approximately the same D. Both A and B

9.

The multimode graded-index fiber that has the best bend performance and will show the least amount of optical degradation if mishandled is what size? A. 50/125 μm B. 62.5/125 μm C. 85/125 μm D. 100/140 μm

10. What are the two basic types of single mode step-index fibers? A. Low NA and high NA B. Solid core and air core C. Enriched clad and depressed D. Matched clad and depressed clad

11. How many wavelength frequency. A. 5 x 108 C. 4 x 108

voice channels can be modulated onto a light carrier at a 1.06 μm? Assume a system BW equal to 1% of the carrier channels channels

B. 8 x 108 channels D. 7 x 108 channels

12. The use of plastic-clad silica and all plastic fibers has what primary drawback? A. Higher NA B. Higher cost C. Higher bandwidth D. Limited optical performance

13. Calculate the midband frequency of color blue. B. 6.33 x 1014 Hz A. 42.5 x 1015 Hz 12 D. 16.8 x 1019 Hz C. 58.33 x 10 Hz 14. Increased extrinsic absorption at 950 nm, 1,250 nm, and 1,383 nm is caused by what impurity in glass optical fibers? A. Phosphorus B. Germanium C. Titanium D. Water

15. The power incident on a detector of light is 100 nW. (a) Determine the number of photons per second incident on the detector if the wavelength is 1550 nm. B. 7.8 x 1012 photons/sec A. 7.8 x 1016 photons/sec D. 7.8 x 1013 photons/sec C. 7.8 x 1011 photons/sec


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16. Which type of scattering loss is proportional to the reciprocal of the fourth power of the wavelength of the light? A. Mie B. Raman C. Rayleigh D. Brillouin

17. For a multi-mode step-index fiber with glass core (n1 = 1.55) and a fused quartz cladding (n2 = 1.46), determine the critical angles. April 2003 A. 70.18 degrees B. 70.38 degrees C. 70.24 degrees D. 7.11 degrees 18. What is the definition of a bound ray? A. A ray that cannot move B. A ray that travels in the air C. A ray that is refracted out of the fiber D. A ray that propagates through the fiber by total internal reflection 19. The A. B. C. D.

fiber NA relates to which of the following characteristics? Physical size of the fiber Tensile strength of the fiber Maximum angle within the fiber acceptance cone Speed of light within the fiber

20. A skew ray is which of the following types of rays? A. An unbound ray B. A meridional ray C. An unbalanced ray D. A ray that propagates without passing through the center axis of the fiber 21. Electromagnetic wave behavior is used to describe the propagation of light along the fiber in what theory? A. Mode theory B. Particle theory C. Maxwell theory D. Rayleigh's theory

22. Calculate the maximum data rate for the 45 km fiber system which has a pulse spreading constant of 100 ns when it is used with a transmitter having a rise time of 50 ns and a receiver having a rise time of 75 ns, if the line code is RZ. A. 0.7 MHz B. 83.1 MHz C. 3.7 MHz D. 7.4 MHz 23. A radius of curvature is larger than the fiber diameter in which of the following types of fiber bends? A. Macrobends B. Microbends C. Gentle bends D. Serpentine bends

24. A fiber optic cable has a loss of 16 dB/Km. What would be its attenuation if the cable is 2000 ft long? April 2003 A. 97.6 dB B. 26 dB C. 2.6 dB D. 9.76 dB


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25. Calculate the velocity of light in a medium which has a relative permittivity of 2.5 B. 5.67 x 10 8 m/s A. 1.89 x 10 8 m/s 7 D. 2.85 x 10 6 m/s C. 3.43 x 10 m/s 26. Only in multimode fibers does which of the following types of dispersion occur? A. Modal B. Material C. Waveguide D. Chromatic 27. When different colors of light travel through the fiber at different speeds, which of the following types of dispersion occurs? A. Modal B. Intramodal C. Intermodal D. Modal delay spreading

28. A single-mode fiber has a numerical maximum core diameter it could have wavelength of 820 nm? A. 2.1 μm B. C. 4.2 μm D. 29. Another name for intramodal dispersion A. Modal B. C. Intermodal D.

aperture of 0.15. What is the for use with infrared light with a 21 μm 42.1 μm Chromatic Rayleigh

30. In a step-index fiber, the refractive index profile of the fiber core has which of the following characteristics? A. It is uniform over the fiber core B. It linearly decreases from a maximum at the fiber center to a minimum at the core-cladding boundary C. It is parabolic with a maximum index of refraction at the center and a minimum index of refraction at the core-cladding boundary D. It linearly increases from a minimum at the fiber center to a maximum at the core cladding boundary

31. Calculate the Rayleigh scattering loss in dB for a 50/125 step-index fiber operating at 1200 nm. A. 5.27 dB B. 3.68 dB C. 2.65 dB D. 3.68 dB 32. In a graded-index fiber, the refractive index profile of the fiber core is best described by which of the following statements? A. It is uniform over the fiber core B. It linearly decreases from a maximum at the fiber center to a minimum at the core-cladding boundary C. It is parabolic with a maximum index of refraction at the center and a minimum index of refraction at the core-cladding boundary D. It linearly increases from a minimum at the fiber center to a maximum at the core cladding boundary 33. Which of the following multimode fiber core sizes is NOT a standard commercial fiber size? A. 25 μm B. 50 μm C. 85 μm D. 100 μm


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34. Elements other than silicon and oxygen are added to glass material by the fiber manufacturer to change its index of refraction. What are these elements called? A. Spices B. Dopants C. Additives D. Impurities

35. Calculate the numerical aperture of an optical fiber with a fractional difference of 0.0258 and a 2.4 relative permittivity at the core A. 1.351 B. 0.351 C. 2.351 D. -0.569 36. Compared to multimode step-index fibers, do multimode graded-index fibers have lower, higher, or approximately equal bandwidths? A. Lower B. Higher C. Approximately equal D. Either A or B

37. To operate properly, a fiber optic receiver requires -34 dBm power. The system losses total 31 dBm from the light source to the receiver. How much power does the light source emit (in mW)? A. 0.05 mW B. 0.005 mW C. 0.5 mW D. 5 mW 38. A T3 system has a 10-9 BER. Compute the number of errors per minute. A. 2.7 errors/min B. 8.16 errors/min C. 4.3 errors/min D. 7.5 errors/min 39. To describe the nature of light, which of the following ways can be used? A. Particles of energy only B. Electromagnetic wave only C. Electromagnetic wave and particles of energy D. Element

40. An optic fiber is made of glass with a surface index of 1.55 and is clad with another glass with a refractive index of 1.51. Launching takes place on air. Determine the numerical aperture of the fiber. April 2003 A. 0.35 B. 0.214 C. 0.0465 D. 0.0305 41. Of the following advantages, which one does NOT apply to fiber optics? A. Improved environmental protection B. Improved signal security C. Increased bandwidth D. Established standards

42. Calculate the travel time of an axial ray in a 10-km multimode step index fiber with a NA of 0.248 and 1.28% fractional refractive index change. A. 15.67 μsec B. 1.67 μsec C. 5.7 μsec D. 51.67 μsec


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43. Of the following factors, which ones are advantages of fiber optic systems? A. Electrical isolation and immunity to noise B. Immunity to noise and low bandwidth C. Signal security and high price D. Electrical isolation and low bandwidth 44. Light exhibits what kind of wave motion? A. Transverse B. C. Longitudinal D.

Turbulent Aperiodic

45. Which of the following factors is a description of transverse wave motion? A. The wave magnitude varies perpendicular to the direction of wave motion B. The wave magnitude varies linearly to the direction of wave motion C. The wave magnitude varies parallel to the direction of propagation D. The wave motion is not predictable 46. What does a transparent substance do to light rays that fall on it? A. Refracts them B. Transmits them C. Reflects them D. Absorbs them

47. Compute the responsivity of a silicon APD operating at 0.82 μm and having a 0.8 quantum efficiency if its gain is 100. How much optical power is needed by this detector to produce 20 nA? A. 3.8 A/W, 5.3 nW B. 53 A/W, 0.38 nW C. 5.3 A/W, 3.8 nW D. 0.38 A/W, 53 nW 48. What does a translucent substance do to light rays that fall on it? A. Reflects and absorbs them B. Refracts and absorbs them C. Transmits and diffuses them D. Transmits and reflects them

49. An erbium-doped fiber amplifier has a noise figure of 6 and a gain of 100. The input signal has a 30-dB signal-to-noise ratio and a signal power of 10 μW. Compute the signal power (in dBm) and a signal-to-noise ratio (in dB) at the amplifier’s output. A. 0 dBm, 22.2 dB B. 22.2 dBm, 2.2 dB C. 22.2 dBm, 0 dB D. 20 dBm, 2.22 dB 50. What does an opaque substance do when light rays fall on it? A. Refracts them B. Reflects or absorbs them C. Transmits them only D. Transmits and diffuses them

51. A single-mode fiber is made with core diameter of 10 micrometers and is coupled to a light source with a wavelength of 1.2 micrometers. Its core glass has a refractive index of 1.55. Determine the cladding index required for producing single-mode propagation. (V cut off = 2.405) April

2003

A. C.

1.504 1.43

B. 1.17 D. 1.547


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52. Calculate the Brewster angle for a parallel-polarized wave incident from air onto paraffin with a relative permittivity of 2. A. 32.5° B. 44.4° C. 54.7° D. 87.5° 53. Fibers are single mode at a particular wavelength only when V < 2.405. Under what condition, if any, will the fiber cease to be single mode? A. When the wavelength of the light is greater than the cutoff wavelength B. When the wavelength of the light is less than the cutoff wavelength C. When the core fiber diameter is extremely small D. Both B and C 54. In making a preform, layers of glass powder are deposited on the inside or outside of a glass rod or tube. What is this glass powder called? A. Soot B. Smoke C. Preform D. Afterburn

55. Calculate the pulse spreading per unit length of a 15-km fiber with an axial propagation time of 53.18μsec and a critical angle meridional propagation time of 54.25μsec A. 1.33 ns/km B. 733 ns/km C. 7.33 ns/km D. 71.33 ns/km 56. The process used in drawing the fiber is best described by which of the following statements? A. The preform is melted and the molten glass is molded, using special fiber molds B. The preform is softened and the glass is pulled into a thin glass filament C. The preform is softened and the glass is rolled into a thin glass filament D. The preform is melted and the fiber is formed by blowing the molten glass through a small hole 57. To protect the fiber from contaminants in the drawing process, what substance is added over the fiber? A. Water B. Coating C. Preform D. Cladding

58. Find the output current of a photodetector whose quantum efficiency is 0.9. The wavelength is 1.3 μm and the incident power level is -37 dBm. A. 18.8 nA B. 1.88 nA C. 0.188 nA D. 188 nA 59. A 45 km length of fiber must not lengthen pulses by more than 100 ns. Find the maximum permissible value for the pulse-spreading constant. A. 1.11 ns/km B. 222 ns/km C. 11.22 ns/km D. 2.22 ns/km 60. The spreading of a light pulse as it travels along the fiber is caused by what mechanism? A. Diffraction B. Attenuation C. Absorption D. Dispersion


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61. A fiber optic system uses a detector with a rise time of 3 ns and a source with a rise time of 2 ns. If an RZ code is used with a data rate of 100 Mbps over a distance of 25 km, calculate the maximum acceptable dispersion for the fiber and the equivalent BW-Distance product. A. 38.6 ns/km, 361 GHz-km B. 0.1386 ns/km, 3.61 GHz-km C. 0.16 ns/km, 31 GHz-km D. 13.86 ns/km, 361 GHz-km 62. A single-mode fiber operating at 1.3 μm is found to have a total material dispersion of 2.2 ns and a total waveguide dispersion of 1.95 ns. Determined the approximate maximum bit rate for the fiber if the transmitted pulse has a width of 0.5 ns. A. 45 Mbps B. 13.2 Mbps C. 170.1 Mbps D. 74 Mbps 63. A photodiode has a responsivity of 0.4 A/W. What optical power is required to produce a current of 500 nA. A. 1.25 μW B. 12.5 μW C. 125 μW D. 0.125 μW 64. Medium 1 is made of silicon and medium 2 is made of glass. Their refractive indexes are 3.2 and 1.2, respectively. For an angle of refraction of 31 degrees, determine the angle of incidence. April 2003 A. 11.14 degrees B. 11.88 degrees C. 11.13 degrees D. 11.53 degrees 65. A cable contains 144 single-mode fiber, many digitized voice messages can be this cable? A. 5175 million B. C. 5.175 million D.

each operating at 2.3 Gbps. How transmitted simultaneously along 51.75 million 517.5 million

66. Calculate the pulse spreading per unit length of a 15-km fiber with a NA of 0.2482 and 1.55 core index of refraction. A. 0.62 ps/km B. 624 ps/km C. 66.24 ns/km D. 6.24 ps/km 67. To operate properly, a fiber optic receiver requires -34 dBm power. The system losses total 31 dB from the light source to the receiver. How much power does the light source emit? A. 0.05 mW B. 5 mW C. 50 mW D. 0.5 mW 68. A silica multimode step-index fiber has core and cladding refractive indices of 1.46 and 1.459, respectively. Compute the RZ rate-length product of this fiber at if the source emits at 1550 nm and has a line width of 120 nm. A. 922 Mbps B. 9.2 Mbps C. 2.2 Mbps D. 92.2 Mbps


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69. Find the total number of propagating modes in a SI fiber having a normalized frequency of 4.5 when the wavelength is 800 nm. A. 10 B. 14 C. 5 D. 18 70. Fiber optic systems use what two types of optical sources? A. PIN diodes and LEDs B. Laser diodes and APDs C. LEDs and APDs D. LEDs and laser diodes 71. The A. B. C. D.

optical source performs which of the following functions? Converts the optical signal to an electrical signal Launches the optical signal into the fiber Amplifies the electrical signal Amplifies the optical signal

72. Given the pulse-spreading constant equal to 10 ns/m and the cable length equal to 100 meters, determine the maximum bit rate in Mbps for UPNRZ transmission. April 2003 A. 500 Mbps B. 1000 Mbps C. 1.0 Mbps D. 0.5 Mbps 73. Find the Brewster angle for the air-to-glass and glass-to-air interface. A. 5.3°, 3.7° B. 56°, 3.7° C. 56.3°, 33.7° D. 5.3°, 33.7° 74. An optic fiber is made of glass with a refractive index of 1.55 and is clad with another glass with a refractive index of 1.51. Launching takes place from air. What numerical aperture and acceptance angle does the fiber have? April 2003 A. 0.35, 20.5° B. 0.512, 10.6° C. 0.532, 10.6° D. 0.352, 25.6° 75. Calculate the pulse spreading per unit length of a 15-km fiber with a core index of 1.55 and 1.53 cladding index of refraction. A. 66.7 ns/km B. 654 ns/km C. 7.54 ns/km D. 6.54 ns/km 76. Calculate the pulse spreading per unit length of a 15-km fiber with a fractional index change of 0.0129 and 1.55 core index of refraction. A. 667 ns/km B. 6.67 ns/km C. 76.67 ns/km D. 66.67 ns/km 77. Find the energy, in electron-volts, in one photon at a wavelength of 1μm. A. 19.9 eV B. 1.24 eV C. 12.4 eV D. 1.99 eV 78. What are the three parts of a fiber optic link? A. Optical fiber, optical connectors, optical splices B. Optical fiber, optical connectors, receiver C. Transmitter, optical fiber, optical connectors D. Transmitter, optical fiber, receiver


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79. A parallel polarized ray is incident at an angle of 85O when traveling from a medium of index 1.48 into a medium having index 1.465. The wavelength is 1300 nm. Compute the reflection coefficient. A. 55.48° B. 34.39° C. 91.2° D. 74.9° 80. The A. B. C. D.

fiber optic transmitter has which of the following functions? Amplifies the output electrical signal Converts the input optical signal to an electrical signal Converts the electrical input signal to an optical signal Amplifies the optical signal

81. What fiber mechanisms weaken and distort the optical signal launched into the fiber? A. Dispersion, radiation, and absorption B. Scattering, radiation, and absorption C. Scattering, reflection, and refraction D. Scattering, absorption, and dispersion

82. A collimated Gaussian beam has a spot size of 1 mm and wavelength of 0.8 μm. Compute the focused spot size when focused by a lens whose focal length is 20 mm. A. 1.44 μm B. 16.71 μm C. 5.09 μm D. 8.54 μm 83. Consider a fiber whose core index is 1.5 and whose cladding index is 1.485. The core radius 100 μm. At what bending radius does a ray traveling along the fiber axis strike the cladding at the critical angle in the bend? A. 1000 cm B. 1 cm C. 100 cm D. 10 cm 84. The A. B. C. D.

fiber optic receiver performs which of the following functions? Amplifies the electrical signal Converts the electrical signal back into an optical signal Converts the optical signal back into an electrical signal Amplifies the optical signal

85. Consider a SI fiber with n1=1.5 and n2=1.485 at 0.82μm. radius is 50 μm, how many modes can propagate? A. 3286 modes B. 328 modes C. 386 modes D. 326 modes 86. What are the two parts of a fiber optic receiver? A. Optical detector and PIN diode B. Fiber and optical detector C. Optical detector and signal conditioning circuits D. Fiber and case 87. What are the two types of optical detectors? A. Laser diodes and PIN diodes B. APDs and laser diodes C. PIN diodes and APDs D. LEDs and APDs


If the core

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88. How many current is produced by a photodetector whose responsivity is 0.5 A/W if the incident optical power level -43 dBm? A. 38 nA B. 46 nA C. 25 nA D. 13 nA 89. Extremely high losses occurred in early fibers because of which of the following conditions? A. Holes in the fiber sides B. Impurities in the fiber material C. Cracks in the fibers D. Core areas too small in the fibers

90. The fiber has zero dispersion at a wavelength of 1310 nm and has a zerodispersion slope of 0.05 ps/(nm2-km). Calculate the total dispersion of 50 km fiber when it is used with a source having a linewidth of 2 nm at a wavelength of 1550 nm. A. 9.49 ps B. 18.98 ps C. 8.44 ps D. 949 ps 91. What was the first light source developed that could be easily coupled into a fiber? A. Lamp B. YAG laser C. PIN diode D. LED 92. What multimode fiber properties help reduce connection losses? A. Smaller core size and higher NA B. Smaller core size and lower NA C. Larger core size and higher NA D. Larger core size and lower NA 93. Light waves that strike a surface but are neither transmitted nor absorbed? A. Diffused B. Refracted C. Reflected D. Diffracted

94. A uniform collimated beam is focused by a lens whose focal length is 20 mm and whose diameter is 10 mm. The wavelength is 0.8 μm. Compute the focused spot size. A. 1.6 μm B. 9.1 μm C. 5.3 μm D. 3.9 μm 95. What is the name of the law that states "The angle of incidence is equal to the angle of reflection"? A. Snell's Law B. Murphy's Law C. Huygen’s Law D. Law of Reflection

96. A receiver has a 10-cm focal length, a 1-cm photodetector diameter, and air between the lens and photodetector. Compute the receiver’s NA. A. 0.55 B. 0.05 C. 0.005 D. 0.523 97. A light wave is incident on a surface. The reflected power is the greatest in which of the following incidences? A. 30º B. 45º C. Perpendicular D. Almost parallel


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98. A light wave passes from one medium into another medium with a different velocity. As the wave enters the second medium, the change of direction is known by which of the following terms? A. Reflection B. Refraction C. Absorption D. Diffusion

99. A certain fiber has a core diameter of 50 μm and is used at a medium light wavelength of 0.80 μm and NA=0.35. Calculate the number of modes it will support. A. 1260 B. 2500 C. 2,361 D. 5345 100. If a light wave passes from a less dense medium to a more dense medium, how does the angle of refraction compare to the angle of incidence? A. Greater than the angle of incidence only B. Equal to the angle of incidence only C. Greater than or equal to the angle of incidence D. Less than the angle of incidence 101. What another word is for diffused? A. Absorbed C. Refracted

B. D.

Scattered Attenuated

102. Calculate the travel time of a critical angle meridional ray in a 10-km multimode step index fiber with a NA of 0.248, n2=1.53 and 1.28% fractional refractive index change. A. 52.34 μsec B. 5.34 μsec C. 0.534 μsec D. 2.4 μsec 103. When light falls on a piece of black paper, what happens to most of the light? A. It is absorbed B. It is reflected C. It is scattered D. It is refracted 104. Light is transmitted along an optical fiber by what two methods? A. Ray theory and mode theory B. Ray theory and photon theory C. Ray theory and quantum theory D. Mode theory and photon theory

105. Compute the rise time of a photodetector if its 3-dB bandwidth is 500 MHz. A. 70 ps B. 7 ps C. 700 ps D. 0.7 ps 106. How does the speed of light in the fiber compare to the speed of light in the air? A. It is slower in the fiber B. It is faster in the fiber C. It is the same in both the fiber and the air D. Both A and B


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107. Compute the pulse spread for a 10-km length fiber when the light source emits at 1320 nm and has a 2-nm spectral width. The fiber has zero dispersion wavelength at 1300 nm and the dispersion is 1.86 ps/nm-km. A. 0.32 ps B. 3.2 ps C. 37.2 ps D. 372 ps 108. Calculate the reflectance for an air to glass interface assuming the refractive index of glass is 1.5 A. 0.4 B. 0.004 C. 4 D. 0.04 109. The relationship between the incident rays and the refracted rays at a boundary between mediums with different indexes of refraction describes what law? A. Bragg's Law B. Snell's Law C. Lenz’s Law D. Law of Reflection 110. Total internal reflection occurs at which of the following angles? A. Obtuse angle B. Fresnel angle C. Right angle D. Critical angle of incidence 111. What are the three basic parts of an optical fiber? A. Core, cladding, and coating B. Inside, middle, and outside C. Fiber, kevlar, and jacket D. Hole, shell, and coating

112. A fiber is rated having a bandwidth-distance product of 500 MHz-km. Find the dispersion in ns/km, and find the rise time of a pulse in a 5 km length if this cable A. 10 ns/km, 5 ns B. 1 ns/km, 0.5 ns C. 10 ns/km, 0.5 ns D. 1 ns/km, 5 ns 113. The cladding performs all except which of the following functions? A. Reduces the loss of light from the core B. Reduces the scattering loss at the surface of the core C. Protects the fiber core from absorbing surface contaminants D. Reduces mechanical strength

114. Calculate the bandwidth and figure of merit of a fiber optic system (50 km) with a total dispersion of 949 ps. A. 52.68 MHz, 2.63 GHz-km B. 526.8 MHz, 2.63 GHz-km C. 526.8 MHz, 26.3 GHz-km D. 52.68 MHz, 26.3 GHz-km 115. What are the two basic types of fibers? A. Small and large B. C. Opaque and diffuse D.

Glass and plastic Single mode and multimode

116. Compute the energy of a photon at 1.3 μm. B. 3.2x10-19J A. 15x10-19J D. 32x10-19J C. 1.5x10-19J


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117. The lowest signal loss and the highest bandwidth are characteristic of which of the following types of fibers? A. Air core B. Multimode C. Single mode D. Plastic core 118. Compared to single mode fibers, multimode fibers have which of the following advantages? A. Ease of making connections only B. Ease of launching light into them only C. Ease of both making connections and launching light into them D. Lower dispersion 119. System performance is affected most by which of the following fiber properties? A. NA and delta B. Core diameter and NA C. Attenuation and dispersion D. Core diameter and cladding

120. There are 1010 photons per second incident on a photo detector at wavelength 0.8 μm. Compute the power incident on the detector. If this detector converts light to current at a rate of 0.65 mA/mW, what current is produced? B. 2.48 x10-9W, 1.6 nA A. 2.48 x10-19W, 1.6 nA D. 2.48 x10-9W, 1.6 pA C. 2.48 x10-19W, 1.6 pA 121. The loss of optical power as light travels along a fiber is called A. attenuation B. absorption C. scattering D. dispersion

122. Determine the refractive index of a coating layer place between fibers whose indices are 1.5 and 1.57 respectively, to produce a zero reflectance. A. 1.534 B. 1.34 C. 1.435 D. 1.4 123. Attenuation is specified in what units? A. dB B. C. μm D.

dB/km μm/km

124. Glass optical fibers have low loss between the infrared and ultraviolet absorptive regions. The approximate wavelength of operation for glass optical fibers is in which of the following ranges? A. 1 nm to 700 nm B. 700 nm to 1600 nm C. 1600 nm to 9000 nm D. 9 μm to 20 μm

125. An optical fiber has a loss of 0.35 dB/km. If an LED with a power output of 25 μW is connected to one end of a 20 km length of this fiber. How much power reaches the detector at the other end? A. -0.23 dBm B. -23 dBm C. -2.3 dBm D. -230 dBm


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126. Optical networks are based on the emergence of the ______ layer in ______ networks. A. optical; integrated B. optical; transport C. transport; optical D. optical; high-capacity 127. DWDM increases the capacity of ____ fiber by first assigning incoming ____ to specific frequencies within a designated frequency band and then ____ the resulting signals out onto one fiber. A. optical fiber; wavelengths; combining B. group; signals; multiplexing C. embedded; optical signals; multiplexing D. dense; wavelengths; multiplexing 128. DWDM enables different formats, such as ______, to be transmitted over the optical layer. A. SONET and SDH B. IP and ATM C. ATM and SONET D. all of the above 129. _____ (DWDM) is a _______ transmission technique. A. Direct wavelength directing medium; telecommunications B. Dense wavelength division multiplexing; fiber-optic C. Direct wave division multiplexing; fiber-optic D. none of the above 130. 1550-nm technology has experienced a significant increase in value since it was deployed in A. 1996 B. 1992 C. 1991 D. 1995 131. Which feature is expected to accelerate the growth of optical Ethernet in the WAN? A. fiber-channel compatibility B. 10.000–Gbps speeds C. improved optical fiber D. SONET OC–192 speeds 132. In current networking practices, optical Ethernet products are required to span distances greater than what? A. 30 meters B. 200 meters C. 2 kilometers D. 50 kilometers 133. When A. in B. in C. in D. in

was the Fast Ethernet over fiber optics first standardized? 1994, as part of the first Fast Ethernet standards 2000, six years after Fast Ethernet over copper 1990, as part of the 10BASE–T specification 1998, six years after Fast Ethernet over copper

134.______ occurs as light travels down single-mode fibers, and when the core of the fiber is asymmetric, the light traveling along one side moves slower or faster than the light traveling along the other side. A. Slope mismatch dispersion B. Chromatic dispersion C. Polarization mode dispersion D. Rayleigh dispersion


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135. The one aspect of SONET that has allowed it to survive during a time of tremendous changes in network capacity needs is its A. versatility B. fiber capacity C. scalability D. functionality 136. One of the great revenue-producing aspects of optical networks is the ability to resell ____ rather than _____. A. bandwidth; fiber B. fiber; bandwidth C. single fiber; double fiber D. wavelength; bandwidth 137. Future DWDM terminals will have the capacity to transmit ____ volumes of an encyclopedia in one second. A. 90 B. 250 C. 1500 D. 90000 138. _____ is based on the principal that different-colored pulses of light travel at different speeds. A. Polarization mode dispersion B. Rayleigh dispersion C. Chromatic dispersion D. Ultraviolet absorption 139. The types of media that can transmit information in the telecommunications world are the following A. copper wire, coaxial cable, fiber, and wireless B. hybrid fiber/coax and copper wire C. copper wire, coaxial cable, fiber, and hybrid fiber/coax D. wireless and copper wire 140. The _____ is a particular set of standards that allows the interworking of products from different vendors. It usually embodies a fiber-optic ring that will permit transmission in both directions. A. LAN B. WAN C. SONET D. CCS 141. Two digital signals whose transitions occur at almost the same rate are A. synchronous B. asynchronous C. plesiochronous D. all of the above 142. SONET systems are _____ technology. A. fiber-optic B. C. wireless D.

twisted-pair, copper-based satellite-based

143. SONET's base signal (STS–1) operates at a bit rate of A. 64 kbps B. 1.544 Mbps C. 51.84 Mbps D. 155.52 Mbps 144. The largest possible optical Ethernet can span what distance? A. 10 kilometers B. global C. 100 kilometers D. 1000 kilometers


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145. Traditional DWDM is capable of multiplexing how many 1550-nm transmitters onto a single fiber? A. 12 B. 2 C. 8 D. 4 146. Line overhead contains _______ bytes of information. A. 18 B. 9 C. 27 D. 90 147. ITU's H.320 standard defines the protocols for transporting voice, data and video over A. PSTN B. the public Internet C. ISDN networks D. SONET 148. _______ occurs in single-mode fibers because dispersion varies with wavelength. This can result in a significant buildup of dispersion, especially at the extremes of a band of wavelength channels. A. Polarization mode dispersion B. Slope mismatch dispersion C. Rayleigh dispersion D. Chromatic dispersion 149.Controlling the ______ at which light waves are transmitted makes it possible to control how efficiently they reach their destination. A. the wavelength above which a single-mode fiber supports only one mode or ray of light B. the inherent curvature along a specific length of optical fiber C. the reduction of signal strength over the length of the light-carrying medium D. the time distortion of an optical signal that results from the time of flight differences of different components of that signal 150. Which of these statements is NOT true for SONET? A. uses frequency division multiplexing (FDM) B. it is a synchronous optical network C. common SONET interface ATM is STS-3c/STM at 155 Mbps D. supports both ATM and BISDN 151. ______ is the degradation of optical signals over distance. A. Capacity B. Distortion C. Dispersion D. Polarization 152. Which of following are not basic SONET network elements? A. switch interface B. digital loop carrier C. add/drop multiplexer D. service control point 153. A physics principle that became the theoretical foundation of optical fiber communications holds that ______ in a ______ medium can carry more information over longer distances. A. electrical signals; glass B. light; coaxial C. light; silica D. electrical signal: glass


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154. To overcome the problem of requiring a large amount of bandwidth for uncompressed analog video, operators should A. install fiber deeper B. use 1550-nm technology C. use dense wave division multiplexing D. increase the number of modes 155. What determines how many fibers a multimedia backbone can run? A. wavelength B. frequency C. spectral efficiency D. spatial multiplexing 156. What are the three primary steps in the optical fiber manufacturing process? A. laydown, compilation, and draw B. laydown, consolidation, and draw C. assembly, consolidation, and pull D. assembly, compilation, and pull 157. Attenuation is which of the following? A. the reduction of signal strength over the length of the light-carrying medium B. the wavelength above which a single-mode fiber supports only one mode or ray of light C. the inherent curvature along a specific length of optical fiber D. smearing an optical signal that results from the many discrete wavelength components traveling at different rates 158.Which of these is NOT an application for fiber channel? A. Multimedia B. backbones C. video conferencing D. all of these 159. What is the term for a switched fiber channel network? A. Packet B. Cell C. Fabric D. Frame 160. Which of these is not an advantage of fiber cables? A. easy to interconnect fiber segments B. secure - difficult to tap C. has a larger usable bandwidth D. freedom from electromagnetic interference


Telephone networks and system

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Section

Telephone Networks

12

and System


DEFINITION. Telephone: An electronic apparatus containing a transmitter and a receiver that is connected to a telecommunication system, enabling the user to speak to and hear other with similar equipment. DEFINITION. Network: A system of two or more communications devices link by wires, cables, or a telecommunication system in order to exchange information. DEFINITION. Telecommunication: The science and technology of transmitting information electronically by means of wire or radio signals with integrated encoding and decoding equipment. ISDN: Integrated Services Digital Network is a multi-use network in which wholly digital transmission is provided between customer locations, with digital telephones and data terminals being used.

DEFINITION.


JOHANN PHILIPP REIS (Germany) build Das Telephon, the 1st telephone to be publicly demonstrated.

1871

ANTONIO MEUCCI (Italy) files a caveat for a telephone that he claims to have built in 1849 though it is not demonstrated and ever patented.

1876

ALEXANDER GRAHAM BELL (Scotland) files a patent for the telephone. Hours later ELISON GRAY files a caveat for the same invention.

1889

ALMOND STROWGER files a patent for the first automatic telephone exchange later called STROWGER SWITCH. WILLIAM GRAY is granted the 1st patent for a coin-operated telephone.

1896

The 1st dial telephone goes into use in Milwaukee.

1946

AT&T inaugurates the 1st commercial mobile radio telephone service.

1981

NORDIC MOBILE TELEPHONE (NMT) system becomes international cellular mobile telephone system serving Denmark, Norway and Sweden.

the 1st Finland,

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Self-Sufficient Guide to ECE by JASON AMPOLOQUIO A. .HIERARCHY OF CLASS OFFICES.

TELEPHONE

LOCAL LOOP

END OFFICE

TOLL OFFICE

TOLL CONNECTING TRUNKS

INTERMEDIATE SWITCHING OFFICE(S)

INTER-TOLL TRUNKS

INTER-TOLL TRUNKS

TOLL OFFICE

END OFFICE

TOLL CONNECTING TRUNKS

TELEPHONE

LOCAL LOOP

1.

Class 5 (End Office) Class 5 offices are the local exchange offices, or central offices, serving end users through local loop connections. There exist approximately 19,000 Class 5 offices in the United States.

2.

Class 4 (Tandem Office) Class 4 offices were tandem toll centers, which served to interconnect Class 5 offices not directly connected. As the lowest class of toll center, these also served as the first point of entry to the long distance, or toll, network. Approximately 1,500 tandem toll centers existed in North America prior to AT&T’s divestiture of the BOCs.

3.

Class 3 (Primary Center) Class 3 offices, or primary toll centers, were higher-order toll centers, generally serving to connect Class 4 offices for intrastate toll calling. Class 4 offices typically served to interconnect independent telephone companies and BOCs. Approximately 200 such offices existed prior to divestiture.

4.

Class 2 (Sectional Center) Class 2 offices were known as sectional toll centers, which served to interconnect primary toll centers, largely for interstate calling within a geographic region such as the Northeast or the Southwest. There were approximately 67 sectional toll centers in the AT&T network prior to divestiture.

5.

Class 1 (Regional Center) Class 1 offices, or regional toll centers, served to interconnect sectional toll centers in support of interregional calling. There were 10 regional toll centers in place in the United States prior to divestiture; seven currently exist, with another two in Canada.



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B. .SWITCHING SYSTEM. 1.

Manual (Cordboard) Switching System These switching system involved operators who manually established the desired connection at the request of the transmitting party. A unique physical and electrical connection was established, on a plug and jack basis, for the duration of the call.

2.

Step-By-Step Switching (SxS) System SxS switches consist of large number of line finders to which groups of individual subscribers are assigned for dial tone. The transmitting party dials a series of numbers, originally with a rotary telephone terminal which causes the making and breaking of an electrical circuit.

3.

Crossbar Switching(Xbar) System In a XBar switch, a request for dial tone is recognized by a marker, which directs a sender to store the dialed digits. A translator is then directed to route the call, reserving a path through a switching matrix. Once the call is connected, these various components are available to serve other calls.


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Electronic Switching System (ESS) ESS switches reflect the marriage of computer technology and telephony. Voice conversations are digitized and switched over highspeed digital circuits, with all processes accomplished through programmed logic. ECC switches are microprocessor controlled, with the total processing power of such a switch often rivaling that of a general purpose, mainframe computer.

C. .COMPONENTS OF TELEPHONE SET. 1.

Transmitter Made up of a flat round diaphragm that converts speech signal into electrical signal. ª

Carbon Transmitter The carbon transmitter is constructed by placing carbon granules between metal plates called electrodes.

ª

Electret Transmitter The electret transmitter is composed of a thin disk of metal-coated plastic and a thicker, hollow metal disk.

Transmitting Objective Loudness Rating (TOLR) TOLR is a measure, in dB, of the efficiency with which a telephone set converts acoustic pressure at the set transmitter to an electrical output voltage. 2.

Receiver The receiver of a telephone set is made from a flat ring of magnetic material with a short cuff of the same material attached to the ring’s outer rim.

Receive Objective Loudness Rating (ROLR) ROLR is a measure, in dB, of the efficiency with which a telephone set converts electrical source voltage to an acoustic pressure output at the receiver.



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3.

Ringer/Alerter The alerter in a telephone is usually called the ringer, because for most of the telephone’s history, a bell was used to indicate a call. The alerter responds only to a special frequency of electricity that is sent by the exchange in response to the request for that telephone number.

4.

Dial Technique ª

Rotary Dial In a rotary dial, the numerals one to nine, followed by zero, are placed in a circle behind round holes in a movable plate. The user places a finger in the whole corresponding to the desired digit and rotates the movable plate clockwise until the user’s finger hits the finger stop.

ª

Push-button dial Push button dialing results in the generation of a unique combination of frequencies or tones when a button is pushed. Push-button telephones usually have a switch on the base that the customer can set to determine whether the telephone will send pulses or tones.

5.

Hybrid Converts the four-wire interface into a two-wire interface, which enables the telephone to operate via a 2-wire connection to the telephone company.

6.

Switch-Hook The switch-hook is place in an open position by pressure from the handset. Thus, when handset is lifted, the switch hook closes, enabling current to flow through the telephone and local loop and signaling the telephone company switch.

7.

Sidetone Generator The sidetone represents a design of the hybrid within the telephone that enables a portion of speech to ‘bleed’ over the earpiece or receiver. The purpose of the sidetone is to enable a person hear themselves talk so they have a better ability to adjust the tones of their conversation.

D. .TELEPHONE SIGNALING. 1.

In-Band Signaling In-band signaling and control functions take place over the same physical path as the conversation, and occupy the same frequency band. As the impact of simultaneous conversation and signaling is most unpleasant, it is seldom used in contemporary networks, with the exception of analog local loops.


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2.

Out-Of-Band Signaling Out-of-band signaling and control, in the simplest analog application, takes place over frequencies separate from those that carry the information. Out-of-band signaling and control is the standard approach in digital networks and, most certainly, in the internal carrier networks. Ex. 3700 Hz, 3825 Hz

3.

Inter-Office Signaling

a.

DC Signaling (Wet & Dry Signaling) Signaling information is indicated by the presence (wet) and absence (dry) of a battery and ground condition on the line at the called end of the trunk.

b.

Loop Reverse-Battery Signaling Loop signaling is accomplished by reversing the polarity of the battery on the line to indicated supervisory conditions.

c.

E and M Signaling Employs two leads to connect the signaling equipment to the trunk circuit where the M lead transMit battery or ground signal to the distant end of the circuit while incoming signals are recEived on the E lead as either a ground or open condition.

d.

A and B Bit Signaling Supervisory and address information may be assigned to specific bits in the pulse stream. These bit positions are derived by time sharing the least significant speech bit during one frame out of six, obtaining a 1.33 kbps signaling bit stream.

e.

Single Frequency Signaling Single Frequency in-band signaling on analog trunks uses a tone frequency of 2600 Hz transmitted at -8dBm0 during pulsing, and 20dBm0 for continuous tone.

f.

Multi Frequency Signaling Multi Frequency signaling is often applied on trunk circuits where digit information is sent by combination of two of five audio frequencies: 700, 900, 1100, 1300, and 1500 Hz.



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4.

Subscriber Loop Signaling

a.

b.

Dial Pulse signaling Consists of series of from one to ten pulses representing the corresponding numerical digits 1 to 9 and 0.

Dialed Digit

Generated Pulse

Duration

1 2 3

1 pulse 2 pulses 3 pulses

100 ms 200 ms 300 ms

0

10 pulses

1 sec

Dual Tone Multi-Frequency (DTMF) signaling DTMF represent numerical digit by one pulse of a specific frequency combination.

An average of 50 to 250 msec for every dialed digit. ------------------------Another 50 to 250 msec for the interdigit interval.


Sample Problem:

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Calculate the time saved in dialing 640-8267 using DTMF system over PD system.

Solution: DTMF system tDTMF ≥ 7(0.25 sec) + 6(0.25sec) = 3.25 s PD system tPD ≥ 6(0.1 sec) + 4(0.1 sec) + 10(0.1 sec) + 8(0.1 sec) 2(0.1 sec) + 6(0.1 sec) + 7(0.1 sec) + 6(0.5 sec) = 7.3 sec Δt = tPD − tDTMF ≥ 4.05 s

Summary of Subscriber loop signaling

Signal

From

To

Function

Off-Hook

User

C.O.

Informs central office that user wants to place a call

On-Hook

User

C.O.

Informs central office that user terminates the connection

Dial tone

C.O.

User

Informs user that central office is ready to accept dialing

Busy signal

C.O.

User

Informs user that destination phone is already in use

Ringback Tone

C.O.

User

Informs user that the destination phone is ringing

Ringing voltage

C.O.

User

Special voltage sent by C.O. to cause a phone’s bell to ring

Touch-tone

User

C.O.

Informs central office of call destination

Flash

User

C.O.

A combination of the on-hook and off-hook signals—the user quickly depresses the switch-hook and release it; often used to pick up a waiting call

Call waiting Tone

C.O.

User

Informs user that another call is waiting on the line



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Frequency parameter

Signaling Dial tone

350

Frequency (Hz) 440 480 620

x

x

Continuous

Busy tone

x

Ringing tone

x

Call-waiting tone

x

Cadence

X

x

0.5 sec ON 0.5 sec OFF 2 sec ON 4 sec OFF Single 500 ms pulse

Summary of Telephone standards

Item

Standards

On hook (idle status) Off hook (busy status) Battery voltage Operating current Subscriber-loop resistance Loop loss Ringing voltage

Line open circuit, min dc resistance 30kΩ Line closed circuit, max dc resistance 200Ω -48 V 20 to 80 mA, 40 mA typical 0 to 1300Ω, 3600Ω (max) 8 dB typical, 17 dB max 90 Vrms, 20 Hz (typical)

Pulse dialing Pulsing rate Duty cycle

10 pulses/sec +10% 58 to 64% break

Touch tone dialing Tone level Maximum level Frequency tolerance Pulse width Time between digits

-6 to -4 dBm +2 dBm +1.5% 50 ms 50 ms minimum

E. .CENTREX & PABX. 1.

Centrex As a concept, Centrex (central exchange), is something of a step back in time, providing PBX-like features through special software loaded into the Central Office Exchange. Generally speaking, each Centrex station is generally connected to the central exchange via an individual twisted pair local loop.


2.

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Private Automatic Branch Exchanges (PABXs/PBXs) Clearly, a better approach would involve effectively moving a partition of the central office exchange to the customer premise. In this fashion, a Private Automatic Branch of the Exchange (PABX) would yield significant benefit to Telco and the user organization.

F. .TELEPHONE LOSSES. 1.

Net Loss The net loss of a transmission channel is the ratio of the signal power at the input and the output of the channel. Net Loss is measured at: American and Canadian Plant European (CCITT)

2.

1004 Hz 1020 (formerly 800) Hz

Insertion Loss The ratio of the power delivered from a source to a load, to the power delivered from the same source to the same load through a transducer.



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3.

Transducer Loss The ratio of the maximum power available from a source to the power delivered by that source to a load through a transducer.

4.

Return Loss A measure of the match between the two impedance on either side of a junction point.

5.

Echo Return Loss The weighted power-average return loss at the reflection point covering the band of approximately 500-2500 Hz.

6.

Singing Return Loss The same as Echo Return Loss, but over a considerably narrow band near an edge of the voice band, e.g., 200-500 Hz or 2500-3200 Hz.

G. .CROSSTALK, SINGING, & ECHO.

1.

2.

Crosstalk Unwanted interference or other signal picked up by one channel of an electronic communications system from other channel. ª

Near-End Crosstalk Measured on a channel at a receiving point near the sending point of the interfering channel.

ª

Far-End Crosstalk Measured on a channel at a receiving point near the receiving point of the interfering channel.

Singing Singing is the result of sustained oscillations due to positive feedback in telephone amplifiers or amplifying circuits.

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Telephone Echo

ª

Talker Echo A signal returned to the talker after making one or more round trips between the talker and a distant reflection point.

ª

Listener Echo A signal first returned down the talker at a distant reflection point and then reflected again toward the listener.

H. .PROPAGATION TIME & VIA NET LOSS (VNL). CCITT recommends the following limitations propagation time when echo source exist. ª ª

ª 1.

on

mean,

0 to 150 ms acceptable. 150 to 400 ms, acceptable, provided that care is exercised on connection when the mean, one-way propagation time exceeds about 300 ms. Above 400 ms, unacceptable.

For Analog Network:

Delay (ms) = 12 + [0.004 x d(km) ]

2.

one-way

For Digital Network:

Delay (ms) = 3 + [0.004 x d(km) ]


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Sample Problem:

Calculate the one-way propagation delay for an analog network and digital network which spans 5 km and 8 km respectively.

Solution: For Analog Network D = 12 + 0.004 x d(km) = 12 + (0.004 x 5) = 12.02 ms

{

}

For Digital Network D = 3 + 0.004 x d(km) = 3 + (0.004 x 8) = 3.032 ms

{

}

Via Net Loss – A concept or method of transmission planning that permits a relatively close approach t an overall zero transmission loss in the telephone network and maintains singing and echo within specified limits.

⎛ L VNL (dB ) = 0.2 ⎜ ⎜ vp ⎝

⎞ ⎟ + 0.4 dB ⎟ ⎠

VNL ( dB ) = 0.2 Δ t + 0.4 dB

L = one - way length of the trunk d = velocity of propagatio n Δ t = one way propagatio n time delay in ms

ECE Board Exam APRIL 2003

A telephone signal takes 11.8 ms to reach its destination. Calculate the via net loss required for an acceptable amount of echo?

Solution: VNL(dB) = 0.2Δt + 0.4 dB = 0.2(11.8) + 0.4 = 2.76

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.SUBSCRIBER LOOP DESIGN. 1.

Two Basic Consideration i.

Resistance Limit

R dc =

0.1095 d2

R dc = loop resistance in Ω / mi d = diameter of the conductor in inches

Table of dc resistance per unit length of Subscriber cable

Gauge

Ω/1000 ft of loop

Ω/mi of loop

Ω/km of loop

26 24 22 19

83.5 51.9 32.4 16.1

440 274 171 85

268 168.5 106 53

Sample Problem:

If we want a 10-mi loop and allow 100Ω/mi of loop. copper wire would we need?

What diameter of

Solution: 100 =

ii.

0.1095 d2

⇒d=

0.1095 = 0.033 in 100

Attenuation Limit The maximum attenuation for a loop is limited to 6 dB. Table of Loss per unit length of Subscriber cable

Gauge 26 24 22 19

dB/1000 ft

Loss dB/mi

dB/km

0.51 0.41 0.32 0.21

2.69 2.16 1.69 1.11

1.61 1.27 1.01 0.46


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Sample Problem:

Calculate the maximum loop length permissible for transmission design consideration if we use a cable gauge of 19.

Solution: L max =

αmax 6 dB = = 28.57 kft = 5.4 mi = 13.04 km 0.21 dB α19 1000 ft

Loading - a method of extending subscriber loop length by simply inserting loading coil into the loop at fixed intervals.

2.

J.

Code Letter

Spacing (ft)

Code Letter

Spacing (ft)

A B C D *most used

700 3000* 929 4500*

E F H Y

5575 2787 6000* 2130

Standard Approach to Subscriber Loop Design ª

Unigauge Design Unigauge is a concept developed by BELL-CORE of North America. This is done by reducing the gauge (diameter) of wire pairs as much as possible while retaining specific resistance and transmission limits.

ª

Resistance Design Resistance design is a method of designing subscriber loops based on establishing a common maximum resistance limit for a switch. This valued is set at 1300Ω primarily to control transmission loss.

ª

Long Route Design The long route design procedure uses several zones corresponding to ranges of resistance in excess of 1300Ω. LRD provides for a specific combination of electronic range extenders and/or fixedgain devices to meet the supervision and loss criteria.

.MULTIPLEXING. Multiplexing is a technique used in communications and input/output operations for transmitting a number of separate signals simultaneously over a single channel or line.

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Time Division Multiplexing Each communication channel is allotted a fixed time slot within a sampling frame, occupying essentially the entire wideband frequency spectrum for the allocated time. i.

T-Carrier System

System

Voice Channel

Data Rate (Mbps)

Overhead (Mbps)

T1 24 1.536 0.008 T2 96 6.144 0.168 T3 672 43.008 1.728 T4 4032 258.048 16.128 T5 8064 516.096 44.064 *BnZS denotes binary n-zero substitution

Line Rate (Mbps)

Line Code

1.544 6.312 44.736 274.176 560.160

BPRZ B6ZS RZ B3ZS RZ Polar NRZ Polar NRZ

T5 = 2T4 = 12T3 = 84T2 = 336T1 Loading ECE SUPERBook

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Telephone networks and system ii.

E-Carrier System

System

Voice Channel

Signaling Channel

Total

Line Rate (Mbps)

E1 E2 E3 E4 E5

30 120 480 1920 7680

2 8 32 128 512

32 128 512 2,048 8,192

2.048 8.192 32.768 131.072 524.288

E5 = 4E4 = 16E3 = 64E2 = 256E1 2.

Frequency Division Multiplexing A unique band of frequencies within the wideband frequency spectrum of the medium is allotted to each communication channel on a continuous time basis.

FDM Process

FDM Demultiplexing Process

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Self-Sufficient Guide to ECE by JASON AMPOLOQUIO i.

AT&T FDM Hierarchy

Hierarchy

JG

MG

SG

G

VB

Group Supergroup Mastergroup Jumbogroup Superjumbogroup

--------3

------6 18

----10 60 180

--5 50 300 900

12 60 600 3600 10,800

Summary of BW and Occupied Spectrum Parameter

Group

SG

Bandwidth

48 kHz

240 kHz

MG 2520 kHz (U600) 2728 kHz (L600)

Occupied spectrum

ii.

60-108 kHz

564-3084 kHz (U600) 60-2788 kHz (L600)

312-552 kHz

CCITT’s FDM Hierarchy

Hierarchy

MG

SG

G

VB

Group Supergroup Mastergroup Supermastergroup

------3

----5 15

--5 25 75

12 60 300 900


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K. .INTEGRATED SERVICE DIGITAL NETWORK. 1.

HDSL HDSL uses two twisted pair (one transmit and one receive) to support 1.544 Mbps at a full duplex at a distance of up to 12,000 ft from the RT. It uses a 2B1Q line code if ISDN or suppressed carrier version of QAM called carrierless amplitude-phase (CAP)

Data

ISDN

Voice

Video

2.

SDSL SDSL is a one pair version of HDSL. It provides full duplex to support 768 kbps in each direction using a hybrid or echo canceller to separate data transmitted from data received.

3.

ADSL ADSL uses one twisted pair to support 6 Mbps sent downstream to the customer and 640 kbps sent upstream over a distance of up to 12,000 ft. The ADSL spectrum is above 25 kHz. The band below 4 kHz is used for a voice frequency POTS signal. Two variations of ADSL called G.DMT and G.Lite.

G.DMT uses discrete multitone modulation with up to 256 carriers and up to 15 bits of data, 32,768 QAM, modulated on each carrier. G.DMT uses a splitter so that the telephone handset will not “short out” the data signal and vice-versa. 4.

VDSL VDSL uses one pair of wires to support 25 Mbps downstream for distances up to 3,000 ft from the RT or 51 Mbps downstream for distances up to 1,000 ft.

5.

ISDN Connections & Interface Units


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Terminal Equipment (TE) TE is the term for a functional device that connects a customer site to ISDN services. Examples include computers, telephones, facsimile machines, and videoconferencing units. TE1 has a builtin ISDN interface, while TE2 devices do not have native ISDN compatibility.

TE1 - Supports standard ISDN interfaces and therefore required no protocol translation. TE2 - This are classified as non-ISDN compatible, physical interface is used like RS 232C and host computer with X.25. ª

Terminal Adapters (TAs) TAs are interface adapters for connecting one or more TE2 (nonISDN) devices to an ISDN network. TAs acts as ISDN DCE, serving a function equivalent to protocol or interface converters.

ª

Network Termination type 2 (NT2s) NT2s are intelligent devices responsible for the user’s side of the connection to the network, performing such functions as multiplexing, switching or ISDN concentration. A NT2 device would likely be in the form of a PABX, LAN router or switching hub.

Used to terminate several S-point connections, provide local switching function and 2-wire to 4-wire conversion & vice-versa. ª

Network Termination type 1 (NT1s) NT1s physically connect the customer site to the carrier side of the connection, performing such functions as signal conversion and maintenance of the local loop’s electrical characteristics.

In a PRI environment, these functions are similar to those provided by Data Service Units (DSUs) and Channel Service Units (CSUs). In a BRI environment, these devices are TE1 devices. ª

Line Termination (LT) LT provides physical interface function between C.O. and local loop lines.

ª

Exchange Termination (ET) ET routes data to an outgoing channel or C.O. users. Reference Points

ª

R reference points - provides interface between a non-ISDN device (TE2) to terminal adapters (TA).


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6.

ª

S reference points - provides interface between an ISDN compatible device (TE1) to network termination 2 (NT2), provide the 2B+D data rate at 192 kbps.

ª

T reference points - separate the network provider’s equipment from the user equipment. Interface between NT2 and NT1.

ª

U reference points - refer to interface between common carrier subscriber loop and the C.O. switch, media interface between NT1 and C.O.

ª

V reference points - media interface between LT and ET.

Data Rates ª

Symmetric

System

Upstream Rate

Downstream Rate

Maximum Distance (ft)

High Bit Rate

1.544 Mbps

1.544 Mbps

12,000

HDSL

2.048 Mbps

2.048 Mbps

12,000

Single Line (SDSL)

1.544 to 2.048 Mbps

1.544 to 2.048Mbps

10,000

128 kbps

128 kbps

18,000

Upstream Rate (kbps)

Downstream Rate

Maximum Distance (ft)

64 kbps

1.544Mbps

8,000

Amplitude Phase (CAP ADSL)

640 kbps

6.312Mbps

12,000

Discrete

176 kbps

1.544Mbps

18,000

Multitone (DMT ADSL) Rate Adaptive (RADSL)

224-260 kbps 128 kbps to 1 Mbps

Very High Bit Rate (VDSL)

1.6 to 2.3 Mbps

ISDN (IDSL) ª

Asymmetric

System Carrierless

6.312Mbps 600 kbps to 7Mbps 12.96Mbps 25.82Mbps 51.84Mbps

12,000 18,000 to 25,000 4,500 3,000 1,000


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Primary Rate Interface (PRI) i.

ii.

ª

American standard 23B+D = 1.544 Mbps (similar to T1 line speed) European standard 30B+D = 2.048 Mbps (similar to E1 line speed)

Basic Rate Interface (BRI)

2B+D = 192 kbps Data rate = 144 kbps Bearer (2B) channel = 128 kbps using circuit switching Signaling (D) channel = 16 kbps using packet switching Overhead = 48 kbps

L.

.TRAFFIC ENGINEERING. 1.

Terminology ª

Concentration: The function associated with a switching network having fewer outlets than inlet terminals.

ª

Coordinate switch: a rectangular array of cross-points in which one side of the crosspoint is multiplied in rows and the other side in columns.

ª

Crosspoint: A two-state switching device containing one or more elements that a low transmission impedance in one state and a very high one in the other.

ª

Full availability: Property of a switch or switching network capable of providing a path from every inlet terminal to every outlet terminal.

ª

Internal blocking: The inability to interconnect an idle inlet to an inlet outlet because all possible paths between them are already in use. Number of crosspoint required = N (N - 1)/2

ª

Busy hour: The continuous one-hour period that, on consecutive days in the busy part of the year, contains the maximum average traffic intensity.

ª

Call: A discrete engagement or occupation of a traffic path.


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2.

ª

Calling rate: The average number of calls placed during the busy hour.

ª

Occupancy: The traffic intensity per traffic path. percent occupancy implies all paths busy.

ª

Traffic Concentration: The average ratio of the traffic quantity during the busy hour to the traffic quantity during the day.

ª

Traffic intensity: The average number of calls present on a group of traffic paths over a period of time.

ª

Traffic path: A channel, time slot, frequency band, line, trunk, switch, or circuit over which individual communications pass in sequence.

ª

Traffic quantity: The aggregate engagement time or occupancy time of one or more traffic paths.

ª

Call Intensity: For many traffic-carrying elements, the number of calls making up the total traffic load is immaterial; the load represented by two calls or ten minutes duration has the same impact as one call of twenty minutes duration.

One hundred

Traffic Unit ª

Erlang The international dimensionless unit of traffic intensity. One erlang is the traffic intensity represented by an average of one circuit busy out of a group of circuits over some period of time.

ª

Traffic Unit (TU) 1 TU is the average intensity in one or more traffic paths carrying an aggregate traffic of 1 call-hour in 1 hour (the busy hour unless otherwise specified).

ª

Equated Busy-Hour Call (EBHC) 1 EBHC is the average intensity in one or more traffic paths occupied in the busy-hour by one 2-min call or an aggregate duration of 2 min.

A European unit of traffic intensity equal to 1/30 of an erlang. ª

Call-Hour The quantity represented by one or more calls having an aggregate duration of 1 hour.

ª

Century or Hundred call-second per hour (CCS) A unit of traffic intensity equal to 1/36 of an erlang.


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ª

Call-minute (Cm) 1 Cm is the quantity represented by one or more calls having an aggregate duration of 1 minute.

ª

Call-second (Cs) 1 Cs is the quantity represented by one or more calls having an aggregate duration of 1 second.

Relation between different traffic units

1Erlang = 1 TU = 1 Ch = 60 Cm = 3600 Cs = 36 CCS = 30 EBHC

Read it till it Hertz…jma 1 Erlang is equivalent to traffic intensity that keeps: ª 1 circuit busy 100% of the time, or ª 2 circuit busy 50% of the time, or ª 4 circuits busy 25% of the time, etc. While 26 Erlangs is equivalent to traffic intensity that keeps: ª 26 circuits busy 100% of the time, or ª 52 circuit busy 50% of the time, or ª 104 circuits busy 25% of the time, etc. 1 ccs is volume of traffic equal to: ª One circuit busy for 100 seconds, or ª 2 circuit busy for 50 seconds, or ª 50 circuits busy for 2 seconds, or ª 100 circuits busy for 1 second, etc. The term “ccs” is unfortunately in common usage when traffic intensity, not traffic volume is describe. Strictly speaking ‘ccs/hr should be used for traffic

intensity and ccs for traffic volume’. 3.

Calling Rate (C) The number of times a route or traffic path is used per unit period, or, more properly defined, “the call intensity per traffic path during the busy hour”

Average number of calls initiated per unit time (e.g. attempts per hour). a.k.a. Arrival Rate Loading ECE SUPERBook


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ª

For n receive calls from a terminal in time t:

C=

ª

n t

For n receive calls from m terminals in time t: a. Group Calling Rate

C group =

b.

n t

Per terminal Calling rate

C=

n mx t

4.

Holding time (T) The duration of occupancy of a traffic path by a call. Sometimes used to mean the average duration of occupancy of one or more paths by calls.

5.

Departure Rate (μ)

μ=

6.

1 T

Traffic Volume (usually in ccs)

V = nx T

7.

T = mean holding time per call

where : V = volume of calls in time t n = # of calls in time period t T = mean holding time per call

Traffic Intensity (A) Traffic intensity or traffic flow is a total traffic volume divided by the duration of time. i.

For a single terminal The traffic in Erlang is the average occupancy of the terminal while the traffic intensity or traffic flow is just the percentage of time the terminal is busy.

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For a group of circuits or terminal a. The average number of circuits simultaneously busy within a group.

b.

The expected number of call arrivals per unit holding time

c.

The number of circuits required to completely carry the offered traffic if each circuit were operating at 100% occupancy.

A=

n xT t

A=

C μ

A = C xT

A=

where: A = Traffic intensity in Erlang T = mean holding time per call C = calling rate V = volume of calls

V t

n = # of calls in time period t t = time period of observations μ = departure rate

Sample Problem:

Suppose that the average holding time is 2.5 min per call and the calling rate in the BH for a particular day is 237. Determine the traffic flow (A) in call-minutes (Cm) and call-hours (Ch).

Solution: A = CxT = 237 x 2.5 min = 592.5 Cm

A = 592 .5 Cm x

1 Ch 60 Cm

= 9.87 Ch


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Sample Problem:

Call established at 2 am between a central computer and a data terminal. Assuming a continuous connection and data transferred at 34 kbit/s what is the traffic if the call is terminated at 2:45 am?

Solution:

A = Cx T ⎛ 1 hr ⎞ ⎟ = 1 call x ⎜⎜ 45 min x 60 min ⎟⎠ ⎝ = 0.75 Erlang

Sample Problem:

A group of 20 subscribers generate 50 calls with an average holding time of 3 minutes, what is the average traffic per subscriber?

Solution: ⎡ ⎛ 1hr ⎞ ⎤ A total = 50 calls x ⎢3 min x ⎜ ⎟⎥ ⎝ 60 min ⎠ ⎦ ⎣ = 2 .5 Erlangs

8.

A sub = =

2.5 Erlangs 20 subscriber 0.125 Erlang subscriber

Offered Traffic (TO) Offered traffic is the traffic intensity that would occur if all traffic submitted to a group of circuits could be processed. The volume of traffic offered to a switch.

9.

Carrier Traffic (TC) Carried traffic is the traffic intensity actually handled by the group of circuits.

The volume of traffic actually carried by a switch. 10. Lost Traffic (TL) Blocked traffic is that portion of traffic that cannot be processed by the group of circuits.

The difference between the offered and carried traffic.

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Self-Sufficient Guide to ECE by JASON AMPOLOQUIO Relation between TO, TC, and TL

TO = TC + TL

TL = T O x P (B )

TC = TO x (1 − P(B))

Where P = Blocking Probability or Call Congestion 11. Circuit Utilization (ρ) The proportion of time a circuit is busy, or average proportion of time each circuit in a group is busy. a.k.a. Circuit efficiency

ρ=

TC #N

where: #N = no. of trunk circuits

Sample Problem:

Calculate the trunk efficiency for a group of 26 trunks that offers 10 Erlangs of traffic and a blocking probability of 0.2%.

Solution: TC ⇒ TC = TO x(1 − P(B)) #N 10(1 − 0.002) = 26 = 0.384 ⇒ 38.4%

ρ=

12. Grade of Service(GOS) or Blocking Probability(P(B)) A measure of probability that, during a specified period of peak traffic, a call offered to a group of trunks of circuits will fail to find an idle circuit at the first attempt. Usually applied to the busy hour of traffic.

GoS = P(B) =

TL TL = TO TL + TC



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Sample Problem:

If we know that there are 354 seizures (lines connected for service) and 6 blocked calls (lost calls) during the BH, what is the grade of service?

Solution: # of lost calls # of offered calls 6 = 354 + 6 = 0.017

GOS =

13. Illustration of Basic Relation between Offered Traffic, Carried Traffic, Grade of Service and Trunk Utilization ª

Dedicated Service

Erlang x150 = 15 Erlang source

Offered Traffic

0.1

Grade of Service (GoS)

Nonblocking (0%)

Carried Traffic

TC (1 − GoS) = 15(1 − 0) = 15 Erlang

Trunk Utilization

15 Erlang x100% = 10% 150 Trunks

Terminal/Trunk ratio

150 ter min al = 1 (Expensive) 150 trunks


Traffic-engineered Service

Erlang x150 = 15 Erlang source

Offered Traffic

0.1

Grade of Service (GoS)

1% blocking (Erlang B condition)

Carried Traffic

TC (1 − GoS) = 15(1 − 0.01) = 14.85 Erlang

Trunk Utilization

14.85 Erlang x100% = 61.875% 24 Trunks

Terminal/Trunk ratio

150 ter min al = 6.25 (Economical) 24 trunks

M. .BLOCKING PROBABILITY.

SCENARIO


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1.

Lost Calls Cleared (Blocked Calls Cleared) The LCC concept, which is used primarily in Europe or those countries accepting European practice, assumes that the user will hang up and wait some time interval before reattempting if the user hears the congestion signal on the first attempt. Such calls, it is assumed, disappear from the system.

The assumption that calls not immediately satisfied at the first attempt are cleared from the system and do not reappear during the period under consideration. Used in the Erlang B Loss-probability equation.

Lost Calls Cleared Call Behavior:

1st call arrived and is served. ----------------------2nd call arrives but server already busy ----------------------2nd call is CLEARED. ----------------------3rd call arrived and is served. ----------------------4th call arrived and is served.

An GOS = PB = n n! Ax x! x=0

∑

where : A = traffic intensity in Erlang n = # of trunks

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Sample Problem:

Suppose we use 5 trunks, and the route offered 1.66 Erlangs of traffic. Calculate the grade of service required to implement this configuration.

Solution:

GoS =

⎡1.665 ⎤ ⎢ ⎥ ⎢⎣ 5! ⎥⎦ ⎡ 1.660 1.661 1.662 1.663 1.664 1.665 ⎤ + + + + + ⎢ ⎥ 1! 2! 3! 4! 5! ⎦⎥ ⎣⎢ 0!

x 100%

= 2%

2.

Lost Calls Delayed (Blocked Calls Wait) The LCD concept assumes that the user is automatically put in queue (a waiting line or pool). For example, this is done when the operator is dialed. It is also done on most modern computer-controlled switching systems, generally referred to under the blanket term stored program control (SPC).

The assumption that calls not immediately satisfied at the first attempt are held in the system until satisfied. Used in the Erlang C delay-probability equation.

Lost Calls Delayed Call Behavior:

1st call arrived and is served. ----------------------2nd call arrives but server already busy ----------------------2nd call WAITS until server is free ----------------------2nd call served. ----------------------3rd call arrives, waits and is served. ----------------------4th call arrives, waits and is served.



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PC =

n −1

∑ x=0

3.

where :

z Ax +z x!

z=

An n n! n − A

Lost Calls Held (Blocked Calls Held) The LCH concept, which is the principal traffic formula used in North America, assumes that the telephone user will immediately reattempt the call on receipt of a congestion signal and will continue to redial. This concept further assumes that such lost calls extend the average holding time theoretically, and in this case the average holding time is zero, and all the time is waiting time.

The assumption that calls not immediately satisfied at the first attempt are held in the system until served or abandoned. Used in the Poisson loss-probability equation.

Lost Calls Held Call Behavior:

1st call arrived and is served. ----------------------2nd call arrives but server already busy ----------------------2nd call is HELD until server is free ----------------------2nd call served. ----------------------3rd call arrives and is served. ----------------------4th call arrives and is served.

Poisson (Molina) Formula

PP = e − A

∞

Ax

∑ x! x =n

Self-Sufficient Guide to ECE by JASON AMPOLOQUIO N. .CHOICE OF FORMULA.

O. .CONGESTION. 1.

Time Congestion Proportion of time a system is congested (all servers busy).

Probability of blocking from point of view of servers. 2.

Call Congestion Probability that an arriving call is blocked.

Probability of blocking from point of view of calls.


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Sample Problem:

On a particular traffic relation the calling rate is 461 (calls in a 1-hr period) and the average call duration is 1.5 minutes during the busy hour. What is the traffic intensity in Erlangs? In ccs?

Solution: 1 hr calls x 1 .5 min x = 11 .525 Erlangs hr 60 min 36 ccs 11 .525 Erlangs x = 414 .9 ccs 1 Erlang

461

Sample Problem:

Company X has 10 employees, each placing an average of 20 minutes of long-distance calls per day. The average call lasts 5 minutes. It has been determined that 20% of the calls are made during the busy hour. A total of 4 external phone lines are used to place the pool of calls. Calculate the traffic intensity in Erlang, during the busy hour.

Solution: Each employee places on average 20 min calls ⇒ = 4 5 min day For 10 employees, a total of calls calls x 10 = 40 day day The traffic during the busy hour is calls calls ⇒ 20% x 40 =8 day hr calls 1 Erlang ⇒8 x 5 min x = 0.6667 Erlangs hr 60 min ⇒ 4


Sample Problem:

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Suppose 100 data terminals are to be connected to a computer by way of leased circuits: 1st plan: The terminals are clustered into four groups that use separate groups of shared circuits. 2nd plan: Traffic from all terminals is concentrated into one group of circuits. Determine the cluster traffic in both cases assuming each terminal is active 10% of the time.

Solution:

a. For the 1st plan; # of terminal for each cluster

⇒

100 ter min als ter min als = 25 4 groups group

The total traffic for each cluster is ⇒ 25 x 10% = 2.5 Erlangs b. For the 2nd plan; # of terminal for each cluster

100 ter min als ter min als = 100 1 group group The total traffic for the cluster is ⇒ 100 x 10% = 10 Erlangs ⇒



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I

H

1.

If there are 2800 maximum calls arriving per hour to a trunk circuits without blocking, calculate the offered traffic for an average calling time of 1.76 minutes and compute the GoS for 2812 simultaneous calls. A. 82.13 Erlangs, 0.43% B. 74.18 Erlangs, 0.22% C. 143.66 Erlangs, 0.22% D. 88.88 Erlangs, 0.34%

2.

If a group of user made 30 calls in one hour, and each call had an average call duration of 5 minutes, this is equivalent to how many Erlangs of traffic? A. 1.5 Erlang B. 8.5 Erlang C. 12.5 Erlang D. 2.5 Erlang

3.

The range of DC current that flows through a telephone is: A. 20 μA to 80 μA B. 200 μA to 800 μA C. 2 mA to 8 mA D. 20 mA to 80 mA

4.

Calculate the blocking probability for a group of 6 trunks that is designed to carry 3 Erlangs of traffic intensity using the Erlang B formula. A. 3.33% B. 2.22% C. 6.33% D. 5.22%

5.

What frequencies are used in pressing 5? A. 770 Hz, 1336 Hz B. C. 770 Hz, 1209 Hz D.

852 Hz, 1336 Hz 852 Hz, 1633 Hz

6.

The separation of control functions from signal switching is known as: A. step-by-step switching control B. crossbar control C. common control D. ESS

7.

Find the holding time if 300 calls registered 13.5 erlang traffic? A. 27 s B. 261 s C. 162 s D. 22.23 s

8.

"Bit-stuffing" is more formally called: A. compensation C. justification

9.

B. D.

rectification frame alignment

If we wish to have a 10-mile long subscriber loop with a total resistance of 1300 ohms, find the diameter of the copper conductor to be used. A. 0.0812 in B. 0.029 in C. 0.0632 in D. 0.03256 in


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10. For a certain telephone, the DC loop voltage is 48 V on hook and 8 V off hook. If the loop current is 40 mA, what is the DC resistance of the local loop? A. 1300 ohms B. 1500 ohms C. 1000 ohms D. 2300 ohms 11. For a certain telephone, the DC loop voltage is 48 V on hook and 8 V off hook. If the loop current is 40 mA, what is the DC resistance of the telephone? A. 48 0hms B. 200 ohms C. 40 ohms D. 1300 ohms 12. The typical voltage across a telephone when on-hook is: A. 48 volts DC B. 48 volts, 20 hertz AC C. 90 volts DC D. 90 volts, 20 hertz AC 13. If a telephone voice signal has a level of 0 dBm, what is its level in dBrn? A. -90 dBrn B. -85 dBrn C. 90 dBrn D. 85 dBrn 14. In DS-1, bits are "robbed" in order to: A. provide synchronization B. C. cancel echoes D.

carry signaling check for errors

15. A telephone test-tone has a level of 80 dBrn at a point where the level is +5dB TLP. If C-weighting produces a 10-dB loss, what would the signal level be in dBrnc0? A. 80 dBrnC TLP B. 65 dBrnC TLP C. 60 dBrnC TLP D. 75 dBrnC TLP 16. What is the approximate data rate for a system using 8 bits per sample and running at 8000 samples per second? A. 56 kbps B. 128 kbps C. 64 kbps D. 384 kbps 17. If bits were "stolen" from every DS-1 frame, what would the useable data-rate be for each channel in the frame? A. 64 kbps B. 8 kbps C. 128 kbps D. 56 kbps 18. POTS stands for: A. Private Office Telephone System C. Primary Operational Test System

B. D.

Primary Office Telephone Service Plain Old Telephone Service

19. Calculate the traffic flow in call-minute and call-hour if 100 calls are generated in 1 hour of 3 minutes average duration. A. 260 call-minute, 4 call-hour B. 600 call-minute, 10 call-hour C. 300 call-minute, 5 call-hour D. 480 call-minute, 8 call-hour 20. DC A. B. C. D.

current flows through a telephone: when it is on hook when it is off hook as long as it is attached to a local loop only when it is ringing



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21. Calculate the dB of VNL required for a channel with a 3 ms delay. A. 1 dB B. 1.4 dB C. 0.4 dB D. 0.8 dB 22. Determine the actual traffic carried by a group of trunks with an offered traffic of 20 Erlangs and 0.98785 blocking probability. A. 1.113 Erlang B. 0.243 Erlang C. 6.743 Erlang D. 13.756 Erlang 23. Signal loss is designed into a telephone system to: A. eliminate reflections B. prevent oscillation C. improve signal-to-noise ratio D. reduce power consumption 24. Calculate the blocking probability for a group of 4 trunks that offers 10 Erlangs of traffic and a trunk efficiency of 88%. A. 0.8666 B. 0.64666 C. 0.9666 D. 0.466 25. Call A. B. C. D.

blocking: cannot occur in the public telephone network occurs on the local loop when there is an electrical power failure occurs only on long-distance cables occurs when the central office capacity is exceeded

26. A single GSM carrier supports 8 (TDM) speech channels. If the traffic volume is 129.6 ccs. How many 3 minute calls does this represent? A. 172 calls B. 7.2 calls C. 72 calls D. 720 calls 27. The typical voltage needed to "ring" a telephone is: A. 48 volts DC B. 48 volts, 20 Hz AC C. 90 volts DC D. 90 volts, 20 Hz AC 28. VNL stands for: A. voltage net loss C. via net loss

B. D.

volume net loss voice noise level

29. If we know that there are 512 seizures (lines connected for service) and 24 blocked calls (lost calls) during the BH, what is the grade of service? A. 0.017 B. 0.032 C. 0.045 D. 0.067 30. SLIC stands for: A. Single-Line Interface Circuit C. Subscriber Line Interface Card

B. D.

Standard Line Interface Card Standard Local Interface Circuit

31. On a particular traffic relation the calling rate is 461 and the average call duration is 1.5 min during the BH. What is the traffic intensity in CCS? A. 345.75 CCS B. 414.9 CCS C. 1,152.5 CCS D. 691.5 CCS 32. The reference noise level for telephony is: A. 1 mW B. C. 1 pW D.

0 dBm 0 dBr


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33. In loading cable 26B66 means _____. A. 26 gauge every 2000 ft with 66 pH inductors B. 26 gauge every 4000 ft with 66 nH inductors C. 26 gauge every 3000 ft with 66 mH inductors D. 26 gauge every 6000 ft with 66 μH inductors 34. The BRI requires _____ kbps of digital transmission capacity without overhead. A. 64 B. 144 C. 128 D. 192 35. PSTN stands for: A. Public Switched Telephone Network B. Private Switched Telephone Network C. Primary Service Telephone Network D. Primary Service Telephone Numbers 36. 0.005 of GoS means_____. A. 1 call in 20 would be lost because of insufficient equipment B. 2 call in 400 would be lost because of insufficient equipment C. 5 call in 100 would be lost because of insufficient equipment D. 10 call in 20000 would be lost because of insufficient equipment 37. How many connections are needed to connect 50 subscribers directly? A. 2125 B. 2450 C. 1225 D. 2521 38. A T-1 multiplexer has _____ channels. A. 12 B. C. 96 D.

24 32

39. In ISDN, the _____ is a 16 kbps or 64 kbps digital channel used to carry signaling information. A. D-channel B. A-channel C. B-channel D. C-channel 40. The number of voice channels in a basic FDM group is: A. 6 B. 12 C. 24 D. 60 41. Calculate the blocking probability for a group of 4 trunks that offers 10 Erlangs of traffic and a trunk efficiency of 2.6%. A. 0.8966 B. 0.966 C. 0.98966 D. 0.866 42. Local loops terminate at: A. a tandem office C. a central office

B. D.

a toll station an interexchange office

43. Determine the actual traffic carried by a group of trunks with an offered traffic of 10 Erlangs and 0.64666 blocking probability. A. 3.533 Erlang B. 0.7193 Erlang C. 8.134 Erlang D. 12.444 Erlang


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44. A particular exchange has been dimensioned to handle 1000 calls during the busy hour. On a certain day during the BH 1100 calls are offered. What is the resulting grade of service? A. 1 B. 10 C. 0.01 D. 0.1 45. Loading coils were used to: A. increase the speed of the local loop for digital data B. reduce the attenuation of voice signals C. reduce crosstalk D. provide C-type conditioning to a local loop 46. The PRI in ISDN consist of 23 or 30B-for user data and a single _____ kbps Dchannel for signaling. A. 16 B. 32 C. 64 D. 48 47. How long does it take to dial 1234560 in sec using pulse dialing? A. 1.5 sec B. 3.4 sec C. 4.8 sec D. 6.1 sec 48. ADSL stands for: A. All-Digital Subscriber Line B. Asymmetrical Digital Subscriber Line C. Allocated Digital Service Line D. Access to Data Services Line 49. _____ refers to the percentage of time a circuit or facility is in use. A. reliability B. traffic C. occupancy D. availability 50. What is the mean holding time if the traffic carried is 4.56 Erlang. The number of calls received is 200 in a given EBHC? A. 273.6 sec B. 41.04 sec C. 14.08 sec D. 82.08 sec 51. The 23 B+D Channel used in North American and Japan uses total transmission capacity of _____ Mbps with overhead. A. 2.048 B. 1.920 C. 1.536 D. 1.544 52. A group of trunks offered 10 Erlangs of traffic, and 0.002 GoS. Calculate the trunk efficiency assuming 20 trunks. A. 49.9% B. 89.9% C. 29.9% D. 69.9% 53. In loading cable 19H88 means _____. A. 19 gauge every 3000 ft with 88 μH inductors B. 19 gauge every 6000 ft with 88 mH inductors C. 19 gauge every 2000 ft with 88 nH inductors D. 19 gauge every 4000 ft with 88 pH inductors


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54. A group of 20 subscribers generate 50 calls with an average holding time of 3 minutes, what is the average traffic per subscriber? A. 0.125 Erlang/subscriber B. 12.5 Erlang/subscriber C. 0.0125 Erlang/subscriber D. 1.25 Erlang/subscriber 55. If there are 3000 calls arriving per hour to a trunk circuits with an average calling time of 1.76 minutes, calculate the traffic intensity. A. 48 Erlangs B. 88 Erlangs C. 68 Erlangs D. 78 Erlangs 56. Traffic _____ is the sum of the holding times of the traffic carried by a pool of resources over a given period of time. A. capacity B. volume C. transit D. intensity 57. Calculate the actual data rate of T3 facilities. A. 45.078 Mbps B. 44.736 Mbps C. 1.728 Mbps D. 44.008 Mbps 58. A switching machine is set to accept pulsed at a rate of 10 pulses/sec with a 60 % break, find the make interval of the switch,. A. 0.4 s B. 0.6s C. 40 ms D. 60 ms 59. The BRI in ISDN consist of _____ B-channel (S) and D-channel. A. 1 B. 23 C. 2 D. 30 60. A data transmission facility is encapsulating packet. Assuming the system is using transmitted per second ignoring overhead? A. 57.12 packets B. C. 67.12 packets D.

4096 bytes of data to form a single T1 line, how many packets are 47.12 packets 77.12 packets

61. A switch could support a maximum of 9800 seizures during the busy hour. Calculate the number of block calls during the BH if the GoS is 0.245%. A. 48 calls will be block B. 12 calls will be block C. 24 calls will be block D. 96 calls will be block 62. On a particular traffic relation, the calling rate is 825. If the traffic intensity is 990 CCS during the BH, what is the average call duration? A. 2 minutes B. 12 minutes C. 6 minutes D. 8 minutes 63. A telephone signal takes 2 ms to reach its destination. Calculate the VNL required for an acceptable amount of echo. A. 0.2 dB B. 0.8 dB C. 0.6 dB D. 0.4 dB


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64. The ff. information was recorded in a 5 trunk network from 9:00-10: am Trunk 1 48 minutes 2 56 min and 28 sec 3 18 min 4 25 min and 24 sec 5 42 min Find the traffic carried by the network in Erlang. A. 18.29 Erlang B. 3.16 Erlang C. 0.781 Erlang D. 5.82 Erlang 65. In ISDN, the _____ is a 64 kbps digital channel that can carry a user data or digitized voice. A. A-channel B. C-channel C. B-channel D. D-channel 66. A duration of traffic occupancy from a call, sometimes referred to as an average duration of occupancy of one or more path from calls. A. traffic B. average call C. Busy hour D. holding time 67. Out of the total capacity required by the BRI _____ kbps is used for overhead. A. 48 B. 144 C. 64 D. 16 68. Find the maximum length of the subscriber loop with the ff. specs. Allowable loss of 6dB using a #24 cable and resistance limit of 1200 ohms a. 7.19 kft B. 13.73 kft c. 14.63 kft D. 11.76 kft 69. One Erlang is equal to _____. A. 360 CCS C. 100 CCS

B. D.

3.6 CCS 36 CCS

70. What is the line rate of T1 and E1 respectively? A. 2.048 Mbps and 1.544 Mbps B. 384 Mbps and 2.048 Mbps C. 1.544 Mbps and 2.048 Mbps D. 1.455 Mbps and 2.408 Mbps 71. ____ is the type of telephone switch having vertical and horizontal paths, and electromagnetically operated means for interconnecting any one of the vertical paths with any of the horizontal paths A. cross bar B. step by step C. strowger D. Hybrid 72. What is a leased line? A. A piece of wire used in local area network in one building B. A piece of wire connecting a telephone set to a PABX C. A temporary connection of one computer to a mainframe via a modern and a telephone line D. A permanent circuit for a private use within a communication network 73. Which of the following devices increases the battery voltage on a loop and extends its signaling range? A. VF repeater B. VF amplifier C. Loop extender D. BORSCHT


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74. One Erlang of traffic is equal to _____ call seconds A. 60 B. 1800 C. 360 D. 3600 75. The 23 B+D Channel used in North American and Japan uses total transmission capacity of ____ Mbps with overhead A. 1.536 B. 1.544 C. 2.048 D. 1.920 76. Blocked calls delayed condition specified delay probability A. Erlang B B. Erlang C C. Erlang D D. Poisson 77. A (an) ___ call attempt is one that cannot be further advanced towards its destination due to an equipment shortage or failure in the network. A. lost B. abandoned C. barred D. terminated 78. In TDM, multiple signals share a channel by transmitting in different A. intervals B. frequency bands C. sampling rates D. amplitudes 79. ____ is a telephone service which enable to user to dial directly telephones outside the user’s local area without the aid of the operator A. ISDN B. DDD C. DID D. Bypass 80. The point between the network termination equipment and the terminal equipment is known as the _____ reference point. A. R B. U C. L D. S 81. Terminal equipment which are non-ISDN compatible can be attached to an ISDN network by using a special A. modem B. Front end processor C. Terminal adapter D. bridge 82. Terminal equipment which are ISDN compatible are generally referred to as A. TE1 B. IBP C. Smart terminals D. Dumb terminals 83. The basic switching element in a PAM multiplexer is a (an) A. ADC B. MOSFET C. DAC D. Bandpass filter 84. In PCM, analog signals are transmitted as A. supergroups B. C. binary codes D.

pulses packets

85. The time period which all channels are sampled once is called A. slot B. domain C. cycle D. frame


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86. ____ is a voice operated device that inserts a high loss in the opposite direction of transmission of the talking party. A. hybrid B. 2 wire circuit C. VNL D. Echo suppressor 87. _____ is a signal returned to the talker after making one or more round trips between the talker and the listener A. singing B. jitter C. echo D. crosstalk 88. The BRI in ISDN consist of ____ B-channel (S) and D-channel. A. 1 B. 2 C. 3 D. 4 89. An even faster class of ISDN interface standards presently being developed is called A. narrowband ISDN B. broadband ISDN C. ATM D. Frame relay 90. Current ISDN standards requires ____ twisted A. 3 B. 6 C. 1 D. 2

pairs for the PRI transmission

91. Traffic ____ is the sum of the holding times of the traffic carried by a pool of resources over a given period of time A. volume B. transit C. intensity D. capacity 92. _____ is a service in which a telephone or PBX location is directly connected to a central office in a distant city via a private line, instead of being connected directly to the central office in that location A. bypass service B. remote switching unit C. foreign exchange D. diversion 93. A signaling method with set combinations of two specific voice-band frequencies, one of which is selected from a group of 4 lower frequencies and the other from a group of either 3 or 4 relatively high frequencies A. Out of Band B. Channel Associated C. In Band D. DTMF 94. PABX means A. private all purpose broadcasting exchange B. public access bi-directional exchange C. private automatic branch exchange D. public automatic branch exchange 95. In ISDN, the ___ is a 64 kbps digital channel that can carry a user data or digitized voice A. A-channel B. C-channel C. B-channel D. D-channel


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96. ISDN is defined as the network that provides an end to end ___ to support a wide range of services including non-voice services A. transparency B. switching C. connectivity D. interfacing 97. The international body which issues recommendations for standards relating to telecommunications is the A. IEEE B. ETSI C. CCIR D. CCITT 98. The multiplexer and demultiplexer are kept in step with one another by a (an) A. clock B. header C. sync pulse D. sequencer 99. A ___ is a local automatic telephone office serving extensions is a business complex and providing access to the public network A. concentrator B. FEX C. PAX D. PABX 100. A T-1 multiplexer has ____ channels A. 24 B. C. 64 D.

32 1024

101. The standard word size in a PCM audio system is ___ bits A. 16 B. 8 C. 32 D. 64 102. The primary benefit of multiplexing is A. redundancy B. C. speed D.

economy error detection

103. What is an echo? A. a signal of the same amplitude but 180 degrees out of phase from the original signal and mixed with the original signal at the transmitter to produce a more negligible output signal B. a wave which has been reflected or otherwise returned with sufficient magnitude and delay for it to perceptible in some manner as a wave distinct from that directly transmitted C. the signal having a higher frequency than the original and transmitted back to earth by a passive satellite D. reflected signal 104. What is a communication link? A. it is a channel or circuit intended to connect other channels or circuits B. it is a piece of wire that is connect to the ground terminals of all communications equipments in one establishment C. it is a cable connecting a transmitter to the antenna D. it refers to a radio link 105. The ISDN network equipment which are installed in the customer premises is called the ___ equipment A. terminal B. DTE C. Adapter D. Server


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106. A A. B. C. D.

Telephone networks and system voice grade channel is suitable for transmission of signals with a frequency from 300 to 3400 Hz VHF signals UHF signals VHF and UHF signals

107. What is trunk? A. the base of the communications tower B. a line connecting the telephone set to a PABX C. a telephone line connecting two central offices D. refers to the body of the tree 108. What is a four-wire circuit? A. is a circuit with three output terminals and one input terminal B. is used between serving central offices for long distance connections, with one being pair being used for each direction of transmission C. is an oscillator that produces simultaneously four different frequencies D. a circuit consisting of four transmission lines 109. For every button pressed on the touch tone telephone, how many signals are transmitted to the CO? A. Two VHF signals B. Two audio frequency tones C. One VHF signal and one audio frequency tone D. Three audible tones 110. The physical connection between the telephone set at the switching equipment is called the A. trunk line B. link C. subscriber loop D. leased line 111. The transmission of information from multiple sources occurring in the same facility but not the same time A. FDM B. TDM C. WDM D. CDM 112. The BRI requires ___ kbps of digital transmission capacity without overhead A. 64 B. 144 C. 128 D. 192 113. What is an Erlang? A. it is equal to the number of simultaneous calls originated during a specific hourly period B. it is a unit of magnetic field intensity measured around a conductor C. it is the number of erroneous bits received per unit of time D. it is unit of electrical energy radiated in space 114. The PRI in ISDN consist of 23 or 30B-for user data and a single ___ kbps D-channel for signaling A. 16 B. 23 C. 32 D. 64


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115. What is the traffic model about block calls cleared condition specified blocking probability? A. Erlang B B. Erlang C C. Erlang D D. Poisson 116. A A. B. C. D.

touch tone telephone generates how many different tones 10 audio frequency tones 8 audible tone frequencies 4 VHF signals 16 audio frequency tones

117. ______ is the operation of the switch selector in searching terminals until anidle one is found A. hunting B. insulation effective C. elective D. loop 118. What is a concentration? A. A system that improves the signal to noise ratio by compressing the volume range of a signal B. A switching system that lets a large number of transmission lines or a narrower bandwidth C. A device that varies the characteristics of a carrier signal in the accordance with the waveform of a modulating signal which contains useful information D. An equipment in the central office 119. Which of the following wire is used to transmit the signal? A. black B. green C. yellow D. red 120. A device converts a 2 wire circuit to a 4 wire circuit A. RS 232 interference B. Hybrid circuit C. Balun D. Stub 121. A passive ____ is an electronic device which reduce signal strength by a specified amount in dB A. splitter B. filter C. himmer D. attenuator 122. What is a multi drop line? A. A line or circuit interconnecting several stations B. A piece of wire with a thick insulating material that serves to protect the conductive material form damage in the event the wire is dropped C. A line designed to withstand high pressure D. A bus line 123. _______ is a trunk between two central offices in the same switching center complex. A. intraoffice B. foreign exchange C. intertoll D. ileline 124. AWG #26 has a typical loss of A. 0.21 dB/1000 ft C. 0.51 dB/1000 ft

B. D.

0.32 dB/1000 ft 0.41 dB/1000 ft


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125. Which the following is commonly used in wire for subscribers loop design? A. AWG #18 B. AWG #30 C. AWG #33 D. AWG #19 126. What is singing? A. An undesired self sustained oscillation in a system, generally caused by excessive positive feedback B. The result of the intermodulating two or more signals of different frequencies to produce a tone having a frequency higher than that of the highest frequency C. the result of intermodulating two or more signals of different frequencies to produce a tone having a frequency equal to the sum of the frequencies of the signals intermodulated D. The art of entertainment that can make one rich 127. _______ is the uninterrupted of 60 minutes for which the average intensity of traffic is at the maximum A. maximum access time B. busy hour C. maximum occupancy D. peak traffic 128. _______ is the probability of the call being blocked during the busy hour because of insufficient equipment or trunks A. contention B. grade of service C. Poisson distribution D. Lockout 129. Presently, this is the “standard” frequency bandwidth for voice transmission A. 300 to 3400 Hz B. 0 to 4000 Hz C. 100 to 3400 Hz D. 300 to 3000 Hz 130. The resistance limit for No. 2 Crossbar Exchange (US) is A. 2000 ohms B. 1250 ohms C. 1300 ohms D. 1200 ohms 131. What is the standard voice channel spacing? A. 44 MHz B. 4 kHz C. 40 kHz D. 40 MHz 132. Which of the following responds to the request of a subscriber by sending a dial tone? A. line finder B. connector C. first selector D. line equipment 133. Combination of modulator, channel and detector A. transceiver B. transponder C. Discrete channel D. T/R channel 134. It A. B. C. D.

is a advantage of side tone transmission efficiency is increased assures the customer that the telephone is working speaker increases his voice resulting in a strengthen signal no dissipation of energy in the balance network


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135. What is a two-wire circuit? A. A circuit usually in the subscribers loop, between the telephone set and the local control office B. A circuit having two terminals, both terminals having the same instantaneous voltage C. A circuit with one input terminal, one output terminal and a common ground D. A circuit consisting of two transmission lines 136. What is the local loop of a telephone system? A. it is a group of wires connecting a telephone set to a modem B. it is a four wire circuit connecting a facsimile machine into a computer C. it is a two-wire or four-wire communication circuit between the customers premise and the central office D. it is a single piece of wire connecting the subscribers telephone set into another set in an adjacent room 137. The corresponding frequency for the digit 7 in the touch tone telephone is A. 770 and 1477 Hz B. 852 and 1209 Hz C. 852 and 1336 Hz D. 770 and 1336 Hz 138. The D-channel messages in an ISDN network are ____ switched A. circuit B. time C. packet D. virtual 139. The standard audio sampling rate in a PCM telephone system is ____ kHz A. 2 B. 4 C. 8 D. 16 140. When human voice and music are transmitted, the type of communications employed is known as A. radio telegraphy B. radio telephony C. audio frequency D. wired radio 141. When one channel picks up the signal carried by another channel A. echo B. party line C. crosstalk D. cross link 142. The circuit connecting a subscriber station with the line terminating equipment in a central office is called A. local loop B. tie line C. trunk D. subscriber drop 143. ______ refers to the percentage of time a circuit or facility is in use A. availability B. traffic C. reliability D. occupancy 144. In an ISDN network, the _____ blank provides only OSI layer functions including the electrical and physical termination of the network on the customer premises A. NT2 B. TE1 C. NT1 D. TE2


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145. ADSL increases existing twisted-pair access capacity by A. threefold B. twofold C. fiftyfold D. thirtyfold 146. The 1948 theorem which is the basis for understanding the relationship of channel capacity, bandwidth, signal-to-noise ratio is known as A. Reeve's Law B. Shannon's Law C. Nyquist Law D. Hartley Law 147. What appears to be the practical limit for analog modems over the standard telephone network? A. 19.2 kbps B. 24 kbps C. 33 kbps D. 28.8 kbps 148. Digital subscriber line (DSL) refers to A. a connection created by a modem pair enabling communications B. a specific length of wire C. a specific gauge of wire used in modem communications D. a modem enabling high-speed communications

high-speed

149. ______ and HDSL are essentially equivalent technologies. A. SONET B. ATM C. T1/ E1 D. ETHERNET 150. What is the source of limitation on the bandwidth of the public switched network? A. local loop B. subscriber line C. telset D. the core network 151. The practical upper limit of length of ADSL is A. 18,000 ft B. 6,000 ft C. 12,000 ft D. 36,000 ft 152. T1 and _____ refer to the same multiplexing system. A. DS-1 B. STM-1 C. OC-1 D. STS-1 153. The correct order of fax call phases is which of the following? A. call setup, pre-image handshake, line testing sequence, image transmission B. call setup, line testing sequence, pre-image handshake, image transmission C. line testing sequence, call setup, pre-image handshake, image transmission D. pre-image handshake, line testing sequence, call setup, image transmission 154. A gigahertz-based coax line is capable of delivering how many times more bandwidth than copper? A. 2 B. 50 C. 1,000 D. 10,000


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155. What is the bandwidth of the telephone system? A. 535 - 1605 kHz B. 340 - 15,000 Hz C. 300 - 3400 Hz D. 88 - 108 Hz 156. Which of these is specified by C-conditioning? A. signal to noise ratio B. harmonic distortion C. gain/attenuation distortion D. phase hits 157. Long distance trunk lines operate as _______ lines. A. simplex B. half duplex C. full duplex D. full/full duplex 158. Pulse spreading is a result of A. relative delay distortion C. phase jitter

B. D.

phase impulse hits amplitude jitter

159. Which measurement compares signal strength to noise level? A. impulse hits B. attenuation/gain distortion C. signal to noise ratio D. amplitude jitter 160. Which telephone line impairment is a change in signal amplitude for a short time duration? A. white noise B. interference C. harmonic distortion D. gain hit 161. Line conditioning is performed on A. dial up lines C. private lines

B. D.

leased lines both B and C

162. Envelope delay distortions a measure of A. relative time delay across the bandwidth B. amplitude variations across the bandwidth C. phase variations in signals across the bandwidth D. signal attenuation across the bandwidth 163. D-conditioning defines specifications of A. propagation delay variations B. C. signal to noise ratio D.

interference limits amplitude variations

164. What is done to a telephone line to tighten gain/attenuation parameters? A. it is replaced with fiber cable B. it is switched to a different line C. it is conditioned D. it is replaced with new copper lines 165. Time division multiplexing A. divides packets into audio cells to be placed on the telephone lines B. assigns channels to different frequencies in the transmission band width C. assigns time slots to each channel's packet sections D. uses asynchronous data transmission only


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166. What is a disadvantage of time division multiplexing (TDM)? A. slower transmission rates B. large packet sizes C. small bandwidth D. more overhead 167. What is the purpose of using alternate mark inversion and return to zero encoding of data? A. faster digital data transmission B. assist in clock recovery C. more efficient transmission of analog data D. reduced susceptibility to noise 168. How many channels are used to make up a group level channel in frequency division multiplexing? A. 600 B. 60 C. 300 D. 12 169. ISDN is A. a transport type protocol that uses the plain old telephone system B. a data link protocol that uses HDLC as a basis C. an integrated syntax network D. digital interface system 170. The U.S. primary rate interface (PRI) data rate is A. 1. 544 Mbps B. 2.048 Mbps C. 192 kbps D. 144 kbps 171. Which ISDN block is responsible for converting non-ISDN format? A. TE 2 B. TE 1 C. TA D. NT 1 172. Which type of media is least susceptible to electrical interference? A. microwave B. unshielded twisted pair cable C. shielded twisted pair cable D. fiber optics 173. What is the composition of a ISDN BRI (basic rate interface)? A. B + 2D B. 2B + D C. 23B + D D. 30B + D 174. What TELCO parameter limits the type of modem that can be used with a given data rate? A. bandwidth B. attenuation C. line impedance D. noise 175. Which of these is NOT a currently used method of multiplexing data channels onto a single transmission medium? A. time division multiplexing B. code division multiplexing C. wave division multiplexing D. encrypted division multiplexing


Section

Facsimile

13

Transmission

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DEFINITION. Facsimile - A form of communication system that copies, sends, and receives documents by way of telephone lines. Also called faxing, this method of communication allows people to share exact copies of important papers by duplicating and sending them on one end, and then receiving and reproducing them on the other. From Latin facere “to do, make” + simile , “similar.” Scanning - A process of directing a light-sensitive device over a surface in order to convert an image into digital or electronic form for further storage, retrieval, and transmission.

DEFINITION.


Scottish physicist Alexander Bain patented the first facsimile device. The device generated electrical impulses by passing a metal rod over a page that contained raised metal letters.

1902

German physicist Arthur Korn transmitted photographs from one device to another over telephone lines, but the basis for modern fax machines did not appear until 1925.

1925

French inventor Edouard Beeline introduced the Belinograph. Beeline’s facsimile machine used a powerful light beam and a photoelectric cell to convert the light, or the absence of light, on an image into electrical impulses.

1966

Xerox Corporation began manufacturing a small facsimile machine that could be connected to any telephone line.


FACSIMILE TRANSMISSION

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A. .SCANNING MECHANISM.

FLYING SPOT SCANNING A finely focused spot of light from the illuminating source is directed at each pixel in turn, and a single broad-angle photocell is used to sense the level of the reflected light from it.

FLYING SENSOR SCANNING The document is completely floodlit and a single photocell is optically focused on the desired pixel.

ELECTRONIC SCANNING The document is floodlit and the entire image is optically focused on a photosensitive surface or photocell array. Scanning is controlled electronically and the source and sensor remain fixed in position.

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Self-Sufficient Guide to ECE by JASON AMPOLOQUIO B. .FACSIMILE RECEIVER. 1.

Synchronization It is necessary for the receiver scanner and the transmitter scanner to run at exactly the same speed. If the receiver runs at slightly faster speed than the transmitter, then each scan line will overlap into the following scan line and delay its start, causing the image to skew diagonally to the right.

2.

Phasing If the receiver and transmitter are in synchronism, but the receiver starts a new line either too early or too late, the image will wrap around.

C. .INDEX OF COOPERATION. The width to height ratio of the document reproduced at the receiver must be the same as that of the originating document if distortion is to be avoided. The IOC is a number derived from width-to-height ratio for proper image reproduction. 1.

IEEE Standard

IOC

General Solution

In terms of Source width & Scan density

IOC ( IEEE ) = S x W

In terms of width-toheight ratio

IOC ( IEEE ) = n x

W L


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FACSIMILE TRANSMISSION 2.

CCITT Standard CCITT developed a slightly different definition of IOC, which is directly applicable to DRUM scanners.

Parameters

General Solution

IOC Drum Circumference Scan Density 3.

IOC ( CCITT ) =

D P

W = πD S=

1 P

Relationship between IEEE and CCITT IOC

IOC ( CCITT ) =

IOC ( IEEE ) π

where: IOC = index of cooperation S = scan density or resolution in lines mm W = widthof the source document or scan stroke length in mm L = document length in mm n = total number of lines in document height D = drum diameter in mm P = scanning pitch in mm line

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Sample Problem:

Calculate the index of cooperation for a facsimile machine with a drum diameter of 70.4 mm and the scanning pitch of 0.2 mm per scan.

Solution:

For CCITT Standard: D IOC(CCITT ) = P 70.4 = = 352 0.2

For IEEE Standard:

IOC(IEEE) = IOC(CCITT ) x π = 352 x π = 1106

Sample Problem:

Calculate the drum diameter in mm and scan density of a fax transmission with an IOC (IEEE) equal to 900 received by a drum scanner that uses 8-in wide paper.

Solution:

For the drum diameter: W 8 in D= = π π = 2.546 in x

2.54 cm 10 mm x 1 in 1cm

= 64.67 mm

For the scan density: IOC(IEEE) 900 S= = W 8 in lines 1 in 1 cm = 112.5 x x in 2.54 cm 10 mm lines = 4.42 mm

Sample Problem:

Find the IEEE and CCITT IOC for a facsimile drum scanner with an FCC specification of 18.85 in for line length and 96 lines/in scan density.

Solution:

For IEEE Standard: IOC(IEEE) = S xW lines = 96 x18.85 in in = 1809.6

For CCITT Standard: IOC(CCITT ) =

IOC(IEEE)

π 1809.6 = = 576 π


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D. .IMAGE DISTORTION DUE TO IOC INCOMPATIBILITY.

Receiver IOC equal to transmitter IOC

Receiver IOC greater than to transmitter IOC

Receiver IOC less than to transmitter IOC

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Self-Sufficient Guide to ECE by JASON AMPOLOQUIO E. .ANALOG FAX TRANSMISSION. Parameters

General Solution

Pixel density for flat-bed scanner

NPX = W x S

Pixel density for drum scanner

NPX = RS =

Line Scan rate

πD P

rpm 60

R PX = NPX x R S

Pixel rate Output frequency

f=

R PX 2

Transmission time

td =

n R PX

Sample Problem:

A certain drum scanner has a pitch of 0.26 mm/line and a diameter of 68.4 mm. The drum rotates at 120 rpm and scans a total of 1075 lines for a standard document page. Find the # of pixel in a scan line, the scan and pixel rate, output frequency of the transmission channel and the length of time required transmitting a page.

Solution:

For the number of pixel in a scan line; D 68.4 NPX = π = π P 0.26 pixels = 826.5 line

For the pixel rate;

For the scan rate;

For the output frequency;

RS

rpm 120 = = 60 60 lines =2 sec

RPX = NPX x R S = 826.5 x 2 = 1653

pixels sec

RPX 1653 = 2 2 = 826.5 Hz

fMAX =

For the transmission time td =

n 1075 = 60RS 60 x 2

= 8.96 min



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F. .CCITT FACSIMILE STANDARDS. Group

Standards

Group 1 G1 Or GI

Group 2 G2 Or GII

Group 3 G3 Or GIII

Group 4 G4 Or GIV

Use analog transmission by FM where WHITE is 1300 Hz and BLACK is 2100 Hz. North American equipment uses 1500 Hz for WHITE and 2300 Hz for BLACK.

The scanning resolution is 96 lines/inch

The average transmission speed is 6 minute/page for an A4 paper. Use analog transmission by using FM or vestigial sideband AM. The vestigial sideband plus part of the upper sideband is transmitted

The resolution is 96 lines/inch

The transmission speed is 3 minutes or less for an A4 page.

Use digital transmission by PCM with black and white only or up to 32 shades of gray.

Either PSK or QAM is used to achieve transmission speeds up to 9600 baud.

The resolution is 200 lines/inch

The transmission speed is less than 1 minute per page with 15 to 30 seconds being typical.

Use digital transmission at 56 kbps.

The resolution is up to 400 lines/inch

The speed of transmission is less than 5 seconds per page.

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Self-Sufficient Guide to ECE by JASON AMPOLOQUIO G. .EIA FACSIMILE STANDARDS. Parameter

Standards

Frequency

300 Hz

Index of Cooperation

576 1500 Hz for Black 2300 Hz for White

Modulation Start Signal

Alternating black and white modulated at a rate of 300 Hz for a period of 5s.

Phasing Signal

25 second transmission (95% of each line in BLACK, then 5% of WHITE)

Stop Signal Scan Spot Size

Alternating black and white interrupted at a rate of 450 Hz for 5 s, then black transmission for a period of 10 s. 0.26 mm x 0.0104 mm

H. .ITU CODING STANDARDS. CCITT DOCUMENTS SCANNED AT 200 x 100 pixel/ square inch Documents

G-II

G-III

G-IV

Time for G-IV (9.6 kbps)

Circuit Diagram

256K

15K

5.4K

4.5 sec

Dense Text

256K

54K

35K

29.17 sec

Graph

256K

23K

8.3K

6.92 sec

Handwriting and Simple Graphics

256K

26K

10K

8.33 sec



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I

H

1.

Calculate the pixel rate of a drum scanner that rotates at 120 rpm if the pixel scan density is 826.5 pixels/line. A. 1653 pixels/sec B. 3615 pixels/sec C. 5163 pixels/sec D. 6351 pixels/sec

2.

Determine the document transmission time of a 1075 scan lines if the line scan rate is 2 lines/sec A. 6.98 minutes B. 9.68 minutes C. 98.6 minutes D. 8.96 minutes

3.

Calculate the index of cooperation for a facsimile machine with a drum diameter of 3.15 in and the scanning pitch of 0.2 mm per scan. A. IOC(CCITT)=400, IOC(IEEE)=1257 B. IOC(CCITT)=600, IOC(IEEE)=1527 C. IOC(CCITT)=200, IOC(IEEE)=1275 D. IOC(CCITT)=800, IOC(IEEE)=1572

4.

Calculate the drum diameter in mm and scan density of a fax transmission with an IOC (IEEE) equal to 1024 received by a drum scanner that uses 8-in wide paper. A. 64.88 mm, 182 lines/in B. 46.88 mm, 182 lines/in C. 46.88 mm, 128 lines/in D. 64.88 mm, 128 lines/in

5.

Find the IEEE and CCITT IOC for a facsimile drum scanner with an FCC specification of 14 in for line length and 115 lines/in scan density. A. 1610, 521.5 B. 1610, 512.5 C. 1160, 512.5 D. 1160, 521.5

Section 14 Pulse Modulation Section 15 Digital Communications Section 16 Data Communications

Computer Communications


4-1


Section

Pulse

14

Modulation


Pulse Amplitude Modulation (PAM) is a method of converting information wherein the amplitude of a constant width, constant position pulse is varied in accordance to the instantaneous amplitude of the modulating signal.

DEFINITION.

Pulse Width/Duration Modulation (PWM/PDM) is a process where the pulse width of a fixed amplitude pulse varies proportionally to the amplitude of the analog signal.

DEFINITION.

DEFINITION. Pulse Position Modulation (PPM) is a form of pulse modulation where the position of a constant width pulse within a prescribed timeslot is varied according to the amplitude of the modulating signal. DEFINITION. Pulse Code Modulation (PCM) is essentially analog-to-digital conversion of a special type where the information contained in the instantaneous samples of an analog signal is represented by digital words in a serial bit stream. A. .PULSE MODULATION. 1.

Pulse Amplitude Modulation (PAM) Designation: (K3E/P3D)


PULSE MODULATION

4-2

2.

Pulse Time Modulation (PTM) PTM is a class of signaling technique that encodes the sample values of an analog signal onto the time axis of a digital signal.

PTM is analogous to angle modulation (FM/PM). i.

Pulse Width/Duration Modulation (PWM/PDM) PWM/PDM is a process where the pulse width of a fixed amplitude pulse varies proportionally to the amplitude of the analog signal. Designation: (L3E/P3E)

ii.

Pulse Position Modulation (PPM) PPM is a form of pulse modulation where the position of a constant width pulse within a prescribed timeslot is varied according to the amplitude of the modulating signal. Designation: (M3E/P3F)

4-3


Pulse Frequency Modulation PFM is a form of pulse modulation where the pulse period and pulse duration are both made proportional to the modulating signal amplitude while keeping the duty cycle of the pulse train constant.

4.

Pulse Code Modulation (PCM) PCM is essentially analog-to-digital conversion of a special type where the information contained in the instantaneous samples of an analog signal is represented by digital words in a serial bit stream.

Vm

t

V t PULSE CODE MODULATION


PULSE MODULATION

4-4

B. .BLOCK DIAGRAM OF PCM.

1.

Band Limiting The anti-alias or bandpass filter limits the frequency of the input analog signal to the standard voice frequency band of 0 to 4 kHz. The purpose is to eliminate any unwanted signal that will result to aliasing or fold-over distortion at the receiver.

2.

Sampling The act of periodically holding a value (sample) of the continually changing analog input signals.


4-5

3 Types of Sampling i.

Ideal Sampling

ii.

Natural sampling (Gating) The natural sampling method retains the natural shape of the sample analog waveform.

iii. Flat-top sampling (Instantaneous sampling) The most common method used for sampling voice signals in PCM where the sample-and-hold circuit convert those samples to a series of constant-amplitude PAM levels

Flat-top sampling alters the frequency spectrum and introduces an error called aperture error.


PULSE MODULATION

4-6

3.

Quantization The process of assigning discrete level to a time-varying quantity in multiples of some fixed unit, at a specified instant or specified repetition rate.

4.

Encoding The process of converting the quantized discrete-signal (PAM samples) to parallel PCM codes.

C. .NYQUIST SAMPLING THEOREM. States that for a sample to be reproduced accurately at the receiver, the sampling frequency must be at least twice of the highest modulating signal.

fs ≥ 2fm

where : fm = highest modulating frequency



4-7

For a sample rate of 30 kHz in a PCM system, determine the maximum analog input frequency.

Solution: fs = 2fm ⇒ fm =

fs 30 kHz = = 15 kHz 2 2

D. .PCM PARAMETERS. 1.

Number of Levels or Code words (M) where:

M=2

M = # of levels, symbols or code words

n

n = # of PCM bits used (sign bit excluded) = # of bits per sample

2.

Bandwidth (Data rate)

BWPCM = fb = nfs

where : fs = sampling rate in Hz fb = bit rate in bps

3.

Dynamic Range (DR)

In terms of ADC/DAC Decoded Voltage

DR =

Vmax Vmin

In terms of # of Bits (excluding the sign bit)

DR = 2n − 1

In dB

DR dB = 20 logDR


PULSE MODULATION

4-8


What is the dynamic range in dB of an 8-bit linear sign-magnitude PCM spectrum whose maximum decode voltage at the receiver is 1.27 Vp?

Solution: DR dB = 20 log(2n − 1) ⇒ n = 7 (sign bit excluded) = 20 log(27 − 1) = 42.076 dB

4.

Resolution (Vmin ) or Quantization Step (in Volts)

R e so lu tio n =

Vm a x 2n


Determine the resolution for an 8-bit linear sign-magnitude PCM for a maximum decode voltage of 2.55 Vp?

Solution: Resolution =

5.

Vmax 2n

=

2.55 27

= 0.0199 ≅ 0.02 V

Scale Factor (SF)

SF =

1 2n

%SF =

100 2n

Sample Problem:

An audio signal with a bandwidth of 3.2 kHz and is converted to PCM signal by sampling at 8 kilosamples/sec and by using a uniform quantizer with a scale factor of 1.5625% and a Vmax of 3V. Determine the number of levels and the resolution of this quantizer.

Solution: 100

;M = 2 n 2n 100 M = = 64 levels 1 . 5625

% SF =

Resolution =

3 = 46 .875 mV 64

4-9


Coding Efficiency (η)

η=

7.

β max = Actual # of bits (including the sign bit)

Maximum Quantization errors (Qe)

Qe =

8.

where : β min = Min # of bits (including the sign bit)

β min x100% β max

Vmin 2

Qe =

Resolution 2

Recovered Signal-to-Quantization Noise ratio (SQNR) i. Ideal SQNR - Only quantization error is present a.

In unitless

S 3 2 = M N 2 b.

S 3 2n ≅ (2 ) N 2

In dB

⎛S⎞ ⎜ N ⎟ = 6.02n + 1.76 ⎝ ⎠dB

Most Used…


PULSE MODULATION

4-10

Sample Problem:

Determine the signal-to-noise ratio in dB, if an audio signal with a bandwidth of 3.2 kHz is converted to PCM signal by sampling at 8 kilosamples/sec and with a data rate of 64 kbps.

Solution: fb = n x fs ⇒ n =

fb fs

64 kbps n = = 8 bits 8 kHz

( )

{

}

3 (2 x8) S 3 2n = = 2 = 98,304 2 2 N 2

S ⎛S⎞ = 10 log = 10 log(98,304) ⎜ ⎟ N ⎝ N ⎠dB = 49.925 dB

Alternate solution: (directly in dB) ⎛S⎞ = 6.02n + 1.76 dB = 6.02(8) + 1.76 dB = 49.92 dB ⎜ ⎟ ⎝ N ⎠dB

ii.

Peak SQNR - Peak Signal-to-Average Quantization Noise Ratio a.

In unitless

General Solution

S 3M2 = N 1 + 4(M2 + 1)Pe b.

Approximate Solution In terms of M

In terms of n

S ≅ 3M2 N

S ≅ 3(22n ) N

In dB

For Non-Linear PCM

For Linear PCM

⎛S⎞ ⎜ ⎟ = 6.02n + α ⎝ N ⎠ dB

⎛S⎞ ⎜ N ⎟ = 6.02n + 4.76 ⎝ ⎠ dB

α = 4.76 − 20 log ⎡⎣ln(1 + μ )⎤⎦

(For μ-law companding)

α = 4.76 − 20 log ⎣⎡1 + ln A ⎦⎤

(For A-law companding)


Sample Problem:

4-11

An audio signal with a passband of 3,200 Hz is converted to PCM signal by sampling at 8 kilosamples/sec and by using a uniform quantizer with 64 steps. Determine the peak SQNR if the probability of error is 10-6.

Solution: S 3M2 3 x 642 = = = 12,089.87 N 1 + 4(M2 + 1)Pe 1 + 4(642 + 1)10−6 S ⎛S⎞ = 10 log = 10 log(12,089.87) = 40.82 dB ⎜ ⎟ N N ⎝ ⎠dB

Alternate solution: (directly in dB) ⎛S⎞ ⎛ ln 64 ⎞ = 6.02n + 4.76 dB = 6.02⎜ ⎜ ⎟ ⎟ + 4.76 dB = 40.88 dB N ⎝ ⎠dB ⎝ ln 2 ⎠


PCM was invented by Alec Reeves and the only digitally encoded modulation technique that is used for digital transmission.

ª

The term PCM is somewhat a misnomer as it is not really a modulation form, but rather a form of source coding.

ª

The sampling process was introduced by Carson.

ª

Sampling theorem is frequently called as Shannon, Whittaker, Kotel’nikov, or Nyquist sampling theorem.

ª

Resolution in ADC/DAC jargon is term LEAST SIGNIFICANT BIT (LSB) which means distance between adjacent quantization levels.

ª

The information to be process is called digital data, not digit data but analog-to-digital conversion is generally called: digitize and digitization, rather than digitalize and digitalization?

ª

The compression parameter μ=255 and A=87.6 curves lies very close to one another, which means that similar and compatible quality is achieved.


PULSE MODULATION

4-12

iii. Average SQNR - Average Signal-to-Average Quantization Noise Ratio a.

In unitless

General Solution

Approximate Solution

S M2 = N 1 + 4(M2 + 1)Pe b.

In terms of M

In terms of n

S ≅ M2 N

S ≅ 22n N

In dB

For Non-Linear PCM

For Linear PCM

⎛S⎞ ⎜ ⎟ = 6.02n + α ⎝ N ⎠ dB

⎛S⎞ ⎜ N ⎟ = 6.02n ⎝ ⎠ dB

Sample Problem: An audio signal with a passband of 3,200 Hz is converted to PCM signal by sampling at 8 kilosamples/sec and by using a uniform quantizer with 64 steps. Determine the number of PCM bits needed and the average SQNR if the probability of error is 10-6. Solution: M = 2n = 64 ⇒ n =

ln 64 = 6 bits ln 2

S M2 642 = = = 4,029.95 N 1 + 4(M2 + 1)Pe 1 + 4(642 + 1)10−6 S ⎛S⎞ ⎜ ⎟ = 10 log = 10 log(4,029.95) = 36.05 dB N N ⎝ ⎠dB

Alternate solution: (directly in dB) ⎛S⎞ = 6.02n = 6.02(6 ) = 36.12 dB ⎜ ⎟ ⎝ N ⎠dB


4-13

E. .CODING METHOD. a.

Level-at-a-time coding This type of coding compares the PAM signal to a ramp waveform while a binary counter is being advanced at a uniform rate. When the ramp waveform equals or exceeds the PAM sample, the counter contains the PCM code.

b.

Digit-at-a-time coding This type of coding determines each digit of the PCM code sequentially. Digit-at-a-time coding is analogous to a balance where known reference weights are used to determine an unknown weight.

c.

Word-at-a-time coding Word-at-a-time coders are flash encoders and are more complex; however, they are more suitable for high-speed applications.

F. .COMPANDING. 1.

Analog Companding Analog compression was implemented in the PCM transmitter prior to the sample-and-hold circuit. While analog expansion was implemented with diodes that were placed just after the receiver low-pass filter. i.

μ-Law Companding

⎡ ⎛ μν in ⎞ ⎤ ν out = Vmax ⎢ln ⎜ 1 + ⎟ ⎥ ÷ ln (1 + μ ) Vmax ⎠ ⎥⎦ ⎣⎢ ⎝

Sample Problem: What proportion of the maximum output voltage is produced if a positive input signal is applied to a μ-law compressor with its voltage three-fourth the maximum value? Solution: ⎡ ⎛ μνin ⎞⎟ ⎤ ⎥ ÷ ln(1 + μ ) ν out = Vo(max) ⎢ln⎜1 + ⎟ ⎜ V ⎢⎣ ⎝ in(max) ⎠ ⎥⎦ ⎡ ⎛ 255 x0.75 Vin(max) ⎞ ⎤ ⎟ ⎥ ÷ ln(1 + 255 ) = Vo(max) ⎢ln⎜1 + ⎟⎥ ⎜ Vin(max) ⎢⎣ ⎝ ⎠⎦ = 0.948 Vo(max)


PULSE MODULATION

4-14

ii.

A-Law Companding

⎡ ν ⎤ ν out = Vmax ⎢ A in ⎥ ÷ (1 + ln A) ⎣ Vmax ⎦

⎡ ⎛ ν ν out = Vmax ⎢1 + ln ⎜ A in ⎢⎣ ⎝ Vmax

2.

0≤

⎞⎤ ⎟ ⎥ ÷ (1 + ln A ) ⎠ ⎥⎦

νin 1 ≤ Vmax A

νin 1 ≤ ≤1 A Vmax

Digital Companding With digital companding, the compression was perform after the input sample has been converted to a linear PCM code and expansion at the receive end prior to PCM decoding.

G. .DELTA MODULATION PCM. Delta modulation uses a single-bit PCM code to achieve digital transmission of analog signals. With delta modulation, only a single bit is transmitted, which simply indicates whether the sample is larger or smaller than the previous sample. Trade-off of DM i. Slope overload Occurs when the input signal changes rapidly and has a large slope (rate of signal amplitude change per unit time) than the DAC can maintain. ii.

Granular Noise Signal variations present in the reconstructed signal if the original analog input signal has a relatively constant amplitude.

Granular noise in DM is analogous to quantization noise in conventional PCM. H. .ADAPTIVE DELTA MODULATION PCM. ADM is a delta modulation system where the step size of the DAC is automatically varied depending on the amplitude characteristics of the analog input signal.

a.k.a. Continuous Variable Sloped Delta Modulation (CVSD)

4-15


I

H

1.

An analog-to-digital converter has an accuracy of 98%. What is the possible range of values it will determine when the analog signal value is 4 V? A. 3.92-4.08 V B. 2.92-5.08 V C. 1.92-3.08 V D. 2.92-3.08 V

2.

In delta modulation, "granular noise" is produced when: A. the signal changes too rapidly B. the signal does not change C. the bit rate is too high D. the sample is too large

3.

The bit rate for each channel in DS-1 is: A. 1.544 Mb/s B. C. 56 kb/s D.

64 kb/s 8 kb/s

4.

Determine the number of quantization step for a 14-bit conversion. A. 28 B. 14 C. 256 D. 16,384

5.

Calculate the dynamic range for an ADC that accept a signal that ranges from 0 to 3 V and convert it to digital format with a quantization error of less than 1 mV. A. 49.5 dB B. 69.5 dB C. 79.5 dB D. 59.5 dB

6.

What is the maximum allowed information signal frequency that will be accepted by an ADC with a Nyquist sampling rate of 16 kHz without producing aliasing distortion? A. 16 kHz B. 32 kHz C. 8 kHz D. 24 kHz

7.

Framing bits in DS-1 are used to: A. detect errors B. carry signaling C. synchronize the transmitter and receiver D. correct burst error

8.

The number of samples per second in DS-1 is: A. 8 K B. 56 K C. 64 K D. 1.544 x 106


PULSE MODULATION

4-16

9.

A voice communication system with a bandwidth of 3300 Hz is to be digitized to a 10-bit resolution, how many bits are generated per second? A. 66 kbps B. 64 kbps C. 54 kbps D. 56 kbps

10. Two clocks operate nominally at 1 MHz and are initially identical, but one clock differs from the other by 0.01% over time. After how many bits do the clocks differ by a full bit period? A. 10,000 bits B. 100,000 bits C. 1,000 bits D. 100 bits 11. So-called "stolen" bits in DS-1 are used to: A. detect errors B. carry signaling C. synchronize the transmitter and receiver D. correct burst error 12. What is companded output voltage of a compander circuit when the input is 0.5 V, for Vmax=1 V and μ=100? A. 0.95 V B. 0.25 V C. 0.85 V D. 0.15 V 13. Calculate the maximum dynamic range for a linear PCM systems using 12-bit quantizing. A. 84 dB B. 74 dB C. 94 dB D. 64 dB 14. Determine the dynamic range for a 10-bit sign-magnitude PCM code.

ECE Board Exam Nov. 2002 A. C.

15. The A. B. C. D.

74 dB 44 dB

B. D.

64 dB 54 dB

dynamic range of a system is the ratio of: the strongest transmittable signal to the weakest discernible signal the maximum rate of conversion to the minimum rate of conversion the maximum bits per sample to the minimum bits per sample none of the above

16. An attempt is made to transmit a baseband frequency of 28 kHz using a digital audio system with a sampling frequency of 44.1 kHz. This will result to an alias audio signal of _____. A. 16.1 kHz B. 32.2 kHz C. 8.05 kHz D. 4.025 kHz 17. In North America, companding uses: A. the Logarithmic Law C. the α Law

B. D.

the A Law the μ Law

18. Natural Sampling does not use: A. a sample-and-hold circuit C. a fixed sample rate

B. D.

true binary numbers an analog-to-digital converter


4-17

19. Which is true about aliasing and foldover distortion? A. They are two types of sampling error B. You can have one or the other, but not both C. Aliasing is a technique to prevent foldover distortion D. They are the same thing 20. Calculate the number of levels used in compact disc (CD) audio system. A. 65,536 B. 16,384 C. 4096 D. 32,768 21. Companding is used to: A. compress the range of base-band frequencies B. reduce dynamic range at higher bit-rates C. preserve dynamic range while keeping bit-rate low D. maximize the useable bandwidth in digital transmission 22. 16-bit quantizing of linear PCM signal result to a dynamic range of _____. A. 83.45 dB B. 88.2 dB C. 91.73 dB D. 98.08 dB 23. What is the minimum data rate needed to transmit audio signal with a sampling rate of 40 kHz and 14 bits per sample? A. 560 kbps B. 760 kbps C. 480 kbps D. 280 kbps 24. Quantizing noise (quantization noise): A. decreases as the sample rate increases B. decreases as the sample rate decreases C. decreases as the bits per sample increases D. decreases as the bits per sample decreases 25. In Europe, companding uses: A. the Logarithmic Law C. the α Law

B. D.

the A Law the μ Law

26. Calculate the raw data rate of CDs before error correction if the coding is linear PCM using 44.1 kHz sampling frequency, 16 bits per sample and two stereo channels. A. 0.705 Mbps B. 2.81 Mbps C. 1.41 Mbps D. 2.049 Mbps 27. Suppose an input signal to a μ-law compressor has a positive voltage and an amplitude 25% of the maximum possible. Calculate the output voltage as a percentage of the maximum input. A. 91.8% B. 75.2% C. 78.5% D. 83.1% 28. The number of bits per sample in DS-1 is: A. 1 B. C. 4 D.

2 8


PULSE MODULATION

4-18

29. Compared to PCM, adaptive delta modulation can transmit voice: A. with a lower bit rate but reduced quality B. with a lower bit rate but the same quality C. only over shorter distances D. only if the voice is band-limited 30. The number of framing bits in DS-1 is: A. 1 B. C. 4 D.

2 8

31. Calculate the number of quantization levels and bit rate if a composite video signal with a baseband frequency of 4 MHz is transmitted by linear PCM, using 8-bits per sample and a sampling frequency of 10 MHz? A. 128, 40 Mbps B. 256, 40 Mbps C. 256, 80 Mbps D. 128, 80 Mbps 32. Two digital clocks differ by 0.3%. After 5 μs this is equal to how many bit difference if the clock and bit rate is 100 kHz? A. 1667 bits B. 1278 bits C. 834 bits D. 239 bits 33. Foldover distortion is caused by: A. noise C. too few samples per second

B. D.

too many samples per second all of the above

34. The immediate result of sampling is: A. a sample alias C. PCM

B. D.

PAM PDM

35. A typical codec in a telephone system sends and receives: A. 4-bit numbers B. 8-bit numbers C. 12-bit numbers D. 16-bit numbers 36. In DS-1, bits are transmitted over a T-1 cable at: A. 1.544 Mb/s B. 64 kb/s C. 56 kb/s D. 8 kb/s 37. The Shannon-Hartley theorem is: A. I = ktB C. C = B log2(1 + S/N)

B. C = 2B log2M D. SR = 2fmax

38. Hartley's Law is: A. I = ktB C. C = B log2(1 + S/N)


39. The Shannon Limit is given by: A. I = ktB C. C = B log2(1 + S/N)


40. The Nyquist rate can be expressed as: A. I = ktB B. C. C = B log2(1 + S/N) D.

C = 2B log2M SR = 2fmax

4-19


Section

Digital

15

Communications

DEFINITION. Information: Information intelligence communicated or received.

is

defined


as

knowledge

or

Data: Information, for example, numbers, text, images, and sounds, in a form that is suitable for storage in or processing by a computer.

DEFINITION.

Digital Transmission is the transmittal of digital pulses between two or more points in a communications system.

DEFINITION.

Digital Radio is the transmittal of digitally modulated analog carrier between two or more points in a communications system.

DEFINITION.

A. .INFORMATION THEORY. 1.

Information Measure The information sent from a digital source when the ith message is transmitter is given by

Ii = logb

2.

1 = − logb Pi Pi

where : Pi = probability of the ith message

Average Information (Entropy) In general, the information content will vary from message to message because the probability of transmitting the nth message will not be equal. Consequently, we need an average information measure for the source, considering all the possible message we can send. N

H=

∑PI

i i

i=0


If the symbols have the same probability of occurrence (P1= P2= P3=… Pn) then the entropy is maximum ( H = logb N ).


DIGITAL communications

4-20

3.

Relative Entropy The ratio of the entropy of a source to the maximum value the entropy could take for the same source symbol.

HR =

4.

H

Hmax = logb N

Hmax

N = total number of symbols

Redundancy

r = 1 − HR

5.

Rate of Information

R =

N

H T

T =

∑tP

i i

i =1

Sample Problem:

A telephone touch-tone keypad has the digits 0 to 9, plus the * and # keys. Assume the probability of sending * or # is 0.005 and the probability of sending 0 to 9 is 0.099 each. If the keys are pressed at a rate of 2 keys/s, compute the entropy and data rate for this source.

Solution: ⎛ 1 ⎞ I0 = I1 = I2 = " I9 = log2 ⎜ ⎟ ⎝ 0.099 ⎠ = 3.34 bits

⎛ 1 ⎞ I∗ = I# = log2 ⎜ ⎟ ⎝ 0.005 ⎠ = 7.64 bits

m

H=

∑ PI

i i

= P0I0 + P1I1 + P2I2 " P#I#

i

Since : P0I0 = P1I1 = P2I2 " P9I9 and P∗I∗ = P#I# = 10{0.099(3.34)} + 2{0.005(7.64)} = 3.38 bits / key H bits keys = H x r = 3.38 x2 T key sec = 6.76 bps

R =


Sample Problem: From the given table Symbol

Probability of Occurrence P(xi)

x1 x2 x3 x4 x5 x6 x7

0.21 0.14 0.09 0.11 0.15 0.18 0.12

Time required to transmit the symbol xi 10 μs 15 μs 20 μs 30 μs 25 μs 15 μs 25 μs

Determine the following: a. Entropy (H) b. Relative Entropy (HR) c. Rate of information

Solution: A. Entropy (H) m

H=

∑ PI

i i

i

B.

⎧ 1 ⎫ ⎧ 1 ⎫ ⎧ 1 ⎫ = 0.21 log2 ⎨ ⎬ + 0.14 log2 ⎨ ⎬ + 0.09 log2 ⎨ ⎬ ⎩ 0.21 ⎭ ⎩ 0.14 ⎭ ⎩ 0.09 ⎭

⎧ 1 ⎫ ⎧ 1 ⎫ ⎧ 1 ⎫ + 0.11 log2 ⎨ ⎬ + 0.15 log2 ⎨ ⎬ + 0.18 log2 ⎨ ⎬ ⎩ 0.11 ⎭ ⎩ 0.15 ⎭ ⎩ 0.18 ⎭ bits ⎧ 1 ⎫ + 0.12 log2 ⎨ ⎬ = 2.76 0 . 12 symbol ⎩ ⎭ Relative entropy (HR) Hmax = log2 N ⇒ max entropy

= log2 7 = 2.81 HR =

C.

H Hmax

=

bits symbol

2.76 = 0.98 2.81

Rate of information 7

T =

∑P t

x x

= 0.21(10) + 0.14(15) + 0.09(20) + 0.11(30)

i=1

+ 0.15(25) + 0.18(15) + 0.12(25) = 18.75μ sec R =

H 2.76 bits = = 147 kbps T 18.75 μ sec


4-21


4-22

6.

Other parameters

Parameter

Equation

Code Word Length

A = log b M

Average Code Word Length

L =

Coding Efficiency

N

∑P A

i i

i =1

η=

L min x 100 % L

γ =1−η

Coding Redundancy

Read it till it Hertz…jma In the equation Ii = logb

1 Pi

ª

If b = 2; the unit of information is bits

ª

If b = 10; the unit of information is dit/Hartley/decit

ª

If b = e; the unit of information is nats/Hepits

ª

1 Hartley = 3.32 bits and 1 nat = 1.443 bits

Sample Problem: Calculate the coding efficiency in representing the 26 letters of the alphabet using a binary and decimal system. Solution: A = log2 26 = 4.7 bits ⇒ 5 bits must be used A = log10 26 = 1.415 dits ⇒ 2 dits must be used

4.7 x 100% = 94% 5 1.415 η= x 100% = 71% 2

η=

This proves the fact that binary coding is more efficient than decimal coding.


4-23

D. .CHANNEL CLASSIFICATIONS.

1.

Lossless Channels A channel described by a channel matrix with only one non-zero element in each column.

2.

Deterministic Channel Describe by a channel matrix with only one non-zero element on each row.

3.

Noiseless Channel A channel which is both lossless and deterministic.

E. .CHANNEL CAPACITY. The maximum rate at which information can be transmitted through a channel. 1.

For Lossless Channels

C = log 2 N The lossless channel capacity is equal to source entropy, and no source information is loss in transmission. 2.

For Deterministic Channel

C 2 = log 2 M The deterministic channel capacity is equal to destination entropy. Each member of the source set is uniquely associated with one, and only one, member of the destination alphabet. Loading ECE SUPERBook


4-24

3.

For Noiseless Channel

C 3 = log 2 M = log 2 N 4.

For Additive White Gaussian Noise Channel (AWGN)

C4 =

1 s⎞ ⎛ log 2 ⎜ 1 + ⎟ 2 N⎠ ⎝

Note: The units of the previous equation are on per sample basis. Since there are 2BW (Nyquist Sampling Theorem) samples per unit time, the capacity per unit time can be written as

C = 2BWC 4 5.

Shannon Limit For Information Capacity

S⎞ ⎛ C = 2BWC 4 = BW log2 ⎜ 1 + ⎟ N⎠ ⎝ 6.

Shannon-Hartley Theorem

C = 2 BWC 2 = 2BW log 2 M where: C = channel capacity in bps N = is the number of input symbols M = is the number of output symbols BW = channel bandwidth in Hz


4-25

ECE Board Exam: April 2003 A binary digital signal is to be transmitted at 10 kbits/s, what is the absolute minimum bandwidth is required to pass the fastest information change undistorted? Solution: From deterministic channel capacity; C = 2BW log2 M ⇒ M = 2 for binary signaling BW =

C 10,000 = = 5 kHz 2 2

ECE Board Exam: April 2003 What is the bandwidth needed to support a capacity of 20,000 bits/s (using Shannon’s theory), when the ratio of power to noise is 200? Also compute the information density.

Solution: S⎞ C ⎛ C = BW log2 ⎜1 + ⎟ ⇒ BW = S⎞ N⎠ ⎛ ⎝ log2 ⎜1 + ⎟ N⎠ ⎝ 20,000 BW = = 2615.54 Hz log2 (1 + 200 ) η=

C S⎞ 20,000 bps bps ⎛ = log2 ⎜1 + ⎟ = = 7.65 BW N ⎠ 2,615.54 Hz Hz ⎝

ECE Board Exam: April 2003 What is the channel capacity for a signal power of 200 W, noise power of 10 W and a bandwidth of 2 kHz of a digital system? Also calculate the spectrum efficiency.

Solution: S⎞ 200 ⎞ ⎛ ⎛ C = BW log2 ⎜1 + ⎟ = 2 x 103 log2 ⎜1 + ⎟ = 8.779 kbps N 10 ⎠ ⎝ ⎠ ⎝ η=

C S ⎞ 8.779 kbps bps ⎛ = log2 ⎜1 + ⎟ = = 4.4 BW N 2 kHz Hz ⎝ ⎠



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Sample Problem: Consider a digital source that converts an analog signal to digital form. If an input analog signal is sampled at 1.25x the Nyquist rate and each sample is quantized into one of 256 equally likely levels. a. What is the information rate of this source? b. Calculate the min bit error rate of this source if transmitted over an AWGN channel with a BW=10 kHz and 20 dB S/N ratio. c. Find the S/N in dB require for error free transmission if the BW=10 kHz d. Find the BW required for error free transmission if the S/N is 20 dB Assume that the successive samples are statistically independent.

Solution: a. Information rate bits ; Nyquist Rate = 2fm(max) = 2(4kHz) = 8 kHz sample kilosample s Actual Rate = 1.25 x R NYQUIST = 1.25(8kHz) = 10 kHz = 10 sec bits ⎧ 3 samples ⎫ Information Rate = H x R actual = 8 ⎨10 x 10 ⎬ = 80 kbps sample ⎩ sec ⎭ H = log2 256 = 8

b.

Minimum BER Channel limitation due to noise : S⎞ ⎛ C = BW log2 ⎜1 + ⎟ = 10 kHz log2 (1 + 100 ) = 66 .88 kbps N⎠ ⎝ BER = R info − C = 80 kbps − 66 .88 kbps = 13 .12 kbps

It cannot be transmitted without errors c.

S/N required for error free transmission Channel capacity ≥ Info Rate S⎞ ⎛ C ≥ R info ⇒ 80 kbps ≥ 10 kHz log2 ⎜1 + ⎟ N⎠ ⎝ S ≥ 255 or 24.1 dB N

d.

BW required for error free transmission Channel capacity ≥ Info Rate C ≥ R info ⇒ 80 kbps ≥ BW log2 (1 + 101 ) BW ≥ 12 .02 kHz

This means to provide an error free transmission the bandwidth must be at least 12.02 kHz or the S/N ratio must be greater than 24.1 dB.

4-27

Self-Sufficient Guide to ECE by JASON AMPOLOQUIO F. .DIGITAL SIGNAL LINE ENCODING FORMAT.

Line encoding is the method used for converting a binary information sequence into a digital signal in a digital communications system.

0

1.

Types of Signaling i.

Unipolar signaling Binary 1 is represented by a high level and binary 0 by a zero level.

ii.

Polar signaling Binary 1’s and 0’s are represented by equal positive and negative levels.

iii. Bipolar (pseudoternary) signaling Binary 1’s are represented by alternately positive or negative values. The binary 0 is represented by a zero level iv. Manchester signaling (Split-phase encoding) Each binary 1 is represented by a positive half-bit period pulse followed by a negative half-bit period pulse. Similarly, a binary 0 is represented by a negative half-bit period pulse followed by a positive half-bit period pulse.



4-28

G. .DEFINITION OF DIGITAL ENCODING FORMAT. 1.

Non-Return to Zero (NRZ) i. Non-Return to Zero-Level (NRZ-L) Where L denotes positive logical level assignment. 1 = High level 0 = Low level ii.

Non-Return to Zero-Level (NRZ-M) Where M denotes inversion on mark 1 = Transition at beginning of interval 0 = No transition

iii. Non-Return to Zero-Level (NRZ-S) Where S denotes inversion on space using negative logic 1 = No change 0 = Transition at beginning of interval 2.

Return To Zero (RZ) 1 = Transition from High to Low in middle of interval 0 = low level

3.

Biphase i. Biphase-Level (Manchester) 1 = Transition from High to Low in middle of interval 0 = Transition from Low to High in middle of interval ii.

Biphase-Mark Always a transition at beginning of interval 1 = Transition in middle of interval 0 = No transition in middle of interval

iii. Biphase-Space Always a transition at beginning of interval 1 = No transition in middle of interval 0 = Transition in middle of interval 4.

Differential Manchester 1 = No transition in middle of interval 0 = Transition at beginning of interval

5.

Delay Modulation (Miller) 1 = Transition in middle of interval 0 = No transition if followed by 1 Transition at end of interval if followed by 0

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Bipolar-AMI (Alternate Mark Inversion) 1 = Pulse in first half of bit level, alternating polarity pulse to pulse 0 = No pulse

Read it till it Hertz…jma The term bipolar has two different conflicting definitions: ª In the space communication industry, polar NRZ is sometimes called bipolar NRZ, or simply bipolar. ª In the telephone industry, the term bipolar denotes pseudoternary signaling, as in the T1 bipolar RZ signaling. So many names… ª Polar NRZ is also called NRZ-L. ª Bipolar NRZ is called NRZ-M. ª Negative Logic Bipolar NRZ is called NRZ-S. ª Bipolar RZ is also called BPRZ, RZ-AMI, BPRZ-AMI, AMI or simply bipolar. ª Manchester NRZ is also called Manchester code, or Biphase-L, for biphase with normal logic level. ª Biphase-M is used for encoding SMPTE time-code data for recording on videotapes. BW and Efficiency of Popular Line Codes

Signaling

Code Type

Bandwidth

Spectral Efficiency (bps/Hz)

NRZ

fb 2

2

RZ

fb

1

NRZ

fb 2

2

RZ

fb

1

Manchester

fb

1

AMI

fb 2

2

UNIPOLAR

POLAR

BIPOLAR



4-30

H. .SIGNAL ELEMENT VS. DATA ELEMENT. One data element per one signal element

One data element per two signal element

Two data element per one signal element

Four data element per three signal element

Relation between Bit Rate and Baud Rate where : n=

fb = n x fB

# of data element # of signal element

fb = bit rate in bps fB = Baud rate in Baud

Sample Problem: A signal is carrying data in which 4 data element is encoded as one signal element. If the bit rate is 100 kbps, what is the average value of the baud rate? Solution: fB =

fb 100 kbps = = 25 kBaud n 4

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.DIGITAL MODULATION. DIGITAL MODULATION

AMPLITUDE SHIFT KEYING (ASK)

FREQUENCY SHIFT KEYING (FSK)

PHASE SHIFT KEYING (PSK)

QUADRATURE AMPLITUDE MODULATION (QAM)

1.

Amplitude Shift Keying (ASK) Digital Amplitude Modulation is simply a double-sideband, full-carrier amplitude modulation where the input modulating signal is a binary waveform. a.k.a. Continuous-wave modulation, On-Off keying (OOK)

Implementation of binary ASK



4-32

2.

Frequency Shift Keying (FSK) Frequency Shift Keying is a form of constant-amplitude angle modulation similar to conventional frequency modulation except that the modulating signal is a binary signal that varies between two discrete voltage levels.

Implementation of binary FSK

i.

Bandwidth Consideration

BW = 2 (fb + Δf ) ii.

FSK Receiver ª Noncoherent FSK demodulator


4-33

Coherent FSK demodulator

Minimum Shift Keying (MSK) With MSK, the mark and space frequencies are selected such that they are separated from the center frequency by an odd exact multiple of one-half of the bit rate.

fm − fs = Δf = n x

fb 2

where : n = positive odd integer

Sample Problem: Calculate the frequency shift (deviation) between mark and space for GSM cellular radio system that uses Gaussian MSK (GMSK) with a transmission rate of 270.833 kbps. Solution: GMSK is a special case of FSK where n=1 fm − fs =

270.833 kbps = 135.4165 kHz 2



4-34

2.

Phase Shift Keying (PSK) Phase Shift Keying is a form of angle-modulated, constant-amplitude digital modulation similar to conventional phase modulation except that with PSK the input signal is a binary digital signal and limited number of output phase are possible.

a.

Binary Phase Shift Keying (BPSK) With BPSK, two output phases are possible for a single carrier frequency.

Implementation of binary PSK

ª

Phasor and Constellation Diagram

4-35


Truth Table & Minimum Nyquist Bandwidth

Binary Input

Output Phase

Logic 0

180°

Logic 1

0°

fN = fb

Sample Problem: Determine the minimum Nyquist bandwidth and the Baud rate for a BPSK modulator with a carrier frequency of 70 MHz and an input bit rate of 10 Mbps. Solution: Nyquist BW fN = fb = 10 MHz

Baud Rate fB = 10 MBaud

M-ary Encoding M-ary is a term derived from the word binary. M is simply a digit that represents the number of conditions or combinations possible for a given number of binary variables.

N = log 2 M

N

M

1

2

2

4

3

8

4

16

5

32

where : N = # of bits per symbol M = # of output conditions or symbols possible w/ N bits



4-36

Sample Problem: How many bits are needed to address 256 different level combinations?. Solution: N = log2M = log2 256 =

b.

log10 256 = 8 bits log10 2

Quarternary Phase Shift Keying (QPSK) QPSK is a form of angle-modulated, constant-amplitude digital modulation where four output phases are possible (M=4)

Implementation of Quarternary PSK

4-37



ª


Binary Input Q

I

QPSK Output

0

0

-135°

0

1

-45°

1

0

+135°

1

1

+45°

fN =

fb 2

Sample Problem: Calculate the minimum double-sided Nyquist BW and the Baud for a QPSK modulator with an input data rate equal to 40 Mbps and a carrier frequency of 110 MHz. Solution: Nyquist BW (QPSK) fb 40 Mbps = 2 2 = 20 MHz

fN =

c.

Baud R ate fb 40 Mbps = 2 2 = 20 MBaud

Baud =

Eight Phase Shift Keying (8-PSK) 8-PSK is another form of angle-modulated, constant-amplitude digital modulation where eight output phases are possible (M=8)



4-38

ª


ª


Binary Input Q I C

8-PSK Output

0

0

0

-112.5°

0

0

1

-157.5°

0

1

0

-67.5°

0

1

1

-22.5°

1

0

0

+112.5°

1

0

1

+157.5°

1

1

0

+67.5°

1

1

1

+22.5°

fN =

fb 3

Sample Problem: Calculate the minimum double-sided Nyquist BW and the Baud for a 8-PSK modulator with an input data rate equal to 25 Mbps and a carrier frequency of 45 MHz. Solution: Nyquist BW (8 - PSK) f 25 Mbps fN = b = 3 3 = 8.33 MHz

Baud Rate fb 25 Mbps = 3 3 = 8.33 MBaud

fB =

4-39

Self-Sufficient Guide to ECE by JASON AMPOLOQUIO d.

Sixteen Phase Shift Keying (16-PSK) 16-PSK is another form of angle-modulated, constant-amplitude digital modulation where sixteen output phases are possible. (N=4 & M=16) Phasor and Constellation Diagram

ª

VCcos t

0100

VCcos t

0100

0011

0101

0010 0001

0110

1011 1100

1111

1001

1110

1010

0000 VCsin t

1000

1111

1001

0001

0111 VCsin t

VCsin t

1000

1010

1101

1011 1100

1110 1101

-VCcos t

-VCcos t


ª

Bit Code

Phase

Bit Code

Phase

0

0

0

0

11.25°

1

0

0

0

191.25°

0

0

0

1

33.75°

1

0

0

1

213.75°

56.25°

1

0

1

0

236.25°

0

0

1

0

0

0

1

1

78.75°

1

0

1

1

258.75°

0

1

0

0

101.25°

1

1

0

0

281.25°

0

1

0

1

123.75°

1

1

0

1

303.75°

1

1

1

0

326.25°

1

1

1

1

348.75°

0

1

1

0

146.25°

0

1

1

1

168.75°

fN =

3.

0010

0110

0000

0111 VCsin t

0011

0101

fb 3

Quadrature Amplitude Modulation (QAM) QAM is a form of digital modulation where the digital information is contained in both the amplitude and phase of the transmitted carrier.



4-40

a.

Eight Quadrature Modulation(8-QAM) 8-QAM is an M-ary encoding technique where M=8. 2 amplitudes and 4 phase are used to give 8 different symbols. ª

b.


Sixteen Quadrature Modulation(16-QAM) 8-QAM is an M-ary encoding technique where M=16.

3 amplitudes and 12 phases are used to give 16 different symbols. 4 amplitudes and 8 phases are used to give 16 different symbols. 2 amplitudes and 8 phases are used to give 16 different symbols. ª



4-41

.SPECTRAL EFFICIENCY OF DIGITAL SYSTEMS. Spectral Efficiency (Bandwidth efficiency) or information density is an indication of how a certain modulation scheme is efficiently utilizing its bandwidth.

In terms of Channel capacity & Bandwidth In terms of Signal to Noise ratio In terms of Bit Rate and Nyquist BW (Baud Rate)

C BW S⎞ ⎛ η = log⎜ 1 + ⎟ N ⎝ ⎠ fb f η= η= b fB BW η=

Sample Problem: Determine the bandwidth efficiency for the following modulation scheme a. BPSK, fb=15 Mbps b. QPSK, fb=20 Mbps c. 8-PSK, fb=28 Mbps d. 8-QAM, fb=30 Mbps e. 16-PSK, fb=40 Mbps f. 16-QAM, fb=42 Mbps Solution: a. BPSK

f 15 Mbps bps =1 η= b = BW 15MHz Hz c. 8PSK f 3(28 Mbps) bps =3 η= b = BW 28MHz Hz e. 16PSK f 4(40 Mbps) bps =4 η= b = BW 40MHz Hz

b. QPSK η=

fb 2(20 Mbps) bps = =2 BW 20MHz Hz

d. 8QAM η=


f.16QAM η=




4-42

Summary of Various Digital Modulation System

System

# of Bits Encoded per Symbol

Minimum Nyquist BW (Baud Rate)

Spectral Efficiency (bps/Hz)

1 2 3 3 4 4 5 6

fb fb/2 fb/3 fb/3 fb/4 fb/4 fb/5 fb/6

1 2 3 3 4 4 5 6

BPSK QPSK 8-PSK 8-QUAM 16-PSK 16-QUAM 32-QUAM 64-QUAM

K.

.ERROR PROBABILITIES FOR DIGITAL COMMUNICATION SYSTEMS.

1.

Coherent Systems with Additive White Gaussian Noise (AWGN) channels

Pe = Q 2 2.

Eb No

Non-Coherent Systems with Additive White Gaussian Noise (AWGN) channels

Pe =

⎛ E −⎜⎜ b 2N 0.5e ⎝ 0

⎞ ⎟ ⎟ ⎠

where : Eb = Energy per bit in J

bps

N 0 = Noise density in W

Hz

Sample Problem: Calculate the probability of error for a non-coherent FSK system if the carrier power is 10-13 W, bit rate of 30 kbps, BW of 60 kHz and noise power of 10-14 W. Solution: Eb =

Po 10−13 W attoJoules = = 3.3 fb 30 kbps bit

Pe =

⎛E ⎞ −⎜⎜ b ⎟⎟ N 0.5e ⎝ O ⎠

=

NO =

N 10 −14 W Watt = = 166.67 x 10-21 BW 60 kHz Hz

⎛ 3.3 x 10 −18 ⎞ ⎟ −0.5⎜ ⎜ 166.67 x 10 −21 ⎟ ⎝ ⎠ 0.5e

= 2.51 x 10−5

This probability of error means that 25 bits will be expected to be corrupted (in error) for every 1 million bits transmitted.


4-43

.ERROR DETECTIONS. 1.

Redundancy Redundancy involves transmitting each character twice. If the same character is not received twice in succession, a transmission error has occurred.

2.

Echoplex Echoplex involves the receiving device echoing the received data back to the transmitting device. The transmitting operator can view the data, as received and echoed, making corrections as appropriate.

3.

Exact-count encoding The number of 1s in each character is the same and therefore a simple count of number of 1s received in each character can determine if a transmission error has occurred.

4.

Parity Checking Parity checking is by far the most commonly used method for error detection and correction, as it is used in asynchronous devices such as PCs. Parity involves the transmitting terminal’s appending one or more parity bits to the data set in order to create odd parity or even parity. Dimensions of parity checking i. Vertical Redundancy Checking (VRC) VRC entails the appending of a parity bit at the end of each transmitted character or value to create an odd or even total mathematical bit value.

ii.

5.

Longitudinal Redundancy Checking (LRC) or Block Checking Character (BCC) LRC adds another level of reliability, as data is viewed in a block or data set, as though the receiving device were viewing data set in a matrix format. LRC/BCC adds a significant measure of reliability. Also known as checksum, the LRC is sent as an extra character at the end of each data block.

Cyclic Redundancy Checking (CRC) CRC validates transmission of a set of data, formatted in a block or frame, through the use of a unique mathematical polynomial known to both transmitter and receiver. The result of that calculation is appended to the block or frame or text as either a 16- or 32-bit value.



4-44

i.

CRC Encoding Procedures

1. Multiply i(x) by xn - k (puts zeros in (n − k) low − order positions). 2. Divide x n - k by g(x). xn − k i(x) = g(x) q(x) = r(x) 3. Add remainder r(x) to xn − k i(x) (puts check bits in the (n − k) low − order positions). b(x) = xn − k i(x) + r(x)

n = number of bits in a codeword k = information bits q(x) = Quotient r(x) = Remainder g(x) = Generator polynomial i(x) = Information polynomial b(x) = Transmitted information ii.

Standard CRC Polynomial Codes

Name

Used In

CRC-8

ATM header error check

CRC-10

ATM CRC

CRC-16

Bisync

CCITT-16

HDLC, XMODEM, V.41

CCITT-32

IEEE 802, V.32

M. .ERROR CORRECTIONS. 1.

Symbol substitution With symbol substitution, if a character is received in error, rather than revert to a high level of error correction or display the incorrect character.

2.

Retransmission (ARQ) Retransmission, as the name implies, is resending a message when it is received in error and the receive terminal automatically calls for retransmission of the entire message.


4-45

Forward Error Correction (FEC) FEC involves the addition of redundant information embedded in the data set in order that the receiving device can detect errors and correct for them without requiring a retransmission. The most commonly employed technique is Hamming code. Hamming Distance The number of bit position in which two codeword differs is called Hamming distance.

2n ≥ m + n + 1

n = Number of Hamming bits m = Number of bits in the data character

ECE Board Exam: APRIL 2003 How many Hamming bits would be added to a data block containing 128 bits? Solution: Number of Hamming bits 2n ≥ m + n + 1 For n = 6 26 = 64 ; m + n + 1 = 128 + 6 + 1 ; 64 < 135 For n = 7 27 = 128 ; m + n + 1 = 128 + 7 + 1;128 < 136 For n = 8 28 = 256 ; m + n + 1 = 128 + 8 + 1 ;256 > 138 (satisfied) Answer : Hamming bits = 8



4-46

Sample Problem:

Calculate the Hamming distance to detect and correct 3 single-bit errors that occurred during transmission. Also compute for the number of Hamming bits for a 23 bit data string.

Solution: Required Hamming distance for error detection Hd = d + 1 = 3 + 1 =4 Required Hamming distance for error correction Hd = 2d + 1 = (2 x 3 ) + 1 =7 Number of Hamming bits 2n ≥ m + n + 1 For n = 4 24 = 16 ; m + n + 1 = 23 + 4 + 1 ; 16 < 28 For n = 5 25 = 32 ; m + n + 1 = 23 + 5 + 1 ; 32 > 29 n=5 Answer : Hamming bits = 5


To detect d single-bit errors, you need a Hamming distance of d+1 code because with such a code there is no way that d single bit errors can change a valid codeword into another valid codeword.

ª

Similarly, to correct d single-bit errors, you need a distance 2d+1 code because that way the legal codeword are so far apart that even with d changes. The original codeword is still closer than any other codeword.

ª

Hamming codes can only correct single bit errors. However there is a trick that can be used to permit Hamming codes to correct burst errors.

4-47


I

H

1.

If a symbol is selected from a group of 200 x’s and 50 y’s. Find the probability if a) any of the x is transmitted b) any of the y is transmitted. A. P(x)=0.7, P(y)=0.3 B. P(x)=0.8, P(y)=0.2 C. P(x)=0.3, P(y)=0.7 D. P(x)=0.2, P(y)=0.8

2.

A TV picture frame contains 211,000 elements and each element may take on 8 possible equiprobable brightness levels. Find the info content of each element and for the whole frame. A. 7 bits/element, 833 kbits/frame B. 6 bits/element, 433 kbits/frame C. 5 bits/element, 333 kbits/frame D. 3 bits/element, 633 kbits/frame

3.

Calculate the maximum data rate without error for a telephone system with a bandwidth of 3.2 kHz and a signal-to-noise ratio if 30 dB, also compute the number of corrupted bits per second if 64 level encoding will be used. A. 31.895 kbps, 3.2 kbits will be corrupted per second B. 38.4 kbps, 6.505 kbits will be corrupted per second C. 31.895 kbps, 6.505 kbits will be corrupted per second D. 38.4 kbps, 3.2 kbits will be corrupted per second

4.

Calculate the Baud and the bit rate for a 64-level modulator that transmit symbols 12,000 times per second A. 12 kBaud, 72 kbps B. 12 kBaud, 62 kbps C. 6 kBaud, 64 kbps D. 6 kBaud, 72 kbps

5.

Determine the channel data rate for the North American digital cellular system that transmit 24.3 kBaud using DQPSK. A. 72.9 kbps B. 97.2 kbps C. 12.15 kbps D. 48.6 kbps

6.

For a BPSK modulation with a carrier frequency of 80 MHz and an input bit rate of 10 Mbps, determine the minimum Nyquist BW. NOV 2002 A. 1 MHz B. 5 MHz C. 10 MHz D. 2.5 MHz

7.

Calculate the bandwidth needed to transmit a DS-1 signal in a channel with a signal-to-noise of 20 dB, and also compute for the spectral efficiency. A. 231.89 kHz, 231.89 bps/Hz B. 1.544 kHz, 6.65 bps/Hz C. 231.89 kHz, 6.65 bps/Hz D. 1.544 MHz, 231.89 bps/Hz



4-48

8.

Calculate the spectral efficiency of an Audio FSK (FSK) that transmits 1200 kbps using an FM signal modulated by tones of 1200 Hz and 2200 Hz, with a 5 kHz frequency deviation. A. 1.37 bps/Hz B. 12.45 bps/Hz C. 0.083 bps/Hz D. 8.3 bps/Hz

9.

A telephone modem uses QAM modulation with four possible phase angles and 2 amplitude levels. Calculate the maximum data rate that could be sent using this modem in a voice telephone channel. (assume 3.2 kHz bandwidth) A. 51.2 kbps B. 19.2 kbps C. 25.6 kbps D. 12.8 kbps

10. 5 Hamming bits are required for a data block containing ____ bits. A. 26 B. 31 C. 21 D. 28 11. Calculate the expected number of error bits for a digital transmission that has an error probability of 10-6 and is 109 bits long. A. 100 bits B. 1000 bits C. 10 bits D. 1 bits 12. 7 error bits out of 5.7 million total bits is equal to ____ error probability. A. 6.81 x 10-6 B. 9.56 x 10-6 -6 C. 4.74 x 10 D. 1.23 x 10-6 13. Determine the number of bits required and the coding efficiency in encoding a system of 50 equiprobable events with a binary code. A. 4 bits, 74 % B. 6 bits, 94 % C. 5 bits, 84 % D. 2 bits, 87.8 % 14. What is the channel capacity for a signal power of 200 W, noise power of 10 W and a bandwidth of 2 kHz of a digital system? April 2003 A. 8.779 kbps B. 9.128 kbps C. 4.751 kbps D. 6.143 kbps 15. Calculate the coding efficiency and redundancy of a discrete system with 105 equiprobable symbols. A. 93.9%, 6.1% B. 98.9%, 1.1% C. 95.9%, 4.1% D. 97.9%, 2.1% 16. 5.54 nats of information is equal to how many bits and Hartley respectively A. 4 bits, 4.4 Hartley B. 6 bits, 1.4 Hartley C. 2 bits, 2 Hartley D. 8 bits, 2.4 Hartley 17. 1 dit is ____ bits A. 0.332 C. 3.32

B. D.

33.2 1.44


4-49

18. A unipolar NRZ line code is converted to a multiple level signal for transmission over a channel. The number of possible values in the multilevel signal is 32, and the signal consists of rectangular pulses that have a pulse width of 0.3472ms. What is the equivalent bit rate for the multilevel signal? A. 28.8 kbps B. 56 kbps C. 7.2 kbps D. 14.4 kbps 19. For quarternary phase shift keying (QPSK) modulation, data with a carrier frequency of 70 MHz, and an input bit rate of 10 Mbps, determine the minimum Nyquist BW. ECE Board Nov 2002 A. 1 MHz B. 5 MHz C. 10 MHz D. 2.5 MHz 20. What is the information density of AMI? A. 2 bps/Hz B. C. 1 bps/Hz D.

3 bps/Hz 4 bps/Hz

21. Calculate the approximate bit rate of a communication system with a minimum Nyquist bandwidth of 5 kHz and using Manchester encoding technique. A. 5 kbps B. 20 kbps C. 10 kbps D. 2.5 kbps 22. How many bits are needed to address 256 different level combinations? A. 5 bits B. 8 bits C. 6 bits D. 7 bits 23. An asynchronous communications system uses ASCII at 9600 bps with eight bits, one start bit, one stop bit and no parity bit. Express the data rate in words per minute. (Assume a word has 5 characters or letters and one space)

April 2003 A. C.

96 wpm 9600 wpm

B. D.

960 wpm 96000 wpm

24. Calculate the phase separation of two adjacent bits constellation in a 16 PSK system? A. 11.25° B. 5.625° C. 22.5° D. 45° 25. 45° maximum phase shift of two adjacent bits combination is possible in ____ system. A. BPSK B. QPSK C. 16 PSK D. 8-PSK 26. 300 Baud is equivalent to what effective bit rate if bits are represented by a single voltage that can have one of 16 distinct values? A. 4.8kbps B. 1.2 kbps C. 9.6 kbps D. 2.4 kbps 27. A BER tester sends bits at 2000 Baud for 10s. There are a total of 50 errors. What is the BER? B. 25x10-3 A. 2.5x10-3 C. 12.5x10-3 D. 21.5x10-3


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28. Using 200 bit/frame at 200 Baud and 10 s of transmission time, what is the percentage of error-free frames when there are 500 errors spread evenly through the 10-s period? A. 50% (1 every 2 frames are corrupted) B. 0% (all frames are corrupted) C. 25% (1 every 4l frames are corrupted) D. 10% (1 every 10 frames are corrupted) 29. Consider a system that sends data in frames of 100 bits, including preamble, message, and CRC bits, and each frame takes 0.5 s. Calculate the BER, percentage of frames with error when 1000 frames are sent. A. 25x10-3, 25% B. 2.5x10-3, 25% D. 2.5x10-3, 50% C. 2.5x10-3, 50% 30. What are the percentage error-free frames when there are 500 errors that occurs at the average rate of 100 errors/sec for the first 5 sec? A. 40% (1st 4 frames has errors, while the next 6 has none) B. 20% (1st 2 frames has errors, while the next 8 has none) C. 50% (1st 5 frames has errors, while the next 5 has none) D. 80% (1st 8 frames has errors, while the next 2 has none) 31. Calculate the number of Hamming bits required for a 16-bit data string A. 5 B. 3 C. 4 D. 2 32. What is the information content for binary 0 in a certain binary transmission system if the probability of binary 1 being transmitted is 0.6 A. 1.32 binits B. 3.2 binits C. 2.13 binits D. 13.2 binits 33. Calculate the Hamming distance to detect and correct 2 single-bit errors that occurred during transmission. A. 4, 7 B. 5, 7 C. 3, 5 D. 2, 4 34. For a ______ channel, we need to use the Shannon capacity to find the maximum bit rate. A. noisy B. noiseless C. bandpass D. low-pass

35. What you see in an eye pattern is the effect of: A. too many bits high B. too many bits low C. intermodulation distortion D. intersymbol interference 36. For QAM, the two dimensions of its symbol space are: A. amplitude and frequency B. amplitude and phase angle C. frequency and phase angle D. I-bits and Q-bits 37. The specs of the old Bell type 103 modem were: A. 300 bps, full-duplex, FSK B. 600 bps, full-duplex, FSK C. 1200 bps, full-duplex, FSK D. 1200 bps, half-duplex, FSK


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38. High-frequency radioteletype systems commonly use: A. FSK B. AFSK C. PSK D. QAM 39. The _________ product defines the number of bits that can fill the link. A. bandwidth-period B. frequency-amplitude C. bandwidth-delay D. delay-amplitude 40. Block coding can help in _______ at the receiver. A. Synchronization B. Error detection C. Attenuation D. A and B 41. In _______ transmission, bits are transmitted simultaneously, each across its own wire. A. Parallel B. Asynchronous serial C. Synchronous serial D. A and B 42. Unipolar, bipolar, and polar encoding are types of _______ encoding. A. Block B. Line C. NRZ D. Manchester 43. How many points will be on the constellation diagram of a QAM system using 8 phase angles and 2 amplitude levels? A. 8 B. 32 C. 16 D. 10 44. _______ encoding has a transition at the middle of each bit. A. RZ B. Manchester C. Differential Manchester D. All the above 45. Which encoding technique attempts to solve the loss of synchronization due to long strings of 0s? A. NRZ B. BnZS C. AMI D. Manchester 46. Suppose a synchronous frame has 16 bits of non-data in the front and a 16-bit BCC at the end. The frame carries 1024 bytes of actual data. Calculate the efficiency of the communication system. A. 80% B. 67.5% C. 85.12% D. 97.0% 47. Baseband transmission of a digital signal is possible only if we have a ____ channel. A. low-pass B. low rate C. bandpass D. high rate 48. Suppose an asynchronous frame holds 8 bits of data, a parity bit, and two stop bits (it could happen). Calculate the efficiency of the communication system. A. 33.3% B. 83.4% C. 66.7% D. 16.67%



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49. One factor in the accuracy of a reconstructed PCM signal is the _______. A. Number of bits used for quantization B. Signal bandwidth C. Carrier frequency D. Baud rate 50. In asynchronous transmission, the gap time between bytes is _______. A. Fixed B. A function of the data rate C. Variable D. Zero 51. How many different characters could be encoded using a six-bit code? A. 64 B. 32 C. 56 D. 6

52. FSK stands for: A. Full-Shift Keying C. Full-Signal Keying

B. Frequency-Shift Keying D. Frequency-Step Keying

53. The ITU is under the auspices of: A. CCITT C. IEEE

B. the U.N. D. ANSI

54. PCM is an example of _______ conversion. A. Analog-to-analog B. Analog-to-digital C. Digital-to-digital D. Digital-to-analog 55. If the frequency spectrum of a signal has a bandwidth of 500 Hz with the highest frequency at 600 Hz, what should be the sampling rate, according to the Nyquist theorem? A. 500 samples/s B. 1000 samples/s C. 200 samples/s D. 1200 samples/s

56. High-speed modems equalize the line to compensate for: A. noise and interference B. uneven phase and frequency response C. low SNR D. inconsistent bit rates at either end of channel 57. The bits sent to allow equalization are called: A. Gaussian bits B. random bits C. a training sequence D. a random sequence 58. PSK stands for: A. Pulse-Signal Keying C. Phase-Signal Keying

B. Pulse-Shift Keying D. Phase-Shift Keying

59. In the equation C = 2Blog2M, M is the: A. margin of noise B. modulation index C. number of possible states per symbol D. maximum number of symbols per second

Self-Sufficient Guide to ECE by JASON AMPOLOQUIO 60. An A. B. C. D.

"eye the the the the

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pattern" shows a good channel when: eye is maximally open eye is maximally closed eye is half open eye alternately opens and closes

61. Which coding scheme requires DC continuity: A. AMI B. Manchester C. unipolar NRZ D. bipolar RZ 62. A T-1 cable uses: A. Manchester coding C. NRZ coding

B. bipolar RZ AMI coding D. pulse-width coding

63. An Ethernet running at 10 Mbits / second uses: A. Manchester encoding B. Three-Level encoding C. NRZ encoding D. AMI encoding 64. Instead of a single bit, a QPSK symbol contains: A. a byte B. 4 bits C. a dibit D. a Q-bit 65. If the available channel is a ____ channel, we cannot send a digital signal directly to the channel. A. low-pass B. bandpass C. low rate D. high rate

66. In the equation I = ktB, I is measured in: A. amperes B. amperes per second C. bits D. bits per second 67. If an asynchronous frame is used to send ASCII characters in the form of bytes (8 bits), what is the shortest time it could take to send 1000 characters if each bit in a frame is 1 msec long? A. 5 seconds B. 0.1 second C. 10 seconds D. 1 second 68. The _____ of a composite signal is the difference between the highest and the lowest frequencies contained in that signal. A. frequency B. period C. bandwidth D. amplitude 69. For a ______ channel, the Nyquist bit rate formula defines the theoretical maximum bit rate. A. noisy B. noiseless C. bandpass D. low-pass

70. To A. B. C. D.

reduce the need for linearity, π/4 DQPSK uses: angles of 0, 90, 180, and 270 degrees angles of 45, 135, 225, and 315 degrees angles of π/4, 2π/4, 3π/4, and 4π/4 double phase-shift angles



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71. For A. B. C. D.

QAM, a "constellation diagram" shows: location of symbols in "symbol space" separation of symbols in "symbol space" effects of noise on symbols all of the above

72. ITU A. B. C. D.

is an abbreviation for: International Telephony Unit International Telephony Union International Telecommunications Union International Telecommunications Units

73. QAM stands for: A. Quadrature Amplitude Modulation B. Quadrature Amplitude Masking C. Quadrature Amplitude Marking D. Quadrature Alternate Modulation 74. Calculate the bits per second capacity of a system sending 1000 symbols per second with 16 possible states per symbol. A. 500 B. 4000 C. 250 D. 16000 75. _______ encoding has a transition at the beginning of each 0 bit. A. Differential Manchester B. RZ C. Manchester D. AMI 76. How many symbols are produced by a QPSK modem? A. 16 B. 4 C. 8 D. 2 77. Which modem is used to handle 9600 bps data rates? A. FSK B. QPSK C. BPSK D. QAM 78. An 8-PSK system has an incoming data stream at 2400 bps. What is the symbol rate of the transmitter? A. 2400 sps B. 600 sps C. 800 sps D. 1200 sps 79. How many different symbols are available from an 8-PSK transmitter? A. 3 B. 4 C. 8 D. 16 80. How many bits are needed to address all the different symbols available in a 64-QAM transmitter? A. 4 B. 5 C. 32 D. 64

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Section

Data

16

Communications


Data: Information, for example, numbers, text, images, and sounds, in a form that is suitable for storage in or processing by a computer.

DEFINITION.

DEFINITION.

Data Communication is the process of transferring digital

information between two or more points.

Compression: The encoding of data so that it requires less disk space for storage and transmission.

DEFINITION.

Encryption: The science of converting computer data and messages to something incomprehensible by means of a key, so that it can be reconverted only by an authorized recipient holding the matching key.

DEFINITION.

A. .STANDARDS ORGANIZATIONS. 1.

International Telecommunications Union (ITU) A voluntary, non-treaty organization founded in 1946. ITU issues standards on vast number of subjects, ranging from nuts and bolts (literally) to telephone pole coatings. Three Main Sector i. Radiocommunications Sector (ITU-R) ii. Telecommunications Standardization Sector (ITU-TSS) iii. Development Sector (ITU-D)

2.

American National Standards Institute (ANSI) Founded in 1918, coordinates and harmonizes private sector standards development. ANSI also serves as the U.S. representative to the International Standards Organization (ISO), the originator of the Open Systems Interconnection (OSI) reference model.

3.

Institute of Electrical and Electronics Engineers (IEEE) IEEE a worldwide professional association dealing with SONET only peripherally. The IEEE has significant responsibility for the development of LAN standards, including FDDI, which have great impact on SONET.


DATA COMMUNICATIONS

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International Standards Organization (ISO) An organization comprised of the national standards organizations (ANSI) of the various nations. The ISO heavily influences international standards set by the ITU-T, and is best known for its involvement in the OSI (Open Systems Interconnection) model.

5.

Exchange Carriers Standards Association (ECSA) Formed in 1984, represents the interests of U.S. IXCs. The ECSA T1 committee addresses issues of functionality and characteristics of interconnection and interoperability.

6.

Electronic Industries Association (EIA) EIA was founded in 1924 as the Radio Manufacturers Association. The EIA is responsible for developing the RS series of standards for data and telecommunications.

B. .NETWORK STANDARDS . 1.

De facto (Latin for “from the fact”) De facto standards are those that have just happened, without any formal plan.

2.

De jure (Latin for “by law”) De jure standards, in contrast, are formal, legal standards adopted by some authorized standardization body.

C. .DATA TRANSMISSION. 1.

Serial Transmission Bits are transmitted one after another. often called serial-by-bit.

This type of transmission is


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Parallel Transmission Bits are transmitted all at once. This type of transmission is called parallel-by-bit or serial-by-character.

D. .DATA COMMUNICATIONS CIRCUITS. 1.

Data Terminal Equipment (DTE) The data equivalent of CPE (Customer Premise Equipment) in the voice world, DTE is the computer transmit and receive equipment, including a wide variety of dumb terminals, or terminals without embedded intelligence in the form of programmed logic.

2.

Data Communications Equipment (DCE) DCE is the equipment that interfaces the DTE to the network, in the process resolving any issues of incompatibility between those domains. Incompatibility issues can include digital versus analog, voltage level, transmission speed and bit density.

E. .DATA COMMUNICATIONS CODES. 1.

Morse Code Morse code is one of the first character code developed. Morse code was design with telegraph operator in mind. The character use the most frequently needs the fewest dots and dashes for transmission.

2.

Baudot Code Baudot code addressed the shortcomings of Morse code by using 5 bits to represent information subsequently known as International Telegraph Alphabet (ITA) #2.


DATA COMMUNICATIONS

4-58 3.

American (National) Standard Code for Information Interchange (ANSCII or ASCII) ASCII employs a 7-bit coding scheme, supporting 128 (27) characters, which is quite satisfactory for most alphabets, punctuations, characters, and so on. CCITT recommendation is International Telegraph Alphabet (ITA) #5.

4.

Extended Binary Coded Decimal Interchange Code (EBCDIC) EBCDIC, developed by IBM in 1962, involves an 8-bit coding scheme, yielding 28 (256) possible combinations and, thereby, significantly increasing the range of expression.

5.

Unicode (Universal Code) UNICODE is an attempt to standardize longer and more complex coding schemes used to accommodate more complex languages such as Japanese and Chinese. UNICODE supports 65,536 (216) characters, thereby accommodating the most complex alphabets; in fact, multiple alphabets can be satisfied simultaneously.

F. .TRANSMISSION METHODS. 1.

Asynchronous Asynchronous or character-framed transmission is a start-stop method of transmission in which each computer value (letter, number, or control character) is preceded by a start bit, succeeded by a stop bit, which advises the receiving terminal that the transmission of that set of information is ended.


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2.

Synchronous Synchronous or message-framed transmission is used for high-speed data communications applications where a relatively large set of data is framed or blocked, with one or more synchronization bits or bit patterns being used to synchronize the receiving terminal on the rate of transmission.

3.

Isochronous Isochronous (Isoc) data is synchronous data transmitted without a clocking source. Bits are sent continuously, with no start-stop bits for timing. Rather, timing is recovered from transitions in the data stream, with a whole number of bit-length intervals between characters.

4.

Pleisiochronous Pleisiochronous communications involves careful synchronization of transmission systems of varying levels of bandwidth through the use of highly accurate clocking devices. The preferred approach involves a master clocking source such as a cesium clock.


Asynchronous comes from Latin and Greek, it translates as not together with time.

ª

Synchronous comes from Latin and Greek origins, synchronous translates as together with time.

ª

Isochronous comes from the Greek word “isochronos”, meaning all bits are of equal importance and are anticipated to occur at regular intervals of time.

ª

Pleisiochronous comes from the Greek word “pleion”, meaning more.


DATA COMMUNICATIONS

4-60 G. .SYNCHRONIZATION.

Synchronization is required at a number of different levels in digital communications systems, these being classified as follows: 1.

Network synchronization - required so that station sharing a network can transmit and receive in an orderly fashion.

2.

Frame synchronization - required to keep track of the individual channels in a time-division multiplexed system.

3.

Codeword and node synchronization - required to keep tracks of block of bits in a bit stream, where each block form a codewords, usually designed for the purpose of error control.

4.

Symbol synchronization - required in order that symbols, which maybe hidden in a noisy waveform, are sampled at the optimum time

5.

Carrier synchronization - required in order to demodulate a carrier modulated wave in the most efficient manner.

H. .DIGITAL INTERFACE STANDARDS. 1.

Serial Interface i. RS 232-C RS 232-C (EIA Recommended Standard) defines exactly how ones and zeroes are to be electronically transmitted, including voltage levels needed as well as the other electronic signals necessary for computer communication.

Electrical Specifications of RS 232 Function

Data Pins

Control Pins

+5 to +15 V

Enable “ON” +5 to +15 V

Disable “OFF” -5 to -15 V

+3 to +25 V

+3 to +25 V

-3 to -25 V

Logic 1

Logic 0

Driver

-5 to -15 V

Terminator

-3 to -25 V


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Terminal Load Capacitance = 2500 pF Terminal Impedance = 3 kΩ Maximum bit rate = 20 kbps Nominal maximum length = 50 ft

RS-232 Pin Function Summary Pins 9,10,11, & 18 Unassigned Pins 1 & 7 Grounds Pins 2, 3, 14, & 16 Data Pins 15, 17, & 24 Timing Remaining pins are used for control and handshaking signals. ii.

RS 449 The RS-449 interface uses a 37-pin connector that provides more functions, faster data transmission, and greater distance capabilities but never embraced by the industry.

iii. RS-530 The RS-530 interface standard is intended to operate at data rates from 20 kbps to 2 Mbps using the same 25-pin DB-25 connector used by the RS-232. iv. RS 422-A (Similar to X.27) Describes a method of balance transmission that can be used for the RS 449 Category I signals. The transmission rate can reach 10 Mbps at approximately 15 m, and 90 kbps is the maximum bit rate that can be transmitted at 1200 m. v.

RS 423-A (Similar to X.26) Describes a method of unbalance transmission that can be used for the RS 449 Category I signals. The unbalanced interface cable will operate at a maximum line speed of 100 kbps and a span distance of 90 m.


V.24 is known in the U.S. as RS 232, which defines only the functional characteristic of the interface while its electrical characteristic is defined in V.28 standards.

ª

V.10 defines unbalance high-speed electrical interface specifications similar to RS-423.

ª

V.11 defines balance high-speed electrical interface specifications similar to RS-422.


DATA COMMUNICATIONS

4-62 2.

Parallel Interface i. Centronics Parallel Interface The Centronics parallel interface became the de facto standard and, today, is the most common interface used for interfacing personal computers to printers and other peripheral devices. ii.

I.

IEEE 488 Bus The IEEE 488 bus uses eight bidirectional data lines connected in parallel to interface up to 15 remote devices. The interface was designed for a maximum distance between adjacent devices of less than 7 ft, and the maximum length of the entire bus is about 65 ft.

.DATA FORMAT. Data formats let the receiving device logically determine what is to be done with the data and how to go about doing it. Data formats include code type, message length, and transmission validation techniques. A data format generally involves a header, text, and a trailer 1.

Header A communications header precedes the data to be transmitted, establishing the fact that the transmission link exists both physically and logically. It also provides for synchronization between the devices and the link, and enables the receiving device to route the data correctly.

2.

Text The text portion of the data set is the information to be communicated. It may contain a fixed or a variable amount of information organized into packets, blocks, frames, or cells.

3.

Trailer The trailer, tail or trace portion of the data set contains information relative to the analysis of the message, including message tracking and diagnostics. Trailing the text, the trailer information may contain the originating ID, the data block number, and total number of blocks being transmitted, and identification of system processing points involved in the transmission.


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.NETWORK PROTOCOLS. Protocol Protocol is a set of rules governing the orderly exchange of data within the network. 1.

Binary Synchronous Protocol (Bisync or BSC) Bisync was developed by IBM in 1966 as a byte-oriented protocol that frames the data (512 characters) with control codes which apply to the entire set of data.

2.

Synchronous Data Link Control (SDLC) SDLC, developed in the mid-1970s, is at the heart of IBM’s System Network Architecture (SNA). SDLC is a bit-oriented protocol that uses bit strings to represent characters. SDLC uses CRC error correction techniques, in this specific case known as Frame Check Sequence (FCS).

3.

High-Level Data Link Control (HDLC) HDLC was developed by the International Standards Organization (ISO) as a superset of IBM’s SDLC and the United States National Bureau of Standard’s (NBS) ADCCP protocols. A version of HDLC is the Link Access Protocol-Balanced (LAP-B), which is used in ITU-TS X.25 packet switched networks.


DATA COMMUNICATIONS

4-64 K. .NETWORK ARCHITECTURE. 1.

Open System Interconnect (OSI) Layer

i.

Physical Layer The lowest layer of the OSI Reference Model is the physical layer. This layer is responsible for the mechanical, electrical, functional, and procedural mechanism required for the transmission of data. Ex. RS-232C Unit of information: BITS

The physical layer is responsible for movements of individual bits from one hop (node) to the next. ii.

Data-Link Layer The data link layer is responsible for the manner in which a device gains access to the medium specified in the physical layer. Ex. HDLC Unit of information: FRAMES

The data link layer is responsible for moving frames from one hop (node) to the next. iii. Network Layer The network layer is responsible for the physical routing of data. To accomplish this task, the network layer performs addressing, routing, switching, sequencing of data packets, and flow control. Ex. Internet Protocol (IP) Unit of information: PACKETS

The network layer is responsible for the delivery of individual packets from the source host to the destination host.


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iv. Transport Layer The transport layer is responsible for ensuring that the transfer of information occurs correctly once a route is establish through a network. Ex. Transmission Control Protocol (TCP) Unit of information: SEGMENTS

The transport layer is responsible for the delivery of a message from one process to another. v.

Session Layer The session layer represents the first of three upper layers of the OSI model. This layer is responsible for establishing and terminating data streams between network nodes.

The session layer is responsible for dialog control and synchronization. vi. Presentation Layer The presentation layer is responsible for the isolating the application layer’s data format from the lower layers in the OSI model. The presentation layer provides data transformation, formatting, and syntax conversion.

The presentation layer is responsible for translation, compression, and encryption. vii. Application Layer The application layer provides support services for user and application tasks. File transfer, interpretation of graphic formats and documents, and document processing are supported at this level. Ex. X.400 e-mail messaging, Telnet, FTP

The application layer is responsible for providing services to the user. 2.

IBM’s System Network Architecture (SNA) IBM’s SNA is a method for unifying network operations. SNA describes a network in terms of a physical network and logical network. The physical network consists of a collection of nodes: host node, frontend communication node, concentration node and terminal node. i.

Host node - The host node is the central processor.

ii.

Front-end Node - The front-end node is concerned with data transmission functions.

iii. Concentration Node - The concentration nodes supervises the behavior of terminals and other peripherals. iv. Terminal Node - The terminal node is concerned with the input and output information through terminal devices.


DATA COMMUNICATIONS

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DECNET’s DNA A DECNET is just a collection of computer (nodes) whose functions include running user programs, performing packet switching or both. The architecture of DECNET is called Digital Network Architecture

Comparison of Different Architecture Control Levels

L.

.DATA SWITCHING. 1.

Circuit Switching In the classic sense, circuit switching provides continuous and exclusive access between physical circuits for the duration of the conversation.

2.

Packet Switching First deployed in 1971, packet switching involves the transmission of data in packets of fixed length across a shared network. Each packet is individually addressed, in order that the packet switches can route each packet over the most appropriate and available circuit.

3.

Frame Switching (Frame Relay) A relative newcomer, frame relay was first offered commercially in 1992 by Wiltel (U.S.). Unlike packet switching, frame relay supports the transmission of virtually any computer data stream in its native form--frames are variable in length (up to 4,096 bytes).

4.

Cell Switching (Asynchronous Transfer Mode) Clearly, cell switching is fundamental to the future of communications. Data is organized into cells of fixed length (53 octets), shipped across very high speed facilities and switched through very high speed, specialized switches.

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Attribute

X.25

Frame Relay

Facilities

Analog Assumed

Digital Assumed

Payload

128B/256B Fixed

< 4,096B Variable

< 56 Kbps-DS1

< 56 Kbps-DS1, DS3

LAP-B

LAP-D/LAP-F

High

Moderate

Connection-Oriented

Connection-Oriented

Network

User

Interactive Data

LAN-to-LAN

Speed Access Link Layer Protocol Latency Orientation Error Control Primary Application 5.

Photonic Switching Still in development, photonic switching are yet another dimension in the evolution of switch technology. Capable of supporting circuitpacket- frame- and cell switching, photonic switches will eliminate the requirement for optoelectric conversion when connected to a fiber optic transmission system.

M. .LOCAL AREA NETWORKING. 1.

LANs defined A LAN is a form of local (limited-distance), shared packet network for computer communications. LANs interconnect computers and peripherals over a common medium in order that users might share access to host computers, databases, files, applications, and peripherals.

2.

LAN Transmission Method ª

Unicast - a single packet is sent from the single source to a specific destination on a network.

A singlecast address is the simple implementation in which one host communicates with another host (or multiple hosts). In this case, a destination and source address is used. ª

Multicast - consists of a single data packet that is copied and sent to a specific subset of nodes on the network.

A multicast address is usually 6 bytes in length. It differs from the singlecast address because only particular hosts understand and receive the multicast address as if the frame were addressed to them with singlecast address. Loading ECE SUPERBook

DATA COMMUNICATIONS

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Broadcast - transmission consists of a single data packet that is copied and sent to all nodes on the networks.

The broadcast address can be used to perform isolated broadcast on networks in which multiple subnets are implemented. This type of address is frequently used when the upper layer protocol is TCP/IP. 3.

LAN Access Method ª

Controlled Access a.

X-on/X-off - the oldest media access control protocols, dating back to the days of the teletype.

b.

Polling - the process of sending signal to a terminal to transmit or asking it to receive it uses “roll call polling” with scheme or sequence.

c.

Token Passing - a channel accessing arrangement that is best suited for ring topology with either a baseband or broadband network. It uses a token and the token circulate in the network. Token is an electrical signal that circulates around the ring network from one station to another. If the station has the token, it can send a message.

ª

Contention - opposite of controlled access. a.

Carrier Sense Multiple Access with Collision Detection (CSMA/CD) - A channel accessing method, a node monitors the line to determine if the line is busy. If the station has a message to transmit but the line is busy, it waits for an idle condition before it transmits its message. Collision is the situation when simultaneously using one link.

two

nodes

transmit

Backoff Time is the ceasing time before the station attempts to transmit again.


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Comparison of Different Access Methods

Access Method

Advantages

Polling

Guaranteed access

Inefficient use of network

Fast in low traffic

Slow in high traffic No access guarantee No priority mechanism

Contention

Token Passing 4.

Disadvantages

Inexpensive Fast in high traffic Guaranteed access Time-critical

Slow in low traffic More expensive

LAN Shared Medium i.

Coaxial Cable

ThickNet, or thick Ethernet (10Base5) Also known as, uses traditional thick coax, often referred to as goldenrod, undoubtedly referring to its high cost and high value.

Twinax, or twinaxial cable Similar to ThinNet coax, but with two conductors, rather than one. Twinax is used in older IBM midrange systems such as Systems 34, 36, and 38, as well as the younger IBM AS/400 and RS/400.


DATA COMMUNICATIONS

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ii.

ThinNet, or thin Ethernet (10Base2) Uses coax of thinner gauge. The thinner cable is less costly to acquire and deploy, although its performance is less in terms of transmission distance.

Twisted Pair (UTP/STP)

UTP

10Base-T, or twisted-pair Ethernet Uses Cat 3, 4, or 5 UTP, 10Base-T actually is a wire hub that serves as a multiport repeater, as well as a central point of interconnection.

CDDI, or Cable Distributed Data Interface Also is known as TPDDI (Twisted-pair Distributed Data Interface). CDDI employs Cat 5 UTP as an inexpensive means of connecting workstations and peripherals to FDDI fiber optic backbone LANs.


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iii. Fiber Optic Cable FDDI The FDDI standard is actually a set of standards as established by the American National Standards Institute. FDDI is a token-based ring access method that allows station to access a cable plant that operates at 100 Mbps.

5.

Summary of Baseband standards. ª

10Base5

Information Category Transmission Speed Access Method Cable Type Max. Segment Length Max. Total Network Length Max. Length between nodes Max. # Segments Max. # of nodes/segment Max. # of nodes in a network Topology Connector Type IEEE specifications Interference Advantages Disadvantages

Specifications 10 Mbps CSMA/CD 50-ohm Thicknet coaxial 500 m (1640 ft) 2500 m (8200 ft) 2.5 m 5, only 3 populated 100 300 Bus Vampire Tap Transceiver 802.3 High resistant Long distances Expensive, difficult to install & troubleshoot


DATA COMMUNICATIONS

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ª

10Base2

Information Category Transmission Speed Access Method Cable Type Max. Segment Length Max. Total Network Length Max. Length between nodes Max. # Segments Max. # of nodes/segment Max. # of nodes in a network Topology Connector Type IEEE specifications Interference Advantages Disadvantages ª

Specifications 10 Mbps CSMA/CD 50-ohm Thinnet coaxial-RG 58A/U 185 m (607 ft) 925 m (3035 ft) 0.5 m 5, only 3 populated 30 90 Bus BNC 802.3 High resistant Simple to install, inexpensive Difficult to troubleshoot

10BaseT

Information Category Transmission Speed Access Method Cable Type Max. Segment Length Max. Total Network Length Max. Length between nodes Max. # of connected Segments Max. # of nodes/segment Max. # of nodes in a network Topology Connector Type IEEE specifications Interference Advantages Disadvantages

Specifications 10 Mbps CSMA/CD UTP 100 m (328 ft) N.A. 2.5 m (8 ft) 1024 1 1024 Star RJ-45 802.3 Low resistant Very inexpensive, simple to connect and easy Difficult to troubleshoot


ª

100-Base T LAN


ª

Specifications 100 Mbps CSMA/CD UTP Category 3-5 100 m (328 ft) N.A. 2.5 m (8 ft) 1024 1 1024 Star RJ-45 802.3 Low resistant Fast, simple to connect and easy troubleshoot Limited distance, expensive

100-Base F LAN


Specifications 100 Mbps CSMA/CD Optical Fiber 2000 m (6561 ft) N.A. N.A. 1024 1 1024 Star Specialized 802.3 Immune to EMI Fast and long distances Very expensive and difficult to install Loading ECE SUPERBook

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DATA COMMUNICATIONS

4-74 ª

Token Ring LAN

Information Category Transmission Speed Access Method Cable Type Max. Segment Length Max. Total Network Length Max. Length between nodes Max. # of connected Segments Max. # of nodes/segment Max. # of nodes in a network Topology Connector Type IEEE specifications Interference Advantages Disadvantages ª

Specifications 4 or 16 Mbps Token Passing IBM cable, Type 1 STP/Cat 3 UTP 45 m with UTP, 101 m with STP N.A. 2.5 m (8 ft) 33 hubs Depends on the hub 72 node with UTP, 260 nodes with STP Star-wired ring RJ-45/ IBM Type A 802.5 Low resistant Fast and reliable More expensive than Ethernet solutions, difficult to troubleshoot

Token Ring LAN (FDDI)

Information Category Transmission Speed Access Method Max. Segment Length Max. Total Network Length Max. Length between nodes Max. # of connected Segments Max. # of nodes/segment Max. # of nodes in a network Topology Cable Type Connector Type IEEE specifications Interference Advantages Disadvantages

Specifications 100 Mbps Token Passing N.A. 100 km N.A. N.A. N.A. 500 Ring Fiber optic Specialized No IEEE specification ANSI X3T9.5 Resistant to EMI Very fast and reliable, long distances, high secure More expensive and difficult to install


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IEEE 802 Standards i.

802.1: Architecture and Internetworking (High-Level Interface). Defines architecture layers and rules for interconnection of disparate LAN protocols. Includes data formatting, network management and internetworking.

ii.

802.2: Defines equivalent of Logical Link Control including protocol for data transfer.

services,

iii. 802.3: Defines CSMA/CD Access Method and Physical Layer specifications. iv. 802.4: Token-Passing Bus Access Method and Physical Layer specifications. v.

802.5: Token-Passing Ring Access Method and Physical Layer specifications.

vi. 802.6: Metropolitan Area Network (MAN) Access Method and Physical Layer specifications. DQDB (Distributed Queue Dual Bus) is defined. vii. 802.7: Broadband Technical Advisory Group. Standards for definition of a broadband cable plant design. Established guidelines for LAN construction within a physical facility such as a building. viii. 802.8: Fiber Optic Technical Advisory Group. Established to assess impact of fiber optics and to recommend standards. Note that this standard is distinct from that of ANSI’s FDDI. ix. 802.9: Integrated Voice and Data Networks (Isochronous Traffic) x.

802.10: Internetwork Security.

xi. 802.11: Wireless LAN. xii. 802.12: High-Speed LANs, Demand Priority Access (e.g., 100VGAnyLAN). xiii. 802.15: A communications specification that was approved in early 2002 by the Institute of Electrical and Electronics Engineers Standards Association (IEEE-SA) for wireless personal area networks (WPANs).

The initial version, 802.15.1, was adapted from the Bluetooth specification and is fully compatible with Bluetooth 1.1. xiv. 802.16: Broadband Wireless Metropolitan Area Network (WMAN).


DATA COMMUNICATIONS

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802.11xx 802.11a (Wireless ATM) 802.11b (Wireless Fidelity or Wi-FI) 802.11g (Short distance WLAN)

Operating Frequency

Logical Topology

Modulation Scheme

Accessing

Bit Rate

5.725 to 5.850 GHz

Ethernet

OFDM*

CSMA/CA

54 Mbps

2.400 to 2.4835 GHz

Ethernet

Complementary Code Keying (CCK)

CSMA/CA

11 Mbps

2.400 to 2.4835 GHz

Ethernet

OFDM

CSMA/CA

11 to 54 Mbps

*Orthogonal Frequency Division Multiplexing

Read it till it Hertz…jma 802.11x refers to a group of evolving wireless local area network (WLAN) standards that are under development as elements of the IEEE 802.11 family of specifications, but that have not yet been formally approved or deployed. As of August 2004, these incomplete standards included the following: ª

802.11e -- Adds Quality of Service (QoS) features to existing

ª

802.11f -- Adds Access Point Interoperability to existing 802.11

802.11 family specifications

ª

ª ª ª ª ª

family specifications 802.11i -- A standard for wireless local area networks (WLANs) that provides improved encryption for networks that use the popular 802.11a, 802.11b (which includes Wi-Fi, and 802.11g standards. The 802.11i standard requires new encryption key protocols, known as Temporal Key Integrity Protocol (TKIP) and Advanced Encryption Standard (AES) 802.11h -- Resolves interference issues with existing 802.11 family specifications 802.11j -- Japanese regulatory extensions to 802.11 family specifications 802.11k -- Radio resource measurement for 802.11 specifications so that a wireless network can be used more efficiently 802.11m -- Enhanced maintenance features, improvements, and amendments to existing 802.11 family specifications 802.11n -- Next generation of 802.11 family specifications, with throughput in excess of 100 Mbps


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N. .NETWORK TOPOLOGY. 1.

Bus Bus topologies are multipoint electrical circuits, which can be implemented using coax, UTP, or STP. Bus networks employ a decentralized method of media access control known as CSMA (Carrier Sense Multiple Access) that allows the attached devices to make independent decisions relative to media access and initiation of transmission.

2.

Star Star network consist of a central node, hub, or switch, to which all other devices are attached directly, generally via UTP or STP.

3.

Ring Rings generally are coax or fiber (FDDI) in nature, operating at raw transmission rates of 4, 16, 20 or 100+ Mbps. Information travels around the ring in only one direction, with each attached station or node serving as a repeater. Token-Passing Ring, IBM Token Ring, and FDDI all are based on ring topologies.


DATA COMMUNICATIONS

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O. .NETWORKING DEVICES. 1.

Bridges Bridges are relatively simple devices that are used to connect LANs of the same architecture (Ethernet-to-Ethernet). Bridges operate at the bottom two layers of the OSI model, providing Physical Layer and Data Link Layer connectivity.

2.

Hubs and Switches Hubs can be either active or passive. Passive hubs act simply as cableconnecting devices, while active hubs also serve as signal repeaters.

3.

Routers Routers are highly intelligent devices that support connectivity between both like and disparate LANs.

4.

Gateways Gateways are at the top of the LAN food chain. Gateway routers perform all of the functions of bridges and routers, including protocol conversion at all seven layers of the OSI Reference Model.

P. .INTERNETWORKING. 1.

Network Domains The Internet is divided logically into domains. Under the terms of IPv4 (Internet Protocol version 4), these are identified as a 32-bit portion of the total address. Addresses follow a standard convention, which is similar to [email protected] Domain types include the following: .com = commercial organizations .edu = educational institutions .gov = government agencies .mil = military .net = network access providers


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World Wide Web (the Web) The Web is a multiplatform web that supports multimedia communications on the basis of a Graphic User Interface (GUI). The GUI provides hypertext, which allows the user to click on a highlighted text word and search related files, across Web servers, through hot links.

For Your Information… The WorldWide Web (WWW), also known as the Web, was developed by Mr. Tim Berners-Lee at CERN, the European Laboratory for Particle Physics in Geneva, Switzerland. 3.

Home Page Home Page is multimedia informational document that may contain graphics, animated graphics, video clips, and audio clips, as well as text.

4.

Universal Resource Locators (URLs) URLs are unique WWW Addresses assigned to each Web site and to each Home Page.

5.

Search Mechanisms and Browsers i.

Archie - A corruption of archive, Archie is a FTP search mechanism first deployed in 1991. Archie allows one to search for a file (exact name unknown) on a file server (name unknown) somewhere on the Net. Archie servers contain directory listings of all such files, updated on a monthly basis through a process of file server polling.

ii.

Gopher - Developed at the University of Minnesota, where the Golden Gopher is the school mascot, Gopher is a user interface that provides easy access to server resources in educational institutions.

iii. Veronica - Veronica (Very Easy Rodent-Oriented Net-wide Index to Computerized Archives) is an Archie variation that supports an index of Gopherspace titles on which a search can be performed. The selected resources are then delivered to the user in the form of a Gopher menu iv. Jughead -Jughead (Jonzy’s Universal Gopher Hierarchy Excavation and Display) is quite similar in operation to Veronica, although it limits the search to a specific organization. Jughead also delivers a custom menu of available resources located on the basis of the keyword search.


DATA COMMUNICATIONS

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Mosaic - Mosaic is a browser developed at the National Center for Supercomputing (NCSA) at the University of Illinois UrbanaChampaign campus. Mosaic provides a consistent user interface supporting Macintosh, Microsoft Windows, and UNIX X-Windows.

vi. Netscape Navigator - Developed by Netscape Communications, Netscape Navigator was built by a team led by Marc Andersen, creator of the original Mosaic. It features simultaneous image loading and continuous document streaming speed performance. vii. Internet Explorer - Developed by Microsoft Corporation whose basic toolbar and features is similar to netscape navigator.

Q. .TCP/IP PROTOCOL SUITE.

1.

Transmission Control Protocol (TCP) TCP was developed to provide a reliable connection-oriented service that supports end-to-end transmission reliability.

2.

User Datagram Protocol (UDP) The UDP was developed to provide an unreliable, connectionless transport service.

3.

Internet Protocol (IP) IP handles software computer addresses.

4.

File Transfer Protocol (FTP) The FTP is a mechanism for moving data files between hosts via a TCP/IP network.

5.

Telnet Telnet is designed to enable client to access a remote computer as though the client were a terminal directly connected to the remote computer

6.

Simple Mail Transfer Protocol (SMTP) The SMTP provides the data transportation mechanism for electronic messages to be routed over a TCP/IP network.

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8.

9.

Hyper Text Transfer Protocol (HTTP) The HTTP represents a relatively recent addition HTTP is the protocol used by the Web browsers Web servers and vice verse. Simple Network Management Protocol (SNMP) The SMTP provides the mechanism to transport statistical information about the operation and devices.

to the TCP/IP family. to communicate with status messages and utilization of TCP/IP

Internet Control Message Protocol (ICMP) ICMP handles errors and sends error messages for TCP/IP

10. Address Resolution Protocol (ARP) ARP relates IP addresses with hardware or MAC addresses. 11. Routing Information Protocol (RIP) RIP find the quickest route between two computers. 12. Open Shortest Path First (OSPF) OSPF is a descendant of RIP that increases its speed and reliability

Read it till it Hertz…jma Tabulated below are the most common port numbers of different TCP/IP family applications.

Application

Port Number

FTP

21

Telnet

23

SMTP

25

TFTP

69

HTTP(WWW)

80

POP3

110

NetBIOS Name (UDP)

137

NetBIOS Datagram (UDP)

138

NetBIOS Session (TCP)

139

SNMP

161

13. Point-To-Point Protocol (PPP) PPP provides for dial up networked connections to networks. PPP is commonly used by ISP to allow customers to connect to their services.


DATA COMMUNICATIONS

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14. Border(Exterior) Gateway Protocol (BGP/EGP) BGP/EGP handles how data is passed between networks. 15. Post Office Protocol Version 3 (POP3) POP3 setup ways for client to connect to servers and collect email. 16. Domain Name Service (DNS) The DNS provides a very important service by enabling “near-English” names (such as ftp.xyz.com to indicate an FTP server operated by the XYZ commercial firm) to be translated into unique IP address that represents the physical address of the interface of the FTP server on a network 17. Real Time Protocol (RTP) The RTP represents a special type of protocol developed to support applications requiring the real-time delivery of data such as audio and video.

R. .TCP/IP ENCAPSULATION.

S. .IP ADDRESSING. 1.

IP address A 32 bit number that identify Internet hosts. These numbers are place in the IP packet header and are used to route packets to their destination. i.

Prefix-based addressing A basic concept of IP addressing is that initial prefixes of the IP address can be used for generalized routing decision. For example, the 1st 16 bits of an address might identify a company and the remaining bits identify a particular host on that network.


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Per-interface assignment IP addresses are assigned on a per-interface basis, so a host might posses several IP addresses if it has several interfaces. This is an important consequence of prefix-based addressing because an IP address doesn’t really refer to a host, it refer to an interface port.

iii. Multiple addresses If a host is know by multiple addresses, every service on this host can be referred to by multiple names. 2.

Types of IP address i.

Class A

ii.

Class B

iii. Class C

iv. Class D


DATA COMMUNICATIONS

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3.

CLASS E

Summary of IP Address Classes

Application

Class A

Class B

Class C

Class D

First few bits

0

10

110

1110

Number of Network bits

7

14

21

28**

Number of Host bits

24

16

8

n.a.

Initial Byte

0-127*

128-191

192-223

224-247

Number of Class

126

16,384

2,097,152

n.a.

Number of Host

16,777,214

65,532

254

n.a.

*0 and 127 are reserved.

** Class Ds are multicast addresses.

T. .COMMON INTERNET APPLICATIONS. 1.

E-mail E-mail clearly is the most popular application. More than one billion e-mail messages of over one trillion bytes (1TB) transverse the Internet each month. It is estimated that over 400 million people used e-mail by the year 2004.

2.

File Transfer File transfer is accomplished through the File Transfer Protocol (FTP). Through any of over 8,000 FTP servers, any of the thousands of Net servers and their resident file resources be accessed and transferred in ASCII or binary code from the server to the user’s computer

3.

Bulletin Board Systems Bulletin Board Systems (BBSs) are specific to a region or area of topical interest. Each supports numbers of subcategories. Network News, or Usenet News, formed in 1979, is a collection of over 6,000 special-interest Bulletin Boards (BBSs).

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Library Catalogs Library catalogs for over 300 libraries are available through the Internet. Such catalogs include the U.S. Library of Congress, the Research Libraries Information Network and many major colleges and universities.

5.

Realtime Applications Realtime applications include collaborative design and development, interactive role-playing, interactive remote education, chat lines (realtime BBSs), voice and video conferencing, network games, gambling, and radio and video broadcasting (cybercasts).

6.

Online Banking Online banking has received a good deal of interest as a result of the failed acquisition talks between Microsoft and Intuit. While security issues abound, there is little doubt that this application has great promise.

U. .MODEM STANDARDS. 1.

Bell Modems

Standard

TX Mode

Baud Rate

Bit Rate

No. of Wire

Modulatio n FSK

Bell 103

FDX

300 baud

300 bps

2-wire

Bell 202

HDX

1200 baud

1200 bps

2-wire

FSK 4-PSK

Bell 212

FDX

600 baud

1200 bps

2-wire

Bell 201

HDX/FDX

1200 baud

2400 bps

2/4wire

4-PSK

Bell 208

FDX

1600 baud

4800 bps

4-wire

8-PSK

Bell 209

FDX

2400 baud

9600 bps

4-wire

16-QAM

2.

ITU-T Modems

TX Mode

Baud Rate

Bit Rate

No. of Wire

Modulation

V.22 bis

FDX

600 baud

1200 to 2400 bps

2-wire

4-DPSK/16QAM

V.32

FDX

2400 baud

9600 bps

2-wire

32-QAM (trellis)

V.32bis

FDX

2400 baud

14400 bps

4-wire

64-QAM

V.32terbo

FDX

2400 baud

19200 bps

4-wire

256-QAM

V.33

FDX

2400 baud

14400 bps

4-wire

128-QAM (trellis)

V.34

FDX

2400 baud

28800 bps

4-wire

4096-QAM

Standard


DATA COMMUNICATIONS

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I

H

1.

When conversing with an individual whose primary language is different from yours, you many need to repeat your words and speak slower. Repeating your words can be compared to _____ and the need to speak slowly can be compared to the _____ functions of the transport layer. A. Flow Control and Reliability B. Flow Control and Transport C. Reliability and Flow Control D. Transport and Acknowledgement

2.

When you mail an unregistered package through the standard mail system you make an assumption that the person to which it is addressed receives it. This is analogous to which protocol? A. IPX B. IP C. TCP D. UDP

3.

Which layer of the OSI model provides connectivity and path selection between two end systems where routing occurs? A. Physical Layer B. Data Link Layer C. Network Layer D. Transport Layer

4.

A file transfer protocol that works over phone lines and is known for its high accuracy and slower speed? A. Veronica B. HTTP C. Kerberos D. Kermit

5.

In telecommunications, what does ICE age mean? A. Information Communication Engineering age B. Informal Communication Engineering age C. Informal Communication Entertainment age D. Information Communication Entertainment age

6.

A text and graphics formatting software that uses a coding to indicate how a received part of a document should be presented by a viewing application such as an internet web browser? A. HTML B. SMTP C. WWW D. FTP


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7.

How is Forward Error Correction implemented? A. By varying the frequency shift of the transmitted signal according to a predefined algorithm B. By transmitting blocks of 3 data characters from the sending station to the receiving station, which the receiving station acknowledges C. By transmitting a special FEC algorithm which the receiving station uses for data validation D. By transmitting extra data that may be used to detect and correct transmission errors

8.

Which layer of the OSI model is responsible for reliable network communication between end nodes and provides mechanisms for the establishment, maintenance, and termination of virtual circuits, transport fault detection and recovery, and information flow control? A. Network Layer B. Transport Layer C. Data Link Layer D. Physical Layer

9.

Which of the following best describes the function of the presentation layer? A. Responsible for reliable network communication between end nodes B. Provides connectivity and path selection between two end systems C. Concerned with data structures and negotiation data transfer syntax. D. Manages data exchange between layer entities

10. What is the maximum segment length of a 10Base2 network in an Ethernet network? A. 100 meters B. 185 meters C. 200 meters D. 500 meters 11. Which of the following UTP cables types is rated for 100 Mbps transmission? A. CAT 3 B. CAT 4 C. CAT 2 D. CAT 5 12. What is a dial-up networking multilink configuration? A. A dial-up networking connection that links your computer to two or more servers simultaneously using a single adapter B. Dial-up networking connection that uses more than physical link to increase bandwidth C. A dial-up networking connection that allows your computer to acts as a link between two servers D. A dial-up networking connection that allows your computer to acts as a link between a server and a network station. 13. What is the principal difference between asynchronous and synchronous transmission? A. The clocking is derived from the data in synchronous transmission B. The clocking is mixed with the data in synchronous transmission C. The bandwidth required is difficult D. The pulse are difficult 14. Which of the following described the very early standard that defines binary digits as space/mark line conditions and voltage levels? A. V.1 B. V.2 C. V.4 D. V.5


DATA COMMUNICATIONS

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15. The SMTP uses _____ protocol to deliver network management information. A. OSPF B. UDP C. TCP D. RIP 16. USB stands for A. Universal System Bus C. Universal Serial Bus

B. D.

Universal System Bios Universal Standard Bus

17. What functions are the data link layer concerned with? A. Physical addressing, network topology, and media access. B. Synchronizes cooperating applications, and establishes agreement on procedures for error recovery and control of data integrity. C. Manages data exchange between presentation layer entities D. Provides mechanisms for the establishment, maintenance, and termination of virtual circuits, transport fault detection, recovery, and information flow control. 18. Some protocols are considered to be technically non-routable. Which of the following statement best describes the most common reason why a protocol would be considered non-routable? A. It defines Physical layer network addresses for internal routing B. It uses advance transport layer service to move across the internet and avoid the routing overhead required by the more primitive networking protocols. C. It does not contain the appropriate datalink information required by the routers D. It does not specify the network addressees required by routers 19. Which of the following OSI layers is responsible for identifying communications partners? A. Application B. Network C. Presentation D. Session 20. Which of the following statement describes a star topology? A. Requires more cabling than a bus topology B. More difficult to troubleshoot than ring topology C. Less reliable than ring topology D. All network computers get equal network access through the use of CSMA/CD 21. What network access method should be implemented to provide centralize control of network transmission? A. Token ring B. CSMA/CD C. Demand priority D. CSMA/CA 22. Which of these hardware devices can be used to boost broadband signal strength on a cable run? A. Multiplexer B. Amplifier C. Repeater D. TDR 23. In a peer to peer network, which of the following may acts as both a client and a server? A. A dedicated server B. A peer computer C. A dedicated workstation D. A standalone computer


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24. What type of protocol is used in the selective calling system for Bell’s System? A. Synchronous Data Protocol B. Asynchronous Link Protocol C. Asynchronous Data Protocol D. Synchronous Link Protocol 25. Ethernet is baseband system using CSMA/CD operating at A. 40 Mbps B. 30 Mbps C. 10 Mbps D. 20 Mbps 26. In a _______ connection, more than two devices can share a single link. A. Secondary B. Multipoint C. Point-to-point D. Primary 27. A port address in TCP/IP is ______ bits long. A. 16 B. 32 C. 48 D. 64 28. The ____ created a model called the Open Systems Interconnection, which allows diverse systems to communicate. A. ISO B. ETSI C. IEEE D. ANSI 29. The seven-layer _____ model provides guidelines for the development of universally compatible networking protocols. A. ANSI B. ISO C. IEEE D. OSI 30. The physical, data link, and network layers are the ______ support layers. A. user B. network C. session D. remote 31. The session, presentation, and application layers are the ____ support layers. A. user B. network C. session D. remote 32. A ________ is a set of rules that governs data communication. A. forum B. protocol C. standard D. timing 33. The _______ layer changes bits into electromagnetic signals. A. Data link B. Transport C. Physical D. Network 34. The physical layer is concerned with the transmission of _______ over the physical medium. A. Bits B. Protocols C. Packets D. Frames 35. Mail services are available to network users through the _______ layer. A. Application B. Transport C. Physical D. Data link


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DATA COMMUNICATIONS

36. As the data packet moves from the lower to the upper layers, headers are _______. A. Rearranged B. Modified C. Subtracted D. Added 37. Communication between a computer and a keyboard involves ________ transmission. A. Simplex B. Half-duplex C. Duplex D. Full-duplex 38. In a network with 25 computers, which topology would require the most extensive cabling? A. Bus B. Star C. Mesh D. Ring 39. Which topology requires a central controller or hub? A. Bus B. Star C. Mesh D. Ring 40. The _______ is the physical path over which a message travels. A. Sink B. Protocol C. Medium D. Signal 41. Which organization has authority over interstate and international commerce in the communications field? A. IEEE B. FCC C. ISO D. ITU-T 42. The information to be communicated in a data communications system is the _______. A. Transmission B. Medium C. Protocol D. Message 43. Frequency of failure and network recovery time after a failure are measures of the _______ of a network. A. Security B. Feasibility C. Reliability D. Performance 44. In _______ transmission, the channel capacity is shared by both communicating devices at all times. A. Simplex B. Half-simplex C. Half-duplex D. Full-duplex 45. An unauthorized user is a network _______ issue. A. Security B. Reliability C. Performance D. Integrity 46. Which topology requires a multipoint connection? A. Ring B. Bus C. Star D. Mesh 47. A _______ connection provides a dedicated link between two devices. A. Primary B. Secondary C. Point-to-point D. Multipoint


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48. A cable break in a _______ topology stops all transmission. A. Bus B. Star C. Primary D. Mesh 49. In the original ARPANET, _______ were directly connected together. A. routers B. networks C. host computers D. IMPs 50. This was the first network. A. ANSNET C. ARPANET

B. D.

NSFNET CSNET

51. _______ are special-interest groups that quickly test, evaluate, and standardize new technologies. A. Regulatory agencies B. Forums C. Standards organizations D. ITU 52. Which agency developed standards for physical connection interfaces and electronic signaling specifications? A. ITU-T B. ISO C. EIA D. ANSI 53. In a ______ connection, two and only two devices are connected by a dedicated link. A. point-to-point B. multipoint C. internet D. ethernet 54. In a ________ connection, three or more devices share a link. A. point-to-point B. multipoint C. internet D. ethernet 55. ______ refers to the physical or logical arrangement of a network. A. Data flow B. Segment C. Mode of operation D. Topology 56. A _______ is a data communication system within a building, plant, or campus, or between nearby buildings. A. LAN B. MAN C. WAN D. WPAN 57. A ______ is a data communication system spanning states, countries, or the whole world. A. LAN B. MAN C. WAN D. WPAN 58. ________ is a collection of many separate networks. A. a LAN B. A WAN C. An internet D. An intranet 59. When data are transmitted from device A to device B, the header from A's layer 4 is read by B's _______ layer. A. Data link B. Network C. Session D. Transport


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DATA COMMUNICATIONS

60. Which layer functions as a liaison between user support layers and network support layers? A. Transport layer B. Application layer C. Physical layer D. Network layer 61. What is the main function of the transport layer? A. Node-to-node delivery B. Process-to-process delivery C. Updating and maintenance of routing tables D. Synchronization 62. The Internet model consists of _______ layers. A. 4 B. 5 C. 6 D. 7 63. The process-to-process delivery of the entire message is the responsibility of the _______ layer. A. Application B. Network C. Transport D. Physical 64. The _______ layer is the layer closest to the transmission medium. A. Network B. Physical C. Transport D. Data link 65. Why was the OSI model developed? A. The rate of data transfer was increasing exponentially B. Standards were needed to allow any two systems to communicate C. Manufacturers disliked the TCP/IP protocol suite D. None of the above 66. The _______ model shows how the network functions of a computer ought to be organized. A. Kennelly-Heaviside B. ANSI C. ISO D. OSI 67. The OSI model consists of _______ layers. A. 4 B. 5 C. 6 D. 7 68. In the OSI model, as a data packet moves from the lower to the upper layers, headers are _______. A. rearranged B. modified C. removed D. added 69. In the OSI model, when data is transmitted from device A to device B, the header from A's layer 5 is read by B's _______ layer. A. transport B. physical C. presentation D. session 70. In the OSI model, encryption and decryption are functions of the ________ layer. A. transport B. session C. application D. presentation


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71. When a host on network A sends a message to a host on network B, which address does the router look at? A. port B. physical C. logical D. MAC 72. To deliver a message to the correct application program running on a host, the _______ address must be consulted. A. port B. physical C. IP D. TCP 73. IPv6 has _______ -bit addresses. A. 32 C. 128

B. D.

64 256

74. The ______ layer is responsible for moving frames from one hop (node) to the next. A. physical B. data link C. network D. transport 75. The ______ layer adds a header to the packet coming from the upper layer that includes the logical addresses of the sender and receiver. A. physical B. data link C. network D. transport 76. The_________ layer is responsible for the delivery of a message from one process to another. A. physical B. data link C. network D. transport 77. The Internetworking Protocol (IP) is a ________ protocol. A. reliable B. connection-oriented C. both A and B D. none of the above 78. _______ is a process-to-process protocol that adds only port addresses, checksum error control, and length information to the data from the upper layer. A. TCP B. IP C. UDP D. RIP 79. __________ provides full transport layer services to applications. A. TCP B. UDP C. ARP D. IP 80. The ________ address, also known as the link address, is the address of a node as defined by its LAN or WAN. A. logical B. port C. physical D. hole 81. Ethernet uses a ______ physical address that is imprinted on the network interface card (NIC). A. 32-bit B. 6-byte C. 64-bit D. 32-byte


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DATA COMMUNICATIONS

82. As the data packet moves from the upper to the lower layers, headers are _______. A. Rearranged B. Modified C. Subtracted D. Added 83. The _______ layer lies between the network layer and the application layer. A. Application B. Transport C. Physical D. Data link 84. Layer 2 lies between the physical layer and the _______ layer. A. Data link B. Network C. Session D. Transport 85. _______ is the protocol suite for the current Internet. A. DOS B. BIOS C. TCP/IP D. UNIX 86. _______ refers to the structure or format of the data, meaning the order in which they are presented. A. Syntax B. Semantics C. Timing D. Flowchart 87. ________ defines how a particular pattern to be interpreted, and what action is to be taken based on that interpretation. A. Syntax B. Semantics C. Timing D. Flowchart 88. _______ refers to two characteristics: when data should be sent and how fast it can be sent. A. Syntax B. Semantics C. Timing D. Flowchart 89. The _______ layer coordinates the functions required to transmit a bit stream over a physical medium. A. network B. transport C. data link D. physical 90. The unit of information at the network layer. A. frames B. packet C. bits D. segment

91. Bit synchronization is a topic associated with which of the following layers? A. Session B. Transport C. Data Link D. Physical 92. A mesh network-________. A. consist of a bus and drop cables connecting all devices in a network B. Has a point-to point connections between every device in the network C. Is relatively easy to install and reconfigure D. Is difficult to troubleshoot


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93. Which is true of asynchronous bit systems? A. They use continuous signals to transmit bits B. They use intermittent signals to transmit bits C. They are commonly used for mainframe-to mainframe communications D. They require an external clocking device 94. Which method is associated with media access? A. asynchronous connection B. dialog control C. message switching D. polling 95. A method that senses the cable prior to transmission, detects collision, and initiates retransmission is ______. A. Contention C. CSMA/CD

B. Polling D. Flow control

96. Data at the Data Link layer is called a ______ A. packet B. datagram C. frame D. token 97. The layer that establishes and maintains the link for transmitting data frames is ___. Session Data Link-MAC

A. C.

B. Network D. Data Link-LLC

98. A message that indicates the reception of a frame is called a(n) ______. A. Handshake B. Datagram C. Reply D. Acknowledgement 99. A method that relies upon transmitting and receiving devices to maintain their own internal clock is_______. synchronous Isochronous

A. C.

B. bisynchronous D. asynchronous

100. The switching technique that connects the sender and the receiver by a single path for the duration of a conversation called _______. Virtual pocket switching B. Datagram packet switching Circuit switching D. Message switching

A. C.

101. Which is a benefit of full –duplex dialog? A. Both ends can transmit at the same time B. Requires only one channel for both transmission and reception C. Requires inexpensive hardware D. Permits broad area coverage 102. When several small computers provide the necessary processing task , the model is called _______. A. Central computing C. Centralized processing

B. Distributive processing D. Distributed computing


DATA COMMUNICATIONS

4-96

103. When two or more computers accomplish the same processing task, the model is called _______. Collaborative computing Central processing

A. C.

B. Distributed processing D. Multitasking

104. A network that spans the earth is called ______. A. An enterprise network B. A wide area network C. A metropolitan area network D. A global network 105. All networks require the following three elements: network services, a transmission media, and _____. Wires and connectors Protocols

A. C.

B. Communications software D. Network operating systems

106. With respect to network computing, clients _____. A. Can request and provide services B. Are only allowed to request services from others C. Are the network users D. Are allowed only to provide services 107. A network, which places restrictions upon a station that may make requests or service the network, is called _____. A. Server-centric B. C. Peer-to-peer D.

User-server Restricted network

108. Protocols are ______. A. A complete set of rules and standards that enable different devices to hold B. C. D.

conversations Computer programs that enable users to access networks in an orderly fashion Rules that describe the hardware configuration of a network Rules required to help entities communicate or understand each other

109. The networking device that extends the maximum distance of a network by connecting separate network segments is called a ______. A. Modem B. Codec C. Router D. Bridge

110. A modem is a _____. A. Device that converts computer digital signals to an analog signals to use B. C. D.

with telephone lines Device that can receive and transmit electromagnetic signals across the transmission media Device that converts analog signals into digital signals Connectivity device between computers

111. A device that combines two or more signals on a single transmission medium is called a _____. A. Multiplexer C. Router

B. Bridge D. Codec


4-97

112. The device that connects two or more logically separate networks is a _____. A. Brouter B. Router C. Bridge D. Repeater 113. The device that prepares electric pulse signals for transmission on WAN transmission media is called _____. A. Modem C. Codec

B. CSU/DSU D. Bridge

114. Which recommendation specifies how messages are stored and forwarded between disparate devices on a computer internetwork? RS-232 B. X.500 X.400 D. ASN.1

A. C.

115. With what layer is the IEEE 802.2 standard associated? A. Physical B. Data link C. Network D. Session 116. Which IEEE 802.3 standard is referred to as Thin Ethernet? A. 10Base5 B. 10BaseT C. 10Base2 D. 10BaseF 117. Which IEEE standard was created to satisfy the LAN needs of industrial automation? A. IEEE 802.4 C. IEEE 802.6

B. IEEE 802.5 D. IEEE 802.7

118. Which specification is based upon the IBM Token-Ring specification? A. IEEE 802.4 B. IEEE 802.5 C. IEEE 802.6 D. IEEE 802.7 119. Which of the following methods is addressed by SDLC? A. Route selection B. Network layer translation C. Token passing D. Flow control 120. What is the maximum number of nodes per segment on a 10BaseT network? A. 3 B. 1 C. 5 D. 1024 121.

The initial MPEG standard (MPEG1) was targeted at A. cellcast-quality video and audio B. VHS–quality video and audio C. DVD-quality video and audio D. broadcast-quality video and audio

122. A. C.

The basic connection unit in an ATM network is known as the virtual public connection B. private connection virtual channel connection D. virtual path connection


DATA COMMUNICATIONS

4-98 123.

In ATM networks all information is formatted into fixed-length cells consisting of _____ bytes. A. 48 B. 53 C. 64 D. 144

124. A. C.

Which method is better suited to handle bursty traffic? circuit switching B. IP TCP D. ATM

125.

_______ is the transmission of computer files from a single site to many sites around the country or around the world. A. File delivery B. FTP C. Store-and-forward D. SMTP

126.

________ applications involve storing or saving a large file and forwarding or delivering that file to the intended audience for use at a later time or date. A. Hold-and-Forward B. Real-time C. Store-and-forward D. Satellite-based

127.

The ______ standard was developed to help minimize congestion when streaming bandwidth-intensive content over local-area networks (LANs) to the desktop. A. Application B. IP multicast C. Satellite D. LAN

128.

_________ applications involve the distribution of time-sensitive content, such as video and/or audio programs and interactive services, to the end user. A. Real-time B. IP–based C. Satellite-based D. Store-and-forward

129. A. C.

Analog signals can be ________ by combining them with a carrier frequency. transported B. carried mixed D. multiplexed

130.

_____ is a high-performance switching and multiplexing technology that utilizes fixed-length packets to carry different types of traffic. A. ATM B. ADSL C. SONET D. SDH

131.

Which feature expanded the maximum span of an Ethernet beyond 3 km? A. Cat-5 unshielded twisted pairs B. bridge ports supporting full-duplex transmission C. 100-megabit (fast-Ethernet) transmission D. high-power lasers

132.

The combination of VDSL and _____ delivers the Internet services of today in an architecture that is capable of supporting tomorrow's applications. A. ATM B. X.25 C. Frame Relay D. IP


Which error detection method uses a quasi-"division" algorithm to generate the error character? A. hamming code B. parity C. cyclic redundancy check (CRC) D. longitudinal redundancy check (LRC)

134. A. C. 135.

4-99

How many Hamming bits are inserted into a 100 bit message? 6 B. 7 8 D. 100

To correct a character with a bad parity bit, the receiver will invert the parity bit ignore the bad character request that the bad character (and possibly, the rest of the message) be retransmitted D. invert the bit that is bad A. B. C.

136. A. C.

Which topology experiences contention for access to the network? ethernet B. bus ring D. star

137.

At which OSI level is a standard description of a modem and its specifications found? A. physical B. application C. data link D. network

138.

In which OSI model level would be found specifications for virtual routing throughout a network? A. application B. network C. datalink D. session

139.

Which of the following encoding types uses the first half of the bit time to hold the complement of the data and the second half to provide for a clock transition? A. non-return to zero mark inverted (NRZ-AMI) B. Bipolar AMI C. Manchester D. return to zero mark inverted (RZ-AMI)

140. A. C. 141.

How many consecutive ones is recognized by SDLC as an abort sequence? 5 B. 6 7 - 14 D. greater than 14

In the SDLC address field the address specifies A. secondary station only B. primary and secondary stations C. destination and source stations D. send and receive stations


DATA COMMUNICATIONS

4-100 142.

An SDLC information control field is recognized by A. a one in the most significant bit B. a SYN sequence C. a zero in the least significant bit D. a DC3 character

143.

The size of a payload for an ATM cell is A. 53 bytes long B. 48 bytes long C. dependent on the upper level protocol transported D. variable

144.

A business housed entirely in a single building would use a ______ type network to interconnect its computers and terminals. A. MAN B. LAN C. WAN D. WMAN A. C.

Which data link protocol is the basis for the Ethernet data link protocol? POP3 B. BISYNC HDLC D. SDLC

A. C.

How long is an untappable Ethernet segment? 1000 m B. 2500 m 1500 m D. 10 m

A. C.

Ethernet sends and receives data at a rate of 2500 Kbps B. 10 Mbps 256 Kbps D. 1500 bps

A. C.

Which service is NOT included on an ISDN line? teleconferencing B. video voice D. all of these

145.

146.

147.

148.

149.

Which term refers to the process of converting a frame or packet into ATM cells? A. reassembly B. adaption C. segmentation D. mapping

150.

Which of these responsibilities does NOT apply to the ATM cell (data link) level? A. flow control B. cell multiplexing C. segmentation and reassembly D. header generation A. C.

Which specification standard incorporates the Ethernet standard? EIA's RS232C B. CCITT's X.25 IBM's Token Ring D. IEEE's 802.3

A. C.

Which type of error detection is used with SDLC/HDLC? Hamming code B. parity checksum D. CRC

151.

152.


4-101

153.

SDLC is a(n) _______ data link protocol A. bit oriented, frame format B. character oriented, packet format C. bit oriented, message format D. character oriented, block format

154.

How many consecutive HDLC frames can be sent without an acknowledgment or response? A. 128 B. 7 C. 56 D. 1544

155.

Which type of infrared connection is used for workstations that are relatively close to each other? A. direct connection B. line of sight C. room roaming D. frequency hopping

156.

Which method is utilized by wireless networks to overcome voice priority in channel use? A. frequency hopping B. direct sequencing C. channel sealing D. code division multiplexing

157.

What is the limitation for parity as an error detection method? a bad parity bit does not guarantee that there is something wrong in the character received B. different parity errors occur if you are using an even or odd parity system C. only reliable if a single error occurs in a character D. none of these apply A.

158. A. C.

Which of the following error methods is used for error correction? Checksum B. LRC Parity D. CRC

159.

How many bit errors within a single transmission can be corrected by using the Hamming code for error correction? A. 2 B. 1 per character C. 1 D. as many errors as are present

160.

Which of these services is provided by TELNET? A. remote database access B. dial up access to Internet C. file transfer D. all of these services are provided by TELNET

161. A. C. 162.

Which OSI level is the TCP part of TCP/IP closely related to? presentation B. transport application D. network

What does the acronym TCP/IP stand for? A. top core protocol / international protocol B. telephone communications policy / internet procedure C. traffic communications protocol / information protocol D. transport control protocol / internet protocol


DATA COMMUNICATIONS

4-102 163.

What is the purpose of Internet Relay Chat (IRC)? A. interactive "real time" access to other users B. video teleconferencing C. e-mail transfers D. web page browsing

164.

Which of these can NOT be used to access the Internet? A. on-line service B. actuated line program grid (ALPG) C. Internet service provider (ISP) D. serial line interface protocol (SLIP)

165.

URL addresses identify all except the following A. e-mail server B. physical location of user C. user's network D. ISP server

166.

What is the unit designation for the area managed by a single cellular phone central office? A. spot beam B. cluster C. cell D. sector

167.

What type of station is connected to both transmission paths of a FDDI network? A. back up station B. dual attached station C. mirrored station D. single attached station

168. A. C.

What is the basic topology of the FDDI network? tree B. star bus D. ring

169.

Who controls the Internet? A. Microsoft B. International Telecommunications Union C. government D. no one

170.

Which of these is NOT a problem with using symmetric private key cryptography? A. brute force attack can be used to uncover key B. transporting key between users C. decrypting long messages takes too much time D. loss or theft of key nullifies security

171.

Firewalls are used for A. creating Intranets B. preventing access by an outside user C. limiting access to certain URLs D. all of these

172. A. C.

The digital URL identifies networks and host nodes location of ISP

B. D.

faulty servers type of protocol used

Section 17 Antenna Fundamentals Section 18 Radio-wave Propagation Section 19 Microwave Engineering Section 20 Satellite Communications Section 21 Cellular Communication System

Wireless Communications Loading ECE SUPERBook

5-1


Section

Antenna

17

Fundamentals


Antenna is basically a transducer that converts electrical alternating current oscillations at an RF frequency to an electromagnetic wave of the same frequency.

DEFINITION.

DEFINITION. Radiation Resistance is a resistance that, if replaced the antenna, would dissipate exactly the same amount of power the antenna radiates. DEFINITION. Effective Isotropic Radiated Power (EIRP) is defined as equivalents transmit power that an isotropic antenna would have to radiate to achieve the same power density in the chosen direction at a given point as another antenna. A. .ANTENNA RECIPROCITY. States that antenna characteristic are essentially the same regardless of whether an antenna is transmitting or receiving electromagnetic energy

B. .ANTENNA FUNDAMENTALS.

A complete antenna system consists of three parts: 1. 2. 3.

Coupling Device - The coupling device (coupling coil) connects the transmitter to the feeder Feeder - The feeder is a transmission line that carries energy to the antenna. Antenna - The antenna radiates this energy into space.


ANTENNA FUNDAMENTALS

5-2

C. .CURRENT AND VOLTAGE DISTRIBUTION ON AN ANTENNA. The principles of radiation of electromagnetic energy are based on two laws: ª

A moving electric field creates a magnetic (H) field.

ª

A moving magnetic field creates an electric (E) field.

D. .ANTENNA PARAMETERS. 1.

Radiation Pattern A polar diagram or graph representing field strengths or power densities at various angular positions relative to an antenna.

2.

Antenna Polarization Polarization refers to the orientation of the electric field vector in space.

3.

Near Field and Far Field Region i. Near Field The term near field refers to the field pattern that is close to the antenna.

FN =

A πD2 = 2λ 8λ


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Far Field The term far field refers to the field pattern that is at great distance from the antenna where power that reach this region continues to radiate outward and is never returned to the antenna.

FF =

2D2 λ

That region of the field of an antenna where the angular field distribution is essentially independent of the distance from a specified point in the antenna region. For Your Information…

Near field is sometimes called the induction field or Fresnel field. Far field is sometimes called the radiation field or Fraunhofer field.

4.

Radiation Resistance Radiation Resistance is a resistance that, if replaced the antenna, would dissipate exactly the same amount of power the antenna radiates. Antenna Efficiency (η)

η=

Prad x 100% Pin

η=

R rad x 100% R rad + R loss

Sample Problem:

Calculate the efficiency of a dipole antenna that has a radiation resistance of 67Ω and a loss resistance of 5Ω, measured at the feed point.

Solution: η =

R rad 67 x 100 % = 93 .05 % x 100 % = 67 + 5 R rad + R loss



5-4

5.

Antenna Gain ª

Directive Gain The ratio of the power density radiated in a particular direction to the power density radiated to the same point by a reference antenna.

D=

ª

Pd(dir) Pd(iso)

Power Gain The ratio between the amounts of energy propagated in these directions compared to the energy that would be propagated if the antenna were not directional is known as its gain.

G = Dxη

Sample Problem:

Calculate the gain of a certain antenna relative to a dipole antenna with a gain of 5.3 dB with respect to an isotropic radiator. Also compute for the power gain if the antenna has an efficiency of 95%.

Solution:

D dBd = D dBi − 2.14 dB = 5.3 − 2.14 = 3.16 dBd

6.

G = Dxη 3.16 ) 10 = 1.97 = 2.94 dB

= 0.95 x log− 1(

Effective Isotropic Radiated Power (EIRP) EIRP is defined as equivalents transmit power that an isotropic antenna would have to radiate to achieve the same power density in the chosen direction at a given point as another antenna.

EIRP = Ptx Gtx

EIRPdB = Ptx(dB) + Gtx(dB)

5-5

Self-Sufficient Guide to ECE by JASON AMPOLOQUIO Effective Radiated Power (ERP)

ERPdB = EIRP(dB) − 2.14 dB

7.

Captured Area & Captured Power i.

Captured Area (Effective Area)

AC =

ii.

Gλ 2 4π

Captured Power (Received Power)

Pr = Pd x A c

⎛ λ ⎞ Pr = Pt Gtx Grx ⎜ ⎟ ⎝ 4π ⎠

2

In dB

Pr(dB) = EIRPdB − FSL dB + Grx(dB)

Sample Problem:

Calculate the captured power 10 km away from a half-wave dipole transmitter with 10 W transmit power for the following antenna at 150 MHz; a. Hertzian dipole b. Half-wave dipole

Solution: a.

For Hertzian dipole (Elementary doublet)

Pr =

b.

Pt G tx 4 πd

2

x

10 (1 . 64 ) λ2 G rx 2 2 (1 . 5 ) x = = 6.23 nW 2 4π 4π 4 π (10 ,000 )

For Half-wave dipole (Hertz antenna) Pr =

Pt G tx

4 π d2

x

10 (1 . 64 ) λ2 G rx 2 2 (1 . 64 ) x = = 6.81 nW 4π 4π 4 π (10 ,000 )2



5-6

Summary for various antennas

Unitless Gain

Effective Area

1

λ2 4π

1.5

1.5λ2 4π

Half-Wave Dipole

1.64

1.64λ2 4π

Turnstile

1.15

1.15λ2 4π

Horn with mouth area A

10A

Parabola with face area A

7A

Antenna Types Isotropic Elementary Doublet or Loop

λ2

λ2

0.81A

0.56A

8.

Front-to-Back Ratio The ratio of the energy radiated in the principal direction compared to the energy radiated in the opposite direction for a given antenna.

9.

Antenna Beamwidth The angular separation between the two half-power points on the major lobe of an antenna’s plane radiation pattern.

E. .ANTENNA FEED MECHANISM. 1.

Center fed antenna If energy is applied at the geometrical center of an antenna, the antenna is said to be center-fed.

2.

Voltage-fed antenna If energy is applied to the point of high voltage along the antenna length, the antenna is voltage-fed.

3.

Current-fed antenna If energy is applied to the point of high circulating current along the antenna length, the antenna is said to be current-fed.


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F. .ANTENNA GROUNDING SYSTEM. The ground screen and the counterpoise are used to reduce losses caused by the ground in the immediate vicinity of the antenna. 1.

Counterpoise Counterpoise consists of a structure made of wire erected at short distance above the ground and insulated from the ground.

2.

Earth Mat (Groundscreen) A network of up to 120 buried wires 15 to 30 cm below the ground under the antenna. Each radial wire has a length which should be at least λ/4, and preferably λ/2.

The ground screen is buried below the surface of the earth. The counterpoise is installed above the ground.

G. .ANTENNA LOADING. Antenna loading is the method used to change the electrical length of an antenna. This keeps the antenna in resonance with the applied frequency. It is accomplished by inserting a variable inductor or capacitor in series with the antenna. 1.

Inserting an inductor in series with a short antenna compensate its capacitive reactance thus effectively increasing its electrical length and bandwidth, while the resonant frequency decreases.

2.

Inserting a capacitor in series with a long antenna compensate its inductive reactance thus effectively decreasing its electrical length and bandwidth, while the resonant frequency increases.

3 General Types a.

Top loading - the loading component is attached at the top of the antenna structure.

b.

Center loading - the loading component is placed along the antenna structure, approximately half-way between the feedpoint and the end.

c.Base loading - the loading component is located at the bottom of the antenna structure.



5-8

H. .BASIC ANTENNAS. 1.

Elementary Doublet The elementary doublet is an electrically short dipole and is often referred to simply as Hertzian dipole. Field strength at a distance

ξ=

60 πI λd

e

where: ξ = induce field strength in V m I = antenna current in A e

= effective antenna length

λ 10 d = distance from the antenna in m ≅

2.

Half-Wave Dipole The half-wave dipole is a resonant antenna, the total length of which is nominally λ/2 at the carrier frequency.

Voltage, Current & Impedance Distribution of Half-wave dipole Voltage is maximum at the ends. ----------------------------------------------Current is maximum at the feedpoint. ----------------------------------------------Impedance is maximum at the ends.

If the dipole is not driven at the centre then the feed point resistance will be higher. If the feed point is distance x from one end of a half wave (λ/2) dipole, the resistance will be described by the following equation. where:

75 Rx = 2 ⎛ πx ⎞ sin ⎜ ⎟ ⎝ λ ⎠

R x = Radiation resistance in Ω λ = signal wavelength in m x = distance from one end of the dipole m

5-9


Sample Problem: Calculate the radiation resistance of a half-wave dipole antenna if the feedpoint is 0.25 m from one end at 300 MHz. Solution: Rx =

75 75 = = 150 Ω ⎛ πx ⎞ 2 ⎛ 0 .2 5 π ⎞ sin ⎜ sin ⎟ ⎜ ⎟ 1 ⎝ λ ⎠ ⎝ ⎠ 2

Sample Problem: Calculate the radiation resistance of a half-wave dipole antenna 0.1 m from one end at 300 MHz. Solution: Rx =

3.

75 75 = = 785 . 4 Ω πx ⎞ 2 ⎛ 0 .1π ⎞ sin ⎜⎜ sin ⎜⎜ ⎟ ⎟ ⎟ ⎟ ⎝ λ ⎠ ⎝ 1 ⎠ 2⎛

Quarter-Wave (Marconi or Long-wire) antenna The quarter wavelength vertical antenna is basically a half-wave dipole placed vertically, with the “other half” of the dipole being the ground (image).

Antenna Types

Exact Formula

Considering the End Effect (F=0.95)

Quarter-Wave Dipole (Marconi or Long wire)

L ft =

246F fMHz

L ft =

234 fMHz

Half-Wave Dipole (Hertz antenna)

L ft =

492 F fMHz

L ft =

468 fMHz

where: F=velocity factor Loading ECE SUPERBook


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ECE Board Exam: APRIL 2005 What is the actual length in feet of one-half wavelength of a coax with velocity factor of 0.63 at 28 MHz? Solution: L ft =

492F f MHz

=

492(0.63) 28

= 11.07 ft

ECE Board Exam: APRIL 2003 What is the actual length in feet of one-quarter wavelength of a coax with velocity factor of 0.695 at 42 MHz? Solution: L ft =

I.

246F f MHz

=

246(0.695) 42

= 4.07 ft

.ANTENNA ELEMENT. 1.

Driven element A driven element obtains its power directly from the transmitter or, as a receiving antenna; it delivers the received energy directly to the receiver.

2.

Parasitic element A parasitic element is located near the driven element from which it gets its power. It is placed close enough to the driven element to permit coupling. ª

When it operates to reinforce energy coming from the driver toward itself, the parasitic element is referred to as a director.

ª

If a parasitic element is placed so it causes maximum energy radiation in a direction away from itself and toward the driven element, that parasitic element is called a reflector. Note:

If all of the elements in an array are driven, the array is referred to as a Driven Array (sometimes as a connected array). If one or more elements are parasitic, the entire system usually is considered to be a Parasitic Array.

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.SPECIAL PURPOSE ANTENNA. 1.

Folded Dipole antenna The folded dipole is essentially a single antenna made up of two elements. One element is fed directly (driven element), whereas the other is conductively (parasitic element) coupled at the ends.

2.

Yagi-Uda antenna A Yagi antenna is a linear array consisting of a dipole and two or more parasitic elements.

0.55λ

Antenna Element

0.5 λ

For your information… Yagi-Uda antenna was invented by Hidetsugi Yagi and Shintaro Uda hence the name YagiUda…

0.45λ

Exact Formula

Considering the End Effect (F=0.95)

Director

0.45λ

0.451λ

Driven element

0.5λ

0.475λ

Reflector

0.55λ

0.49875λ ≈ 0.5λ



5-12

Sample Problem: A Yagi-Uda antenna is designed to receive signals centered at 174 MHz. Calculate the length of the driven element, reflector, and director. Solution: a. For the driven element Length

=

⎧⎪ 3 x 10 8 λ = 0 .5 ⎨ 2 ⎪⎩ 174 x 10 6

⎫⎪ ⎬ = 0 . 86 m ⎪⎭

Considering End effect Length = 0.5λ − 5%(0.5λ) = 0.475λ = 0.86 − (0.05 x 0.86) = 0.817 m

b.

For the reflector ⎧⎪ 3 x 108 ⎫⎪ Length = 0.55λ = 0.55⎨ ⎬ = 0.948 m ⎪⎩174 x 106 ⎪⎭

Considering End effect Length = 0.475 λ + 5%(0.475λ) = 0.49875 λ ≈ 0.5λ = 0.817 + (0.05 x 0.817) = 0.8599 m

c.

For the director ⎧⎪ 3 x 108 ⎫⎪ Length = 0.45λ = 0.45⎨ ⎬ = 0.776 m ⎪⎩174 x 106 ⎪⎭

Considering End effect Length = 0.475λ − 5%(0.475 λ) = 0.451λ = 0.817 − (0.05 x 0.817) = 0.776 m

3.

Turnstile antenna

The turnstile antenna is formed by placing two dipoles at right angles to each other, 90° out of phase. The radiation pattern is the sum of the radiation patterns from the two dipoles, which produces a nearly omnidirectional pattern.

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Loop antenna A closed-circuit antenna, that is, one in which a conductor is formed into one or more turns so its end are close together.

Parameter

Equation

ν = 2 πfBAN

Induced Voltage Induced Voltage (Tuned Loop Antenna)

ν max = Qν

B = magnetic flux density in Teslas A = loop area in m2 N = number of turns in the loop Q = qualitiy factor

5.

Helical antenna A circularly polarized antenna that is wound into a helix. S D

Two Principal Modes

1.

In the normal (broadside) mode or, the dimensions of the helix are small compared with the wavelength. The far field radiation pattern is similar to an electrically short dipole or monopole.

These antennas tend to be inefficient radiators and are typically used for mobile communications where reduced size is a critical factor. Loading ECE SUPERBook


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2.

In the axial (endfire) mode, the antenna produces true circular polarization.

These antennas are best suited for space communication, where the orientation of the sender and receiver cannot be easily controlled, or where the polarization of the signal may change. i. Gain of a helical antenna

G=

N = # of turns, 3 to 20 turns

15NS(πD)2 λ3

S = pitch or spacing D = helix diameter

ii. 3dB Beamwidth (ϕ)

ϕ=

52λ λ πD NS

Sample Problem: Calculate the gain and beamwidth of a helical antenna if the optimum diameter is 80 mm, pitch of 62.5 mm, with eight turns and will operate at 1.2 GHz. Solution: ⎧⎪ 15 NS ( π D ) 2 G = 10 log ⎨ ⎪⎩ λ3 ϕ =

52 λ πD

⎫⎪ ⎧⎪ 15 ( 8 )( 0 . 0625 )( π x 0 . 08 ) 2 ⎬ = 10 log ⎨ ⎪⎭ 0 . 25 3 ⎩⎪ λ NS

=

52 ( 0 . 25 ) π x 0 . 08

0 . 25 8 x 0 . 0625

⎫⎪ ⎬ = 14 . 8 dB ⎪⎭

= 36 . 6 °

K. .MULTI-ELEMENT ARRAYS. 1.

Multi-element arrays frequently are classified according to their directivity. i.

Bidirectional array A bidirectional array radiates in opposite directions along the line of maximum radiation.


2.

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Unidirectional array A Unidirectional array radiates in only one general direction. Arrays can be described with respect to their radiation patterns and the types of elements of which they are made.

In Terms of Direction of Radiation i.

Collinear array The collinear array antenna usually uses two or more wire halfwave dipoles mounted end-to-end. The pattern radiated by the collinear array is similar to that produced by a single dipole. The addition of two or more radiator, however, tends to intensify the pattern.

ii.

Broadside array A broadside array designates an array in which the direction of maximum radiation is perpendicular to the plane containing these elements. In actual practice, this term is confined to those arrays in which the elements themselves are parallel, with respect to each other.



5-16

iii. End-fire array An end-fire array is one in which the principal direction of radiation is along the plane of the array and perpendicular to the elements. Radiation is from the end of the array, which is the reason this arrangement is referred to as an end-fire array.

iv. Log-Periodic antenna The log-periodic antenna derives its name from the fact that the feedpoint impedance is a periodic function of the operating frequency.

γ=

L1 L 2 L 3 = = =… L2 L 3 L 4

γ=

d1 d2 d3 = = =… d2 d3 d4

⎛ L ⎞ α = 2 tan−1 ⎜ 1 ⎟ ⎝ 2d1 ⎠ γ = design constant (typically between 0.7 to 0.9) α = design angle in degrees L1 ,L2

= element length from shortest to longest

d1 , d2

= spacing between elements from the apex


.PARABOLIC REFLECTOR ANTENNA. A form of a dish antenna with a reflecting surface that is a geometric paraboloid. Parabolic reflectors resemble the shape of a plate or dish; therefore, they are sometimes called parabolic dish antennas or simply dish antennas.

Parameter

Equation

Aperture Number

f D

Angular Aperture

2ψ

Relation between Angular aperture and Aperture number

f ⎛ψ⎞ = 0.25 cot ⎜ ⎟ D ⎝2⎠ A eff = η x A face

Effective Area

⎛λ⎞ ϕ = 70 ⎜ ⎟ ⎝D⎠ ⎛λ⎞ ϕo = 2ϕ = 140 ⎜ ⎟ ⎝D⎠

Beamwidth Beamwidth between nulls

Gain of a Parabolic Reflector antenna

G=

Areaeffective(parabolic ) Areaeffective(isotropic )

=

πd2 2 4 = η ⎛ πD ⎞ ⎜ ⎟ 2 λ ⎝ λ ⎠ 4π

ηx

where: η = aperture or illumination efficiency = 0.5 to 0.75 (0.55 typical) D = mouth or face diameter in m


5-17

5-18


When η=55% ⎛D⎞ G = 5.4 ⎜ ⎟ ⎝λ⎠

When η=60%

2

⎛D⎞ G = 6⎜ ⎟ ⎝λ⎠

2

GdB = −42.22 + 20 log ( fMHz x Dm )

GdB = −52.6 + 20log ( fMHz x Dft )

GdB = 17.78 + 20 log ( fGHz x Dm )

GdB = 7.5 + 20 log ( fGHz x D ft )

Sample Problem: Calculate the directive gain and beamwidth between nulls for a paraboloidal reflector antenna with a mouth diameter of 2.4m and the illumination efficiency is 0.55 operating at 6 GHz. Solution: Directive Gain (D) D =

4π λ

2

xA

eff

=

4π λ

2

⎛ πD 2 x η⎜ ⎜ 4 ⎝

⎛ π x 2 .4 2 ⎞ 4π ⎟ = x 0 . 55 ⎜ 2 ⎟ ⎜ 4 0 . 05 ⎠ ⎝

⎞ ⎟ ⎟ ⎠

= 12 ,507 D dB ≅ 10 log D = 10 log 12 ,507 = 41 dB

Beamwidth between nulls

ϕo = 2ϕ = 140

λ ⎛ 0.05 ⎞ = 140⎜ ⎟ = 2.92° D ⎝ 2.4 ⎠

Sample Problem: To minimize interference, a 500-MHz dish needs to have a 1° beamdwith. What diameter dish is required, in wavelength and meters? Solution: ϕ =

⎡ 3 x 10 8 ⎤ 70 λ 70 λ 70 λ ⇒ D = = = 70 λ = 70 ⎢ ⎥ = 42 m D 1° ϕ ⎢⎣ 500 x 10 6 ⎦⎥


5-19

M. .HORN ANTENNA. Horn radiator is a tapered termination of a length of waveguide that provides the impedance transformation between waveguide and free space impedance.

dE

dH

Parameter Gain

General Solution G=

7.5dEdH λ2

H-plane Beamwidth

θH =

70λ dH

E-plane Beamwidth

θE =

56λ dE

Sample Problem: Calculate the gain, beamwidth in the E and H plane of a pyramidal horn antenna that has an aperture of 60 mm in the E-plane, 80 mm in the Hplane and operating at 6 GHz. Solution: 7 . 5 x 0 . 06 x 0 . 08 = = 14 . 4 ⇒ 11 . 58 dB λ2 0 . 05 2 56 x 0 .05 70 x 0 .05 56 λ 70 λ θE = = = 46 .67 ° θH = = = 43 .75 ° dE 0 .06 dH 0 .08 G =

7 . 5 dE dH


5-20


N. .SUMMARY OF VARIOUS ANTENNAS. ª

Half-wave dipole antenna

Antenna

Radiation Pattern

.Elevation Pattern.

.Azimuth Pattern.

ª

Parameter

Typical Values

Gain

2.14 dB

Bandwidth

10% (1.1:1)

Frequency Limit

10 Hz to 8 GHz (due to size)

Polarization

Linear (vertical as shown)

Half-Power Beamwidth

80º x 360º

Vertical Whip antenna

Antenna

Radiation Pattern

Parameter

Typical Values

Gain

2 to 6 dB

Bandwidth

10% (1.1:1)

Frequency Limit

None

Polarization



45º x 360º

.Elevation Pattern.

.Azimuth Pattern.

5-21


ª

Linear Dipole Array antenna

Antenna

Radiation Pattern

.Elevation Pattern.

.Azimuth Pattern.

ª

Parameter

Typical Values

Gain

Dependent on the number of elements

Bandwidth

Narrow

Frequency Limit

10 MHz to 10 GHz

Polarization



Related to gain

Yagi-Uda antenna

Antenna

Radiation Pattern

.Elevation Pattern.

.Azimuth Pattern.

Parameter

Typical Values

Gain

5 to 15 dB

Bandwidth

5% (1.05:1)

Frequency Limit

50 MHz to 2 GHz

Polarization

Linear (horizontal as shown)


50º x 50º


5-22


ª

Log-Periodic antenna

Antenna

Radiation Pattern .Elevation Pattern.

.Azimuth Pattern.

ª

Parameter

Typical Values

Gain

6 to 8 dB

Bandwidth

163% (10:1)

Frequency Limit

3 MHz to 18 GHz

Polarization



60º x 80º

Circular Loop

Antenna

Radiation Pattern

.Elevation Pattern.

.Azimuth Pattern.

Parameter

Typical Values

Gain

-2 to 2 dB

Bandwidth

10% (1.1:1)

Frequency Limit

50 MHz to 1 GHz

Polarization


Half-Power eamwidth

80º x 360º

5-23


Square Loop

Antenna

Radiation Pattern

.Elevation Pattern.

.Azimuth Pattern.

ª

Parameter

Typical Values

Gain

1 to 3 dB

Bandwidth

10% (1.1:1)

Frequency Limit

50 MHz to 1 GHz

Polarization



100º x 360º

Alford Loop

Antenna

Radiation Pattern

.Elevation Pattern.

.Azimuth Pattern.

Parameter

Typical Values

Gain

-1 dB

Bandwidth

67% (2:1)

Frequency Limit

100 MHz to 2 GHz

Polarization



80º x 360º


5-24


ª

Axial Mode Helical antenna

Antenna

Radiation Pattern

Parameter

Typical Values

Gain

10 dB

Bandwidth

52% (1.7:1)

Frequency Limit

100 MHz to 3 GHz Circular (Lefthand as shown )

Polarization Half-Power Beamwidth

ª

50 º x 10º

Normal Mode Helical antenna

Antenna

Radiation Pattern

.Elevation Pattern.

.Azimuth Pattern.

Parameter

Typical Values

Gain

0 dB

Bandwidth

5% (1.05:1)

Frequency Limit

100 MHz to 3 GHz

Polarization

Circular (with an ideal pitch-todiameter ratio)


60º x 360º

5-25


Horn antenna

Antenna

Radiation Pattern

.Elevation Pattern.

.Azimuth Pattern.

ª

Parameter

Typical Values

Gain

5 to 20 dB

Bandwidth

120% (4:1) --ridged 67% (2:1) nonridged

Frequency Limit

50 MHz to 40 GHz

Polarization



40º x 40º

Horn antenna (with Polarizer)

Antenna


.Azimuth Pattern.

Parameter

Typical Values

Gain

5 to 10 dB

Bandwidth

60% (2:1)

Frequency Limit

2 to 18 GHz

Polarization

Circular (depends on polarizer)


40º x 40º


5-26


ª

Discone antenna

Antenna


.Azimuth Pattern.

ª

Typical Values

Gain

0 to 4 dB

Bandwidth

100% (3:1)

Frequency Limit

30 MHz to 3 GHz

Polarization



20 to 80º x360º

Parabolic antenna

Antenna

ª

Parameter

Radiation Pattern

Parameter

Typical Values

Gain

10 to 60 dB

Bandwidth

33% (1.4:1)

Frequency Limit

400 MHz to +13 GHz

Polarization

Polarization of feed


1 to 10º

V-antenna

Antenna

Radiation Pattern

Parameter

Typical Values

Gain

2 to 7 dB

Bandwidth

Broadband

Frequency Limit

3 to 500 MHz

Polarization



60º x 60º

5-27


Rhombic antenna

Antenna

ª

Radiation Pattern

Typical Values

Gain

3 dB

Bandwidth

Broadband

Frequency Limit

3 to 500 MHz

Polarization



60º x 60º

Biconical antenna

Antenna

Radiation Pattern

Parameter

.Elevation Pattern.

.Azimuth Pattern.

ª

Parameter

Typical Values

Gain

0 to 4 dB

Bandwidth

120% (4:1)

Frequency Limit

500 MHz to 40 GHz

Polarization



20 to 100º x360º

Corner Reflector antenna

Antenna

Radiation Pattern

Parameter

Dependent upon feed emitter

Typical Values

Gain

10 dB above feed

Bandwidth

Narrow

Frequency Limit

1 GHz to 40 GHz

Polarization

Feed dependent


40º x variable


5-28


ª

Biconical antenna (with Polarizer)

Antenna

Radiation Pattern

.Elevation Pattern.

.Azimuth Pattern.

ª

Parameter

Typical Values

Gain

-3 to 1 dB

Bandwidth

100% (3:1)

Frequency Limit

2 GHz to 18 GHz

Polarization

Circular (Direction depends on polarization)


20 to 100º x360º

Guide Fed Slot antenna

Antenna


.Azimuth Pattern.

Parameter

Typical Values

Gain

0 dB

Bandwidth

Narrow

Frequency Limit

2 GHz to 40 GHz

Polarization



45º x50º (elevation) 80º (azimuth)

5-29


Cavity Backed Circuit Fed Slot and Microstrip Patch antenna

Antenna

Radiation Pattern .Elevation and. .Azimuth Pattern.

Parameter

Typical Values

Gain

6 dB

Bandwidth

Narrow

Frequency Limit

50 MHz to 18 GHz

Polarization



80º x 80º

O. .RADIATION PATTERN OF VARIOUS DIPOLE ANTENNA. Notice the development of lobes as the antenna length increases…

0.5λ

1.0λ

0.75λ

1.25λ


5-30


1.5λ

1.75λ

2.0λ

2.25λ

2.5λ

2.75λ

3.0λ

5-31


I

H

1.

What aperture number correspond to an angular aperture of 55° A. 0.35 B. 0.72 C. 0.66 D. 0.48

2.

What are the theoretical element lengths and overall boom length for a threeelement Yagi designed for 220 MHz operation with element spacing of 0.18λ? A. driver=0.682 m, director=0.648 m, reflector=0.716, boom=0.49 m B. driver=0.648 m, director=0.682 m, reflector=0.49, boom=0.716 m C. driver=0.716 m, director=0.49 m, reflector=0.648, boom=0.682 m D. driver=0.49 m, director=0.648 m, reflector=0.682, boom=0.49 m

3.

Calculate the antenna power gain for a beamwidth of 10 degrees. APRIL 2005 A. 20 dB B. 2.6 dB C. 26 dB D. 2.0 dB

4.

A helix antenna has its pitch halved while everything else remains unchanged. What is the change in gain, in dB? In beamwidth? In axis length? A. G=-3dB, φ=0.5x, L=0.5 B. G=3dB, φ=1.4x, L=1.5 C. G=-3dB, φ=1.4x, L=1.5 D. G=-3dB, φ=1.4x, L=0.5

5.

Calculate the gain and beamwidth of a parabolic antenna that has a diameter of 3 m, and an efficiency of 60%. A. 40 dB, 1.75° B. 45 dB, 4.91° C. 32 dB, 3.33° D. 38 dB, 6.12°

6.

In high frequency radio transmission, the lower the radio frequency the _________of the antenna. A. smaller the diameter B. shorter the length C. bigger the diameter D. longer the length

7.

Determine from the following an advantage of using a trap antenna. A. Has high gain B. Can be used for multiband operation C. Minimizes harmonic radiation D. It has directivity in the high-frequency bands

8.

Type of antenna which is normally used for satellite tracking services. A. Omni B. Yagi C. Helical D. Dipole


5-32

9.

ANTENNA FUNDAMENTALS Find the approximate width of a pyramidal horn antenna that will operate at 10,000 MHz intended to have a beamwidth of 10°. April 2000 A. 14 cm B. 34 cm C. 24 cm D. 44 cm

10. Calculate the approximate physical length for a half wave dipole operating on a frequency of 7.25 MHz. A. 13.66 ft B. 73.86 ft C. 64.55 ft D. 33.81 ft 11. An antenna with a length of 6 cm radiates a 12-cm wavelength signal. Calculate the near field distance. A. 2 cm B. 6 cm C. 12 cm D. 16 cm 12. An increase in the effective power radiated by an antenna in a certain desired direction at the expense of power radiated in other directions. A. Antenna efficiency B. Antenna gain C. Antenna total ratio D. Antenna back lobe ratio 13. Two wires that are bent 90 degrees apart. A. Dipole B. C. Hertz D.

Log-periodic Rhombic

14. Gain of Hertz antenna. A. 1.64 C. 1.76

1.5 4.24

B. D.

15. A TVRO installation for use with a C-band satellite (downlink frequency at 4 GHz), has a diameter of about 3.5 meters and an efficiency of 60%. Calculate its beamwidth. NOV 2003 A. 5.05° B. 11.7° C. 1.5° D. 3.2° 16. Electric circuit designed specifically to radiate the energy applied to it in the form of electromagnetic waves, normally used to transmit and receive radio waves. A. Bandwidth B. Antenna C. Wavelength D. Radiator 17. What is the reason why 5/8 whip vertical antenna is a advantage than quarter wave vertical whip antenna? A. Can handle high power signal B. Flexible C. More gain D. Less than interference 18. Which of the following antenna where its beamwidth is determine by the dimensions of its horn, lens or reflector? A. Long wire antenna B. Aperture antenna C. Whip antenna D. Aperiodic antenna


5-33

19. What is the effective radiated power (ERP) of a repeater with 450 W transmitting power output, 4 dB feedline loss, 6 dB duplexer loss, 7 dB circulator loss, and feedline antenna gain of 25 dB? APRIL 2004 A. 3980 W B. 2839.31 W C. 3141.98 W D. 1139.8 W 20. What is meant by the term radiation resistance for an antenna? A. An equivalent resistance that would dissipate the same amount of power as that radiated from an antenna B. The resistance in the trap coils to received signal C. Losses in the antenna elements and feed line D. The specific impedance of the antenna 21. Where does the maximum current and minimum voltage values on a resonant Hertz dipole exist? A. Ends of the antenna B. Near the center of the antenna C. Near the end of the antenna D. Center of the antenna 22. In a VHF mobile radio system, the base station transmits 100 W at 160 MHz frequency using half-wave dipole antenna 20 meters above ground. Calculate the field strength at a receiving antenna at a height of 4 meters and a distance of 30 km. (Gain of antenna is 1.64) Nov 2004 A. 161.58 μV/m B. 15.45 μV/m C. 41.78 μV/m D. 34.13 μV/m 23. Radiation pattern of a discone. A. Bidirectional C. Omnidirectional

B. D.

24. Which is properly terminated antenna? A. Rhombic B. C. Marconi D.

Figure eight Unidirectional Dipole Hertz

25. A 400 feet antenna is to be operated at 6200 kHz. What is the wavelength at this frequency? NOV 2004 A. 52.16 m B. 41.58 m C. 48.39 m D. 33.56 m 26. Which of the following antenna feedline can easily be buried underground for a distance without adverse effects? A. Twisted pair B. Twin lead C. Coaxial cable D. Waveguide 27. Top loading is used in an antenna in order to increase its A. Beamwidth B. Bandwidth C. Input capacitance D. Effective height 28. Unity gain antenna. A. Dummy C. Isotropic

B. D.

Half-wave dipole Rhombic


5-34


29. In order to fully utilize the _____ a quarter-wave vertical antenna should require an excellent ground system. A. troposcatter effect B. ground effect C. mirror image principle D. reflected wave 30. A microwave communications uses plane reflectors as passive repeaters. The diameter of the parabolic antenna is 18 ft while the effective area is 310 ft2. Determine the reflector coupling factor. APRIL 2005 A. 0.76 B. 0.92 C. 0.906 D. 0.706 31. One of the following is a type of radio antenna used in a transmission and reception of microwave signals. A. whip antenna B. horn antenna C. l-type antenna D. Discone antenna 32. What is the phase separation of two antennas 3/8 wavelength apart? NOV

2004 A. C.

135 degrees 165 degrees

B. D.

185 degrees 195 degrees

33. The antenna property of interchangeability for transmitting and receiving electromagnetic energy under the same radio frequency is referred to as A. back-to-back B. reciprocity C. maximum energy D. re-transmission 34. What is the term for the ratio of the radiation resistance of an antenna to the total resistance of the system? A. Antenna efficiency B. Beamwidth C. Radiation conversion loss D. Effective radiated power 35. ______ is a device that detects both vertically and horizontally polarized signals simultaneously. A. Light transducer B. Orthomode transducer C. Optoisolator D. Crystal 36. What is the effective height of a broadcast antenna if the voltage induced in the antenna is 2.7 volts and the measured field intensity received in the antenna site is 27 mV/m? APRIL 2005 A. 29.7 meters B. 72.9 meters C. 100 meters D. 127 meters 37. An antenna which is not resonant at a particular frequency and so can be used over a wide band of frequencies is called A. boresight B. aperiodic C. top-loaded D. cassegrain 38. A helical antenna has the following dimensions: pitch is 1/3 wavelength, diameter is ¼ wavelength and turns is 27. What is the gain in dB? NOV 2004 A. 10.34 dB B. 16.8 dB C. 19.2 dB D. 13.5 dB


5-35

39. What is the usual electrical length of a driven element in a HF beam antenna? A. ½ wavelength B. ¼ wavelength C. ¾ wavelength D. 1 wavelength 40. What is an antenna element that is not directly connected to the feed line, but which affects the radiation pattern and feed-point impedance of the antenna? A. Balun B. Driven element C. Parasitic element D. Radiator 41. What is the purpose of the two yagi beam antenna on a VHF TV receiver? A. For receiving low and high band stations B. To increase selectively C. To increase selectively on both hands D. Used to receive VHF and UHF stations 42. A 500 kHz antenna radiates 500 W of power. The same antenna produces field strength equal to 1.5 mV/m. If the power delivered by the antenna is increases to 1kW, what would be the expected field intensity? NOV 2003 A. 1.66 mV/m B. 2.12 mV/m * C. 8.45 mV/m D. 5.58 mV/m 43. What is a parasitic element of an antenna? A. An element that receives its excitation from mutual coupling rather than from a transmission line B. An element polarized 90 degrees apposite the driven element C. An element dependent on the antenna structure for support D. A transmission line that radiates radio frequency energy 44. Determine the approximate length of a simple whip quarter wave antenna at UHF frequency of 450.25 MHz. April 2000 A. 3.47 ft B. 13.69 ft C. 1.89 ft D. 0.55 ft 45. What is the electric field strength 10 km away in the direction of maximum radiation from a half-wave dipole that is fed, by means of lossless, matched line, by a 15 W transmitter? A. 27.1 mV/m B. 271 mV/m C. 0.271 mV/m D. 2.71 mV/m 46. Calculate the power density 25 km away from a 90% efficient half-wave dipole if the transmit power is 125 W. A. 23.5 nW B. 318.5 nW C. 51.7 nW D. 931.35 nW 47. A dipole antenna requires to be feed with 20 kW of power to produce a given signal strength to a particular distant point. If the addition of reflector makes the same field strength available with an input power of 11 kW. What is the gain in dB obtained by the use of the reflector? (Gain referred to this particular dipole) Nov 1996 A. 2.6 dB B. 4.12 dB C. 1.6 dB D. 1.76 dB


5-36


48. Calculate the operating frequency for a helical antenna that consists of 10 turns with a spacing of 10 cm and a diameter of 12.7 cm. A. 0.75 MHz B. 750 MHz C. 7.5 MHz D. 75 MHz 49. A type of antenna that has physical length equivalent to one half wavelength of the radio frequency being used. A. Yagi B. Hertz C. Rhombic D. Marconi 50. Calculate the gain, with respect to an isotropic antenna, of a half-wave dipole with an efficiency of 90% that is placed one-quarter wavelength from a plane reflector that reflects 100% of the signal striking it. A. 12.46 dBi B. 4.8 dBi C. 7.68 dBi D. 8.39 dBi 51. Due to the presence of parallel LC networks in the trap antenna, one of the following is a disadvantage of using this kind of antenna. A. Reduce power B. Radiate harmonics C. Allow entry of interference D. Reduced beamwidth 52. How can the approximate beamwidth of a rotatable beam antenna be determined? A. Measure the ratio of the signal strengths of the radiated power lobes from the front and near of the antenna B. Note the two points where the signal strength of the antenna is down 3 dB from the maximum signal point and compute the angular difference C. Measure the ratio of the signal strengths of the radiated power lobes from the front and side of the antenna D. Draw two imaginary lines through the ends of the elements and measure the angle between the lines 53. What is the approximate length of a half-wave dipole antenna radiating at 6450 kHz? Nov 2000 A. 72.5 ft B. 136.6 ft C. 13.73 ft D. 112.47 ft 54. What is the length of a simple quarter wave antenna using a VHF frequency of 150.55 MHz? April 2000 A. 5.82 ft B. 1.55 ft C. 4.97 ft D. 3.52 ft 55. What happens to gain and beamwidth if the number of turns is doubled when the pitch is halved? A. 0.5x B. unchanged C. 4x D. 2x 56. This is flexible vertical rod antenna commonly used on mobiles. A. Hertz B. Whip C. Ground plane D. Marconi


5-37

57. Calculate the distance from the apex of the 1st and 2nd dipole element of a logperiodic antenna to cover the low VHF TV band, from 54-88 MHz. Use α=30°, γ=0.7 and L1=1.5 m A. 2.8 m, 4 m respectively B. 4 m, 2.8 m respectively C. 4.8 m, 2 m respectively D. 2 m, 8 m respectively 58. When testing transmitter to prevent interfering with other stations, which type of antenna must be used? A. Dipole B. Hertzian antenna C. Void antenna D. Dummy antenna 59. In consists of a number of dipoles of equal size, equally spaced along a straight line with all dipoles fed in the same phase from the source. A. Broadside array B. Log-periodic antenna C. Yagi antenna D. End-fire array 60. A Yagi antenna has a gain of 10dBi and a front-to-back ratio of 15 dB. It is located from a transmitter with an ERP of 100 kW at a frequency of 10 MHz. The antenna is connected to a receiver via a matched feedline with a loss of 2 dB. Calculate the signal power supplied to the receiver if the antenna is pointed directly away from the transmitting antenna. A. 82.8 μW B. 8.28 μW C. 828 μW D. 0.828 μW 61. Radio wave concentration in the direction of the signal emitted by a directional antenna. A. Major lobe radiation B. Back lobe radiation C. Transmitted signal D. Side lobe radiation 62. Referred as an effect of parasitic elements in an antenna system. A. Makes the antenna isotropic B. Makes the antenna omnidirectional C. Increase its directivity D. Decrease its directivity 63. A dish designed for operation at 150 MHz is operated at twice that frequency. By what factors do gain and beamwidth change? A. G=2x, φ=2.5 B. G=2x, φ=0.5 C. G=4x, φ=1.5 D. G=4x, φ=0.5 64. What is the gain of four identically polarized antennas stacked one above the other and fed in phase? A. 10 dB the gain of one antenna B. 4 dB over the gain of one antenna C. 3 dB over gain of one antenna D. 6 dB over the gain of one antenna 65. What do you mean by the outward flow of an energy from any source in the form radio waves. A. Encoding B. Radiation C. Emission D. Tracking


5-38


66. What is the driven element of an antenna? A. Always the forward most element B. The element fed by the transmission line C. Always the rearmost element D. The element connected to the rotator 67. Theoretical gain of a hertzian dipole. A. 1.76 dB C. 0 dB

B. D.

2.14 dB 1 dB

68. Best description of a collinear and broadside antenna radiation pattern. A. Unidirectional B. Perfect circle C. Omnidirectional D. Bidirectional 69. To minimize interference, a 500-MHz dish needs to have a 1° beamdwith. What diameter dish is required, in wavelength and meters? A. 50λ B. 70λ C. 60λ D. 80λ 70. One of the following refers to the usage of a small loop antenna in radio communications. A. For mobile communications B. For spread spectrum communications C. For direction finding purposes D. For satellite use 71. Determine the dB gain of a receiving of antenna which delivers a microvolt signal to a transmission line over that of an antenna that delivers a 2 microvolt signal under identical circumstances. A. 3 dB B. 6 dB C. 8 dB D. 10 dB 72. How can the antenna efficiency of an HF grounded vertical antenna be made comparable to that of a half-wave antenna? A. By shortening the vertical B. By length the vertical C. By installing a good ground radial system D. By isolating the coax shield from ground 73. A horizontal antenna is A. perpendicularly polarized C. centrally polarized

B. D.

horizontally polarized vertically polarized

74. A parabolic dish has a diameter of 10λ. What is the gain, in dB? What is the beamwidth? If the dish is built for operation at 200 MHz, what would be the actual diameter A. G=27.8 dB, φ=0.7°, D=15 m B. G=27.8 dB, φ=7°, D=1.5 m C. G=27.8 dB, φ=7°, D=15 m D. G=2.8 dB, φ=7°, D=15 m 75. The A. B. C. D.

capture area of a antenna is directly proportion to the Power density of the signal gain of the antenna distance between transmitter and receiver frequency of the received signal


5-39

76. What property of an antenna which is a reason for its efficiency in receiving and transmitting the intended electromagnetic energy and is physically identical to its physical configuration? A. Resonance B. Direction C. Orientation D. Polarization 77. What is the effective area of a Hertzian dipole operating at 30 MHz. A. 16.4 m2 B. 11.9 m2 C. 17.38 m2 D. 27.18 m2 78. Known as the technique for adding a series inductor at or near the center of an antenna element in order to cancel the capacitive reactance of an antenna A. Center loading B. Reflector C. Loading coil D. Dipole 79. A helical antenna is used for satellite tracking because of A. maneuverability B. good front-to-back C. circular polarization D. broad bandwidth 80. In a VHF mobile radio system, the base station transmits 100 W at 150 MHz, and the antenna is 20 m above ground. The transmitting antenna is a halfwave dipole. Calculate the field strength at a receiving antenna of height 2m at a distance of 40 km. A. 1.1 μV/m B. 101 μV/m C. 0.11 μV/m D. 11 μV/m 81. What is the estimated medium wind loading in the Philippines for antenna tower design? A. 160 kph B. 180 kph C. 200 kph D. 240 kph 82. The standard reference antenna for the directive gain is A. infinitesimal dipole B. elementary doublet C. half-wave dipole D. isotropic antenna 83. One of the following is referred to as a way of increasing the electrical length of an antenna. A. By adding a capacitor in series B. By adding an inductor in parallel C. By adding an inductor in series D. By adding a resistor in parallel 84. Find the power received by a Hertz antenna if the 150 MHz transmitter output power is 50 kW. The transmitter is 100 km distant and also uses a Hertz antenna. A. 574.3 nW B. 340.6 nW C. 810.9 nW D. 22.16 nW 85. Determine the dB gain of a receiving antenna which delivers a microvolt signal to a transmission line over that of an antenna that delivers a 2 microvolt signal under identical circumstances. Nov 1998 A. -5 dB B. -8 dB C. -6 dB D. -3 dB


5-40


86. What makes an antenna physically long but electrically short? A. Adding C in series B. Adding L in series C. Top loading D. All of these 87. What is the front-to-back ratio for an antenna with a forward gain of +20 dB and a backward gain of +15 dB? A. -5 dB B. 0 dB C. 5 dB D. 10 dB 88. A type of a grounded antenna is a/an _______ antenna. A. Parabolic B. Marconi C. Hertz D. Isotropic 89. What is the term for an antenna element which is supplied power from a transmitter through a transmission line? A. Parasitic element B. Director element C. Driven element D. Reflector element 90. In the propagation of radio signal through an aerial, at what angle does magnetic field are positioned with reference to the direction of the propagation? A. 180° B. 90° C. 0° D. 45° 91. Determine the gain of a certain antenna that will produce an effective area 199 m2 at 30 MHz? A. 38.47 dB B. 26.37 dB C. 13.98 dB D. 71.16 dB 92. Calculate the open-circuit voltage induce in a half-wavelength dipole when 10 W at 150 MHz is radiated from another half-wavelength dipole 50 km distant. The antennas are positioned for optimum transmission and reception. A. 282 μV B. 2.82 μV C. 0.282 μV D. 28.2 μV 93. How do you compare the length of the reflector element of a parasitic beam antenna with that of its driven element? A. Reflector element is 15% shorter B. Reflector element is 2% shorter C. Reflector element is half shorter D. Reflector element is 5% longer 94. What do you call the LC networks inserted in an antenna element to provide multiband operation? A. Loading coil B. Reflector C. Traps D. Director 95. Where does the voltage node of a half-wave antenna exist? A. At feed point B. Near the center C. At the end D. Near the feed point 96. The minimum number of turns a helix antenna must have A. 3 B. 4 C. 5 D. 6


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97. Which of the following type of radio communication antenna does not deliver an omni-directional radiation pattern? A. Whip B. Discone C. Yagi D. vertical dipole 98. A helical antenna has 20 turns of diameter λ/4, pitch of λ/4. What are the gain and beamwidth? If the frequency of operation is 250 MHz, what are the actual dimensions of turn diameter, pitch, and length of the helix axis? A. G=16.6 dB, φ=2.6°, D=0.3 m. S=0.5 m, L=6 m B. G=16.6 dB, φ=29.6°, D=0.3 m. S=0.3 m, L=6 m C. G=16.6 dB, φ=29.6°, D=0.5 m. S=0.3 m, L=5 m D. G=166 dB, φ=2.6°, D=0.3 m. S=0.3 m, L=6 m 99. Two half-wave dipoles are separated by 50 km. They are aligned for optimum reception. The transmitter feeds its antenna with 10 W at 144 MHz. Calculate the received power. A. 296 pW B. 38 pW C. 26 pW D. 96 pW 100. A. C.

Antenna impedance is similar to ______ in fiber optics numerical aperture B. refractive index critical angle D. acceptance angel

101.

A directional variation of the dipole with parasitic elements added with functionality similar to adding a reflector and lenses (directors) to focus a filament lightbulb A. Loop antenna B. Patch antenna C. Yagi-uda antenna D. Beverage antenna

102.

Two wires pointed in opposite directions arranged either horizontally or vertically, with one end of each wire connected to the radio and the other end hanging free in space A. quad antenna B. helical antenna C. loop antenna D. dipole

103. A. C. 104.

An array of square loops that vary in size log-periodic antenna B. patch antenna broadside antenna D. quad antenna

A one A. C.

105. A. C.

form of directional long-wire antenna which uses a resistive termination at end and feed from the other Hertz antenna B. Beverage antenna Helical antenna D. Phased array antenna

Antenna that is used by submarines when submerged synthetic aperture antenna B. phased array antenna Beverage antenna D. trailing wire antenna


Radio-wave propagation

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Section

Radio-Wave

18

Propagation


Free-Space Path Loss is often defined as the loss incurred by an electromagnetic wave as it propagates in a straight line through a vacuum with no absorption or reflection of energy from nearby objects.

DEFINITION.

DEFINITION. Power Density is the rate at which the energy passes through a given surface area in free space. A. .ELECTRO-MAGNETIC WAVES. 1.

Transverse Electro-Magnetic Waves (TEM)

Quantities of interest: ª The period, which is the time (T) in which one complete vibratory cycle of events occurs, ª The frequency of vibration (f), which is the number of cycles taking place in one second, and ª The wavelength, which is the distance the disturbance travels during one period of vibration 2.

Characteristic Impedance i.

Free-space

Ζ=

ξ Η

Ζ=

μ0 ε0

Ζ = 120πΩ ≅ 377Ω

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Non-magnetic medium

Ζ=

377

Ζ=

εr

120 π k

where: ε r = relative permittivity k = dielectric constant Relation ∴ ε r = k

Sample Problem:

Find the characteristic impedance of polyethylene, which has a dielectric constant of 2.3.

Solution: Ζ =

2.

120 π 2 .3

= 248 . 6 Ω

Relation between Electric & Magnetic Field Strength

ξ = 120 π Η

where: Η = magnetic field strength in A

m

B. .FREE SPACE PROPAGATION PARAMETERS. 1.

Power Density Rate at which the energy passes through a given surface area in free space. i.

For isotropic source (omnidirectional source)

Pd =

Ptx

4 πd2

where: d = distance from the source in m



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ii.

For anisotropic source (directional source)

Pd =

Ptx Gtx

where :

4 πd2

G tx = gain of the TX antenna source

iii. Relation between power densities and distances 2

⎛d ⎞ Pd2 = Pd1 x ⎜ 1 ⎟ ⎝ d2 ⎠


Calculate the power density 25 km away from a 90% efficient half-wave dipole if the transmit power is 125 W.

Solution: Pd = =

Ptx G tx

∴ G tx = η G = 0 .9 x 1 .64 = 1 .476

4 πd2 125 x 1 .476

4 π(25,000 )2

= 23 .49

nW m2

Sample Problem:

At 20 km in free-space from a point source, the power density is 200μW/m2. What is the power density 25 km away from this source?

Solution: ⎛d Pd2 = Pd1 x ⎜⎜ 1 ⎝ d2

⎞ ⎟ ⎟ ⎠

2

⎛ 20 ⎞ = 200 x 10 − 6 ⎜ ⎟ ⎝ 25 ⎠

2

= 128

μW m2

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Electric Field Intensity The intensity of the electric and magnetic fields of an electromagnetic wave propagating in free space.

In terms of Power density and impedance

Pd Ζ

ξ=

In terms of Power density and Magnetic field Intensity

ξ=

In terms of output power and distance

ξ=

30Ptx

Pd Η

30Ptx Gtx

ξ=

d

d

For VHF Propagation (Line-Of-Sight) where:

⎡ 4 πhtxhrx ⎤ ξ = ξo ⎢ ⎥ 2 ⎣ λd ⎦

ξo =

30Ptx Gtx 1m

ECE Board Exam: Nov 2003

Determine the electric field strength 10 km away from a half-wave dipole transmitter with an input power of 100 W?

Solution: For Anisotropic antenna ξ =

30PtxGtx d

=

30(100)(1.64) 10 x 103 m

=7

mV m


In a VHF mobile radio system, the base station transmits 100 W at 160 MHz frequency using half-wave dipole antenna 20 meters above ground. Calculate the field strength at a receiving antenna at a height of 4 meters and a distance of 30 km. (Gain of antenna is 1.64)

Solution: ⎡ 4 πhtxhrx ⎤ ξ = ξo ⎢ ⎥ 2 ⎣ λd ⎦

∴λ =

c = 1 .875 m , ξ o = f

30Ptx G tx 1m

= 70 .143

V m

, G tx = 1 .64

⎡ ⎤ 4 π x 20 x 4 = 70 .143 ⎢ ⎥ 2 ⎣⎢ 1 .875 x (30,000 ) ⎦⎥ = 41 .786

nV m



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Free-Space Path Loss FSL is often defined as the loss incurred by an electromagnetic wave as it propagates in a straight line through a vacuum with no absorption or reflection of energy from nearby objects.

Sample Problem:

Calculate the power delivered to the receiver, assuming free-space propagation for a transmitter that has a power output of 12W at a carrier frequency of 0.28 GHz. It is connected by 10m of a transmission line having a loss of 3dB/100m to an antenna with a gain of 12dBi. The receiving antenna is 20 km away and has a gain of 8dBi. There is negligible loss in the receiver feedline, but the receiver is mismatched; the antenna and line are designed for a 50Ω impedance, but the receiver input is 75Ω.

Solution:

Prx(dB ) = Ptx(dB ) + G tx(dB ) + G rx(dB ) − (FSL dB + L tl + L mismatch )

Computing all the losses; FSL = 92 .4 + 20 log( 0 .28 x 20) = 107 .36 dB; α 3dB xA = x 10 m = 0 .3 dB; A 100 m ⎡ ⎛ Z − ZO = 10 log(1 − Γ 2 ) = 10 log ⎢1 − ⎜⎜ L ⎢ ⎝ ZL + Z O ⎣

L tl = Lm

Prx(dB )

⎡ = 10 log ⎢1 − ⎢ ⎣ = 10 .8 + 12

⎞ ⎟ ⎟ ⎠

2⎤

⎥ ⎥ ⎦

2 ⎛ 75 − 50 ⎞ ⎤ ⎜ ⎟ ⎥ = 0 .18 dB ⎝ 75 + 50 ⎠ ⎥⎦ + 8 − (107 .36 + 0 .3 + 0 .18) = −77 .04 dB

C. .POLARIZATION. The physical orientation of electric field vector in space. 1.

Linear The electric field vectors are oriented in the same direction in space. i.

Vertical Polarization The electric field progress in the direction of propagation in the vertical plane.


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The vertical or y-component of electric field vector is not equal to zero while the horizontal or x-component is equal to zero.

ξ y ≠ 0, ξ x = 0

ii.

Horizontal Polarization The electric field progress in the direction of propagation in the horizontal plane.

The horizontal or x-component of electric field vector is not equal to zero while the vertical or y-component is equal to zero.

ξ x ≠ 0, ξ y = 0

2.

Circular The electric field vector rotates in circular fashion as it progress in the direction of propagation.

The horizontal or x-component of electric field vector is equal to the vertical or ycomponent but 90° out-of-phase forcing the electric field to rotate in the direction of propagation.

ξx = ξy = 0

3.

i.

Right-Handed Circular Polarization The electric field vector rotates in a clockwise manner as it recedes in the direction of propagation.

ii.

Left-Handed Circular Polarization The electric field vector rotates in a counter-clockwise manner as it recedes in the direction of propagation.

Elliptical A variation of circular polarization where the electric field rotates in elliptical fashion in the direction of propagation.



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The horizontal or x-component of electric field vector is not equal to the vertical or y-component but 90° out-of-phase forcing the electric field to rotate in the direction of propagation.

ξx ≠ ξy ≠ 0

4.

i.

Right-Handed Elliptical Polarization The electric field vector elliptically rotates in a clockwise manner as it recedes in the direction of propagation.

ii.

Left-Handed Elliptical Polarization The electric field vector elliptically rotates in a counter-clockwise manner as it recedes in the direction of propagation.

Random The electric field radiation has no definite pattern or orientation.

D. .EFFECTS OF THE ENVIRONMENT. 1.

Attenuation Reduction in power density w/ distance.

In terms of Power Densities

α = −10 log

2.

℘1 ℘2

In terms of their relative distance from the source

α = 20 log

d2 d1

Absorption The reduction in the intensity of radiated energy within a medium caused by converting some or all of the energy into another form.

Some of the energy of the electromagnetic waves is transferred to the atoms and molecules of the atmosphere thereby causing some radio waves to be absorbed. 3.

Interference Phenomena that occurs when two radio waves that left one source and traveled different path arrive at a point.


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E. .OPTICAL PROPERTIES OF RADIO WAVES. 1.

Reflection Phenomenon of wave motion, in which a wave is returned after impinging on a surface. When wave energy traveling from one medium encounters a different medium, part of the energy usually passes on while part is reflected.

Bouncing of radio waves as it hit a reflective surface. 2 General Types of Reflection i.

Specular Reflection (Mirror-like) Reflection phenomenon where the wave hits a smooth surface and obeys the law of reflection.

θinc = θref ii.

Diffuse The reflected signal are scattered in different direction after it hits a rough surface.

Rayleigh Criterion A wave impinging on a semi-rough surface satisfy the Rayleigh criterion the reflected wave will be reflected as if the wave phenomenon is specular reflection.

cos θinc >

2.

λ 8h

where: h = irregularities height in m

Refraction The change in direction that occurs when a wave of energy such as light passes from one medium to another of a different density, for example, from air to water.

Bending of electromagnetic waves as it passes through from one medium to another w/ different density. 3.

Diffraction Modulation or redistribution of energy w/in a wavefront when it passes near the edge of an opaque object.

The bending or spreading out of waves as they pass around the edge of an obstacle or through a narrow aperture.


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Radio-wave propagation Huygen’s Principles States that every point on a given spherical wavefront can be considered as a secondary point source of emw from w/c other secondary wave (wavelets) are reradiated.

Read it till it Hertz…jma 1690 1801 1815

The wave theory of light was propounded by Huygens. Interference of light was discovered by British physicist Thomas Young Refraction of light was explained by French physicist Augustin Fresnel

F. .TERRESTRIAL PROPAGATION OF EMW.

1.

Ground-Wave (Surface Wave) Propagation The ground wave or surface wave reaches the receiving site by traveling along the surface of the ground. A surface wave can follow the contours of the Earth because of the process of diffraction. i.

Field strength at a distance

ξ=

120 πht I λd

where: ht = transmitter height in m I = antenna current in A d = distance in m λ = signal wavelength in m


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Induced Antenna Voltage

Va = ξ x hr

hr = receiver height in m

Advantages of ground-wave propagation 1. Given enough transmit power; ground waves can be used to communicate between any two locations in the world. 2. Ground waves are relatively unaffected by changing atmospheric conditions. Disadvantages of ground-wave propagation 1. Ground waves require a relatively high transmission power. 2. Ground waves are limited to VLF,LF, and MF that requires large antennas 3. Ground losses vary considerably with surface materials.

Sample Problem:

A 125-m antenna, transmitting at 1.5 MHz has an antenna current of 8A. What field strength and voltage is received by a receiving antenna 40 km away, with a height of 2 m?

Solution: ξ =

120 πht I 120 π(125 )(8) mV = = 47 .12 λd m 200 (40 x 10 3 m)

Va = ξ x hr = 47.12

2.

mV x 2 m = 94.24 mV m

Space-Wave (Trophosperic Wave) Propagation The space wave follows two distinct paths from the transmitting antenna to the receiving antenna—one through the air directly to the receiving antenna, the other reflected from the ground to the receiving antenna.


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Parameter

In miles

In km

Radio Horizon

d = 2ht(ft)

d = 4 ht(ft)

Radio Range Equation

D = 2ht(ft) + 2hr(ft)

D = 4 ht(ft) + 4 hr(ft)


A microwave transmitting antenna is 600 feet high. The receive antenna is 240 feet high. The maximum transmission distance is ______.

Solution: D=

3.

2ht +

2hr =

2(600 ) +

2(240 ) = 56 .55 mi

Sky-Wave Propagation (Ionospheric Wave Propagation) EMW are beam to the upper portion of the Earth’s atmosphere. i.

Refraction in the ionosphere The amount of refraction that occurs depends on three main factors: a. the density of ionization of the layer, b. the frequency of the radio wave, and c. the angle at which the wave enters the layer

ii.

Skip Distance The skip distance is the distance from the transmitter to the point where the sky wave is first returned to Earth. The size of the skip distance depends on the frequency of the wave, the angle of incidence, and the degree of ionization present.

iii. Skip Zone Skip zone is a zone of silence between the point where the ground wave becomes too weak for reception and the point where the sky wave is first returned to Earth. iv. Ionospheric Layers a.

D layer The D layer ranges from about 30 to 55 miles. Ionization in the D layer is low because it is the lowest region of the ionosphere. This layer has the ability to refract signals of low frequencies. High frequencies pass right through it and are attenuated.


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b.

E Region or E-layer ª The E layer limits are from about 55 to 90 miles. ª This layer is also known as the Kennelly- Heaviside layer, because these two men were the first to propose its existence. ª The rate of ionic recombination in this layer is rather rapid after sunset and the layer is almost gone by midnight. ª This layer has the ability to refract signals as high as 20 megahertz.

c.

F layer ª The F layer exists from about 90 to 240 miles. During the daylight hours, the F layer separates into two layers, the F1 and F2 layers. ª The ionization level in these layers is quite high and varies widely during the day. The F layers are responsible for high-frequency, long distance transmission.

Read it till it Hertz…jma 1902

Oliver Heaviside discovered the existence of an ionized layer of air in the upper atmosphere which was verified by Kennelly and was later known as Kennelly- Heaviside.

1924

Appleton made his study of Kennelly-Heaviside layer.

v.

Variations In The Ionosphere a.

Regular Variations The regular variations that affect the extent of ionization in the ionosphere can be divided into four main classes: ª

Daily Daily variations in the ionosphere are a result of the 24hour rotation of the Earth about its axis.

ª

Seasonal Seasonal variations are the result of the Earth revolving around the sun; the relative position of the sun moves from one hemisphere to the other with changes in seasons.


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b.

ª

11-year sunspot cycle Sunspots are responsible for variations in the ionization level of the ionosphere. This cycle has both a minimum and maximum level of sunspot activity that occurs approximately every 11 years.

ª

27-day sunspot cycle As the sun rotates on its own axis, these sunspots are visible at 27-day intervals, the approximate period required for the sun to make one complete rotation. The 27-day sunspot cycle causes variations in the ionization density of the layers on a day-to-day basis.

Irregular Variations Irregular variations are irregular and unpredictable; they can drastically affect communications capabilities without any warning. ª

Sporadic E Irregular cloud-like patches of unusually high ionization, called sporadic E, often form at heights near the normal E layer.

ª

Sudden Ionospheric Disturbances SIDs are caused by gigantic emission of hydrogen from the sun. These disturbances may occur without warning and may prevail for any length of time, from a few minutes to several hours.

ª

Ionospheric Storms Scientists believe that ionospheric storms result from particle radiation from the sun. Particles radiated from a solar eruption have a slower velocity than ultraviolet light waves produced by the eruption.

vi. Frequency Parameters a.

Index of Refraction

n= 1−

81N f2

N = electron density per m3

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Self-Sufficient Guide to ECE by JASON AMPOLOQUIO b.

Critical frequency Highest frequency at a given ionization density that will be returned down to earth when beam vertically upward.

fc = 9 Nmax

Radio waves transmitted at frequencies higher than the critical frequency of a given layer will pass through the layer and be lost in space. Radio waves of frequencies lower than the critical frequency will also be refracted back to Earth unless they are absorbed or have been refracted from a lower layer. c.

Maximum Usable Frequency (MUF) Highest frequency that will be returned down to the earth at a given distance when beam at an specific angle other than the normal.

MUF =

d.

fc cos θinc

MUF = fc sec θinc

Optimum Working Frequency (OWF)

OWF = 85%MUF

Sample Problem:

The critical frequency at a particular time is 11.6MHz. What is the OWF for a transmitting station if the required angle of incidence for propagation to a desired destination is 70°?

Solution: MUF = fc sec θ =

fc 11.6 MHz = = 33.92 MHz cos θ cos 70°

OWF = 0.85 x MUF = 28.83 MHz


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G. .SCATTER PROPAGATION. 1.

Ground-Scatter The signals are reflected from the ionosphere obliquely to a ground region, scattered from the ground, and propagated again by ionospheric reflection to the receiver.

If the TX and RX are collocated, the process is referred to as backscatter. 2.

Tropospheric-Scatter When a radio wave passing through the troposphere meets a turbulence, it makes an abrupt change in velocity. This causes a small amount of the energy to be scattered in a forward direction and returned to Earth at distances beyond the horizon. This phenomenon is repeated as the radio wave meets other turbulences in its path. The total received signal is an accumulation of the energy received from each of the turbulences.

This scattering mode of propagation enables vhf and uhf signals to be transmitted far beyond the normal line-of-sight. 3.

Ionospheric-Scatter Same principles with Troposcatter, except that the scattering medium is the E region of the ionosphere, with some help of the D and F layers.

4.

Meteor-Burst Scatter A type of propagation of VHF and UHF waves utilizes the phenomenon of scattering of a radio signal from the ionization trails caused by meteors entering the earth’s atmosphere.

5.

Auroral Scatter Ionization associated with auroral disturbances gives radio reflections at HF and VHF.

6.

Equatorial F Scatter Based from the fact that the ionosphere over the equatorial region is higher, thicker, and denser than elsewhere because of its more constant exposure to the solar radiation, the equatorial belt has high nighttime MUF possibilities.

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I

H

1.

The dielectric strength of air is about 3 MV/m. What is the maximum power density of an electromagnetic wave in air? A. 0.45 GW/m2 B. 167 GW/m2 2 C. 884.4 GW/m D. 23.9 GW/m2

2.

A transmitter has a power output of 150 W at a carrier frequency of 325 MHz. It is connected to an antenna with a gain of 12 dBi. The receiving antenna is 10 km away and has a gain of 5 dBi. Calculate the power delivered to the receiver. A. 815 nW B. 327 nW C. 409 nW D. 134 nW

3.

Radio-frequency waves cannot be seen for which of the following reasons? A. Because radio-frequency energy is low powered B. Because radio-frequency waves are below the sensitivity range of the human eye C. Because the human eye detects only magnetic energy D. Because radio-frequency waves are above the sensitivity range of the human eye

4.

Which of the following statements about a wave is the law of reflection? A. The angle of incidence is equal to the refracted wave B. The angle of incidence is not equal to the refracted wave C. The angle of incidence is equal to the angle of reflection D. The angle of incidence is not equal to the angle of reflection

5.

If a wave passes first through a dense medium and then through a less dense medium, which of the following angle-of-refraction conditions exists? A. The angle of refraction is greater than the angle of incidence B. The wave will pass through in a straight line C. The angle of refraction is equal to the angle of incidence D. The angle of refraction is less than the angle of incidence

6.

What is the distance to the radio horizon for an antenna that is 40 ft above the top of a 4000 ft mountain peak? A. 89.88 mi B. 32 mi C. 72 mi D. 45.22 mi

7.

A radio signal travels ______ yards per microseconds. APRIL 2003 A. 618 B. 123.6 C. 273 D. 328


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8.

The antenna for a television station is located at the top of a 1,500-ft transmission tower. Compute the LOS coverage for the TV station if the receiving antenna is 20 ft above ground. A. 71.1 mi B. 91.1 mi C. 61.1 mi D. 81.1 mi

9.

An ionospheric disturbance caused by solar activity resulting in severe attenuation or loss of radio signals, also known as Mogul-Delinger fadeout is known as ______ disturbance. APRIL 2003 A. traveling ionospheric B. sudden ionospheric (SID) C. sporadic E D. magnetic STORM

10. After the radiation field leaves an antenna, what is the relationship between the E and H fields with respect to (a) phase and (b) physical displacement in space? A. (a) Out of phase (b) 90 degrees B. (a) Out of phase (b) 180 degrees C. (a) In phase (b) 90 degrees D. (a) In phase (b) 180 degrees 11. At a certain time, the MUF for transmission at an angle of incidence of 75° is 18 MHz. What is the critical frequency? A. 43.91 MHz B. 4.66 MHz C. 6.77 MHz D. 0.45 MHz 12. Electrically charged particles that affect the propagation of radio waves are found in what atmospheric layer? A. Ionosphere B. Stratosphere C. Troposphere D. Chronosphere 13. Long range, surface-wave communications are best achieved when the signal is transmitted over seawater with (a) what polarization at (b) what relative frequency? A. (a) Vertical (b) High B. (a) Horizontal (b) High C. (a) Vertical (b) Low D. (a) Horizontal (b) Low 14. Energy radiated from an antenna is considered horizontally polarized under which of the following conditions? A. If the induction field is in the horizontal plane B. If the electric field is in the horizontal plane C. If the wavefront is in the horizontal plane D. If the magnetic field is in the horizontal plane 15. Determine the radio horizon for a transmit antenna that is 100 ft high and a receiving antenna that is 50 ft high A. 21.3 mi B. 33.2 mi C. 24.14 mi D. 32 mi 16. A space wave (a) is primarily a result of refraction in what atmospheric layer and (b) extends approximately what distance beyond the horizon? A. (a) Troposphere (b) One-tenth farther B. (a) Troposphere (b) One-third farther C. (a) Ionosphere (b) One-third farther D. (a) Ionosphere (b) One-tenth farther


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17. The signal of a space wave is sometimes significantly reduced at the receiving site because of which of the following interactions? A. Ground-wave diffraction B. Ground-wave reflections C. Space-wave refraction D. Space-wave reflections 18. For long-range communications in the HF band, which of the following types of waves is most satisfactory? A. Space wave B. Reflected ground wave C. Sky wave D. Surface wave 19. Ionization in the atmosphere is produced chiefly by which of the following types of radiation? A. Alpha radiation B. Cosmic radiation C. Infrared radiation D. Ultraviolet radiation 20. A 10-MHz wave entering the ionosphere at an angle greater than its critical angle will pass through the ionosphere and be lost in space unless which of the following actions is taken? A. The ground wave is reinforced B. The frequency of the wave is increased C. The frequency of the wave is decreased D. The ground wave is canceled 21. The polarity of a radio wave is determined by the orientation of (a) what moving field with respect to (b) what reference? A. (a) Electric (b) antenna B. (a) Magnetic (b) earth C. (a) Electric (b) earth D. (a) Magnetic (b) antenna 22. Radio communications can be diffracted to exceptionally long distances through the use of (a) what frequency band at (b) what relative power level? A. (a) Very high frequency (b) High power B. (a) Very low frequency (b) High power C. (a) Very high frequency (b) Low power D. (a) Very low frequency (b) Low power 23. Which of the following ionospheric variation causes densities to vary with the position of the earth in its orbit around the sun? A. Seasonal variation B. 11-year sunspot cycle C. Daily variation D. 27-day sunspot cycle 24. A sudden and intense burst of ultraviolet light is especially disruptive to communications in which of the following frequency bands? A. MF B. VLF C. HF D. LF 25. The most consistent communications can be expected at which of the following frequencies? A. Critical frequency B. Maximum working frequency C. Maximum usable frequency D. Optimum working frequency 26. If the optimum working frequency for a communications link is 4,250 kHz, what is the approximate maximum usable frequency? A. 5,500 kHz B. 5,000 kHz C. 4,500 kHz D. 6,000 kHz


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27. Under certain conditions, such as ducting, line-of-sight radio waves often propagate for distances far beyond their normal ranges because of which of the following factors? A. Ionospheric storms B. Temperature inversions C. Low cloud masses D. Frequency fluctuations 28. The distance between the transmitter waves return to earth is referred to as A. skip distance C. reception distance

and the nearest point at which refracted the ______. B. return distance D. ground-wave distance

29. For a radio wave entering the atmosphere of the earth at a given angle, the highest frequency at which refraction will occur is known by which of the following terms? A. Usable frequency B. Critical frequency C. Maximum usable frequency D. Optimum working frequency 30. If the critical frequency is 10 MHz, what is the OWF at an angle if incidence of 60°? A. 17 MHz B. 19 MHz C. 25 MHz D. 20 MHz 31. When ground-wave coverage is LESS than the distance between the transmitter and the nearest point at which the refracted waves return to earth, which of the following reception possibilities should you expect? A. Strong ground wave B. A zone of silence C. No sky-wave D. Weak ground wave 32. If the signal strength of an incoming signal is reduced for a prolonged period, what type of fading is most likely involved? A. Absorption B. Multipath C. Selective D. Polarization 33. Radio waves that arrive at a receiving site along different paths can cause signal fading if these waves have different A. velocities B. amplitudes C. phase relationships D. modulation percentages 34. Which of the following descriptions of tropospheric scatter signal reception is NOT true? A. Receiver signal strength decreases as the turbulence height is increased B. The level of reception depends on the number of turbulences causing scatter C. The energy received is the portion of the wave reradiated by the turbulence D. Increased communications distance enables more turbulence to act on the signal, thereby raising the received signal level 35. The technique of reducing multipath fading by using several receiving antennas at different locations is known as what type of diversity? A. Space B. Receiver C. Frequency D. Modulation


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36. Which of the following ionospheric variation causes densities to vary with the axial rotation of the sun? A. Daily variation B. 11-year sunspot cycle C. 27-day sunspot cycle D. Seasonal variation 37. The tropospheric scatter signal is often characterized by very rapid fading caused by which of the following factors? A. Extreme path lengths B. Turbulence in the atmosphere C. Multipath propagation D. Angle of the transmitted beam 38. The critical frequency at a particular time is 11.6MHz. What is the MUF for a transmitting station if the required angle of incidence for propagation to a desired destination is 70°? A. 3390 MHz B. 33.9 MHz C. 339MHz D. 3.39 MHz 39. The greatest amount of absorption takes place in the ionosphere under which of the following conditions? A. When sky wave intensity is greatest B. When the density of the ionized layer is greatest C. When collision of particles is least D. When precipitation is greatest

40. The A. B. C.

"attenuation of free space" is due to: losses in the characteristic impedance of free space losses due to absorption in the upper atmosphere the decrease in energy per square meter due to expansion of the wavefront D. the decrease in energy per square meter due to absorption of the wavefront

41. Ground waves are most effective: A. below about 2 MHz B. above about 20 MHz C. at microwave frequencies D. when using horizontally polarized waves 42. Radio waves would most strongly reflect off: A. a flat insulating surface of the right size B. a flat dielectric surface of the right size C. a flat metallic surface of the right size D. a flat body of water 43. Which polarization can be reasonably well received by a circularly polarized antenna: vertical circular

A. C.

B. horizontal D. all of the above

44. The number of circular polarization modes (directions) is: A. 1 B. 2 C. 3 D. many


MICROWAVE ENGINEERING

5-62

Section

Microwave

19

Engineering


Microwaves: A type of electromagnetic wave whose wavelength ranges from 1.0 mm to 30 cm, used in radar, to carry radio transmissions, and in cooking or heating devices.

DEFINITION.

Microwave Link: A widely employed broadband transmission medium commonly used to transport the analog FDM or digital PCM.

DEFINITION.

Line Of Sight: Straight path, unobstructed by the horizon, between a transmitting and receiving antenna.

DEFINITION.

A. .ADVANTAGES OF MICROWAVE SYSTEMS. 1.

The gain of an antenna is proportional to its electrical size. Therefore, one can construct high-gain antennas at microwave frequencies that are physically small.

2.

A 1% bandwidth provides more frequency range at microwave frequencies than that of HF.

3.

Microwave signals travel predominantly by LOS. Plus they don’t reflect off the ionosphere like RF signals.

4.

At microwave frequencies, the electromagnetic properties of many materials are changing with frequency. This is due to molecular, atomic, and nuclear resonance. This behavior is useful for remote sensing and other applications.

5.

There is much less background noise at microwave frequencies than at RF.

6.

Microwave systems do not require a right-of-way acquisition between stations.

7.

Fewer repeaters are necessary for amplification.

8.

Underground facilities are minimized.

9.

Increased reliability and less maintenance.

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Self-Sufficient Guide to ECE by JASON AMPOLOQUIO B. .LINE OF SIGHT MICROWAVE SYSTEMS. 1.

Microwave Frequency Bands

Millimeter Wave Region

Microwave Region (30 cm to 8 mm)

Region

2.

Band Designation

Frequency (GHz)

L

1-2

S

2-4

C

4-8

X

8-12

Ku

12-18

K

18-27

Ka

27-40

U

40-60

V

60-80

W

80-110

Mm

110-300

Procedures of Microwave (Radio Link) Engineering i. ii. iii. iv. v. vi. vii. viii. ix.

Selection of sites (radio equipment plus tower locations) that are in line-of-sight of each other. Selection of an operating frequency band. Development of path profiles to determine radio tower heights. Path calculations Making a path survey Establishment of a frequency plan and necessary operational parameters. Equipment configuration to achieve the fade margins set in step 4 most economically. Installation. Beam alignment (Boresighting), equipment lineup, checkout, and acceptance by a customer.

For Your Information… Band L S C X

Origin of Names… Meaning Band Long wave Ku Short wave K Compromise Ka Cross V

Meaning Kurz-under Kurz Kurz-above Very



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3.

Types Of Path i.

Line of Sight The straight path (LOS) between a transmitting and receiving antenna unobstructed by the horizon.

ii.

Grazing Path The microwave beam barely touches the obstruction (zero clearance).

iii. Obstructed Path The microwave beam is hindered by an obstruction. 4.

The K-Curve A numerical figure that considers the non-ideal condition of the atmosphere resulting to atmospheric refraction that causes the ray beam to be bent toward the earth or away from the earth.

k =

Effective earth radius re = True earth radius ro

k<1 k=∞

k>1

K-curve Conditions i. Sub-standard condition

k <1

ii.

Under this condition, the microwave beam is bent away from the earth, it is as if we expanded the earth curvature (bulge) or raised it up toward the beam above its true value.

Standard condition

4 k= 3

Under this condition, the fictitious earth radius appears to the microwave beams to be longer than the true earth radius.

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Self-Sufficient Guide to ECE by JASON AMPOLOQUIO iii. Super-standard condition

k>

This condition results in an effective flattening of the equivalent earth’s curvature.

4 3

iv. Infinity Condition (Flat Earth Condition)

This condition results to zero curvature (as if the earth is flat) and the microwave beam follows the curvature of the earth.

k=∞

5.

Earth Bulge (eb) The number of feet or meters an obstacle is raised higher in elevation (into the path) owing to earth curvature or earth bulge.

eb(ft) =

d1(mi)d2(mi) 1.5

eb(m) =

d1(km)d2(km) 12.75

Sample Problem:

Calculate the earth bulge 2 mi, 10 mi, 22 mi away from a transmitter for a 25 mi terrestrial microwave link.

Solution:

For 2 miles; eb =

d1d2 2(25 − 2) = = 30.67 feet 1.5 1.5

eb =

d1d2 10(25 − 10) = = 100 feet 1.5 1.5

For 10 miles;

For 22 miles; eb =

d1d2 22(25 − 22) = = 44 feet 1.5 1.5

For Your Information… If we consider atmospheric refraction that may cause the ray beam to be bent toward or away from the earth we must include the k-factor to the earth bulge equation. Loading ECE SUPERBook


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Effective Earth Bulge Equation (Earth Curvature)

eb(ft) =

d1(mi)d2(mi)

eb(m) =

1.5k

d1(km)d2(km) 12.75k

Sample Problem:

Calculate the effective height of a 100 ft obstruction situated 10 mi from the receiving end of a 25 mi radio link for the following values of k; a. 4/3 b. ½ c. 5/2

Solution: He = Hactual + Effective Earth Bulge When k=4/3 He = 100 +

10(25 − 10) = 175 feet ⎛4⎞ 1.5⎜ ⎟ ⎝3⎠

When k=½ He = 100 +

10(25 − 10) = 300 feet ⎛1⎞ 1.5⎜ ⎟ ⎝2⎠

When k=5/2 He = 100 +

6.

10(25 − 10) = 140 feet ⎛5⎞ 1.5⎜ ⎟ ⎝2⎠

Effective Earth Radius

re(km) =

where:

ro

1 − 0.04665 e(0.005577NS )

NS = Surface refractivity ro = true earth radius = 6370 km

7.

Sea Level Refractivity

NS = N0 x e −0.1057hS

where: N0 = Sea level refractivity hs = height of potential site in km

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Sample Problem:

Determine the surface refractivity for a potential microwave site 250 m above sea level with a sea level refractivity of 312 and also calculate the effective earth radius.

Solution:

For the surface refractivity

NS = Noe −0.1057ht = 312 x e

−0.1057 x

250 1000

= 303.86

For the effective earth radius ro 6370 km re = = 0.005577NS 1 − 0.04665e 0.005577(3 03.86) 1 - 0.04665e = 8,539 km

8.

Antenna Height

hTX(ft) =

9.

d1(mi)d2(mi) 2

hTX(m) =

d1(km)d2(km) 17

Radio Range (Standard condition, K=4/3)

D(mi) = 2hT(ft) + 2hR(ft)

D(km) ≅ 4 hT(m) + 4 hR(m)



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Sample Problem:

Calculate the maximum range for a microwave link for which the antenna heights are 100 and 60 ft.

Solution: D = 2hT + 2hR = 2(100) + 2(60) = 25.1 mi

10. Approximate relation between Path length and frequency As a general rule, the lower the frequency, the farther the microwave link. Thus, a general estimate can be made as to the approximate range of frequencies which may be used for specific distance: Frequency

Approximate Path Length

8 GHz 10.5 GHz 18 GHz 23 GHz

30 miles maximum 25 miles maximum 18 miles maximum 10 miles maximum

B. .FRESNEL CLEARANCE. It derives from electromagnetic wave theory that a wavefront has expanding properties as it travels through space. This factor must be added to the obstacle height to obtain an effective obstacle height.

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Fresnel Zone Radius The amount of additional clearance that must be allowed to avoid problems with the Fresnel phenomenon is expressed in Fresnel zones. i.

1st Fresnel Zone Radius

F1(ft) = 72.1

d1(mi)d2(mi) f(GHz)D(mi)

F1(m) = 17.3

d1(km)d2(km) f(GHz)D(km)

60% of the 1st Fresnel Zone Radius (0.6F1) A situation when there is no net change in attenuation or “no gain, no loss” condition occurs when 60% of the first Fresnel radius clears a path obstruction in microwave systems. ii.

Higher Fresnel Zone Radius

Fn = F1 n

n = nth Fresnel zone

Sample Problem: Solve for the total height extended in feet for an obstacle situated 27-mi away fro a 35-mi microwave system assuming if the tree growth exists, add 40 ft for the trees and 10 ft for additional growth (use 6 GHz and 0.6F1). Solution:

H = Earth Curvature + Fresnel Clearance + Vegetation

⎧⎪ 27(35 − 27) 27(35 − 27) ⎫⎪ + 0.6⎨72.1 ⎬ + 50 4 6 x 35 ⎪⎩ ⎪⎭ 1.5 x 3 = 108 + 43.87ft + 50 ≅ 202 ft

H=



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Read it till it Hertz…jma A French physicist, Augustin Fresnel, defined the propagation of a radio wave as a three-dimensional elliptical path between the transmitter and receiver. ª

Fresnel divided the path into several zones based on the phase and speed of the propagating waves.

ª

As frequency decreases, the size of the Fresnel Zone increases.

ª

As the length of the path increases, the size of the Fresnel Zone also increases.

ª

A Fresnel Zone’s radius is greatest at the midpoint of the path. Therefore, the midpoint requires the most clearance of any point in the path.

Sample Problem: Calculate the 5th Fresnel zone radius to clear a 35 mi radio link operating at 12 GHz if the 1st Fresnel zone radius is 61.57 ft. Solution: Fn = 72.1

nd1(mi)d2(mi) f(GHz)D(mi)

= 72.1

5(17.5 x 17.5) = 137.67 ft 12 x 35

Alternate Solution; Fn = F1(ft) n = 61.57 5 = 137.67ft

C. .MICROWAVE LINK CALCULATIONS. Path Profile A Path Profile is a graphical representation of the path traveled by the radio waves between the two ends of a link. The path profile determines the location and height of the antenna at each end of the link, and it insures that the link is free of obstructions, such as hills, and not subject to propagation losses from radio phenomena, such as multipath reflections.

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Self-Sufficient Guide to ECE by JASON AMPOLOQUIO Power Level Diagram

1.

Transmit Parameters i.

ii.

Transmitter Power

In dBW

In dBm

⎛ P ⎞ PT(dBW) = 10 log ⎜ T ⎟ ⎝1W ⎠

⎛ PT ⎞ PT(dBm) = 10 log ⎜ ⎟ ⎝ 1mW ⎠

Transmitter Transmission Line Loss (dB)

L T (dB) =

α (dB) length

x total length + L M

iii. Transmitter Antenna Gain

General Solution

⎛ πD ⎞ GT = η ⎜ ⎟ ⎝ λ ⎠

2

⎛D⎞ GT = 6 ⎜ ⎟ ⎝λ⎠

2

Metric system

GT(dB) = 17.8 + 20 log fGHz + 20 logDm

English system

GT(dB) = 7.5 + 20 log fGHz + 20 logDft

In dB


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iv. Effective Isotropic Radiated Power (EIRP) a.

In Watts

System

General Solution

Ideal

EIRP = PT x GT

Practical

b.

2.

EIRP =

PT x GT LT

In dB

System

General Solution

Ideal

EIRP(dBW) = PT(dB) + GT(dB)

Practical

EIRP(dBW) = PT(dB) + GT(dB) − L T(dB)

Path Parameters i.

ii.

Free Space Loss The loss incurred by an electromagnetic wave as it propagates in a straight line through vacuum with no absorption or reflection of energy from nearby objects.

System

In dB

Metric

FSL (dB) = 92.4 + 20 log fGHz + 20 logDkm

English

FSL (dB) = 96.6 + 20 log fGHz + 20 logDmi

Miscellaneous Noise ª

Atmospheric Noise

Na = 65.5 + En − 20 log fMHz iii. Isotropic Receive Level (IRL)

IRL (dBW) = EIRPdBW − FSL dB

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Receive Parameters i.

Receiver Antenna Gain (dB) Typically the same with TX antenna gain otherwise specified.

ii.

Receiver Transmission Line Loss (dB) Typically the same with TX line losses otherwise specified.

iii. Net Path Loss (NPL)

General Solution

NPL (dB) = Total Losses − Total Gains

In dB

NPL (dB) = L TX(dB) + FSL dB + LRX(dB) − GTX(dB) + GRX(dB)

(

) (

iv. Receive Signal Level (RSL)

In term of Output Power and Net Path Loss In terms of fade margin and Improvement threshold v.

RSL (dBW) = PO(dBW) − NPL dB RSL (dBm) = PO(dBm) − NPL dB RSL (dBW) = FMdB + ITdB RSL (dBm) = FMdB + ITdBm

Noise Threshold

N(dBW) = 10 log(kTB) + NFdB In dBW

N(dBW) = −204 + 10 logB + NFdB

In dBm

⎛ kTB ⎞ N(dBm) = 10 log ⎜ ⎟ + NFdB ⎝ 1mW ⎠

N(dBm) = −174 + 10 logB + NFdB


)


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vi. FM Improvement Threshold (IT) The point where “capture effects” takes place and the output signal-to-noise ratio suddenly jumps to 30 dB.

System

In dB

In dB

IT(dBW) = NdBW + 10 dB

In dBm

IT(dBm) = NdBm + 10 dB

vii. Carrier-to-Noise Ratio The ratio of the minimum wideband carrier power at the input to a receiver that will provide a usable baseband output to the wideband noise power present at the input of a receiver, the noise introduce within the receiver, and the noise sensitivity of the baseband detector.

C = RSL dBm − NdB N (dB)

For Your Information… AM: C/N = S/N FM: S/N about 30 dB larger than C/N ("FM IT")

viii. Fade Margin (dB) A “fudge factor” included in the system gain equation that considers the non-ideal and less predictable characteristics of radio-wave propagation, such as multipath propagation and terrain sensitivity. Rayleigh Table

Propagation Reliability (%) 90 99 99.9 99.99 99.999 99.9999

Required Fade Margin (dB) 8 18 28 38 48 58

Self-Sufficient Guide to ECE by JASON AMPOLOQUIO In terms of RSL and IT

In dB

FM(dB) = RSL dBW − ITdBW

In dBm

FM(dBm) = RSL dBm − ITdBm

In terms of Propagation Parameters

FMdB = 30 log dkm + 10 log ( 6abfGHz ) − 10 log (1 − R ) − 70 where:

Surface Factor (a) 4.0 1.0 0.25

Climate Factor (b)

Description For very smooth terrain, over water, desert For average terrain with some roughness For mountainous, very rough, or very dry terrain

Description

0.5

For hot, humid coastal areas

0.25

For normal, interior temperature

0.125

For mountainous or very dry but not reflective terrain


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5-76

Sample Problem: An FM LOS microwave link operates at 6.15 GHz. The required receiver IF bandwidth is 20 MHz. The transmitter output power is 30dBm. The receiver front end’s first active stage is a mixer with a noise figure of 9 dB. The path length is 21 mi; the antennas at each end have a 35-dB gain and the transmission line losses at each end are 3 dB. If the FM Improvement threshold is used as the unfaded reference, what is the reliability of the radio link? Solution: 1. EIRP EIRP = Pt(dBm) − L tx(dB) + Gtx(dB) = 30 dBm - 3 dB + 35 dB = 62 dBm

2.

FSL FSL = 96.6 + 20log(fGHz xDmi )

3.

= 96.6 + 20log(6.15x21) = 138.82 dB

IRL

IRL dBm = EIRPdBm - FSL dB = 62 dBm - 138.82 dB

4.

= -76.82 dBm

RSL

RSL dBm = IRL dBm + Grx(dB) − L rx(dB) = −76.82 dBm + 35dB − 3 dB

5.

Noise Threshold

= −44.82 dBm

NdBm = −174 + 10log(BW) + NF

6.

= −174 + 10log(20x106 ) + 9 dB = −91.99 dBm

FMIT

FM ITdBm = NdBm + 10 dB 7.

= −81.99 dBm Carrier-to-Noise Ratio ⎛C⎞ = RSL dBm − NdBm ⎜ ⎟ ⎝ N ⎠ dB = −44.82 dBm - (-91.99 dBm)

8.

Fade Margin

= 47.17 dB

FMdB = RSL dBm − FM ITdBm

9.

= −44.82 dBm - (-81.99 dBm) = 37.17 dB Reliability (By interpolation) Reliability = 99.982%

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Self-Sufficient Guide to ECE by JASON AMPOLOQUIO D. .MICROWAVE SYSTEM PERFORMANCE. 1.

System Gain The difference between the nominal output power of a transmitter and the minimum input power required by a receiver.

GS(dB) = P0(dBW) − Threshold(dBW)

2.

GS(dB) = P0(dBm) − Threshold(dBm)

System Reliability The percentage of time a system or link meets performance requirements. ª

Reliability

R = (1 − Outage) x 100%

ª

For Multi-hop link

R s = R1 xR 2 xR 3 … xR n where: Outage = is the amount of time that the requirements will not be meet R1 ,R2 …Rn = individual reliability

Reliability vs. Outage time table

Reliability (%) 0 50 80 90 95 98 99 99.9 99.99 99.999

Outage Time (%)

Per Yr.

100 50 20 10 5 2 1 0.1 0.01 0.001

8760 hr 4380 hr 1752 hr 876 hr 438 hr 175 hr 88 h 8.8 hr 53 min 5.3 min

Outage Time Per Month (avg.)

Per Day (avg.)

720 hr 360 hr 144 hr 72 hr 36 hr 14 hr 7 hr 43 min 4.3 min 26 s

24 hr 12 hr 4.8 hr 2.4 hr 1.2 hr 29 min 14 min 1.4 min 88.6 s 0.86 s


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Availability

ª

A=

MTBF x 100% MTBF + MTTR

Unavailability

ª

U=

MTTR x 100% MTBF + MTTR

U = (1 − A) x 100%

Sample Problem:

What fade margin is required for a microwave LOS link with a time availability requirements of 99.997%?

Solution:

From Rayleigh table

Propagation Reliability (%) 99.99 99.997 99.999

Required Fade Margin (dB) 38 ? 48

By interpolation method; FM − 38 0.99997 − 0.9999 = ⇒ FM = 45.77 dB 48 − 38 0.99999 − 0.99997


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E. .DIVERSITY TECHNIQUE. Diversity suggests that there is more than one transmission path or method of transmission available between a transmitter and a receiver. 1.

Frequency Diversity Frequency diversity is simply modulating two different RF carrier frequencies with same IF intelligence, then transmitting both RF signals to a given destination.

2.

Space Diversity With space diversity, the output of a transmitter is fed to two or more antennas that are physically separated by an appreciable number of wavelengths.

where:

3λ R e S= L

3.

R e = fictitious earth radius in km L = path length in km λ = signal wavelength in m

Polarization Diversity A single RF carrier is propagated with two different electromagnetic polarizations.



5-80

I

H

1.

To provide a reliable 1 GHz cellular phone service in the ground floor dining room of a 52-story (160 m) building in the center of a metropolis, a passive link via coaxial cable was installed between the top of the building and the dining room. Without the link, signals via multipath reflections were unreliable. The top of the building antenna is a λ/2 dipole. The dining room antenna is a λ/4 stub (probe) projecting down from the ceiling. The handheld cell phone with λ/4 antenna projecting up has equivalent effective area. If the dining room cellular phone is 10m from the ceiling antenna and the top of the building antenna is 1.2-km LOS from the cell tower antenna, which has a gain of 10dB and the cable loss is 2dB/100m. What is the total path loss in dB? A. -154 dB B. -134 dB C. -174 dB D. -114 dB

2.

When drawing a path profile, we calculated the appropriate k-factor as 0.85. How will such a k-factor impact tower height compared to a k-factor of 1.0? A. It will decrease by 7.5% B. It will increase by 15% C. It will increase by 7.5% D. It will decrease by 15%

3.

What is the effective earth radius if Ns = 301? A. 8393 km B. 7270 km C. 6370 km D. 8493 km

4.

The effects of fading due to multipath reception are often reduced using: A. diversity B. low power C. high-gain antennas D. repeater

5.

Repeaters are used in a microwave system: A. always B. when distance exceeds line-of-sight C. above 10 GHz D. below 10 GHz

6.

Calculate the maximum range for a microwave link for which the antenna heights are 100 and 60 ft. A. 23.5 mi B. 44.5 mi C. 25.1 mi D. 21.8 mi

7.

What reliability is equal to an average outage time of 88.6 seconds per day? A. 99.99% B. 96.97% C. 90.91% D. 98.19%


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8.

What is the system reliability of a microwave network consisting of three hops with the following individual reliability? April 1992 Hop 1: 99.99% Hop 2: 99.95% Hop 3: 99.8% A. 96.52% B. 97.52% C. 99.52% D. 98.52%

9.

A certain radio station is transmitting at 0 dBW. The transmission line losses at the transmitting and receiving ends are both 2 dB, total propagation path loss of 138 dB, and the receiver noise threshold is -124 dBW. What should be the combined gain of the transmitting and receiving antenna to obtain a C/N ratio of 10 dB at the receiver front end? April 1989 A. 18 dB B. 48 dB C. 28 dB D. 38 dB

10. Satisfactory performance of an analog microwave system is defined as: A. a carrier-to-noise ratio that exceeds a given value B. an ERP level that exceeds a given value C. an energy-per-hertz level that exceeds a given value D. a power density that exceeds a given value 11. A point-to-point communication system consists of a transmitter operating at 400 MHz with an RF output of 90 W. It is fed to a 6 dBi antenna through a 150 ft coaxial cable which has an attenuation of 1.7 dB per 100 ft. The receiving stations, located several miles away, has an antenna system consisting of a 9 dB gain antenna, with a 100-ft coaxial cable with an attenuation of 1.5 dB per 100 ft. If the path loss between the transmitter and the receiving stations is 124 dB, determine the signal level in dB at the receiver. A. -95.6 dBm B. -63.5 dBm C. -148.1 dBm D. -88.2 dBm 12. Fading is caused by: A. multipath reception C. ducting

B. D.

attenuation due to weather all of the above

13. MMDS stands for: A. Multichannel Microwave Distribution System B. Multipoint Microwave Distribution System C. Multichannel Multipoint Distribution System D. Multiple Microwave Distribution Systems 14. For a system gain of 112 dB, a total NF of 6.5 dB, an input noise power of -104 dBm, and a minimum output S/N of the FM demodulator of 32 dB, determine the minimum received carrier power and the minimum transmit power. A. -55.2 dBm, 33.5 dBm B. 29.5 dBm, -82.5 dBm C. -33.5 dBm, 55.2 dBm D. -82.5 dBm, 29.5 dBm 15. An A. B. C. D.

advantage of digital techniques over analog in a microwave system is: less bandwidth is required accumulation of noise is reduced it requires less power lower cost



5-82

16. 90% reliability is equivalent to how many hours per year that a certain microwave system will not meet the desired propagation requirements. A. 876 hrs/yr B. 568 hrs/yr C. 347 hrs/yr D. 239 hrs/yr 17. A frequency diversity microwave system operates at an RF=8 GHz. The IF is a low-index frequency-modulated subcarrier. The baseband signal is a single mastergroup FDM system (BW=2520 kHz). The antennas are 2.4-m parabolic dishes. The feeder lengths are 12 0m at one station and 80 m at the other station with a feeder loss of 6.5-dB/100m and a total 3 dB branching loss for each station. The reliability objective is 99.995%. The system propagates over an average terrain that has a very dry climate. The distance between stations is 40 km. The minimum C/N ratio at the receiver input is 28 dB. Determine the system gain and the minimum transmit power. A. 93. 52 dB, 14.2 mW B. 93. 52 dB, 41.2 mW C. 39. 52 dB, 41.2 mW D. 39. 52 dB, 14.2 mW 18. A microwave receiver receives –60 dBm of signal. The noise power is –100 dBm. What is the carrier-to-noise power ratio? A. 160 dB B. 40 dB C. -160 dB D. -40 dB 19. A typical microwave system uses a transmitted power of about: A. 2 watts B. 20 watts C. 200 watts D. 2000 watts 20. Consider a receiver with an FM IT of –114 dBW, a free space attenuation of 140 dB, an antenna gain of 20 dB each, and 2 dB transmission line losses. What would the transmitter output have to provide a –114 dBW input level to the receiver? A. 3.5 W B. 0.1 W C. 5.3 W D. 8.2 W 21. _____ fading is a partial isolation of the transmitting and receiving antennas because of intrusion of the earth’s surface or atmospheric layers into the propagation path. A. voltage B. current C. Rayleigh D. power 22. What Fade Margin is required for 99.95 % single hop propagation reliability? A. C.

25 dB 18 dB

B. D.

33 dB 45 dB

23. LMDS stands for: A. Local Microwave Distribution System B. Local Multipoint Distribution System C. Local Multichannel Distribution System D. Low-power Microwave Distribution System 24. Calculate the half-power beamdwidth of a parabolic reflector with a gain 44 dB. A. 1° B. 23° C. 15° D. 21°


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25. Calculate the required clearance for an obstacle located 10 km from the transmitting end of a 40-km line of sight radio link operating at frequency of 6 GHz. A. 11.6 m B. 22.8 m C. 5.5 m D. 8.14 m 26. A transmitter has a power output of 10W at a carrier frequency of 250 MHz. It is connected by 10 m of a transmission line having a loss of 3 dB/100m to an antenna with a gain of 6 dBi. The receiving antenna is 20 km away and has a gain of 4 dBi. There is negligible loss in the receiver feedline, but the receiver is mismatched; the antenna and line are designed for a 50-Ω impedance, but the receiver input is 75-Ω. Calculate the power delivered to the receiver, assuming free-space propagation. A. 204 nW B. 2040 nW C. 2.04 nW D. 20.4 nW 27. A 10 hops LOS microwave system has a system time availability equal to 99.85%. What would the per hop time availability be? A. 99.85% B. 90.95% C. 99.985% D. 97.85% 28. The use of redundant system to reduce the effects of multipath fading is _____. A. combining B. diversity C. modulation D. multiplexing 29. In digital microwave systems, additional repeaters increase the: A. reliability B. noise level C. jitter D. security 30. The A. B. C. D.

microwave signal path should clear obstacles by at least 60% of the Faraday zone 60% of the Fresnel zone 60% of the height of the antenna tower 60% of the highest obstacle height

31. What is the availability of a system where the MTBF is 40,000 and the MTTR is 10 hr? A. 98.5% B. 99.975% C. 99.918% D. 95.95% 32. The extra strength needed in order to assure that enough signal reaches the receiving antenna and must be made available to compensate for fades; computed as the difference between the received signal strength and the threshold level A. RSL B. Noise Figure C. Threshold Level D. Fade Margin 33. A microwave path over which radio waves barely touches the obstruction is called _____. A. Grazing Path B. Obstructed Path C. Line of Sight D. Crooked Path


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34. Calculate the system propagation reliability given the following individual link availability: Link 0-1 = 99.85% Link 1-2 = 99.99% Link 2-3 = 99.9% Link 3-4= 99.95% A. 99% B. 99.45% C. 99.95% D. 99.845% 35. A line-of-sight microwave link operating at 4 GHz has a separation of 40 km between antennas. An obstacle in the path is located midway between the two antennas. By how much must the beam clear the obstacle? A. 12.4 meters B. 14.4 meters C. 16.4 meters D. 18.4 meters 36. What fade margin is required for a microwave LOS link with a time availability requirements of 99.997%? A. 42.5 dB B. 45.77 dB C. 43.5 dB D. 46 dB 37. In analog microwave systems, additional repeaters increase the: A. reliability B. noise level C. jitter D. security 38. Consider a space-diversity microwave radio operating at an RF carrier frequency of 1.8 GHz. Each station has a 2.4-m diameter parabolic antenna that is fed by 100-m of air filled coaxial cable, which has a branching loss of 2 dB/100 m and a feeder loss 5.4 dB/100m. The terrain is smooth and the area has a humid climate. The distance station is 40 km and a reliability objective of 99.99% is desired. Calculate the system gain. A. 113.35 dB B. 77.45 dB C. 105 dB D. 145 dB 39. Calculate the value of k-factor that will, as if effectively give an earth bulge of 200 ft for a 25 mi radio link system. A. 1.33 B. 0.521 C. 0.75 D. 0.92 40. A transmitter has a power output of 150 W at a carrier frequency of 32 MHz. It is connected to an antenna with a gain of 12 dBi. The receiving antenna is 10 km away and has a gain of 5 dBi. Calculate the power delivered to the receiver, assuming free-space propagation. A. 4.04 nW B. 404 nW C. 40.4 nW D. 44 nW 41. Determine the power delivered to the receiver in dBm of a transmitter located 40-km away with an output power of 2 W using a 20 dBi antenna, assuming the receive antenna gain is 25 dBi at 6 GHz. A. -37 dBm B. -153 dBm C. -62 dBm D. -85 dBm 42. A situation when there is no net change in attenuation or “no gain, no loss” occurs when _____% of the first fresnel radius clears a path obstruction in microwave systems. A. 60 B. 85 C. 45 D. 75

Self-Sufficient Guide to ECE by JASON AMPOLOQUIO 43. Calculate the noise temperature of the the receiver input of a microwave temperature of 120 K, and the antenna A. 94 K C. 271 K

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antenna/feedline system, referenced to system if the antenna sees a sky feedline loss of 2 dB. B. 182 K D. 144 K

44. A component of a microwave station that samples signal traveling in one direction down to the transmission line. A. Attenuator B. Combiner C. Directional coupler D. Modulator 45. Calculate the thermal noise power in dBm, referred to the receiver input of a microwave antenna/feedline system with a combined noise temperature of 182 K connected to a receiver with a NF of 2 dB and 20 MHz of bandwidth. A. -80 dBm B. -120 dBm C. -100 dBm D. -150 dBm 46. How far from the transmitter could a signal be received if the transmitting and receiving antennas were 40m and 20m, respectively, above level terrain? A. 44.5 km B. 87.6 km C. 32.7 km D. 15.8 km 47. If the line-of-sight distance for an optical beam is 12 km, what would it be, approximately, for a microwave beam? A. 15 km B. 16 km C. 12 km D. 8 km 48. Satisfactory performance of a digital microwave system requires a: A. low level of transmitted power B. high level of ERP C. good energy per bit per transmitted Watt ratio D. good energy per bit per noise density ratio 49. A transmitter and receiver operating at 1 GHz are separated by 10 km. How many dBm of power gets to the receiver if the transmitter puts out 1 Watt, and both the sending and receiving antennas have a gain of 20 dBi? A. -22.4 dBm B. 32.4 dBm C. -42.4 dBm D. 42.4 dBm 50. A microwave system has a feed-line loss of 2 dB and sees a sky temperature of 150 K. Calculate the noise temperature of the antenna/feed-line system referenced to the receiver input. A. 201 K B. 300 K C. 178 K D. 290 K


SATELLITE COMMUNICATIONS

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Section

Satellite

20

Communications


DEFINITION. Satellite: (Telecommunications) An object put into orbit around Earth or any other planet in order to relay communications signals or transmit scientific data. Angle of Elevation is the angle subtended by the direction of travel of an electromagnetic wave radiated from an earth station antenna pointing directly toward a satellite and the horizontal plane.

DEFINITION.

DEFINITION.

Apogee: Point in a satellite orbit located farthest from Earth.

DEFINITION.

Perigee: Point in a satellite orbit located closest to Earth.


Arthur C. Clarke proposes a station in geosynchronous orbit to relay communications and broadcast television. (Coined the term ‘satellite’)

1957

Russia launched Sputnik I that becomes the 1st active satellite. Sputnik I transmit telemetry information for 21 days.

1958

Explorer I was launched. The 1st American satellite, which also transmits telemetry information’s for nearly five months.

1958

NASA launched Score. Score was the 1st artificial satellite used for relaying terrestrial communications that rebroadcast President Eisenhower’s 1958 Christmas message. (1ST communications satellite)

1960

Echo I was launched. Echo is consisted of an aluminized plastic balloon 30 m (100 ft) in diameter. (1st passive reflector satellite) Courier 1B- (1st active repeater satellite)

1962

AT&T launched Telstar I, the 1st satellite capable to transmit and receive simultaneously but eventually destroyed by the new discovered Van Allen radiation belts. (1st duplex satellite)

1963

Synchronous communication launched Syncom I, The first attempt to place a satellite in geostationary orbit but unfortunately was lost during orbit injection. (Considered as the 1st geo-stationary satellite)

1965

Intelsat or Early Bird was launched and was the first commercial telecommunications satellite which used 2 transponder, 25 MHz


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bandwidth which could simultaneously carry 1 TV signal and 480 voice channels. (1st satellite for commercial service) 1997

Aguila II, launched in China becomes the first Filipino satellite.

A. .GENERAL TYPES OF SATELLITES. 1. Astronomical satellites -- are satellites used for observation of distant planets, galaxies, and other outer space objects. 2. Communications satellites -- are artificial satellites stationed in space for the purposes of telecommunications using radio at microwave frequencies. Most communications satellites use geosynchronous orbits or near-geostationary orbits, although some recent systems use low Earth-orbiting satellites. 3. Weather satellites -- are satellites that primarily are used to monitor the weather and/or climate of the Earth. 4. Navigation satellites -- are satellites which use radio time signals transmitted to enable mobile receivers on the ground to determine their exact location accurately on the order of a few metres in real time. 5. Reconnaissance satellites -- are Earth observation satellite or communications satellite deployed for military or intelligence applications. 6. Earth observation satellites -- are satellites specifically designed to observe Earth from orbit, similar to reconnaissance satellites but intended for non-military uses such as environmental monitoring, meteorology, map making etc. 7. Solar power satellites -- are proposed satellites built in high Earth orbit that use microwave power transmission to beam solar power to very large antenna on Earth where it can be used in place of conventional power sources. 8. Biosatellites -- are satellites designed to carry living organisms, generally for scientific experimentation. 9. Miniaturized satellites -- are satellites of unusually low weights and small sizes. New classifications are used to categorize these satellites: minisatellite (500–200 kg), microsatellite (below 200 kg), nanosatellite (below 10 kg).



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B. .KEPLER’s LAW. The laws concerning the motions of planets formulated by German astronomer Johannes Kepler early in the 17th century. 1.

First Law The orbit of a planet around the sun is an ellipse.

2.

Second Law (Law of Areas) A straight line from the planet to the center of the sun sweeps out equal areas in equal time intervals as it goes around the orbit; the planet moves faster when closer to the sun and slower when distant. \

Sun Aphelion Perihelion A1=A2 if t1=t2

3.

Third Law (Law of Periods or Harmonic Law) The square of the period (in years) for one revolution about the sun equals the cube of the mean distance from the sun's center, measured in astronomical units.

The square of the periodic time of orbit is proportional to the cube of the mean distance between the primary and the satellite. Astronomy P2 =k MD3 MD = mean distance P = orbital period A = constant

Satellite Communication

α = A xP

2

3

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Calculate the approximate height of a GEO satellite using Kepler’s Third law (A=42241.0979; P = 0.9972)

Solution: 2 AP 3

h

2 0.9972 3

α = = 42241 .0979 x = 42, 162 .21 km hkm = α − R earth = 42, 162 .21 − 6378 = 35, 784 .21 km

α

Rearth

E

C. .SATELLITE ORBITAL PATTERNS.

1.

Polar Orbit Satellite rotates in a path that takes it over the North and South poles in an orbit perpendicular to the equatorial plane.

2.

Equatorial Orbit Satellite rotates in an orbit directly above the equator, usually in a circular path.

3.

Inclined Orbit Virtually all orbit except those that travel directly above the equator or directly over the North and South poles.



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D. .SATELLITE ELEVATION CATEGORY.

Comparison between Different Elevation Categories Orbit

Typical Height (mi)

Operating Frequency (GHz)

Orbital Period

Availability

LEO

100 - 300

1.0 – 2.5

1.5 hrs.

0.25 hrs.

1.22 – 1.66

6–12 hrs.

2–4 hrs.

2 - 18

24 hrs.

24 hrs.

MEO GEO

6,000 to 12,000 19,000 to 25,000

E. .GEOSTATIONARY ORBIT.


Parameter

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Value 19,322 nmi, 22, 300 smi, 36,000 km

Altitude

23 hr, 56 min, 4.091 s

Period Orbit Inclination

0° 6879 mi/h

Velocity Coverage Number of satellites Subsatellite point Area of no coverage

42.5% of earth’s surface 3 for global coverage (120° apart) On the equator Above 81° north and south latitude

F. .EARTH STATION DESIGN PARAMETERS. 1.

Antenna Look Angles ª

Angle of Elevation (AOE) The angle subtended by the direction of travel of an electromagnetic wave radiated from an earth station antenna pointing directly toward a satellite and the horizontal plane.

⎡ 1 ⎤ R β = tan−1 ⎢ − ⎥ φ + φ tan R h sin ( ) ⎣⎢ ⎦⎥



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ª

Azimuth Angle The horizontal pointing angle of an earth station antenna generally referred to true north.

⎛ tan ϕ ⎞ Az = cos −1 ⎜ − ⎟ ⎝ tan φ ⎠

2.

β

Slant Distance The Line-of-Sight (LOS) distance between an earth station antenna and the satellite

d = −R sin β +

( R + h)

2

− R 2 cos2 β

Where: β = Angle of elevation in degrees φ = cos −1 ⎣⎡cos ϕ cos λ ⎦⎤ ϕ = Latitude of Earth station antenna λ = Difference in longitude between an Earth station antenna and the sub-satellite point h = Satellite HEIGHT in km R = Earth's radius = 6378km Az = Azimuth angle in degrees

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Sample Problem:

Calculate the elevation angle, azimuth and slant range between the TVRO site (38.8°N latitude, 77°W longitude) and Hughes Galaxy satellite that is in a geo-stationary orbit at 134°W longitude above the equator.

Solution:

Elevation Angle ⎡ 1 ⎤ R β = tan−1 ⎢ − ⎥ ( ) tan R h sin φ + φ ⎣ ⎦ φ = cos −1 [cos(ϕ ) cos(λ )]

= cos −1 [cos(38.8° ) cos(134 ° − 77° )]

= 64.9° ⎡ ⎤ 1 6400 km β = tan−1 ⎢ − ⎥ ( ) ( ) ( ) tan 64 . 9 6400 36 , 000 sin 64 . 9 ° + ° ⎣ ⎦ = 16.8° Azimuth Angle ⎡ − tan ϕ ⎤ −1 ⎡ − tan(38.8° ) ⎤ Az = cos −1 ⎢ ⎥ ⎥ = cos ⎢ tan φ ⎣ ⎦ ⎣ tan(64.9° ) ⎦ = 247 .9° Slant Distance d = −R sin β +

(R + h)2 − R 2 cos2 β

= −6400 sin(16.8° ) +

(6400 + 36,000 )2 − 64002 cos2 (16.8°)

= 40, 105 .2 km Answer : β = 16.8°, Az = 247 .9°, d = 40, 105 .2 km

3.

Orbital Height (Vertical Height) The average distance of a satellite above the surface of the earth.

General Solution h=

3

gR 2 T 2 −R 4 π2

Alternate Solution h=

3

(4 x 105 ) T 2 −R 4 π2

2

h = 21.64 T 3 − R



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4.

Orbital Period The period of time that it takes a satellite to rotate around the earth.

General Solution T = 2π 5.

Alternate Solution

3

(h + R) gR 2

T = 2π

(h + R)3 4 x 105

Orbital Velocity The apparent velocity of a satellite as it rotates around the earth.

Alternate Solution km/s m/s

General Solution ν=

2 π(h + R) T

ν=

4 x 105 (h + R)

ν=

4 x 1011 (h + R)

ν = orbital velocity

T = orbital period in sec g = acceleration due to gravity

h = satellite Height in km R = earth's radius

= 9.81 m

= 0.00981 km 2 s2 s gR2 = gravitational constant

= 6378 km

3

= 4 x 105 km

s2

Sample Problem:

Determine the orbital period and orbital velocity of a satellite located 15,000 km above the surface of the earth.

Solution:

Orbital Period T = 2π

(R + h)3 gR 2

= 2π

= 31, 100 .72 sec x

1 hr 3600 s

= 8 .64 hrs Orbital Velocity ν =

4 x 1011 (6400 + 15,000 )

= 4,323 .38 m

s

(6400 + 15,000 )3 4 x 105

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Propagation Time (One-way) The amount of time it takes a signal to reach the satellite after it leaves the earth station antenna or vice-versa.

Ptime =

7.

Slant distance d = Speed of light c

Propagation Delay (Two-way) The amount of time that elapsed after the signal reaches the receiving earth stations after it was transmitted by an earth station.

Ptime = Propagation time in sec

Pdelay = 2xPtime

d = Slant distance in m c = Speed of light = 3 x 108 m

s

Reminder… In some books propagation delay is the same as round-trip propagation time.

8.

Free Space Loss FSL = Free Space Loss in dB

FSL = 92 .4 + 20 log( f x d)

f = Frequency in GHz d = Distance in km



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Sample Problem:

Calculate the propagation time, propagation delay, and FSL for a geostationary satellite located directly above an earth station antenna with an operating frequency of 12 GHz.

Solution:

Propaga tion time d 36 ,000 km Ptime = = c 3 x 105 km s = 120 ms

Propagation delay Pdelay = 2 x Ptime

= 2(120 ms ) = 240 ms

Free Space Loss

FSL = 92.4 + 20 log(f x d)

= 92.4 + 20 log(12 x 36,000) = 205.1 dB

For Your Information… ª The average distance of an earth station from a geostationary satellite is between 36,000 to 42,000 km ª The average propagation time for a geostationary satellite is between 120 ms to 140 ms ª The average round-trip propagation delay for a geostationary satellite is between 240 ms to 280 ms ª The approximate free space loss (at 6 GHz) for a geostationary satellite is between 199.1 dB to 200.4 dB This means that your solution must be between these specified limits!

G. .SATELLITE FOOTPRINTS. Footprint or footprint map is the geographical representation of a satellite antenna’s radiation pattern.

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Self-Sufficient Guide to ECE by JASON AMPOLOQUIO Footprint Summary Beam Spot Beam

Coverage Area 10% of earth’s surface

Domestic coverage

Hemispheric Beam

20% of earth’s surface

Regional coverage

Global Beam

42% of earth’s surface (using 17° beamwidth)

Earth coverage

Zonal Beam

H. .EARTH STATION SYSTEM PARAMETERS. 1.

Bit Energy The amount of energy carried by a single bit of information.

Expressed in

General Solution

Joule/bps

Eb = P t xTb

dBJ or dBW/bps

Eb = P t(dBW) +10 log Tb

Alternate Solution

Eb =

Pt fb

Eb = P t(dBW) −10 log fb

ECE Board Exam: NOV 2002 In satellite communications system, for a total transmit power of 500 watts, determine the energy per bit for a transmission rate of 50 Mbps expressed in dBW. Solution: Eb =Pt(dBW)−10logfb = 10log(500) − 10log(50x106 ) = −40

dBW bps


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SATELLITE COMMUNICATIONS 2.

Noise Density Noise density is the noise power normalized or present in a 1-Hz bandwidth.

Expressed in

General Solution No =

Watts/Hz

Alternate Solution

N BW

No = kTe

dBW/Hz

No = NdBW − 10 logBW

No = −228.6 + 10 log Te

dBm/Hz

No = NdBm − 10 logBW

No = −198.6 + 10 log Te

ECE Board Exam: NOV 2002 For an equivalent noise bandwidth of 10 MHz in a satellite system and noise power of 0.0280 pW, determine the noise density in dBW. Solution: No = NdB − 10logBW = 10log(0.02 80x10-12 ) - 10log(10x1 06 ) = -205.53

3.

dBW Hz

Carrier-to-Noise Density Ratio The ratio between the average wideband carrier power to noise density.

Expressed in

General Solution

unitless

C C = No kTe

dB

C = C dBW − No(dBW) No

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Self-Sufficient Guide to ECE by JASON AMPOLOQUIO System Carrier-to-Noise Density Ratio

1 1 1 = + ⎛ C ⎞ ⎛ C ⎞ ⎛ C ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ No ⎠S ⎝ No ⎠ UL ⎝ No ⎠DL

Sample Problem:

For a satellite communication channel, the uplink C/No ratio is 80 dB/Hz. And the downlink value is 90 dB/Hz. Calculate the overall C/No ratio in dB/Hz

Solution: ⎛ C ⎞ ⎟ ⎜ ⎜N ⎟ = ⎝ o ⎠S

1 1 = = 90.9x106 1 1 1 1 + + ⎛ C ⎞ ⎛ C ⎞ (108 )UL (109 )DL ⎟ ⎜ ⎟ ⎜ ⎜N ⎟ ⎜N ⎟ ⎝ o ⎠DL ⎝ o ⎠UL

= 10log(90.9 x106 ) = 79.586 dB/Hz

4.

Energy Per Bit-to-Noise Density Ratio

Expressed In Unitless

dB

General Solution

C Eb fb = No N BW

Eb C BW = x No N fb

⎛ Eb ⎞ ⎛ BW ⎞ ⎛C⎞ ⎜ ⎟ =⎜ ⎟ +⎜ ⎟ ⎝ No ⎠ dB ⎝ N ⎠ dB ⎝ fb ⎠ dB

ECE Board Exam: NOV 1997 A coherent binary phase shift keyed BPSK transmitter operates at a bit rate of 20 Mbps with a carrier-to-noise ratio C/N of 8.8 dB. Find the Eb/No. Solution: For BPSK system fb = BWNyquist ⎛ Eb ⎜ ⎜N ⎝ o

⎛ BW ⎞ ⎞ 20 MHz ⎛C⎞ ⎟ ⎟ = 8.8 dB + 10log = 8.8 dB = ⎜ ⎟ + ⎜⎜ ⎟ ⎟ 20 Mbps ⎠dB ⎝ N ⎠dB ⎝ fb ⎠dB



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5.

Gain-to-Equivalent Noise Temperature Ratio A figure of merit used to represent the quality of a satellite or an earth station receiver. i.

Usual Approach

ii.

Expressed in

General Solution

unitless

G A r + ALNA = Te Te

dB

G = A r(dB) + ALNA(dB) − Te(dBK ) Te

Considering the antenna noise temperature

Expressed in

General Solution

unitless

G A r + ALNA = T Ta + Te

dB

G = A r(dB) + ALNA(dB) − 10 log(Ta + Te ) Te

A LNA = Low Noise Amplifier (LNA) gain A r = receiving antenna gain A = loss in feedline & antenna Ta = effective noise temperature of antenna & feedline = 290(A − 1) + Tsky A

Sample Problem:

A receiving antenna with a G/T of 25 dB is used to receive signals from a satellite 38,000 km away. The satellite has a 100-watt transmitter and an antenna with a gain of 30 dBi. The signal has a bandwidth of 1 MHz at a frequency of 12 GHz. Calculate the C/N at the receiver.

Solution: EIRP = 10logPtx + Gtx(dB) = 50 dBi

∴ FSLdB = 92.4 + 20 log{12 x 38,000} = 205.6 dB

C G = RSLdB − NdB = EIRPdBW − FSLdB + + 228.6 dB − 10 log BW N dB T dB = 50 − 205.6 + 25 + 228.6 − 10 log(1 x 106 ) = 38 dB


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Sample Problem: Calculate the G/T of a receiving antenna with a gain of 38 dB and looks at the sky with a noise temperature of 15 K if the loss between the antenna and the LNA input, due to feedhorn, is 0.5 dB, and the LNA has a noise temperature of 38 K. Solution: G = 38 dBi - 0.5 dB = 37.5 dBi

290(A − 1) + Tsky

⎛ 0.5 ⎞ ⇒ A = log−1 ⎜ ⎟ ⎝ 10 ⎠ 290(1.12 − 1) + 15 = = 50 °K 1.12

Ta =

A

G = Ar(dB) + ALNA(dB) − 10log(Ta + Te ) Te = 37.5 + 0 - 10log(50 + 38) = 18.16 dB

I.

.SATELLITE ACCESSING TECHNIQUES. 1.

Frequency Division Multiple Access (FDMA) A method of multiple earth stations accessing technique where a given RF bandwidth (typically 500 MHz) is divided into smaller frequency bands (36 MHz) called subdivisions.

i.

Single Channel Per Carrier (SCPC) Each subdivision carries only one 4-kHz voice band channel.

ii.

Multiple Channel Per Carrier (MCPC) Several voice band channels are frequency-division multiplexed together to form a wider subdivision.



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ª

Fixed-assignment multiple access (FAMA) The assignment of capacity is distributed in a fixed manner among multiple stations.

ª

Demand-assignment multiple access (DAMA) Capacity assignment is changed as needed to respond optimally to demand changes among the multiple stations.

2.

Time Division Multiple Access (TDMA) Each earth station is allotted a fixed time slot(called EPOCH) within a TDMA frame, occupying essentially the entire wideband frequency spectrum for the allocated time.

3.

Code Division Multiple Access (CDMA) Referred to as Spread Spectrum Multiple Access; transmission can spread throughout the entire allocated bandwidth. Each earth station’s transmission is encoded with a unique binary word called CHIP code.

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CDMA Techniques (Spread Spectrum) ª Direct Sequence (DSSS) DS spread spectrum is produced when a bipolar data-modulated signal is linearly multiplied by the spreading signal in a special balanced modulator called spreading correlator.

J.

ª

Frequency Hopping (FHSS) A form of CDMA where a digital code is used to continually change the frequency of the carrier.

ª

Hybrid DS/FH Combination of direct sequence and frequency hopping.

.COMPARISON BETWEEN NAVSTAR GPS AND RUSSIAN GLONASS.

Parameter

NAVSTAR/GPS

GLONASS

21 + 3

21 + 3

Number of orbit

6

3

Orbital Altitude

10,898 nmi

10,313 nmi

12 hrs

11 hrs 15 min

Planned Constellation

Orbital Period

55°

64.8°

CDMA

FDMA

1,023 bits

511 bits

C/A Code BW

2 MHz

1 MHz

Bit Rate

50 bps

50 bps

Orbital Inclination Access Method C/A Code



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Personal Communication Service (PCS) Satellite System Summary

Parameter

IRIDIUM

ICO

GLOBALSTAR

Motorola

ICO Global

Qualcomm

Orbit

LEO

MEO

LEO

Altitude

485

6,450

880

66

10

48

6

2

6

Inclination of planes

86.4°

45°

52°

No. of VF channels

1,110

4,500

2,400

Access technique

TDMA

TDMA

CDMA

Owner(s)

No. of satellites No. of orbit planes

Timeline of 1st Orbital Launches by Nationality

Country

Year Launched

First Satellite

Soviet Union

1957

Sputnik 1"

United States

1958

Explorer 1

Canada

1962

Alouette 1

France

1965

Astérix

Japan

1970

Osumi

China

1970

Dong Fang Hong I

United Kingdom

1971

Prospero X-3

European Union

1979

Ariane 1

India

1980

Rohini

Israel

1988

Ofeq 1

Pakistan

1990

Badr-1 1

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I

H

1.

It is often claimed that the KU-band (12 GHz) is better than the C-band (4 GHz) for TVRO reception because a parabolic antenna of a given diameter has higher gain at the higher frequency. Though the gain is undoubtedly higher, what is the difference in path loss to find out if there is really a net improvement in the signal strength obtain for a given receiving antenna? A. 10.542 dB B. 9.542 dB C. 6.542 dB D. 8. 542 dB

2.

Consider a bent pipe satellite system where the uplink C/NO= 105 dB and the downlink C/NO= 95 dB. What is the system C/NO? A. 94.586 dB B. 85.935 dB C. 81.478 dB D. 98.153 dB

3.

Calculate the required C/No for a digital satellite link if the desired Eb/No ratio is 9.6 dB and the bit rate is equal to that of T1 carrier. A. 61.5 dB B. 51.5 dB C. 71.5 dB D. 81.5 dB

4.

A satellite receiver has a G/T ratio of -7 dB/K, and the receiver feeder loss are 1 dB. The earth station transmits an EIRP of 50 dBW, and the transmission path loss amount to 205 dB. Calculate the C/No at the receiver. A. 83.8 dB B. 65.6 dB C. 38.5 dB D. 56.5 dB

5.

Calculate the EIRP of a satellite with a TWTA output power of 10 W, the satellite antenna gain is 40 dB, and the transmit feeder loss are 1 dB. A. 49 dB B. 79 dB C. 59 dB D. 69 dB

6.

A satellite is put into final geosynchronous orbit from its transfer orbit by firing the _____? A. Jet Thruster B. Orbit stabilizer C. Apogee Kick Motor D. Booster Rocket

7.

What balance the gravitational pull of the earth to allow the satellite to stay on its orbit? A. Satellite self power B. Atmospheric condition C. Wind velocity D. Centripetal force


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8.

Calculate the Doppler frequency shift for a satellite with a relative velocity of 5000 km/s as receive by an earth station antenna operating at 6 GHz. A. 5.5 GHz B. 3.5 GHz C. 6.1 GHz D. 8.1 GHz

9.

Determine the EIRP of a satellite used for downlink transmission required to produce a C/N ratio of 26 dB at the earth station. The bandwidth is 36 MHz, the transmission path loss is 203 dB, and the earth station receiver feeder loss is 2 dB. A. 55.96 dB B. 88.96 dB C. 66.96 dB D. 77.96 dB

10. For the downlink of a digital satellite circuit, the transmission path loss is 207 dB and the G/T at the receiver is -5 dB/K. Calculate the satellite EIRP required to maintain a transmission rate of 65 Mbps at a Eb/No of 9 dB. A. 70.53 dB B. 66.38 dB C. 88.25 dB D. 83.57 dB 11. Calculate the propagation delay for a satellite and an earth station if the angle of elevation is 30°. A. 260 ms B. 75 ms C. 130 ms D. 520 ms 12. Calculate the signal strength in dBm and the time it will take for a signal that emanates from an interstellar space probe Voyager 1 with a transmit power 10 W, 45 dBi antenna gain to reach the earth when it passed the orbit of Pluto. (Assume Pluto’s location equal to 3.7x109 miles when the probes passed by and the Earth station antenna gain is 55 dBi operating at 1.8 GHz). A. 1.13 hrs -83.1 dBm B. 5.53 hrs -83.1 dBm C. 5.53 hrs -93.1 dBm D. 1.13 hrs -93.1 dBm 13. In a transoceanic satellite conversation, how much is the typical delay before a reply is heard? A. 200 ms B. 900 ms C. 600 ms D. 50 ms 14. Under a circular satellite orbit, how high is a certain satellite located above the surface of the earth if the total satellite height is 9,869 miles? A. 3,960 miles B. 6,000 miles C. 4,984.50 miles D. 3,000 miles 15. At what apogee in an elliptical orbit must a geosynchronous satellite be initially stationed before it is finally fired into its final geostationary orbit? A. 22,300 miles B. 10,500 miles C. 50,000 miles D. 30,000 miles 16. A positional tolerance of 0.1 on a geostationary orbit is equivalent to an arc distance of? A. 200 km B. 74 km C. 240 km D. 300 km


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17. Telephone communications takes place between two earth stations via a satellite that is 38,000 km from each earth station. Suppose Tamyboy, at station 1, asks a question and Jason, at station 2, answer immediately, as soon as he hears the question. How much time elapses between the end of Tamyboy’s question and the beginning of Jason’s reply, as hear by Tamyboy? A. 253.33 ms B. 87.88 ms C. 158.45 ms D. 224.12 ms 18. A satellite receiving system has a figure of merit of -8dB/K. The satellite antenna is a paraboloidal reflector type with an illumination efficiency of 70% and a -3dB beamwidth of 18. Calculate the total noise temperature at the receiver input. A. 435.6 K B. 659.2 K C. 345.6 K D. 235.6 K 19. Calculate the C/No at the earth receiving station, from a satellite transmitting an EIRP of 49.5 dBW on a frequency of 12 GHz. The earth station antenna angle of elevation is 7° and the receiving figure of merit is 40.7 dB A. 11.2 dB B. 151.2 dB C. 112.5 dB D. 100.3 dB 20. Determine the signal strength at the satellite if an earth station is transmitting 7 W of power using a parabolic antenna with a gain of 45 dBi, operating at 4 GHz, the path length is 40,000 km and the satellite receiver gain is 40 dBi. A. -79 dBm B. -73 dBm C. -70 dBm D. -77 dBm 21. Calculate the orbital velocity of a satellite located 500 km above the earth’s surface. A. 3.6 km/s B. 5.6 km/s C. 7.6 km/s D. 9.6 km/s 22. Determine from the following the basic technique used to stabilize a satellite. A. Hub B. Solar panel orientation C. Spin D. Gravity forward motion balance 23. How many satellites does the GPS system consist? A. 60 satellites B. 3 satellites C. 12 satellites D. 24 satellites 24. What is the equivalent noise temperature for a receiving installation that has a G/T of 30 dB. If its antenna gain is 45 dB and the antenna noise temperature is 22 K. A. 9.62 K B. 29.62 K C. 39.62 K D. 19.62 K 25. Find the noise temperature of an antenna on a satellite if it looks at the earth (giving it a “sky” noise temperature of 290 K) and is coupled to the reference plane by a waveguide with loss of 0.3 dB. A. 290 K B. 310 K C. 210 K D. 300 K



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26. The IRL from a satellite is -155 dBW; the earth station receiving system has an antenna gain of 47 dB, an antenna feed loss of 0.1 dB, a waveguide loss of 1.5 dB, a directional coupler insertion loss of 0.2 dB, and a bandpass filter loss of 0.3 dB; the system noise temperature is 117 K. What is the C/NO? A. 93.51 dB B. 95.9 dB C. 857.88 dB D. 97.82 dB 27. Sputnik 1 is the first active satellite which transmits _____ information. A. data B. weather C. telemetry D. celestial 28. Also known as the “Law of Areas” A. 1st Law of Kepler C. 3rd Law of Kepler

B. D.

4th Law of Kepler 2nd Law of Kepler

29. Calculate the power density in W/m2 as received from a 10-W satellite source that is 22,000 mi away from earth. A. 8.77x10-16 W/m2 B. 3.33x10-16 W/m2 -16 2 C. 1.48x10 W/m D. 6.35x10-16 W/m2 30. In this type of satellite transponder, a single mixer converts all channels within the 500-Mhz bandwidth simultaneously to their downlink frequency. A. Broadband transponder B. Double-conversion C. Channelized transponder D. Regenerative transponder 31. The A. B. C. D.

two factors that keep a satellite in orbits are? Centrifugal force created by earth’s rotation and satellite centrifugal force Gravitational pull of the earth and satellite velocity Satellite velocity and centrifugal force Gravitational pull of the earth and centripetal force of the revolving satellite

32. Calculate the average distance of the moon from the earth if the moon has an orbital period of approximately 28 days. A. 34x106 m B. 84x106 m C. 384x106 m D. 38x106 m 33. In satellite communication what does TTC means? A. Telemetry, Tracking, and Control B. Telecommand, Telemetry, and Control C. Telecommand, Tracking, and Communication D. Telemetry, Telecommand, and Communication 34. An amateur radio hobbyist is communicating using the moon as a reflector. Calculate the amount of time it will take the signal from the hobbyist transmitter to reach the moon surface and be returned and receiver back here on earth. Also calculate the total path loss at 1 GHz. A. 2.56 seconds, 48.2 dB B. 2.56 seconds, 408.2 dB C. 5.88 seconds, 408.2 dB D. 5.88 seconds, 48.2 dB 35. Satellite bandwidths are typically ____ wide and are divided into ____ segments, each _____ wide. A. 500-MHz, 12, 24-MHz B. 500-MHz, 36, 12-MHz C. 500-MHz, 24, 70-MHz D. 500-MHz, 12, 36-MHz


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36. Calculate the orbital velocity of satellite located 10,000 km above the surface of the earth. A. 4,931.16 m/s B. 4,656.34 m/s C. 4,578.9 m/s D. 4,456.56 m/s 37. The FDMA technique wherein voice band channels are assigned on “as needed” basis. A. CDMA B. DAMA C. PAMA D. SSMA 38. Calculate the received signal strength at the satellite if the earth station uplink transmitter operates at 6 GHz with a transmitter power of 10 kW and an antenna gain of 50 dBi. The geostationary satellite receiver is using a parabolic antenna with a gain of 40 dBi, the elevation angle to the satellite as viewed from the earth station is 45°. A. -29.5 dBm B. -39.5 dBm C. -59.5 dBm D. -49.5 dBm 39. A form of CDMA where a digital code is used to continually change the frequency of the carrier. A. Store and Forward B. SPADE C. Spread Spectrum D. Frequency Hopping 40. The satellite frequency re-use method which sends different information signals using vertical or horizontal electromagnetic polarization. A. Dual polarization B. Spread spectrum C. Multiple coverage areas D. Spatial polarization 41. The line that connects the apogee and the perigee. A. line of apsides B. line of nodes C. line of gee D. line of shoot 42. Calculate the EIRP in dBm of an earth terminal with an antenna with a gain of 30 dB, transmitter output of 200 W, and transmission line losses of 2.3 dB. A. 20.5 dBm B. 36.67 dBm C. 80.7 dBm D. 56.89 dBm 43. What is the isotropic receive level (IRL) at an earth terminal antenna where the free -space loss to the associated satellite is 196.4 dB and other link losses are 2.6 dB? The satellite EIRP is + 34 dBW. A. -163.4 dBW B. -110.6 dBW C. -143.2 dBW D. -165 dBW 44. What is the receiving system noise temperature (Tsys) if the antenna noise temperature is 300 K and the receiver noise temperature is 100 K? A. 200 K B. 200 K C. 316.22 K D. 400 K 45. Calculate No of a receiving system with a noise temperature of 100 K. A. -123.3 dBW B. -208.6 dBW C. -39.3 dBW D. -210.2 dBW


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46. Determine C/No at an earth terminal of a satellite downlink where the RSL is 139 dBW and the system noise temperature is 400 K. A. 34.4 dB B. 12.45 dB C. 63.58 dB D. 56.3 dB 47. Referred to as a function and/or a design of a double conversion satellite. A. Has individual LNA, HPA, mixer and band pass filter for each channel B. Demodulate its up-link signal to recover its baseband signals and use them to remodulate a downlink transmitter C. Equipped with a single mixer that converts all satellite channels within the bandwidth simultaneously to their downlink frequency D. Equipped with two mixers 48. How does spatial isolation technique in satellite communications avoid interference? A. Use of different polarity antennas B. Use of low gain antennas C. Employment of highly directional spot-beam antennas D. Use of different types of antennas 49. An area on earth covered by a satellite radio beam. A. Footprint B. Bandwidth C. Identity D. Beamwidth 50. Which type of satellite transponder improves the S/N ratio which demodulates the uplink signal to recover the baseband signals and use them to remodulate the downlink transmitter? A. Regenerative B. Baseband C. Double conversion D. Broadband 51. What spatial separation is needed for 72 satellites to be parked in a geosynchronous orbit? A. 7 B. 4 C. 6 D. 5 52. What is the typical IF of satellite receivers? A. 70 MHz B. 455 MHz C. 30 MHz D. 10.7 MHz 53. Calculate the transmission path loss at vertical incidence between a geostationary satellite and a ground station operating at a frequency of 4 GHz, allowing 0.04 dB for atmospheric attenuation and 0.1 dB for rain attenuation. A. 155.2 dB B. 212.5 dB C. 160.8 dB D. 195.7 dB 54. One of the following devices in satellite transponder serves as output of the receive antenna: A. mixer; B. low noise amplifier; C. power amplifier; and D. local oscillator. A. A B. B C. C D. D


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55. In a re-use frequency technique of increasing satellite channel capacity of corresponding transponders, how do you control the antenna used to prevent interference? A. Different polarity B. Low gain antenna C. High directional antenna D. Different type antenna 56. Refers to a mobile earth station in the mobile-satellite service located on board ship. A. Mobile satellite station B. Ship earth station C. Mobile satellite D. Ship station 57. A technique in satellite communications which uses a highly directional spotbeam antenna to prevent interference from frequency sharing. A. Directional technique B. Beaming technique C. Spatial isolation technique D. Frequency re-use technique 58. How do you determine the satellite location in latitude and longitude measurements? A. Designate a point on earth directly below the satellite B. Designate south or North Pole as reference C. Designate any reference point on the surface of the earth D. Designate any reference point on the earth along the equator 59. A point in the satellite orbit known to be the closest location to the surface of the earth. A. Zenith B. Azimuth C. Perigee D. Apogee 60. Which satellite transponder has the most number of mixers? A. Broadband B. Regenerative C. Double-conversion D. Channelized 61. The height of the geosynchronous orbit above the equator is about: B. 35,780 km A. 3,578 km D. depends on satellite velocity C. 357,800 km 62. The high and low points of a satellite's orbit are called, respectively,: B. perigee and apogee A. apogee and perigee D. downlink and uplink C. uplink and downlink 63. The area on the earth that is "covered" by a satellite is called its: B. downlink A. earth station D. plate C. footprint 64. The velocity required to stay in orbit: A. is constant B. is zero (freefall) C. is lower close to the earth than far from the earth D. is higher close to the earth than far from the earth



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65. An antenna is aimed by adjusting the two "look angles" called: B. azimuth and declination A. azimuth and elevation D. apogee and perigee C. declination and elevation 66. The power per transponder of a typical Ku-band satellite is in the range: B. 50 to 250 watts A. 5 to 25 watts D. depends on its orbit C. 500 to 2500 watts 67. The power level for an earth station to transmit to a satellite is on the order of: B. 102 watts A. 101 watts D. 104 watts C. 103 watts 68. The "payload" on a communications satellite consists of: B. batteries A. transponders D. all of the above C. solar cells 69. "Station-keeping" refers to: A. antenna maintenance C. orbital adjustments

B. power-level adjustments D. antenna alignment

70. DBS stands for: A. decibels of signal C. direct-broadcast system

B. down-beam signal D. direct-broadcast satellite

71. LNA stands for: A. low-noise amplifier C. low-noise amplitude

B. low north angle D. low-noise array

72. A reduction in TWT power for linearity is called: B. backoff A. backdown D. EIRP drop C. power-down 73. TVRO stands for A. television receive only C. television remote origin

B. television repeater only D. television random origin

74. VSAT stands for: A. video satellite B. video signal antenna terminal C. very small antenna terminal D. very small aperture terminal 75. What is the first Japanese and Canadian satellite?

A. C.

Alouette 1, Explorer 1 Osumi, Astérix

B. Astérix, Explorer 1 D. Osumi, Alouette 1

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Section

Cellular

21

Communication System

DEFINITION.

Cell:

Read it till it Hertz

A cell is the basic geographic unit of cellular system.

Cellular Communications: A cellular communications system uses a large number of low-power wireless transmitters to create a complex and interconnected network of cells.

DEFINITION.

DEFINITION. Frequency reuse is the process in which the same set of frequencies can be allocated to more than one cell, provided the cells are at a certain distance apart. A. .MODERN CELLULAR SYSTEM.

B. .ANALOG CELLULAR. Analog cellular was the first approach in wireless telephony, given its widespread implementation in the early years. Analog standards include AMPS, N-AMPS, TACS and NMT. 1.

AMPS (Advanced Mobile Phone System) Developed by Motorola and AT&T and is used in the United States. It is an analog technology operating on 50 MHz in the 800 MHz band and supports 666 channels split into 30-kHz voice channels, one each for forward and reverse communications.

2.

N-AMPS (Narrowband AMPS) NAMPS is also developed by Motorola, enhances the performance of an analog AMPS system. System capacity is improved by splitting the 30kHz channels into three 10-kHz channels, thereby tripling AMPS capacity.


Cellular communication system

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3.

TACS (Total Access Communications System) TACS supports up to 1,000 channels, compared with the 666 supported by AMPS.

4.

NMT (Nordic Mobile Telephone) Developed and placed into service in the early 1980s in Scandinavian countries including, Denmark, Finland, Norway, and Sweden. ª

NMT 450 - Operates in the 450 MHz range, which yields excellent signal propagation.

ª

NMT 900 - Operates in the 900 MHz range, and is appropriate for more densely populated areas.

C. .DIGITAL CELLULAR. Current digital standards include JDC, USDC, TDMA, CDMA, and GSM, with GSM clearly being the dominant. Most digital standards, including GSM, are based on TDMA and TDD. 1.

JDC (Japanese Digital Cellular) It has not found acceptance outside Japan, with the exception of a few Asian countries under Japanese economic influence.

2.

USDC (United States Digital Cellular) Developed as a standard by the TIA (Telecommunications Industry Association) subcommittee TR 45.3 operating in the 800 MHz band. a.k.a. D-AMPS, Interim Standard 54 or IS-54

3.

GSM (Global System for Mobile Communications) GSM operates in the 800 MHz and 900 MHz frequency ranges and is ISDN-compatible.

4.

DECT (Digitally Enhanced Cordless Telephone) DECT is the European standard for a Digital Cordless Electronic Telephone. It is designed for use as an in-office wireless phone, where a single cell is sufficient, as a wireless link between a computer and a nearby printer, or as a link to an email system.

5.

TDMA (Time Division Multiple Access) The North American digital cellular standard for personal communication service operating in the 1900 MHz band similar to European GSM. a.k.a. Interim Standard 136 or IS-136

6.

CDMA (Code Division Multiple Access) CDMA is a direct-sequence spread-spectrum system developed in the United States by Qualcomm and known as IS-95, or by its tradename, CDMAoneTM. a.k.a. Interim Standard 95 or IS-95

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Self-Sufficient Guide to ECE by JASON AMPOLOQUIO D. .COMPARISON OF FAMOUS CELLULAR STANDARDS. 1.

Analog Cellular

Parameter

AMPS

NMT-450

TACS

Base Station (MHz)

824-849

453-458

890-915

Mobile Station (MHz)

869-894

463-468

935-960

Spacing between TX & RX (MHz)

45

10

45

Channels Spacing

30 kHz

25 kHz

25 kHz

Bandwidth

20 to 25 MHz

4 to 5 MHz

25 MHz

Number of channels

666/832

180/200

1000

21x2

21x2

---

Audio signal (kHz)

FM Dev=+12

FM Dev=+5

FM Dev=+9.5

Control signal (kHz)

FSK Dev=+8

FSK Dev=+3.5

FSK Dev=+6.4

FDMA

FDMA

FDMA

FDD

FDD

FDD

Radius

2-20 km

1.8-40 km

2-20 km

Data Transmission

10 kbps

1.2 kbps

8 kbps

Frequency Spectrum

Control Channel Modulation Technique

Accessing Technique Duplex Method

AMPS Forward and Reverse Communications



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AMPS FDMA Architecture

Digital-AMPS TDMA-FDMA Architecture


GSM TDMA-FDMA

Read it till it Hertz…jma Details…details…details… System NMT AMPS ETACS (Europe) JTACS (Japan) GSM TDMA (IS-136) NAMPS NTACS CDMA (IS-95)

Year Introduced 1981 1983 1985 1988 1990 1991 1992 1993 1993


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IS-95 Forward Transmission (CDMA-FDMA)

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IS-95 Reverse Transmission (CDMA-FDMA)




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2.

Digital Cellular

D-AMPS (TDMA)

GSM

CDMA

DECT

824-849

890-915

824-849

1897-1913

869-894

935-960

869-849

1897-1913

Duplex Distance

45 MHz

45 MHz

45 MHz

---

Channels BW

30 kHz

200 kHz

1.23 MHz

1.728 MHz

Bandwidth

25 MHz

25 MHz

25 MHz

---

Number of channels

832

125

19-20

24 slot channels

Control Channel

21x2

---

---

---

π/4 DQPSK

GMSK

BPSK or QPSK

GFSK

Accessing Technique

TDMA

TDMA

CDMA

TDMA/TDD

Radius

32 km

32 km

13 mi

500 m

2.2 mW to 6 W

3.7 mW to 20 W

10 nW to 1W

250 mW

48.6 kbps

280.833 kbps

1.2288 Mbps

1.152 Mbps

Parameter Frequency Spectrum

Uplink (MHz) Downlink (MHz)

Modulation

Mobile Output Power

Data Transmission

E. .BASIC CELLULAR CONCEPTS. A cellular mobile communications system uses a large number of lowpower wireless transmitters to create an interconnected complex network of cells.

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Cell A cell is the basic geographic unit of cellular system. i.

Macrocell Cover a relatively large area. One (1) macrocell might support 12 channels and only 12 simultaneous conversations. In a 7-cell reuse pattern (which is typical) with each cell covering a radius of about 11 miles, no improvement is realized; only 12 conversations can be supported.

ii.

Microcell Cover a smaller area. If a macrocell were divided into microcells, in a 7-cell reuse pattern, a reuse factor of 128 is realized. The same 12 channels could support 1,536 simultaneous conversations.

iii. Picocell Picocells are quite small, covering only a few blocks of an urban area or, perhaps, a tunnel, walkway, or parking garage. In a 7-cell reuse pattern, with each cell covering a radius of approximately 1/2 mile, the reuse factor climbs to 514. The same 12 channels could theoretically support up to 6,168 simultaneous conversations. Required Number of Cells for a given area (N)

N=

A 3.464r 2

where: A = total area to be covered r = cell radius


A province in the Philippines has an area of 2000 sq. kms. It has to be covered by cellular mobile telephone service using cells with a radius of 2 kms. Assuming hexagonal cells, find the number of cellsites needed.

Solution: N=

A

3.464r 2 = 145 cells

2.

=

2000

rcell

3.464 x 22

Cluster Cluster is a group of cells. No channels are reused within a cluster.



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3.

Frequency Reuse Frequency reuse is the process in which the same set of frequencies can be allocated to more than one cell, provided the cells are at a certain distance apart. When the same channel is to be reused in two cells, the two cells are called co-channel cells. The distance D is the separation of the two co-channel cells. i. Radio Channel Available

R = nC

ii.

where: C = number of cells in a cluster n = number of full-duplex channels in a cell

Total Number of Full-Duplex Channels

F = mR

where: m = number of clusters

Sample Problem:

Calculate the theoretical number of full duplex channels available in a cluster and the total capacity for a cellular system where there are 20 clusters, each consisting of 10 cells with 16 channels each cell.

Solution:

F = mR

R = nC channels cells = 16 x10 cell cluster = 160 full - duplex channels

4.

= 20 clusters x160

channels cluster

= 3200 full - duplex

Cellular Interference i. Co-channel interference Two cells using the same set of frequencies are called co-channel cells and the interference between them is called co-channel interference. Co-channel Reuse ratio

Q = 3n

Q=

d r


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where: n = number of cells in a cluster d = distance to the nearest co-channel cell r = radius of cell ii.

Adjacent channel interference Occurs when transmission from adjacent channels interfere with each other.

Sample Problem:

Determine the co-channel reuse ratio for a cluster with 25 cells.

Solution: Q=

5.

3n =

3 x 25 = 8.66

Handoff Handoff occurs when the mobile telephone network automatically transfers a call from radio channel to radio channel as mobile crosses adjacent cells.

2 Types of Handover ª Break Before Make – hard handover ª Make Before Break – soft handover


How often will hand-offs occur when vehicle travels through a CMTS at 100 kph speed if the distance between cellsites is 10 kms?

Solution: ν=

d d 10 km 3600 s ⇒t= = = 0.1hr x = 360 s t ν 100 km 1 hr hr


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Cellular communication system 6.

Antenna separation The antenna separation for cellular systems is determined by the formula

h = 11 d

where: h = antenna height d = spacing between 2 antennas

F. .CELLULAR SYSTEM COMPONENTS. 1.

Public Switched Telephone Network (PSTN) The PSTN is made of local networks, the exchange area networks, and the long-haul network that interconnect telephones and other communication devices on a worldwide basis.

2.

Mobile Telephone Switching Office (MTSO) The MTSO is the central office fro mobile switching. It houses the mobile switching center (MSC), field monitoring and relay station for switching calls fro cell sites to wireline central offices (PSTN)

3.

Cell Site The term cell site is used to refer to the physical location of radio equipment that provides coverage within a cell. Cell sites includes power sources, interface equipment, RF transmitter, receivers, and antenna system

4.

Mobile Subscriber Unit (MSU) The mobiles subscriber unit consists of a control unit and a transceiver that transmits and receives radio transmissions to and from a cell site. Three type of MSU i. The mobile telephone (typical transmit power is 4 watts) ii. The portable (typical transmit power is 0.6 watts) iii. The Transportable (typical transmit power is 1.6 watts)


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G. .ACCESSING TECHNIQUE.

1.

Frequency Division Multiple Access (FDMA) A method of multiple mobile stations accessing technique where a given RF bandwidth is divided into smaller frequency bands.

2.

Time Division Multiple Access (TDMA) Each mobile station is allotted a fixed time slot (called epoch) within a TDMA frame, occupying essentially the entire wideband frequency spectrum for the allocated time.

3.

Code Division Multiple Access (CDMA) Referred to as Spread Spectrum Multiple Access; transmission can spread throughout the entire allocated bandwidth. Each mobile station’s transmission is encoded with a unique binary word called CHIP code.


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Cellular communication system Frequency Hopping Spread Spectrum (FHSS)

Frequency-hopping systems are the simpler of the two systems available.

A frequency generator is used that generates a carrier that changes frequency many times a second according to a programmed sequence of channels known as pseudo-random (PN) noise sequence.

It is called this because if the sequence is not known, the frequencies appear to hop about unpredictably.

Direct Sequence Spread Spectrum (DHSS)

Direct-sequence systems inject pseudo-random noise (PN) into the bit stream that has a much higher rate than the actual data to be communicated.

The data to be transmitted is combined with the PN. The PN bits are inverted when real data is represented by a one and leave the bit stream unchanged when a data zero is transmitted. The extra bits transmitted this way are called chips. Most direct-sequence systems use a chipping rate of at least ten times the bit rate.


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H. .GSM NETWORK.

1.

Network Switching System (NSS) The switching system is responsible for performing call processing and subscriber-related functions. i.

Home Location Register (HLR) The HLR is a database used for storage and management of subscriptions. The HLR is considered the most important database, as it stores permanent data about subscribers, including a subscriber’s service profile, location information, and activity status.

ii.

Mobile-service Switching Center (MSC) The MSC performs the telephony switching functions of the system. It control calls to and from other telephone and data system. It also performs such functions as toll ticketing, network interfacing, and CCS.

iii. Visitor Location Register (VLR) The VLR is a database that contains temporary information about subscribers that is needed by the MSC in order to service visiting subscribers. The VLR is always integrated with the MSC. When a mobile station roams into a new MSC area, the VLR connected to the MSC will request data about the mobile station for the HLR. iv. Authentication Center (AuC) The AuC provides authentication and encryption parameters that verify the user’s identity and ensure the confidentiality of each call. The AuC protects network operators from different types of fraud found in today’s cellular world. v.

Equipment Identity Register (EIR) The EIR is a database that contains information about the identity of mobile equipment that prevent calls form stolen, unauthorized, or defective mobile stations. The AuC and EIR are implemented as stand-alone nodes or as a combined AuC/EIR node.


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3.

I.

Base Station System (BSS) All radio related functions are performed in the BSS, which consists of Base Station Controllers (BSC) and the Base Transceiver Stations (BTS). i.

Base Station Controllers (BSC) The BSC provides all the control functions and physical links between the MSC and BTS. It is a high capacity switch that provides functions such as handover, cell configuration data, and control of RF power levels in the BTS

ii.

Base Transceiver Stations (BTS) The BTS handles the radio interface to the mobile stations. The BTS is the radio equipment (transceivers and antennas) needed to service each cells in the network.

Operation & Maintenance Center (O&MC) The O&MC is connected to all equipment in the switching system and to the BSC. The implementation of O&MC is called the operation and support system (OSS). The purpose of OSS is to offer the customer cost-effective support for centralized, regional and local operational and maintenance activities that are required for a GSM network.

.GSM DIGITAL LOGICAL CHANNELS. 1.

2.

Broadcast Channels i. Frequency Correction Channel (FCCH) Used to tell the mobile that this is the BCCH carrier and to enable the mobile synchronize to the frequency being broadcast by the BTS. ii.

Synchronization Channel (SCH) Used for sending BSIC and give TDMA frame number to the mobile.

ii.

Broadcast Control Channel (BCCH) Used for sending information to the mobile like CGI, LAI.

Common Control Channels i. Paging Channel (PCH) Used for paging the mobile subscriber, reason could be an incoming call or an incoming short message ii.

Random Access Channel (RACH) Used for responding to the paging, location updating or to make call access by asking for a signaling channel.

iii. Access Grant Channel (RACH) Used to allocate SDCCH to the mobile.


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Dedicated Control Channels i. Stand-alone Dedicated Control Channel (SDCCH) Used for allocating voice channel (TCH) to the mobile and location updating and sending short text message to idle mobile. ii.

Slow-Associated Control Channel (SACCH) Used for sending information to the mobile like CGI, LAI, BCCH of the neighboring cells, signal strength, bit error measurement of the serving cell and for sending short text message to busy mobile.

iii. Fast-Associated Control Channel (FACCH) Used for handover

J.

.GSM NUMBERING PLAN. 1.

Mobile Station ISDN number (MSISDN) The mobile number used in a GSM PLMN. Ex: 0917-1234567

2.

International Mobile Subscribers Identity (IMSI) The subscriber number used over radio path for all signaling in the GSM PLMN. This number is stored in SIM, HLR and VLR.

3.

Temporary Mobile Subscribers Identity (TMSI) Used for the subscriber confidentiality. Each time a mobile request for location updating or call setup, MSC/VLR allocates to the IMSI a new TMSI, so the TMSI is used on the signaling path, protecting the IMSI identity.

4.

Location Area Identity (LAI) Used to uniquely identify each location area in the GSM PLMN.

5.

Cell Global Identity (CGI) Used for cell identification within the GSM network.

6.

Base Station Identity Code (BSIC) Used to distinguish co-channel frequency used in the neighboring cell.


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K. .COMPARISON OF GSM 900, 1800(DCS), & 1900(PCS). GSM 900

GSM 1800 (DCS)

GSM 1900 (PCS)

Uplink

890-915 MHz

1710-1785 MHz

1850-1910 MHz

Downlink

935-960 MHz

1805-1880 MHz

1930-1990 MHz

Bandwidth

25 MHz

75 MHz

60 MHz

BW/channel

200 kHz

200 kHz

200 kHz

125

375

300

45 MHz

95 MHz

80 MHz

270.833 kbps

270.833 kbps

270.833 kbps

GMSK

GMSK

GMSK

TDMA

TDMA

TDMA

Parameter

No. of channels Duplex Distance Bit Rate Modulation Technique Accessing Technique

L.

.GSM SUBCRIBER SERVICES. 1.

Advice of charge (AoC) The AoC service provides the mobile subscribers with an estimate of the cal charges. There are two types of AoC information: one that provides the subscriber with an estimate of the bill and one that can be used for immediate charging purposes.

2.

Barring of incoming calls This function allows the subscriber to prevent incoming calls. The following two conditions for incoming call barring exist: baring of all incoming calls and barring of incoming calls when roaming outside the home PLMN.

3.

Barring of outgoing calls This service makes it possible for a mobile subscriber to prevent all outgoing calls.

4.

Call Forwarding This service gives the subscriber the ability to forward incoming calls to another number of the called mobile unit is not reachable, if it is busy, if there is no reply, or if call forwarding is allowed unconditionally.


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5.

Call Hold This service enables the subscriber to interrupt an ongoing call and then subsequently re-establish the call. The call hold service is only applicable to normal telephony.

6.

Calling Line Identification Presentation/Restriction These services supply the called party with the integrated service digital network number of the calling party. The restriction service enables the calling party to restrict the presentation. The restriction overrides the presentation.

7.

Call Waiting This service enables the mobile subscribers to be notified of an incoming call during a conversation. The subscriber can answer, reject, or ignore the incoming call. Call waiting is applicable to all GSM telecommunications services using circuit-switched connection.

8.

Cell Broadcast A variation of the short message service is the cell broadcast facility. A message of maximum of 93 characters can be broadcast to all mobile subscribers in a certain geographic area.

9.

Close User Groups (CUGs) CUGs are generally comparable to a PBX. They are group of subscriber who are capable of only calling themselves and certain numbers.

10. Dual Tone Multi-Frequency (DTMF) DTMF is a tone signaling used for various control purposes via the telephone network, such as remote control of an answering machine. GSM supports full-originating DTMF. 11. Facsimile Group III GSM supports CCITT Group 3 facsimile. 12. Fax Mail With this service, the subscriber can receive fax messages at any fax machine. The messages are stored in a service center from which they can be retrieved by the subscriber via a personal security code to the desired fax number. 13. Multiparty service The multiparty service enables a mobile subscriber to establish a multiparty conversation-that is, a simultaneous conversation between three and six subscribers. 14. Short Message Service A message consisting of maximum of 160 alphanumeric characters can be sent to or form a mobile station.


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Cellular communication system 15. Voice Mail This service is actually an answering machine within the network, which is controlled by the subscribers. Calls can be forwarded to the subscriber’s voice-mail box and the subscriber checks for messages via a personal security code.

M. .CELLULAR TRAFFIC ENGINEERING. 1.

Blocking Probability i.

Erlang B Loss-probability equation.

An GOS = PB = n n! Ax x! x=0

∑

where: A = traffic in Erlang n = number of trunks/channels

Sample Problem:

Calculate the blocking probability for 5-channel cell with an offered traffic of 1.66 erlangs.

Solution: P(B ) =

1.660 0! = 2%

1.665 5! x100% 1.661 1.662 1.663 1.66 4 1.665 + + + + + 1! 2! 3! 4! 5!

Sample Problem:

From the following data, determine the offered traffic of a cell. Maximum calls per hour in one cell = 3000 Average calling time = 1.76 minutes Blocking probability = 2%

Solution: A=

n 3000 ⎛ 1hr ⎞ xT = x ⎜ 1.76minx ⎟ = 88 Erlangs t 1hr 60min ⎠ ⎝

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Peak and Average Traffic where:

A = NT

N = # of subscribers T = peak or average holding time

Sample Problem:

A particular cellular telephone system consisting of 120 cells and catering 20,000 subscribers is using a 12-cell repeating pattern. If each subscriber on this network uses his/her phone on average 45 minutes per day, and also on average 20 of those minutes are used during the peak hour, calculate the peak and average traffic for one cell, assuming callers are evenly distributed over the system.

Solution:

For the entire system: 20 min Apeak = 20,000 x 60 min = 6666.67 Erlang

⎛ 1hr 1 day ⎞ ⎟ A ave = 20,000 x ⎜⎜ 45min x x 60 min 24 hr ⎟⎠ ⎝ = 625 Erlang

For only one cell: 6666.67 120 cell = 55.55 Erlangs

Apeak =

625 120 cell = 5.21 Erlangs

A ave =

Sample Problem:

Assuming the average duration of each call be 1.76 minutes and from Erlang B tables for 2% blocking

# of channels

Offered Traffic

600 587.2 666 ? 700 688.2 How many customers can be served with 666 voice channels if all the channels are used in a single cell?

Solution: By interpolation 666 − 600 A 666 = x(688.2 − 587.2 ) + 587.2 700 − 600 = 653.86 Erlangs

A T 653.86 E 60min N= x 1.76 min 1E

A = NT ⇒ N =

= 22,291 calls


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N. .3G GLOBALLY.

O. .MIGRATION TO 3G.

1.

International Mobile Telecommunication or IMT-2000 Radio Standards i.

IMT-SC or Single Carrier (UWC-136): EDGE GSM evolution (TDMA); 200 KHz channels; sometimes called “2.75G”.

ii.

IMT-FT or FDMA/TDMA Digital European Cordless Telephone (DECT) legacy.

iii. IMT-MC or Multi Carrier CDMA: CDMA2000 Evolution of IS-95 CDMA, i.e. cdmaOne

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Self-Sufficient Guide to ECE by JASON AMPOLOQUIO iv. IMT-DS or Direct Spread CDMA: W-CDMA New from 3GPP; UTRAN FDD v.

2.

IMT-TC or Time Code CDMA New from 3GPP; UTRAN TDD New from China; TD-SCDMA

Data Evolution

Summary of Data Evolution

Parameter

1G

2G

2.5G

2.75G

3G

System

NMT AMPS TACS

GSM TDMA CDMA

GPRS WAP

EDGE

UMTS

---

9.6 kbps

115 kbps

384 kbps

Up to 2 Mbps

Analog Voice

Digital Voice

Packet Data

Intermediate Multimedia

Multimedia

Data Rate Service


5-136

Cellular communication system i.

General Packet Radio Service GPRS is a new nonvoice service that is being added to existing IS-136 TDMA networks in the United States and GSM works in the United States and Europe. It provides for the transmission of IP packets over existing cellular networks, bringing the Internet to the mobile phone. GPRS Time Slots Allocation Type 2+1 - two time slots for download, one for upload Type 3+1 - three time slots for download, one for upload Type 4+1 - four time slots for download, one for upload GPRS phones classification Class A - capable of simultaneous voice and data transmission Class B - automatic switching (according to phone settings) between voice and data Class C - hand-operated switching between voice and data

ii.

3.

Enhance Data rate for Global Evolution (EDGE) EDGE takes the cellular community one step closer to UMTS. It provides higher data rates than GPRS and introduces a new modulation scheme called 8-PSK.

2G Multimedia Message Service (MMS) Architecture


2.5G (GPRS) Architecture

5.

3G (IMT-2000) Architecture


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3rd Generation CDMA Standards

Technology

W-CDMA (FDD)

TD-SCDMA

Wideband Code Division Multiple Access Frequency Division Duplex Korea, Japan, Europe, USA, other Asian countries

Time DivisionSynchronous Code Division Multiple Access China; other countries deploying W-CDMA (TDD)

Introduction

2001

2004 or later

2001

Frequency range

1920-1980 MHz 2110-2170 MHz

1900-1920 MHz 2010-2015 MHz

IS-95 bands plus

Multiple access technology

CDMA

TDMA/CDMA

CDMA

Symbol rate/chip rate

3.84 Mcps

1.28 Mcps

1.2288 Mcps

Channel spacing

5 MHz

1.6 MHz

1.23-1.25 MHz

QPSK downlink

QPSK initially, 8PSK later

QPSK

HPSK uplink

QPSK initially, 8PSK later

HPSK

Primary service

High mobility data and voice



Maximum data rate

Up to 384 kbps (up to 2 Mbps indoor)

Up to 384 kbps (phase one); 2 Mbps (phase two)

Up to 307.2 kbps

Switching

Packet switched or circuit switched



Description

Geography

Modulation

cdma2000 (1xRTT) 1x Radio Telephone Technology Korea first; USA and Japan later

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3G IP Multimedia Service (IMS) Architecture

8.

Comparison of 3G and 4G

Parameter

3G

4G

Frequency

1.8-2.5 GHz

2-8 GHz

Bandwidth

5-20 MHz

5-20 MHz

Up to 2 Mbps (384 kbps deployed)

Up to 20 Mbps

W-CDMA

MC-CDMA or OFDM

Convolutional rate 1/2, 1/3

Concatenated coding scheme

Switching

Circuit/Packet

Packet

Mobile top speeds

200 kph

200 kph

Data rate Access Technique Forward Error Correction


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P. .VARIOUS WIRELESS COMMUNICATIONS SPECIFICATIONS. 1.

Analog Cordless Telephones

Standard

CT0 Cordless Telephone 0

CT1/CT1+ Cordless Telephone 1

2/48 (U.K.) 26/41 (France) Mobile Frequency Range (MHz)

30/39 (Australia) 31/40 (The Netherlands/Spain)

CT1: 914/960 CT1+: 885/932

46/49 (China, Taiwan, U.S.A.) 48/74 (China) Multiple Access Method

FDMA

FDMA

Duplex Method

FDD

FDD

Number of Channels

10, 12, 15, 20 or 25

CT1: 40 CT1+: 80

Channel Spacing

1.7, 20, 25 or 40 kHz

25 kHz

FM

FM

Modulation 2.

Digital Cordless Telephones

Standard

Cordless Telephone 2 (CT2)

Digital Enhanced Cordless Telephone (DECT)

Personal Handy Phone System (PHS)

Mobile Frequency Range

864-868 MHz

1880-1900 MHz

1895-1918 MHz

Multiple Access Method

TDMA/FDM

TDMA/FDM

TDMA/FDM

Duplex Method

TDD

TDD

TDD

40

10

300

1

12

4

100 kHz

1.728 MHz

300 kHz

GFSK

GFSK

π/4 DQPSK

72 kb/s

1.152 Mb/s

384 kb/s

Number of Channels Users Per Channel Channel Spacing Modulation Channel Bit Rate



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Data Rates Supported by Cellular Standards

Wireless LAN/ Wireless PAN/ Wireless MAN

Technology Application

Technology

Date Rates

Typical Range

Bluetooth

Wireless Personal Area Network (WPAN)

0.723 Mbps

10 m

802.15.4 Zigbee

Wireless Personal Area Network (PAN)

0.250 Mbps

10 m to 100 m

802.11a (Wireless ATM)

Wireless Local Area Network (WLAN)

54 Mbps

300

802.11b (WiFi)

Wireless Local Area Network (WLAN)

11 Mbps

100 m

802.16a (WiMax)

Broadband Wireless Metropolitan Area Network (WMAN)

70 Mbps

<3 km range

Wireless Mobile Standard

Cellular Family

Standard

Peak Data Typical Rate Connection Modulation Data Rate (kbps)

CSD

9.6 kbps 14.4 kbps

9.6 kbps

Circuit Switched

GMSK

HS-CSD

28.8 kbps 43 kbps

28.8 kbps

Circuit Switched

GMSK

GPRS

115 kbps 171 kbps

50 kbps

Packet Switched

GMSK

EDGE

384 kbps 513 kbps

115 kbps

Packet Switched

8-PSK

FDD

384 kbps 2 Mbps

144 kbps

Packet Switched

QPSK

TDD

384 kbps 2 Mbps

144 kbps

Packet Switched

QPSK

IS-95A

14.4 kbps

14.4 kbps

Circuit Switched

QPSK

IS-95B

64 kbps 115 kbps

56 kbps

Packet Switched

QPSK

CDMA2000

IX

144 kbps 307 kbps

130 kbps

Packet Switched

QPSK

TDMA

CSD

9.6 kbps

9.6 kbps

Circuit Switched

DQπ/4PSK

PDC

i-mode

9.5 kbps

9.6 kbps

Packet Switched

DQπ/4PSK

GSM

UMTS

CDMAOne


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CSD -- Circuit-Switched Data FDD -- Frequency-Division Duplexing HS-CSD --High Speed CSD TDD -- Time-Division Duplexing PDC -- Personal Digital Cellular IS -- Interim Standard GPRS -- General Packet Radio Service GMSK --Gaussian Minimum Shift Keying EDGE -- Enhanced Data rate for Global Evolution UMTS -- Universal Mobile Telecommunication System DQπ/4PSK-- Differential Quadrature π/4 Phase Shift Keying

Q. .BLUETOOTH SPECIFICATIONS. Bluetooth is a universal radio interface in the 2.4 GHz frequency band that enables electronic devices to connect and communicate wirelessly via short-range (10-100 m), ad-hoc networks. The Bluetooth specifications are developed and licensed by the Bluetooth Special Interest Group. Key Features: Peak data rate : 1 Mbps Low power : peak TX power < 20 dBm Low cost : target is $5-10 per piece Ability to simultaneously handle both voice and data Line of sight not required History: 9 9 9

Invented in 1994 by L. M. Ericsson, Sweden Named after Harald Blaatand “Bluetooth”, king of Denmark 940981 A.D. Bluetooth SIG founded by Ericsson, IBM, Intel, Nokia and Toshiba in Feb 1998

Technical Features:

Connection Type: Spectrum: Modulation: Supported Stations: Transmission Power: Data Rate: Range: Module size:

FHSS & TDD (1600 hops/sec) 2.4 GHz ISM Band (79 channels) GFSK 8 devices 1 mW – 100 mW 1 Mbps 30 ft 9x9 mm


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Class

Maximum Power

Range (Approximate)

Class 1

100 mW

20 dBm

~100 meters

Class 2

2.5 mW

4 dBm

~10 meters

Class 3

1 mW

0 dBm

~1 meter

Typical Bluetooth Scenario:

Point to Point Link 9 Master - slave relationship 9 Bluetooth devices can function as masters or slaves

Piconet

9 9 9

It is the network formed by a Master and one or more slaves (max 7) Each piconet is defined by a different hopping channel to which users synchronize. Each piconet has max capacity of 1 Mbps.

A group of overlapping piconets is called a SCATTERNET

Self-Sufficient Guide to ECE by JASON AMPOLOQUIO 9 9

9

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Users in a piconet share a 1 Mbps channel – individual throughput decreases drastically as more units are added The aggregate and individual throughput of users in a scatternet is much greater than when each user participates on the same piconet Collisions do occur when 2 piconets use the same 1 MHz hop channel simultaneously. As the number of piconets increases, the performance degrades gracefully

Time-Division Duplex Scheme:

Channel is divided into consecutive slots (each 625 μs) One packet can be transmitted per slot Subsequent slots are alternatively used for transmitting and receiving 9 Strict alternation of slots between the master and the slaves 9 Master can send packets to a slave only in EVEN slots 9 Slave can send packets to the master only in the ODD slots

Connection Modes:

Active Mode: Device actively participates on the piconet channel Power Saving modes

Sniff Mode: Slave device listens to the piconet at a reduced rate . Least power efficient.

Hold Mode: The ACL link to the slave is put on hold. SCO links are still supported. Frees capacity for inquiry, paging, participation in another piconet.

Park Mode: The slave gives up its active member address. But remains synchronized (beacon channel). Listens to broadcasts. Most power efficient.


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I 1.

The dimension of the chip-sized SIM card. A. 15 x 20 mm B. C. 10 x 20 mm D.

H 20 x 30 mm 15 x 25 mm

2.

In the third generation of cellular phones, _______ uses W-CDMA. A. IMT-TC B. IMT-MC C. IMT-DS D. IMT-SC

3.

A transmission method that uses only one channel for transmitting and receiving, separating them by different time slots. No guard band is used. This increases spectral efficiency by eliminating the buffer band, but also increases flexibility in asynchronous applications. A. TDM B. TDD C. FDM D. FDD

4.

A secret number issued to a cellular phone that is used in conjunction with a subscriber's shared secret data information for authentication A. MIN B. A-key C. IMSI D. Cipher key

5.

_______ is a first-generation cellular phone system. A. GSM B. D-AMPS C. IS 136 D. AMPS

6.

A type of advanced smart antenna technology that continually monitors a received signal and dynamically adapts signal patterns to optimize wireless system performance A. Phased array antennas B. Helical antenna C. Adaptive array antennas D. Parabolic antenna

7.

Technique employed by wireless infrastructure systems that lowers the power of a signal in a cell site whenever the site detects that the user's phone is close to the source of the signal A. Adaptive power control B. Pre-emphasis C. Level down D. Power attenuation

8.

In wireless technology, _____ refers to transporting voice and data traffic from a cell site to the switch A. Backbone B. Backhaul C. Routing D. Switching


A Bluetooth "piconet" has: A. 2 nodes C. 2 to 8 nodes

B. D.

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2 to 4 nodes 2 to 16 nodes

10. A protocol designed for advanced wireless devices allowing the easy transmission of data signals, particularly Internet content, to micro-browsers built into the device's software A. Bisync B. Wireless Application Protocol C. TCP D. SMTP 11. In the third generation of cellular phones, ________ uses CDMA2000. A. IMT-TC B. IMT-MC C. IMT-DS D. IMT-SC 12. The monitoring, manipulating and troubleshooting of computer equipment through a wireless network A. Wireless ethernet B. Wireless IT C. WiFi D. Radio sounding 13. The part of the wireless system's infrastructure that controls one or multiple cell sites' radio signals, thus reducing the load on the switch A. Home Location Register B. Visitor Location Register C. Base station controller D. Base Transceiver Station 14. _______ is a second-generation cellular phone system. A. NMT B. TDMA C. AMPS D. IS 54 15. An uncompleted call made from a wireless phone A. Dropped call B. Attempt C. Blocked calls D. Seizures 16. The code name for a new wireless technology being developed by Ericsson Inc., Intel Corp., Nokia Corp. and Toshiba A. WiMax B. Zigbee C. Bluetooth D. WiFi 17. Refers to planned developments in mobile communications. Increased bandwidth, from 128 Kbps while moving at high speeds to 2Mbps for fixed stations, will enable multimedia applications and advanced roaming features. A. 4G B. 2.75G C. 3G D. 2.5G 18. Facility offered by a few handsets enabling calls to be made by using voice commands rather than pressing the numbers. The memory can be programmed to store and identify names spoken into the handset and call numbers associated with them. A. Automatic Dialing B. Abbreviated Dialing C. Voice Calling D. Voice Recognition 19. Air interface standard, which will enable Third Generation mobile phones to carry new mobile multimedia services. A. WCDMA B. TDMA C. UMTS D. TDD


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20. A framework from the ITU for third-generation (3G) wireless phone standards throughout the world that deliver high-speed multimedia data as well as voice. A. GSM 1900 B. TACS C. IMT-2000 D. IS-136 21. The total bit rate of GSM RF channels. A. 250.833 kbps C. 270.833 kbps

B. D.

290.833 kbps 210.833 kbps

22. A measure of the number of subscribers who leave or switch to another carrier's service A. Offered traffic B. Churn C. Carried traffic D. Population 23. A wireless phone programmed with stolen or duplicated electronic serial and mobile identification numbers A. RAID B. Backup C. Clone D. Party line 24. An automobile travels at 60kph. Find the time between fades if the car uses a PCS phone at 1900 MHz. A. 4.7 ms B. 2.35 ms C. 9.14 ms D. 1.12 ms 25. A metropolitan area of 1000 km2 is to be covered by cells with a radius of 2km. How many cell sites would be required, assuming hexagonal cells? A. 70 B. 73 C. 76 D. 78 26. A designation of the American National Standards Institute--usually followed by a number--that refers to an accepted industry protocol A. IMSI B. Access Code C. Protocol Stack D. IS (Interim Standard) 27. Synonymous with smart card A. Flash card C. Prepaid card

B. D.

Network interface card SIM

28. Also known as 802.11b A. Wireless ATM C. Wi-Fi

B. D.

Bluetooth Zigbee

29. A cellular radio transmitter has a power output of 3-W at 800 MHz. It uses an antenna with a gain of 12 dBi. The receiver is 5-km away, with an antenna gain of 12 dBi. Calculate the received signal strength in dBm, ignoring any losses in transmission lines. A. -14.7 dBm B. -24.7 dBm C. -45.7 dBm D. -84.7 dBm


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30. A cell site has 57 voice channels. Assume a blocking probability of 2%. From the Erlang B table, the offered load A=46.8 Erlangs. Assume the average talking time per call is 100 sec; calculate the number of calls per busy hour and the number of calls per radio. A. 1658 calls/b-h, 92.5 calls/radio B. 1586 calls/b-h, 25.9 calls/radio C. 1865 calls/b-h, 95.2 calls/radio D. 1685 calls/b-h, 29.5 calls/radio 31. The basic range of a Bluetooth device is: A. 10 cm to 1 meter B. C. 10 cm to 100 meters D.

10 cm to 10 meters within 10 feet

32. IRDA stands for: A. Infrared Data Association C. Infrared Restricted Data Area

Infrared Digital Association Infrared Roaming Data Area

B. D.

33. The ERP of a typical handheld AMPS cell phone is: A. less than 600 μW B. less than 600 mW C. between 1 and 2 watts D. 4 watts 34. The combination of the mobile cell phone and the cell site radio equipment is called the: A. BSC B. MTSO C. RF interface D. air interface 35. One way to increase the capacity of a cell phone system is: A. increase the number of cells B. decrease the number of cells C. increase the ERP D. decrease the ERP 36. The frequency band designated for PCS in North America is: A. 800 MHz B. 900 MHz C. 1.9 GHz D. 12 GHz 37. In CDMA: A. all frequencies are used in all cells B. each cell uses half the available frequencies C. each cell is assigned a frequency by the base D. the frequency is selected by the mobile phone 38. Bluetooth uses: A. CDMA C. QPSK

B. D.

frequency hopping direct sequence

39. An improvement in signal gain can be obtained by raising the height of the base station antenna. On the assumptions that the received radio signals is 110 dBm and the height of the base station antenna is 100 ft, how much higher must the base station antenna be raised to obtain an increase from -110 to 100 dBm in the received signal? A. 116 ft B. 216 ft C. 416 ft D. 316 ft 40. Calculate the wavelength of Bluetooth signal. A. 2.4 m B. 1.25 m C. 0.125 m D. 0.24 m


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41. A cellular system uses 12-cell repeating pattern. There are 120 cells in the system and 20,000 subscribers. Each subscriber uses the phone on average 30 minutes per day, but on average 10 of those minutes are used during peak hour. Calculate the average and peak traffic in erlangs for the whole system. A. 416 E & 3333 E B. 41.6 E & 33.33 E C. 4.16 E & 3.333 E D. 41.6 E & 3.333 E 42. The __________ cellular phone system will provide universal personal communication. A. first-generation B. second-generation C. third-generation D. fourth-generation 43. In a ______ handoff, a mobile station only communicates with one base station. A. medium B. soft C. hard D. make before break 44. From the following data, determine the average holding time of a cell. Maximum calls per hour in one cell = 28000 Blocking probability = 2% Offered traffic A = 821 erlangs. A. 1.76 minutes B. 0.176 minutes C. 176 minutes D. 17.6 minutes 45. Calculate the number of full duplex channels required in each cell for a cellular system 12 clusters, 15 cells per cluster, and a total capacity of 3960 full duplex channel. A. 22 B. 88 C. 44 D. 11 46. A cooperation of standards organizations that specifications for IMT-2000, or 3G. A. ITU B. TIA C. UMTS D. 3GPP 47. The number of channel of GSM 1800. A. 300 C. 125

B. D.

develops

the

technical

150 375

48. Calculate the co-channel reuse ratio for a cluster when the distance to the center of the nearest co-channel cell is 4 km and the cell radius is 1.2 km. A. 1.33 B. 4.33 C. 3.33 D. 2.33 49. GSM is using a modulation technique which is a variant of FSK called GMSK using a frequency deviation of _____. A. 167.708 kHz B. 67.708 kHz C. 16.708 kHz D. 617.708 kHz 50. IMSI contain the following number code. A. MIC+MNC+MSIN B. C. CC+MNC+MSIN D.

MCC+MNC+MSIN MCC+MIC+MSIN

Self-Sufficient Guide to ECE by JASON AMPOLOQUIO 51. The duplex distance of GSM 1800. A. 20 MHz C. 60 MHz

B. D.

80 MHz 95 MHz

52. Two or more connected piconets forms a: A. micronet B. C. TDD net D.

multinet scatternet

53. The range of an IRDA system is: A. 1 meter C. 1 foot

10 meters 10 feet

B. D.

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54. The International Telecommunication Union's name for the new third generation global standard for mobile telecommunications A. UMTS B. IMT-2000 C. 3GPP D. HSDPA 55. ________ is a second-generation cellular phone system based on CDMA and Direct Sequence Spread Spectrum (DSSS). A. GPRS B. IS-95 C. EDGE D. IMT-MC 56. In a ______ handoff, a mobile station can communicate with two base stations at the same time. A. medium B. break before make C. hard D. make before break 57. In AMPS, each band is divided into ______ channels. A. 666 B. 800 C. 900 D. 1000 58. Wi-Fi is capable of transmitting data normally at distances up to about A. 300 feet at a data rate of 11 megabits per second B. 3000 feet at a data rate of 11 megabits per second C. 300 feet at a data rate of 54 megabits per second D. 3000 feet at a data rate of 54 megabits per second 59. A CDMA mobile measures the signal strength from the base as -100dBm. What should the mobile transmitter power be set to? A. 320 mW B. 43 mW C. 250 mW D. 54 mW 60. Calculate the maximum and minimum hopping rate for the Bluetooth system. A. 600 & 30 Hz B. 1600 & 320 Hz C. 160 & 4320 Hz D. 100 & 20 Hz 61. The technical organs of the United Nations specialized agency for telecommunications. They functions trough international committees of telephone administration and private operating agencies. A. CEPT B. ETSI C. CCITT D. ANSI


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62. _______ is an analog cellular phone system using FDMA. A. AMPS B. D-AMPS C. GSM D. IS 136 63. Used to uniquely identify each location area in the GSM PLMN. When the system receives an incoming call it knows in which location area it should page the mobile and does not page the entire network. A. CGI B. BSIC C. IMSI D. LAI 64. Used for cell identification within the GSM network. A. CGI B. MCC C. PLMN D. TMSI 65. The number of channel of GSM 1900. A. 125 C. 300

B. D.

375 150

66. Channel used for sending signal strength & bit error rate measurement of the serving cell and signal strength of the BCCHs of the neighboring cells and used for sending short text message to busy mobile. A. SDCCH B. RACH C. SACCH D. PCH 67. D-AMPS uses ______ to divide each 25-MHz band into channels. A. FDMA B. TDMA C. CDMA D. Both A and B 68. IS-95 uses the _______ satellite system for synchronization. A. Teledesic B. Iridium C. Globalstar D. GPS 69. Bluetooth is operating on what frequency band. A. 2.4 GHz B. 1.4 GHz C. 5.4 GHz D. 3.4 GHz 70. The bandwidth of GSM 1800. A. 75 MHz C. 60 MHz

B. D.

25 MHz 80 MHz

71. In the Bluetooth technology, the SIG is the core pioneer of this technology, what does SIG mean? A. Special Information Guide B. Super Intelligent Group C. Super Interest Group D. Special Interest Group 72. The downlink frequency of GSM 1900. A. 1910-1990 MHz C. 1910-1970 MHz

B. D.

1930-1990 MHz 1905-1980 MHz


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73. When the signal from the mobile cellular unit drops below a certain level, what action occurs? A. the MSTO increases power level B. the units is “handed off” to a closer cell C. the cell site switches antennas D. the call is terminated 74. What is the bandwidth of a cellular CDMA system? A. 1.23 GHz B. 1.23 KHz C. 1.23 MHz D. 1.23 THz 75. Cellular mobile system was first operated in _____. A. 1983 B. 1985 C. 1979 D. 1981 76. TACS is a cellular system with ____ channels. A. 832 B. 1000 C. 666 D. 200 77. CDMA technology was invented by: A. AT&T C. Bell Labs

B. D.

Lucent Qualcomm

78. Suppose we are designing picocells for an indoor cellular system in an underground mall. If we assume people using the system are walking at a speed of 5m/s or less, what is the smallest cell radius we could have if we wanted to have handoffs occur no more than once every 45 seconds? A. 115 m B. 1.125 m C. 11.25 m D. 112.5 m 79. AMPS has a frequency reuse factor of _______. A. 1 B. 3 C. 5 D. 7 80. GSM allows a reuse factor of _______. A. 1 B. C. 5 D.

3 7

81. The duplex distance of GSM 900. A. 80 MHz C. 45 MHz

75 MHz 25 MHz

B. D.

82. The Bluetooth original core SIG is composed of these companies except ____. A. Ericsson B. Sony C. Nokia D. Toshiba 83. A cellular system is designed to operate reliably with traffic of 2000 E. If each customer on average uses the phone for 3 minutes during the busiest hour, how many customers can be accommodated, assuming an even distribution of customers? A. 30,000 B. 20,000 C. 40,000 D. 50,000


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84. The bandwidth of GSM 1900. A. 60 MHz C. 75 MHz

B. D.

25 MHz 80 MHz

85. The "forward" PCS channel is: A. from the base to the mobile C. from mobile to mobile

B. D.

from the mobile to the base same as the uplink

86. In an IS-95 system, the frequency-reuse factor is normally _____. A. 1 B. 3 C. 5 D. 7 87. In the third generation of cellular phones, ______ uses a combination of WCDMA and TDMA. A. IMT-TC B. IMT-MC C. IMT-DS D. IMT-SC 88. In GSM, voice channels are called: A. traffic channels C. bearer channels

B. D.

voice channels talking channels

89. A vehicle travels through a cellular system at 100 kph. often will handoffs occur if the cell radius is 500 meters? A. 36 s B. 4.8 s C. 9.6 s D. 8.3 s 90. AMPS operates in the ISM _____ band. A. 800-MHz B. C. 1800-MHz D.

Approximately how

900-MHz 2484-MHz

91. A handshaking procedure where the cell site sends a control message on the forward voice channel confirming the channel. A. CMAC B. SAT C. SID D. ESN 92. CDMA uses a set of PN sequences that are: A. common B. unique C. rotating D. orthogonal 93. BSS stands for: A. Basic Service Set C. Bluetooth Service System

B. D.

Basic Service System none of the above

94. The downlink frequency of GSM 1800. A. 1810-1880 MHz C. 1710-1785 MHz

B. D.

1710-1770 MHz 1805-1880 MHz

95. The kHz A. C.

analog cellular standards which allows voice messages to travel through 30 voice channels. NAMPS B. DAMPS AMPS D. NMT


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96. A means of increasing the capacity of a cellular system by subdividing cells onto two or more smaller cells. A. cell splitting B. channel splitting C. cell allocation D. channel allocation 97. The uplink frequency of GSM 900. A. 890-920 MHz C. 890-915 MHz

B. D.

890-900 MHz 890-935 MHz

98. Bluetooth is operating in an unlicensed ISM band, what does ISM stand for? A. Industrial Scientific Medicine B. Infinite Scientific Mechanism C. Integrated System Modulation D. Indefinite Social Market 99. Channel that enable the mobile synchronize to the frequency being broadcast by the BTS. A. FCCH B. RACH C. PCH D. SCH

A. C.

______ is a digital version of AMPS. GSM B. D-AMPS D.

A. C.

_______ is a second-generation cellular phone system used in Europe. IS 54 B. IS-95 GSM D. IS 136

A. C.

AMPS uses ______ to divide each 25-MHz band into channels. TDMA B. CDMA FDMA D. SSMA

A. C.

In the third generation of cellular phones, _______ uses TDMA. IMT-TC B. IMT-MC IMT-DS D. IMT-SC

100.

101.

102.

103.

104.

IS 54 IS-95

The SID is used by a cell phone to: A. identify the type of system (analog or digital) B. recognize an AMPS system C. set its transmitted power level D. recognize that it is "roaming" A. C.

In an AMPS system, voice is sent using: AM B. FSK D.

A. C.

In an AMPS system, control-channel signals are sent using: AM B. FM FSK D. CDMA

A. C.

An industrial specification for wireless personal area networks GPRS B. WAP WiFi D. Bluetooth

105.

106.

107.

FM CDMA


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A. C.

The name Bluetooth was born from the 10th century king of ______. Finland B. Sweden Denmark D. France

A. C.

What is the approximate range and power output of class I Bluetooth device? 100 m, 20 dBm B. 10 m, 4 dBm 10 cm, 0 dBm D. 100 mm, 5 dBm

108.

109.

110.

What is the approximate range and power output of class II Bluetooth device? A. 100 m, 20 dBm B. 10 m, 4 dBm C. 10 cm, 0 dBm D. 100 mm, 5 dBm

111.

A. C. 112.

Bluetooth is also known as IEEE 802.12.1 IEEE 802.11b

B. D.

IEEE 802.15x IEEE 802.15.1

What is the approximate range and power output of class III Bluetooth device? A. 100 m, 20 dBm B. 10 m, 4 dBm C. 10 cm, 0 dBm D. 100 mm, 5 dBm Bluetooth specification allows connecting 2 or more piconets together to form

113.

a A. C.

nanonet femtonet

B. D.

scatternet broadnet

114.

In order to avoid interfering with other protocols which use the 2.45 GHz band, the Bluetooth protocol divides the band into A. 49 channels (each 1 MHz wide) and changes channels up to 1600 times per second B. 59 channels (each 5 MHz wide) and changes channels up to 1600 times per second C. 79 channels (each 1 MHz wide) and changes channels up to 1600 times per second D. 89 channels (each 10 MHz wide) and changes channels up to 600 times per second

115.

Bluetooth function as _______, whereas Wi-Fi is _______ A. wireless Ethernet, wireless USB B. wireless network, wireless computer C. wireless USB, wireless Ethernet D. wireless computer, wireless network

116.

What makes the wireless channel more unpredictable than the wireline channel? A. Doppler spread B. Multipath C. Shadow fading D. All of the above

117.

A. C.

Rayleigh fading occurs when there is a LOS component present B. Intercell interference D.

no LOS component present Multipath fading


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118.

Spread spectrum systems employ a form of diversity called _____, where the signal is spread over a much larger bandwidth than is needed A. Radio diversity B. Spatial diversity C. Frequency diversity D. Bandwidth diversity

119.

The use of equalizers adds to the complexity and costs of the ______ systems, because equalization requires significant amounts of signal-processing power. A. OFDM B. CDMA C. TDMA D. FDMA

120.

With ______, fast and accurate power-control loops must be used so that the signal from one user does not drown out the signals from other users, and each cell must be tightly synchronized. A. CDMA B. TDMA C. OFDM D. FDMA

121.

A. C.

How many conversations per channel can TDMA digital cellular carry at once? 1 B. 2 3 D. 10

122.

Larger cells are more useful in A. rural areas B. mountainous areas C. densely populated urban areas D. lightly populated urban areas

123.

Interference effects in cellular systems are a result of A. the power of the transmitters B. the distance between areas C. the height of the antennas D. the ratio of the distance between areas to the transmitter power of the areas

124.

A. C. 125.

Which is one method of securing cellular transmissions? frequency hopping B. packet coding cell identification D. cell jamming

Which of these measures provides security to mobile phone transmissions by varying the transmitted frequency? A. direct sequencing B. frequency division multiplexing C. frequency hopping D. code division multiplexing


Section 22 Navigation System Section 23 RADAR Fundamentals

RADAR RCS

Navigational Aids Loading ECE SUPERBook

6-1


Section

22

Navigation

System


DEFINITION. Navigation: Art and science of maneuvering safely and efficiently from one point to another. The word navigation (Latin navis, “boat”; agire, “guide”) traditionally meant the art or science of conducting ships and other watercraft from one place to another. DEFINITION. Bearing: The Angular distance of any terrestrial object from an observer, or from another object measured clockwise from a reference north pointing through 360° on the arc of the horizon. DEFINITION. Course: The angular distance of a ship’s intended line of movement measured clockwise from a reference north point through 360° on the arc of the horizon. A. .MAJOR NAVIGATION AGENCIES. 1.

Federal Aviation Administration (FAA), Washington, D.C. Operates navigational aids and air traffic control systems for both civil and military aircraft in the US and its possessions.

2.

Federal Communications Commission (FCC), Washington, D.C. The agency that licenses transmitters and operators in the United States and aboard US registered ships and aircraft.

3.

International Air Transport Association (IATA). Montreal, Canada The international association representing scheduled airlines.

4.

International Civil Aviation Organization (ICAO), Montreal, Canada A United Nations agency that allocates standards and recommended practices, including navigational aids, for all civil aviation.

5.

International Telecommunication Union (ITU), Geneva, Switzerland An agency of the United Nations that allocates frequencies for best use of the radio spectrum.


NAVIGATION system

6-2

B. .METHODS OF NAVIGATION. METHODS

CELESTIAL NAVIGATION

NAVIGATION BY COASTAL PILOTING

NAVIGATION BY DEAD RECKONING

ELECTRONIC NAVIGATION

PRINCIPLES OF OPERATION When no landmarks or aids to navigation are visible, navigators may use the Sun, the Moon, or other celestial bodies to fix the craft’s position. In celestial navigation, navigators measure the altitude of a celestial body to derive a circle of position. Altitude of a celestial body refers to its angle, in degrees, above the horizon. From every point on the circle of position, the altitude of the celestial body is the same. Coastal piloting means navigating within sight of land. In coastal piloting, navigators determine their position more accurately by taking compass bearings —angular measurements of the line of sight between the craft and nearby landmarks or aids to navigation. Dead Reckoning, basic method of navigation in which the position of a ship or aircraft is determined by calculation from a previous position of the craft, the direction of travel from the previous position, the speed of the craft, and the time traveled. Modern navigators rarely rely exclusively on their own measurements and calculations. They often use position calculations derived by high-tech electronic navigational instruments. These instruments usually can determine positions faster and more accurately than humans. They function in nearly all weather conditions, day or night, and have a range far beyond that of the human senses alone.


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C. .RADIO FREQUENCIES USED IN ELECTRONIC NAVIGATION. SYSTEM

FREQUENCY BAND

1. Omega 2. VLF 3. Decca 4. Loran-C/D 5. LFR 6. ADF/NDB 7. Loran A 8. Marker Beacon 9. ILS Localizer 10. VOR 11. ILS Glide Slope 12. DME 13. TACAN 14. ATCRBS 15. GPS 16. Altimeter 17. MLS 18. Weather Radar 19. Doppler Radar

10 to 13 kHz 16 to 24 kHz 70 to 130 kHz 100 kHz 200 to 300 kHz 200 to 1,600 kHz 2 MHz 75 MHz 108 to 112 MHz 108 to 118 MHz 329 to 335 MHz 960 to 1215 MHz 960 to 1215 MHz 1030 to 1090 MHz 1227 to 1575 MHz 4200 MHz 5 GHz 5, 9 GHz 10 to 20 GHz

D. .MAJOR ELECTRONIC NAVIGATION SYSTEM. 1.

Automatic Direction Finding/Radio Direction Finding (ADF/RDF) The oldest navigational system employing a combined rotating loop and sense antenna. ADF Basic Operation The ground station transmits an amplitude modulated signal in an omnidirectional pattern. The receiver at the aircraft receives the transmitted signal in a combined loop and sense antenna whose output is then calculated to give relative station bearing The relative station bearing output from the ADF receiver drives the bearing pointers in Radio Magnetic Indicator (RMI) and EHSI’s (Electronic Horizontal Station Indicator) of the Electronic Flight Instrument System (EFIS) The pointer position with respect to the fixed lubber line is the relative bearing while the position with respect to the North of the compass is the magnetic bearing. Direction Finding Errors i. Coastline refraction (Land effect) Radio waves crossing from a land to a water area, at any angle rather than 90° will refract, or bend towards the coastline.


NAVIGATION system

6-4

ii.

Non-opposite minimums (Antenna effect) This is a form of loop unbalance due to stray signal pickup through earphone cords or power lines or improper shielding or grounding of the loop.

iii. Reradiation Signals striking the loop may also strike nearby metal objects, inducing current in them that produce reradiated signals and also pickup by the loop. iv. Great Circle error Errors produce because of earth’s curvature. v.

Polarization Error (Night effect) During nighttime hours the sky wave refracts from the ionosphere and is receivable for long distance. When the sky wave amplitude approaches the ground wave amplitude, the null of a loop antenna will no longer be reliable.

Read it till it Hertz…jma Are you familiar with the Main antenna Switch mode? ª

Main transmitter--In this position, the main antenna is connected to the main receiver and to the main transmitter through a breakin relay.

ª

Emergency transmitter--The main antenna is connected to the emergency receiver and to the emergency transmitter through a break-in relay.

ª

AA--The main antenna is connected to the auto-alarm receiver, and power is connected to the AA equipment. All transmitters are disabled by interconnecting circuits.

ª

DF (Direction Finder)--The main antenna is open-circuited, or an auxiliary antenna is connected to the main receiver, depending on the antenna conditions when the DF is calibrated. The transmitters are disabled.

ª

AA-DF--The main antenna or an auxiliary antenna is connected to the auto alarm, as when the DF was calibrated. Power is applied to the AA equipment.

ª

HF--The main antenna is connected to the HF transmitter.


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VHF Omnidirectional Range (VOR) Navigation system that provides omnidirectional bearing, establishes the airways for airplane, and provides station identification. VOR Basic Operation The VOR beacon is a ground station that transmits signals in all directions (omnidirectional radials). The transmitted radials all contain different information, which makes it possible to separate the radial from each other.

The VOR beacon transmits the fist radial in the direction magnetic north of the station we call it the zero degrees radial.

The VOR beacon transmits on its carrier frequency two modulated signals, a 30-Hz reference signal and a 30-Hz variable signal. The radial is measured from the difference in phase between the reference and variable signal.

The 30-Hz reference-phase signal always has the same phase whatever the aircraft’ position is with respect to the station.

The 30-Hz amplitude modulated variable-phase signal is generated by a goniometer, a mechanically rotatable variable RF transformer device.

The phase of the variable signal varies with the aircraft’s position.

VOR Deviation VOR magnetic bearing sine and cosine signals are combined with selected course information. The difference between these two parameters is the VOR deviation.

For a difference of 5° left the receiver output is standardized at 75μA which causes a VOR deviation bar deflection of one dot.

For a difference of 20°, the receiver output is 300μA produces a four dot deflection.

FOR ADF/RDF: Relative Bearing = Pointer Position w.r.t. the fixed Lubberline (Aircraft Heading) Magnetic Bearing = Pointer position w.r.t. the North of the compass card FOR VOR: Radial = Phase of Reference Signal – Phase of Variable Signal VOR Magnetic Bearing = Received Radial + 180° VOR Deviation = Selected Course – VOR MB


NAVIGATION system

6-6

VOR Summary

Parameter

3.

Description

Guidance information

Omnidirectional bearing

Operating Frequency

108 to 118 (117.95) MHz

Service Area

200 mi for high- flying aircraft (Enroute VOR) 25 mi for low-flying aircraft (Terminal VOR)

RF output

25 to 200 W

No of channels

20 or 40

Spacing between adjacent channels

100 kHz

Antenna pattern

Cardioid pattern rotates at a 30 Hz rate

Accuracy

+ 1.4°, + 3° with site error

Distance Measuring Equipment (DME) Distance Measuring Equipment is an interrogator-transponder two-way distance ranging system. It became an ICAO standard in 1959. The airborne equipment (interrogator) generates a pulsed signal that is recognized by the ground equipment (transponder). DME Summary

Parameter

Description


Slant distance and Time-to-Station

Operating Frequency

960 to 1215 MHz 1kW pulses of 3.5μs duration, 30 times per second

Interrogator RF output No of channels

126


1 MHz

Ground Transponder interrogation capacity

100 aircraft simultaneously

Time Delay

50 μs

Transponder reply

63 MHz above or below the interrogating channel

Accuracy

+ 0.2 mi


6-7

4.

Tactical Air Navigation (TACAN) TACAN is a NATO military system that adds a bearing function to DME, on the same frequencies, allowing greater portability of the ground station than with ICAO’s VOR/DME, particularly on aircraft carriers.

5.

Collocated VOR/TACAN station (VORTAC) In NATO countries having a common air traffic control system for the civil and the military, the ICAO Rho-theta system is implemented. Rho-theta system is a generic term for navigation system that derives position by measurement of distance and bearing from a single collocated VOR/DME or VOR/TACAN station.

6.

Instrument Landing System (ILS) An ILS normally consists of two or three marker beacons, a localizer, and a glide slope to provide both vertical and horizontal guidance information. i.

Localizer Basic Operation The localizer signal comes from a transmitter located at the end of the runway that operates in the frequency range from 108 to 111.95 MHz.

The localizer transmits two beams, one on the right side of the runway center line amplitude modulated by a 150-Hz audio signal and one on the left side of the runway center line amplitude modulated by a 90-Hz audio signal.

When the airplane position is on the left of the on-course path, the strength of the 90-Hz audio signal predominates and localizer deviation indicators are deflected to the right and a yellow visual lamp indicator are on, indicating that the runway centerline is to the right.

While when the airplane position is on the right side of the oncourse path, the strength of the 150-Hz audio signal predominates and localizer deviation indicators are deflected to the left and a blue visual lamp indicator are on, indicating that the runway centerline is to the left.


NAVIGATION system

6-8

ILS Summary

Parameter

Description


Azimuth (horizontal or lateral)

Operating Frequency

108 to 112 MHz band

RF power

25 W

Location No of channels

1000 feet beyond the approach end of the runway. 40; where each is paired with one of 40 Glideslope channel

Spacing between adjacent channel

50 kHz

Left-Hand Pattern

Amplitude-modulated by 90 Hz

Right-Hand Pattern


Accuracy

+ 0.1° + 35° from the centerline of the runway

Lateral threshold ii.

Glide-Slope Basic Operation The glide-slope signal comes from a transmitter at the beginning of the runway that operates in the UHF band with a frequency range of 328.6-MHz to 335.4-MHz.

The glide-slope receiving circuits are tune automatically whenever a localizer frequency is tuned or selected.

Two antenna patterns are produced: one above the normal 2.5° ascent angle to the runway’s surface at an audio modulated tone of 90-Hz and one below the normal 2.5° ascent angle to the runway’s surface at an audio modulated tone of 150-Hz.

If the airplane is located above the 2.5° glide-path, the 90-Hz audio signal modulates the glide slope carrier forcing the glide-slope pointer to deflect downward and turn-on the yellow visual indicator lamp, indicating that the airplane is above the glide-path and the pilot must maneuver the airplane below to obtain the desired glide-path.

While if the airplane is located below the 2.5° glide-path, the 150-Hz audio signal modulates the glide-slope carrier forcing the glide-slope pointer to deflect upward and turn-on the blue visual indicator lamp, indicating that the airplane is below the glide-path and the pilot must maneuver the airplane above to obtain the desired glide-path.


6-9

Glideslope Summary

PARAMETER

DESCRIPTION


Vertical

Operating Frequency

328.6 to 335.4 MHz band

RF power

7W

Location No of channels

Approach end of the runway and up to 500 ft to the side 40; where each is paired with one of 40 Localizer channel

Spacing between adjacent channel

50 kHz

Pattern above the Glideslope


Pattern below the Glideslope


Accuracy

+ 0.1°

Vertical threshold

+ 3° above the horizontal

iii. Marker Beacons The marker beacon system gives information about the distance to the runway. Marker Beacons Three marker beacon are strategically positioned at fixed distance from the runway having the same carrier frequency of 75-MHz. If the airplane is 7.2-km away from the runway center line, the pilot will hear a Morse-coded sound of 2 dashes/sec and the Blue (Purple in other books) light indicator will turn on. If the airplane is 1.05-km away from the runway center line, the pilot will hear an alternating Morse-coded sound of dots and dashes and the Amber (Orange in other books) light indicator will turn on. And finally, if the airplane is 75-m away from the runway center line, the pilot will hear a continuous dots and the White light indicator will turn on, an indication that the airplane is very near the runway threshold.


NAVIGATION system

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Marker Beacon Summary

Marker

Modulating Frequency

Typical Location

Visual (Light ) Indicator

Aural (Sound) Indicator

Outer

400 Hz

7200 m

Blue (Purple)

2 dashes/sec

Middle

1300 Hz

1050 m

Amber (Orange)

Alternating dots and dashes

Inner

3000 Hz

75 m

White

Continuous dots

ICAO Categories for Minimum Approach Ceiling and Forward Visibility.

Category

7.

Description

I

200 ft ceiling and ½ mile visibility

II

100 ft ceiling and ¼ mile visibility

III-A

50 ft ceiling and 700 ft visibility

III-B

35 ft ceiling and 150 ft visibility

III-C

0 ceiling and 0 visibility

Microwave Landing System (MLS) The MLS is the ICAO-approved replacement for the current ILS. The system is based on time-reference scanning beams, reference to the runway, which enables the airborne unit to determine precise azimuth and elevation angle.

Parameter Operating Frequency Azimuth (lateral) scan Elevation (vertical) scan

Description 5000 to 5250 MHz 60° either side of the runway centerline 0 to 30°

Scanning rate

50μs/degree

No of channels

200


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Air Traffic Control Radar Beacon System (ATCRBS) In 1958 IFF (Identification Friend or Foe) became an ICAO standard known as SSR (Secondary Surveillance Radar) or ATCRBS.

Parameter

Description


IFF, Range, Bearing, and Altitude information

Operating Frequency

1030 to 1090 MHz

Ground Interrogator

1030 MHz

Airborne Transponder

1090 MHz


1 MHz

Power output

500 W

Available Pulse code

4096

Interrogator Pulse

400 pulses pair/s

Reply Pulse

14 pulses lasting 21μs

E. .HYPERBOLIC NAVIGATION SYSTEM. 1.

Decca This system has been developed in the United Kingdom starting 1939. It is a continuous-wave hyperbolic system operating in the 70 to 130 kHz.

Position information is obtained by measuring the relative phase differences of the received signals. A typical chain comprises four stations, 1 Master and 3 Slaves. 2.

Omega The Omega system (8-station only) is a worldwide VLF navigation system used for marine and enroute air navigation. The system 8 CW transmitting stations sequentially transmit long, but precisely timed, pulses at four frequencies: 10.2 kHz, 11.3 kHz, 13.6 kHz, and 11.05 kHz.

Position information is obtained by measuring the relative phase differences of the received signals. A typical chain comprises two stations, 1 Master and 1 Slave station Loading ECE SUPERBook

NAVIGATION system

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3.

Loran-C Loran-C is a long-range hyperbolic radio navigation system that possesses an inherent high degree of accuracy at ranges of 800 to 1000 nautical miles.

Position information is obtained by measuring the relative time differences of the received signals. A typical chain comprises three stations, 1 Master and 2 Slaves station. F. .SATELLITE NAVIGATION SYSTEM. 1.

TRANSIT Transit is a satellite navigation system consisting of four or more satellites in approximately 600-nmi polar orbit.

2.

Navigation System using Time And Ranging/Global Positioning System (NAVSTAR/GPS) Global Positioning System (GPS), space-based radionavigation system, consisting of 24 satellites and ground support. GPS provides users with accurate information about their position and velocity, as well as the time, anywhere in the world and in all weather conditions.


Three satellites are needed to determine latitude and longitude, while a fourth satellite is necessary to determine altitude.

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GPS Services a. Standard Positioning Service (SPS) The SPS is a positioning and timing service that is available to all GPS users on a continuous, worldwide basis with no direct charge. SPS provides a horizontal position that is accurate to about 100 m. b.

Primary Positioning Service (PPS) The PPS is a highly accurate military positioning, velocity, and timing service that is available on a continuous, worldwide basis to users authorized by the DoD.

Accuracy

3.

SPS

PPS

Horizontal

100 m

22 m

Vertical

156 m

27.7 m

3-D

185 m

35.4 m

Time transfer

340 ns

200 ns

Global Orbiting Navigation Satellite System (GLONASS) Satellite navigation system developed by Russia similar to Navstar GPS. GLONASS Services a.

Civil Service The specified positioning accuracies are 100-m in the horizontal plane and 150-m in the vertical plane. Civil velocity accuracy is specified at 0.15 m/sec. The time dissemination capability is within 5 msec of UTC.

b.

Military Service Military service yields accuracies comparable to the GPS PPS.


NAVIGATION system

6-14

Comparison between GPS and GLONASS

Parameter Planned Constellation

NAVSTAR/GPS

GLONASS

21 + 3

21 + 3

Number of orbit

6

3

Orbital Altitude

10,898 nmi

10,313 nmi

Orbital Period

12 hrs

11 hrs 15 min

Orbital Inclination

55°

64.8°

Access Method

CDMA

FDMA

C/A Code

1,023 bits

C/A Code BW

2 MHz

1 MHz

Bit Rate

50 bps

50 bps

511 bits

P code is a long precision code with 10.23-MHz chip rate used by the military. C/A code stands for Course/Acquisition code for public used.

G. .CARDINAL DIRECTIONS OF THE EARTH.

Lines of Latitude The parallel lines of latitude are horizontal, running from east to west. The equator is the imaginary line from which latitude is measured; it is equidistant from the poles, dividing the globe into the northern and southern hemispheres.

Meridians of Longitude Lines running from top to bottom (north to south) on a map, chart, or globe. A longitude coordinate indicates distance from the prime meridian.


6-15

H. .AXIS AND DIRECTIONS.

1.

True North Axis (Geographic Axis) Axis around which Earth rotates, also called the North Pole. i.

2.

Magnetic North Axis The fluid motion of the earth’s outer core generates magnetism such that its magnetic field within the earth creates a magnetic axis which is an angle away from the geographic or true axis. ii.

3.

True North Directions Courses and bearings reckoned from TRUE north.

Magnetic North Directions Courses and bearings reckoned from MAGNETIC north.

Compass North Axis Any magnetic substance for instance, the steel in a ship can induce magnetism. The axis of reference of compass direction is called compass meridian, the magnetic field of which is the sum total of the ship’s magnetism and all other magnetism on board. iii. Compass North Directions Courses and bearings reckoned from COMPASS north.


NAVIGATION system

6-16

Read it till it Hertz…jma

I.

ª

The north magnetic pole is situated in the vicinity of Ellef Ringnes Island in northern Canada, about 1300 km (about 800 mi) from the North Pole. This point lies at approximately latitude 74° north, longitude 101° west—about 1,600 km (about 1,000 mi) from true north.

ª

The south magnetic pole is located just off of the coast of Wilkes Land, Antarctica. It is 2550 km (1600 miles) from the geographic South Pole.

ª

In 1730, English mathematician John Hadley and American inventor Thomas Godfrey independently invent the sextant, an optical instrument used for the measurement of angular distance between any two objects.

.COMPASS ERRORS. 1.

Variation

An error of the compass indicated by the angle between the meridian of true north and the meridian of magnetic north.


6-17

Deviation

An error of the compass indicated by the angle between the meridian of magnetic north and the meridian of the compass north.

J.

.COURSES & BEARINGS. 1.

Course (Heading) Course is the angular distance of a ship’s direction of movement on the surface of the earth, measured clockwise from three reference points of the earth –true north, magnetic north and compass north – through the 360° system of the compass.

2.

Bearing Bearing is the angular distance of an object from an observer, or from another object, measured clockwise from the same three points of references –true, magnetic and compass north –through the 360° system of the compass, giving rise to the three bearings in one direction. i.

Relative Bearing Relative Bearing is the angular distance of an object measured clockwise through 360° from the ships’ bow (intended line of movement).

ii.

True Bearing True Bearing is the angular distance of an object from an measured clockwise from the true north. Conversion from TB to RB and vice versa

TB = H + RB RB = TB − H H = TB − RB


NAVIGATION system

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iii. 4-Points Bearing 4-points bearing are eye approximations of relative bearings measured clockwise or anti-clockwise from the ship’s bow, stern or beams.

Sample Problem:

Find the compass course to steer in order to make a true course of 270°. Variation is 15° W and the deviation is 12° E.

Solution: True course Variation

270 ° 15 ° W

Magnetic course Deviation

255 ° 12 °E

Compass course

267 °

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I

H

1.

What is the frequency range of the ground-based Very High Frequency Omnidirectional Range (VOR) stations used for aircraft navigation? A. 108.00 kHz to 117.95 kHz B. 108.00 MHz to 117.95 MHz C. 329.15 MHz to 335.00 MHz D. 329.15 kHz to 335.00 kHz

2.

What is the main underlying operating principle of the Very High Frequency Omnidirectional Range (VOR) aircraft navigational system? A. A definite amount of time is required to send and receive a radio signal B. The difference between the peak values of two DC voltages may be used to determine an aircraft's altitude above a selected VOR station C. A phase difference between two AC voltages may be used to determine an aircraft's azimuth position in relation to a selected VOR station D. A phase difference between two AC voltages may be used to determine an aircraft's distance from a selected VOR station

3.

Choose the only correct statement about the effective range of a Very High Frequency Omnidirectional Range (VOR) station used for aircraft navigation. A. Its reception range is based on both the aircraft's altitude and the aircraft's line-of-sight to the VOR station B. Its reception range is not a function of the aircraft's altitude C. Its reception range is not a function of the aircraft's longitude and latitude position in relation to the VOR station's position D. Its reception range is greatly affected by atmospheric effects and propagation anomalies

4.

What is the name of the mechanically rotatable variable RF transformer device of a Very High Frequency Omnidirectional Range (VOR) station that is used to generate the amplitude modulated variable phase signal used in aircraft navigation? A. A ghandimeter B. A gondolameter C. A goniometer D. A gorgonzolameter

5.

Are courses and bearings reckoned from magnetic north? A. Magnetic directions B. Compass directions C. Deviation D. Variation


NAVIGATION system

6-20

6.

What is the frequency range of the localizer beam system used by aircraft to find the centerline of a runway during an Instrument Landing System (ILS) approach to an airport? A. 108.10 kHz to 111.95 kHz B. 329.15 MHz to 335.00 MHz C. 329.15 kHz to 335.00 kHz D. 108.10 MHz to 111.95 MHz

7.

An error of the compass indicated by the angle between the meridian of true north and the meridian of magnetic north. A. Variation B. Compass error C. PPI error D. Deviation

8.

Courses and bearings reckoned from compass north. A. Magnetic directions B. PPI direction C. Compass directions D. Gyrocompass error

9.

The angular distance of any object measured clockwise from the ship’s bow through 360° on the arc of the horizon. A. Course bearing B. Four point bearing C. Compass bearing D. Relative bearing

10. Are eye approximations of relative bearings measured clockwise or anticlockwise from the ship’s bow, stern or beams. A. Relative bearing B. Four point bearing C. Course bearing D. Compass bearing 11. What is the frequency range of the used to indicate an aircraft's slant based navigation station? A. 962 MHz to 1213 MHz C. 108.00 MHz to 117.95 MHz

Distance Measuring Equipment (DME) range distance to a selected groundB. 108.10 MHz to 111.95 MHz D. 329.15 MHz to 335.00 MHz

12. A device use for determining magnetic directions by using a magnetic needle that is free to turn around to align itself with the magnetic meridian of the earth then point to magnetic north. A. PPI B. GCA C. Magnetic compass D. Gyrocompass 13. Choose the only correct statement about the glideslope beam system used by aircraft to maintain the proper ascent angle to the surface of a runway during an Instrument Landing System (ILS) approach to an airport. A. Two antenna patterns are produced; one to the left of the runway's centerline at an audio modulated tone of 90 Hz and one to the right of the runway's centerline at an audio modulated tone of 150 Hz B. Two antenna patterns are produced; one to the left of the runway's centerline at an audio modulated tone of 150 Hz and one to the right of the runway's centerline at an audio modulated tone of 90 Hz C. Two antenna patterns are produced; one above the normal 2.5 degree ascent angle to the runway's surface at an audio modulated tone of 150 Hz and one below the normal 2.5 degree ascent angle to the runway's surface at an audio modulated tone of 90 Hz


6-21

D. Two antenna patterns are produced; one above the normal 2.5 degree ascent angle to the runway's surface at an audio modulated tone of 90 Hz and one below the normal 2.5 degree ascent angle to the runway's surface at an audio modulated tone of 150 Hz 14. The angular distance of a ship’s intended line of movement measured clockwise from a reference north point through 360° on the arc of the horizon. A. Directions B. Bearing C. Axis D. Course 15. An error of the compass indicated by the angle between the meridian of magnetic north and the meridian of the compass north. A. Variation B. Deviation C. Magnetic directions D. Compass error 16. M.V. Princess Lorena is proceeding on course 060° C. Deviation for the ship’s compass heading is 9° E Variation for the place is 19° W. What is the true course? A. 050° T B. 28° T C. 5° T D. 10° T 17. What is the distance to a selected Distance Measuring Equipment (DME) station if an aircraft receives the ground station's reply 175 microseconds after it transmits its airborne interrogation signal? Use the standard 50 microsecond DME reply delay. A. 20.2 nautical miles B. 11.6 statute miles C. 14.2 nautical miles D. 10.1 statute miles 18. What is the frequency range of the marker beacon system used to indicate an aircraft's position during an Instrument Landing System (ILS) approach to an airport's runway? A. The outer, middle, and inner marker beacons' UHF frequencies are unique for each ILS equipped airport to provide unambiguous frequency protected reception areas in the 329.15 MHz to 335.00 MHz range B. The outer marker beacon's carrier frequency is 400 MHz, the middle marker beacon's carrier frequency is 1300 MHz, and the inner marker beacon's carrier frequency is 3000 MHz C. The outer, the middle, and the inner marker beacon's carrier frequencies are all 75 MHz but the marker beacons are 95% tonemodulated at 400 Hz (outer), 1300 Hz (middle), and 3000 Hz (inner) D. The outer marker beacon's carrier frequency is 3000 kHz, the middle marker beacon's carrier frequency is 1300 kHz, and the inner marker beacon's carrier frequency is 400 kHz


6-22

NAVIGATION system

19. What is the main underlying operating principle of an aircraft's Distance Measuring Equipment (DME)? A. A measurable amount of time is required to send and receive a radio signal through the earth's atmosphere B. The difference between the peak values of two DC voltages may be used to determine an aircraft's distance to another aircraft C. A measurable frequency compression of an AC signal may be used to determine an aircraft's altitude above the earth D. A phase inversion between two AC voltages may be used to determine an aircraft's distance to the exit ramp of an airport's runway 20. A type of radar that provides the aircraft ID and altitude to the A TC ground controller. A. PSR B. SSR C. LADAR D. IFF 21. A(n) _______ is a transmitter receiver (transponder) that replies to an interrogating aircraft signal and identifies itself. A. radar beacon B. ILS C. VOR D. Localizer 22. What type of information is derived from an aircraft's mode C transponder transmission operating in the Air Traffic Control Radar Beacon System (ATCRBS)? A. Range, aircraft weight, and fuel aboard information B. Range, aircraft weight, and altitude information C. Range, fuel aboard, and altitude information D. Range, bearing, and altitude information 23. What type of encoding is used in an aircraft's mode C transponder transmission to a ground station of the Air Traffic Control Radar Beacon System (ATCRBS)? A. Differential phase shift keying B. Pulse position modulation C. Doppler effect compressional encryption D. Amplitude modulation at 95% 24. At a certain location, the magnetic variation is 11o 16' W. The ships compass has a deviation of 0.50oE. What must the compass leading be if the ship wishes to steer due E? A. 100o 46’ B. 101°46' C. 99046' D. 102o46' 25. What type of encryption is used in the P6 informational pulse of an aircraft's mode S transponder transmission to a ground station of the Air Traffic Control Radar Beacon System (ATCRBS)? A. Differential phase shift keying B. Pulse position modulation C. Doppler effect compressional encryption D. Amplitude modulation at 95%


6-23

26. What is the frequency range of an aircraft's radio altimeter? A. 962 MHz to 1213 MHz B. 329.15 MHz to 335.00 MHz C. 4250 MHz to 4350 MHz D. 108.00 MHz to 117.95 MHz 27. What type of transmission is radiated from an aircraft's radio altimeter antenna? A. An amplitude modulated continuous wave B. A pulse position modulated UHF signal C. A differential phase shift keyed UHF signal D. A frequency modulated continuous wave 28. What is the frequency range of an aircraft's Automatic Direction Finding (ADF) equipment? A. 190 MHz to 1750 MHz B. 108.10 MHz to 111.95 MHz C. 190 kHz to 1750 kHz D. 108.00 MHz to 117.95 MHz 29. Choose the only correct statement about an aircraft's Automatic Direction Finding (ADF) equipment. A. An aircraft's ADF transmission exhibits primarily a line-of-sight range to the ground-based target station and will not follow the curvature of the earth B. Only a single omnidirectional sense antenna is required to receive an NDB transmission and process the signal to calculate the aircraft's bearing to the selected ground station C. All frequencies in the ADF's operating range except the commercial standard broadcast stations (550 kHz to 1660 kHz) can be utilized as a navigational Non Directional Beacon (NDB) signal D. An aircraft's ADF antennas can receive transmissions that are over the earth's horizon (sometimes several hundred miles away) since these signals will follow the curvature of the earth 30. A ship’s gyrocompass is not working. Find the compass course of 327° T. Variation is 7° E and deviation is 9° W. A. 20° C B. 329° C C. 320° C D. 30° C 31. What is the cause of quadrantal error when using an aircraft's Automatic Direction Finding (ADF) equipment? A. Quadrantal error is caused by the altitude of an aircraft in relation to the curvature of the earth B. Quadrantal error is caused by the geographic position of an aircraft in relation to the North, South, East, or West quadrant of the earth C. Quadrantal error is caused by the presence of the aircraft in the electromagnetic field of the NDB transmission D. Quadrantal error is caused when the aircraft's ADF transmission is not attenuated sufficiently and is then received at an elevated power level


NAVIGATION system

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32. What is meant by the term, "night effect", when using an aircraft's Automatic Direction Finding (ADF) equipment? A. Night effect refers to the fact that all Non Directional Beacon (NDB) transmitters are turned-off at dusk and turned-on at dawn B. Night effect refers to the fact that Non Directional Beacon (NDB) transmissions can bounce-off the earth's ionosphere at night and be received at almost any direction C. Night effect refers to the fact that an aircraft's ADF transmissions will be slowed at night due to the increased density of the earth's atmosphere after sunset D. Night effect refers to the fact that an aircraft's ADF antennas usually collect dew moisture after sunset which decreases their effective reception distance from an NDB transmitter 33. An aircraft has a magnetic heading of 150 degrees and a relative bearing to a NDB station of 8O degrees. Determine the magnetic bearing of the station in degrees. A. 180o B. 210° C. 230o D. 120° 34. A navigational error which is caused by the location of navigational aid having bearing. A. Propagation error B. C. Polarization error D.

reflection of obstruction close to a normal concern on directional Site error Instrument error

35. What is the nominal glide path angle in an ILS? A. 5 o B. 1 o C. 3 o D. 30 o 36. What is the frequency range of the glideslope beam system used by aircraft to maintain the proper ascent angle to the surface of a runway during an Instrument Landing System (ILS) approach to an airport? A. 108.00 MHz to 117.95 MHz B. 329.15 kHz to 335.00 kHz C. 329.15 MHz to 335.00 MHz D. 108.10 kHz to 117.95 kHz 37. Choose the only correct statement about the localizer beam system used by aircraft to find the centerline of a runway during an Instrument Landing System (ILS) approach to an airport. A. If the ratio of the 90 Hz audio signal strength to the 150 Hz audio signal strength of the antenna patterns is equal; the aircraft is on the proper 2.5 degree approach glidepath B. If the strength of the 90 Hz audio signal is greater than the strength of the 150 Hz audio signal of the antenna patterns; the aircraft is to the left of the centerline of the runway C. If the strength of the 150 Hz audio signal is greater than the strength of the 90 Hz audio signal of the antenna patterns; the aircraft is to the left of the centerline of the runway


6-25

D. If the strength of the 150 Hz audio signal is greater than the strength of the 90 Hz audio signal of the antenna patterns; the aircraft is above the proper 2.5 degree approach glidepath 38. If the aircraft is flying off-path in the glide slope pattern, which modulation signal is prevailing or has the higher strength? A. 75 Hz B. 75 MHz C. 150 Hz D. 90 Hz 39. What is the slant range distance of an aircraft's Distance Measuring Equipment (DME)? A. It is the distance between two aircraft of different altitudes B. It is the distance between two ground-based navigation stations having differences in their elevations above mean sea level C. It is the line-of-sight distance between an aircraft and a selected ground-based navigation station D. It is the radius-of-the-earth distance between two ground-based navigation stations having the same elevations above mean sea level 40. The Angular distance of any terrestrial object from an observer, or from another object measured clockwise from a reference north pointing through 360° on the arc of the horizon. A. Course B. Four point bearing C. Relative bearing D. Bearing 41. What are the transmission and the reception frequencies of an aircraft's mode C transponder operating in the Air Traffic Control Radar Beacon System (ATCRBS)? A. Transmit at 1090 MHz and receive at 1030 MHz B. Transmit at 1030 kHz and receive at 1090 kHz C. Transmit at 1090 kHz and receive at 1030 kHz D. Transmit at 1030 MHz and receive at 1090 MHz 42. The compass error is 16° W and deviation 8° E. If M.V. Princess Alma is steering on course 245° C, what is the true course by which she is proceeding? A. 24° T B. 253° T C. 229° T D. 261° T 43. The use of telecommunication for automatic indicating and recording measurement at the distance from measuring instrument. A. Telemetry B. Telecomand C. Tracking D. Monitoring 44. Choose the only correct statement about the localizer beam system used by aircraft to find the centerline of a runway during an Instrument Landing System (ILS) approach to an airport. A. The localizer beam system operates within the assigned frequency range of 108.10 GHz to 111.95 GHz


NAVIGATION system

6-26

B.

The localizer beam system produces two amplitude modulated antenna patterns; one pattern above and one pattern below the normal 2.5 degree approach glidepath of the aircraft C. The localizer beam system's frequencies are automatically tuned-in when the proper glideslope frequency is selected on the aircraft's Navigation and Communication (NAV/COMM) transceiver D. The localizer beam system produces two amplitude modulated antenna patterns; one pattern with an audio frequency of 90 Hz and one pattern with an audio frequency of 150 Hz 45. Is a device, which, when activated by electricity, aligns itself with the geographic meridian and cause its needle to point to true north. A. PPI B. Gyrocompass C. Compass north D. Magnetic compass 46. A PPI cathode-ray tube as used in radar set A. is used to check the percentage of modulation B. indicates only the range of the target C. is used for receiver alignment D. indicates both the range and azimuth of a target 47. Courses and bearings reckoned from a certain reference north point. A. Deviation B. Directions C. Variation D. Compass bearing 48. The sum or difference of variation between the meridians of true north A. Compass error C. Magnetic error

and deviation indicated by the angle and compass north. B. Variation error D. Deviation error

49. The amplitude modulated variable phase signal and the frequency modulated reference phase signal of a Very High Frequency Omnidirectional Range (VOR) station used for aircraft navigation are synchronized so that both signals are in phase with each other at: A. 180 degrees South, true bearing position of the VOR station B. 360 degrees North, magnetic bearing position of the VOR station C. 180 degrees South, magnetic bearing position of the VOR station D. 0 degrees North, true bearing position of the VOR station 50. Find the variation A. 257° C. 272°

compass course to steer in order to follow the course 250° T if is 15°W and deviation is 7° W. C B. 22° C C D. 275° C

51. TACAN is a distance indication A. instrument-landing glide paths C. speed and height indication

B. D.

bearing and weather information bearing and distance indication


6-27

52. Which of the following is a navigational equipment which utilizes a ground radar system to determine the position of a plane during its approach? A. DME B. ODR C. GCA D. Sonar 53. What is the operating frequency of ILS Glideslope? A. 1030 to 1090 MHz B. 70 to 130 kHz C. 329 to 335 MHz D. 1227 to 1575 MHz 54. Refers to an effect of selective fading A. A fading effect caused by large changes in the height of the ionosphere, as experienced at the receiving station B. A fading effect caused by small changes is beam heading at the receiving and transmitting stations C. A fading effect caused by phase differences between radio wave components of the same transmission, as experienced at the receiving station D. A fading effect caused by phase differences between radio wave components of the same transmission, as experienced at the transmitting station 55. Which band in the radio spectrum does the radio navigational system Omega transmit? A. UHF B. VLF C. VHF D. HF 56. Term used to describe a process of approaching a desired point by directing the vehicle towards the point A. ILS B. ILS reference point C. Homing D. Heading 57. What is the operating frequency of ILS Localizer? B. 108 to 112 MHz A. 329 to 335 MHz C. 70 to 130 kHz D. 960 to 1215 MHz 58. An aircraft deviation measured by an ILS localizer A. Ground speed B. Altitude C. Vertical D. Horizontal 59. Refers to the vertical deviation of an airplane upon takeoff from the horizontal plane of the runway A. Distance of airplane from the runway B. A horizontal deviation of an airplane from its optimum path of descent along the axis of the runway C. Angle of landing base from horizontal plane it the runway D. Foremast 60. What navigational system technique uses an antenna directivity to reduce an undesired multipath signals? A. Pulse transmission B. Space diversity C. Frequency diversity D. Loran-C


NAVIGATION system

6-28

61. What is being measured in radar theory to complete the determination of a distance to a target or object, after a high radio frequency signal is transmitted to a target? A. Transmission time from source to target B. Transmission time from target to source C. Echo time off the object to the source D. Complete signal transmission from source and back 62. LORAN is a navigational system used primarily for A. blind landing B. obtaining your fixed location over large distances C. automatic collision warning D. approach control 63. What is the frequency used by Global Positioning System? B. 960 to 1215 MHz A. 70 to 130 kHz C. 1227 to 1575 MHz D. 1030 to 1090 MHz 64. Frequencies most affected by knife-edge refraction A. VHF and UHF B. VLF C. HF D. 100 kHz to 3 MHz 65. RADAR means A. Radio Distance and Ranging C. Radio Detection and Ranging

B. D.

Radio Delay and Ranging Radio and detection and Rating

66. What do you call a circuit that controls the magnetron output? A. Converter B. Modulator C. Inverter D. Impeller 67. How do you account the effect of selective fading on the transmitted signal terms of its bandwidth? A. Equally affirmative to both narrow and wide bandwidth B. More affirmative at wider bandwidth C. More affirmative at narrow bandwidth D. Dependent on receiver bandwidth 68. Which of the following systems is not used in radio detection and ranging? A. Frequency shift B. Pulse radar C. Amplitude modulation D. Frequency modulation 69. Referred to as the cycle time difference between the master and the slave signals to reach the receiver in the operation of loran navigational equipment A. Cycle time B. Station delay C. Echo delay D. Time delay 70. In radio navigation, one of the following deviations of an airplane is determined by an ILS localizer A. Curvature of path B. Horizontal C. Vertical D. Slope

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Section

Radar

23

Fundamentals


Radio - Relating to electromagnetic waves or electromagnetic phenomena with frequencies between 10 kHz and 300,000 MHz

DEFINITION.

Detection - The act of noticing or discovering the existence of something, or the state of having been detected.

DEFINITION.

DEFINITION. Ranging - The distance between something, especially a gun or a tracking device and the object it is aimed at RADAR - The use of reflected radio waves to determine the presence, location, and speed of distant objects. The system has military, lawenforcement, and navigational applications. Examples of its uses include the locating of enemy aircraft or ships and the monitoring of vehicle speeds.

DEFINITION.

A. .BASIC RADAR CONCEPTS. The electronics principle on which radar operates is very similar to the principle of sound-wave reflection. The radio-frequency (rf) energy is transmitted to and reflects from the reflecting object. A small portion of the energy is reflected and returns to the radar set. This returned energy is called an ECHO, just as it is in sound terminology. Radar sets use the echo to determine the direction and distance of the reflecting object.


RADAR FUNDAMENTALS

6-30

B. .RADAR REFERENCE COORDINATES. The locations of an object with respect to the antenna are described by three coordinates: 1.

Range The line from the radar set directly to the object is referred to as the line of sight (LOS). The length of this line is called range.

2.

Elevation Angle The angle between the horizontal plane and the LOS is the elevation angle.

3.

True bearing or Azimuth angle The angle measured clockwise from true north in the horizontal plane is called the azimuth angle.

C. .RADAR RANGE EQUATION . 1.

RADAR Range ª

In terms of received echo

To obtain RANGE In any units In nautical miles In yards

where: Δ t = time elapsed c = speed of light

General Solution cΔt 2 Δt (μs) = 12.36

R=

R (nmi)

R = 164 Δt

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Sample Problem:

Calculate the range of a steady target that produced an echo and is received by the radar equipment once every 385μs.

Solution: cΔt 162,000(385 x 10 −6 ) = = 31.185 mi 2 2 Alternate solution 385 R(nmi) = = 31.15 nmi 12.36 R =


Electromagnetic energy travels through air at approximately the speed of light, which is 186,000 statute miles per second.

ª

The Navy uses nautical miles to calculate distances; 186,000 statute miles is approximately 162,000 nautical miles.

ª

While the distance of the statute mile is approximately 5,280 feet, the distance for a nautical mile is approximately 6,080 feet.

2.

Minimum Range

To obtain RANGE In any units

In nautical miles

In yards

General Solution R min =

R min =

cPW 2

PW(μs) 12.36

R min = 164 x ( PW + R T )

PW = pulse width R T = Recovery Time


RADAR FUNDAMENTALS

6-32

Sample Problem:

Calculate the minimum range in yards for a radar system that transmit pulses for 38 μs and a recovery time of 0.1 μs.

Solution: c(PW + tr ) 186,000 (38 μs + 0.1 μs) = 2 2 smi 5280 ft yard = 3.54 x x ≈ 6,230.4 yards s smi 3 ft

Rmin =

Alternate Solution R min = 164(38 + 0.1) = 6,248.4 yards

3.

Maximum Unambiguous Range

To obtain RANGE

General Solution

In any units

R max =

In nautical miles

R max =

In yards

c x PRT 2

PRT(μs) 12.36

R max = 164 x PRT

6-33

Self-Sufficient Guide to ECE by JASON AMPOLOQUIO where: PRT = Pulse Repetition Time or Period PRT =

1 PRF

Sample Problem:

Calculate the maximum unambiguous range for radar facilities with a PRF of 1000 pulses per second.

Solution: c x PRT 1 1 ⇒ PRT = = = 1x10− 3 s 2 PRF 1000 162,000 = x 1x10− 3 s 2 = 81 nmi

Rmax =

(

ª

)

In terms of RADAR parameter

R max =

4

Pt λ 2 Gt2 (RCS) (4 π )3 Pr(min)

Rmax = maximum RADAR range Pt = peak pulse power, W Gt = gain of RADAR antenna RCS = RADAR cross section, m2 Pr(min) = min receive power

Sample Problem: A radar facility operates at 10 GHz with a peak pulse power of 450 kW, and 40dBi antenna gain. If the target RCS is 20 m2 and the minimum receivable power is 1 pW, determine the maximum range. Solution: R max =

4

Pt λ2Gt2 (RCS) 3

(4π) Pr(min)

=

4

(450 x 103 )(0.03)2(104 )2(20) (4π)3(1 x 10−12 )

= 142.14 km


RADAR FUNDAMENTALS

6-34

Read it till it Hertz…jma RADAR Cross Section – The RADAR cross section of a target is the (fictional) area intercepting that amount of power which, when scattered equally in all directions, produces an echo at the RADAR equal to that from the target.

Typical Target Dimensions & RCS

4 πA 2 λ2

ª

πD2 4

Considering the effect of noise

⎡ P D 4 (RCS) ⎤ ⎥ R max(km) = 48 ⎢ 4 2t ⎢⎣ λ BW(F − 1) ⎥⎦ R max(km) = maximum RADAR range, km D = antenna diameter, m BW = noise bandwidth, Hz F = noise factor

πwh2 λ

6-35


Sample Problem: Calculate the peak transmitted pulse power of a low power, short-range radar with an overall NF of 5 dB, 500 kHz IF bandwidth, 10 GHz operating frequency, 1.2 m antenna diameter if the target is a sphere with a radius of 2.5 m located 12 km away. Solution: ⎡ ⎤ P D 4 (RCS ) ⎥ R max( km ) = 48 ⎢ 4 2t ⎢ λ BW (F − 1) ⎥ ⎣ ⎦ ⎛ R max( km ) ⎞ ⎟ ⇒ ⎜⎜ ⎟ 48 ⎠ ⎝ pt =

ª

4

λ2BW (F − 1) 4

256 D (RCS )

= =

PtD 4 (RCS )

⎛ 12 ⎞ =⎜ ⎟ 2 λ BW (F − 1) ⎝ 48 ⎠

4

=

1 256

(0 .03 )2 (500 x 10 3 )(3 .16 − 1) = 0 .373 W 2 ⎞ ⎛ 4 ⎜ π(2 .5) ⎟ 256 (1 .2) ⎜ ⎟ 4 ⎝ ⎠

RADAR Beacon Range Equation The presence of a beacon on a target increases enormously the distance over which a target may be tracked.

RANGE

General Solution

Interrogation link

R max(int) =

Reply link

R max(rep) =

A oTPtT A oB

λ 2N(FB − 1)

A oBPtB A oT

λ 2N(FT − 1)

For Your Information… All symbols have their previous defined meanings, except that the subscript T is now used for quantities pertaining to the Transmitter of the main radar, and B is used for the beacon functions


RADAR FUNDAMENTALS

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Sample Problem: To communicate from earth to Voyager 2, ground station used 100-kW transmitters, 70-m antenna operating at 8 GHz frequency, 1.5 noise factor and 5-kHz bandwidth. The dish antenna on Voyager 2 has a 3.66-m diameter, 13 dB noise figure, and the actual transmitted power is just 22 W. Determine the interrogation link range of the Earth tracking stations. Also compute for the reply link range of Voyager 2 beacon and maximum active tracking range of the system. Solution: Capture area for the earth tracking station; 0.65πDT 2 0.65π(70)2 = = 2,501 .5 m2 4 4 Capture area for the Voyager 2 beacon; A oT = kA T =

A oB = kAB = Noise Power;

0.65πDB2 0.65π(3.66)2 = = 6.84 m2 4 4

{

}

N = kTB = (1.38 x 10-23 )(290)(5 x 103 ) = 20.01 x 10 −18 W Noise Fact or of the Voyager 2 beacon; FB = 101.3 = 20 Interrogat or link tracking range; R max(int) =

A oTPtT A oB 2

λ N(FB − 1)

=

(2,501 .5)(100 x 103 )(6.84) (0.037)2 (20.01 x 10 −18 )(20 − 1)

= 5.67 x 1013 m = 56.7 billion km Re ply link range R max(rep) =

A oBPtB A oT

λ2N(FT − 1)

=

(6.84)(22)(2,501 .5) (0.037)2 (20.01 x 10 −18 )(1.5 − 1)

= 5.24 x 1012 m = 5.24 billion km

Since the Voyager 2 reply link (5.24 billion km) is lesser than the earth station tracking range (56.7 billion km), the MAXIMUM ACTIVE TRACKING RANGE IS 5.24 billion km.


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Duty Cycle (DC) The duty cycle is a ratio of the time on to the time off of the transmitter

In terms of

General Solution

Pulse Width & Pulse Repetition Frequency

DC = PW x PRF

Pulse Width & Pulse Repetition Time

DC =

PW PRT

Average & Peak Power

DC =

Pave Ppk

where: Ppk = Peak Power in, W PW = Pulse Width PRF = Pulse Repetition Frequency or Rate(PRR),Hz

Sample Problem: Calculate the duty cycle and pulse width of a pulse magnetron with an average power of 1.2 kW and a peak power of 18.5 kW if one pulse is generated every 10 ms. Solution: 1 .2 x 100% = 6.5% 18.5 PW DC = ⇒ PW = DC x PRT = 0.065 x (10 ms) = 0.65 ms PRT

DC =


6-38

RADAR FUNDAMENTALS 5.

RADAR Receive Power

Pr =

Pt G t 2 λ 2

(4 π )3 R 4

x RCS

Sample Problem: A radar transmitter has a power of 20 kW and operates at a frequency of 10 GHz. Its signal reflects from a rectangular target 25 km away with a dimension of 3.5x3.5-m. The gain of the antenna is 30 dBi. Calculate the received signal power. Solution: Pr =

Pt λ2Gt2

(4π)3R 4

x RCS =

(20 x 103 )(0.03)2(1000)2 (4π)3(25,000)4

x

4π(3.5 x 3.5)2 0.032

= 48.65 nW


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D. .RADAR ANTENNA HORIZON.

d (nm i) = 1.25 h t

where: d = RADAR antenna horizon, nmi h T = Antenna height, ft

Sample Problem: What is the service range of a radar antenna mounted 85 ft above the platform of a battleship if the platform is 25 ft above sea level. Solution: d(nmi) = 1.25 ht = 1.25 (85 + 25) = 13.1 nmi

E. .RADAR HITS PER SCAN. Hits Pulses per revolution = Scan 360 °

Sample Problem: Calculate the number of pulses for each degree of azimuth for a radar antenna that rotates at 20 rpm with a PRF of 200 pulses/s. Solution: Hits Pulses per revolution 60s = ⇒ Re volution = = 3s Scan 360° 20 rpm No. of pulses 200 pulses 600 pulses = x3s = revolution s revolution Hits 600 = = 1.67 pulses / deg Scan 360°


6-40

RADAR FUNDAMENTALS

F. .RADAR TARGET RESOLUTION. The target resolution of radar is its ability to distinguish between targets that are very close together in either range or bearing.

1.

Range Resolution Range resolution is the ability of a radar system to distinguish between two or more targets on the same bearing but at different ranges.

2.

Bearing Resolution Bearing, or azimuth, resolution is the ability of a radar system to separate objects at the same range but at different bearings.

G. .EFFECTS OF ATMOSPHERIC CONDITIONS TO RADAR RANGE. 1.

Atmospheric Refraction The path followed by electromagnetic energy in the atmosphere, whether direct or reflected, usually is slightly curved; and the speed is affected by temperature, atmospheric pressure, and the amount of water vapor present in the atmosphere, which all affect the refractive index.

2.

Ducting The temperature and moisture content of the atmosphere normally decrease uniformly with an increase in altitude. However, under certain conditions the temperature may first increase with height and then begin to decrease. Such a situation is called a temperature inversion that produces atmospheric ducts.


6-41

H. .RADAR SYSTEM & COMPONENTS.

I.

1.

Synchronizer (Timer) The synchronizer ensures that all circuits connected with the radar system operate in a definite timed relationship.

2.

Transmitter The transmitter generates powerful pulses of electromagnetic energy at precise intervals.

3.

Duplexer A duplexer is essentially an electronic switch that permits a radar system to use a single antenna to both transmit and receive.

4.

Antenna System The antenna system routes the pulse from the transmitter, radiates it in a directional beam, picks up the returning echo, and passes it to the receiver with a minimum of loss.

5.

Receiver The receiver accepts the weak echo signals from the antenna system, amplifies them, detects the pulse envelope, amplifies the pulses, and then routes them to the indicator.

6.

Indicator The indicator uses the received signals routed from the radar receiver to produce a visual indication of target information.

.RADAR SCANNING MECHANISM. Scanning is the systematic movement of a radar beam in a definite pattern while searching for or tracking a target.

1.

Mechanical Scanning The most common type of mechanical scanning is the rotation of the antenna through 360 degrees to obtain azimuth coverage. i.

2.

Conical Scan A conical scan can be generated by nutation of the waveguide. In this process the axis of the waveguide itself is moved through a small conical pattern.

Electronic Scanning Electronic scanning can accomplish lobe motion more rapidly than, and without the inherent maintenance disadvantages of, the mechanical systems.


6-42

RADAR FUNDAMENTALS i.

Monopulse (Simultaneous) Lobing All range, bearing, and elevation-angle information of a target is obtained from a single pulse

D. .DOPPLER EFFECT. A perceived change in the frequency of a wave as the distance between the source and the observer changes.

1.

Doppler Frequency of Sound Waves

ν ± νo fo = fs ν ∓ νs

fo = Observed frequency in Hz ν o = Velocity of observer in m/s ν s = Velocity of source in m/s fs = Source frequency in Hz


Sample Problem:

An ambulance travels down a highway at a speed of 75.0 mi/h with its siren emitting a sound with a frequency of 400 Hz. What frequency is heard (a) by someone standing still when the ambulance approaches? (b) by a passenger in a car traveling at 55 mi/h in the opposite direction as it approaches the ambulance? (c) by a passenger in a car traveling at 55 mi/h in the opposite direction as it moves away from the ambulance? 75 mi/h = 33.5 m/s, 55 mi/h = 24.6 m/s.

Solution: (a) ν ≈ 345 m s , ν s = 33.5 m s, ν o = 0 (observer is not moving) fo = fs

ν + νo ⎛ 345 + 0 ⎞ = 400⎜ ⎟ = 443 Hz ν − νs ⎝ 345 − 33.5 ⎠

(b) ν ≈ 345 m s , ν s = 33.5 m s , ν o = 24.6 m s fo = fs

ν + νo ⎛ 345 + 24.6 ⎞ = 400⎜ ⎟ = 475 Hz ν − νs ⎝ 345 − 33.5 ⎠

(c) ν ≈ 345 m s , ν s = 33.5 m s , ν o = 24.6 m s fo = fs

ν − νo ⎛ 345 − 24.6 ⎞ = 400⎜ ⎟ = 339 Hz ν + νs ⎝ 345 + 33.5 ⎠


6-43

Doppler Frequency of Electromagnetic Waves

fo = fs

c ± νr c ∓ νr

fo = Observed frequency in Hz c = Speed of light = 3 x 10 8 m / s ν r = Velocity of source relative to observer in m/s fs = Source frequency in Hz 3.

Doppler Frequency (RADAR) fo = observed frequency in Hz

2 νr fo = λ

ν r = relative velocity in m/s λ = RADAR signal wavelength in m

Sample Problem:

Find the Doppler shift caused by a vehicle moving toward a radar at 60 mph, if the radar operates at 10 GHz.

Solution: 2νr fo = = λ

4.

⎡ mi 1.609 km 1000 m 1 hr ⎤ 2⎢60 x x ⎥ h 1 mi 1 km 3600 s ⎦ ⎣ = 1.78 kHz 0.03

RADAR Blind Speed

νb =

n λ PRF nλ = 2 2 PRT

n = any integer PRF = pulse repetition rate in Hz PRT = pulse repetition time in s


6-44

RADAR FUNDAMENTALS

Sample Problem:

Calculate the 3rd blindspeed for an MTI RADAR that operates at 6 GHz, with a pulse repetition rate of 1000 Hz.

Solution: νb =

J.

nλPRF 3(0.05)1000 = = 75 m s ⇒ 270 kph 2 2

.MILITARY CLASSIFICATION OF RADAR SYSTEMS. Function Search Track

Installation Vehicle Ground Or Land Based Airborne

Height Finder

Shipboard

Radar Equipment Indicator 1ST Symbol – Installation (1.1) (1.2) (1.3) (1.4) (1.5) (1.6) (1.7) (1.8) (1.9) (1.10) (1.11) (1.12) (1.13) (1.14)

Piloted Aircraft Underwater mobile, submarine Pilotless carrier Fixed ground General ground use Amphibious Ground, mobile Portable S—Water Ground, transportable General utility Ground, vehicular Water surface and under water combination Piloted and pilotless airborne vehicle combination

A B D F G K M P S T U V W Z

2ND Symbol – Type of Equipment (2.1) (2.2) (2.3) (2.4) (2.5)

Invisible light, heat radiation No modulating signal Carrier Telegraph or Teletype Interphone and public address Electromechanical or Inertial wire covered

A C G I J

Self-Sufficient Guide to ECE by JASON AMPOLOQUIO (2.6) (2.7) (2.8) (2.9) (2.10) (2.11) (2.12) (2.13) (2.14) (2.15) (2.16) (2.17) (2.18)

Telemetering Countermeasures Meteorological Sound in air Radar Sonar and underwater sound Radio Special types, magnetic, etc., or combinations of types Telephone (wire) Visual and visible light Armament (peculiar to armament, not otherwise covered) Facsimile or television Data processing

6-45

K L M N P Q R S T V W X Y

3RD Symbol – Purpose (3.1) (3.2) (3.3) (3.4) (3.5) (3.6) (3.7) (3.8) (3.9) (3.10) (3.11) (3.12) (3.13) (3.14) (3.15) (3.16) (3.17)

Bombing B Communications (receiving and transmitting Telegraphy (manual) C Direction finder reconnaissance and/or surveillance D Ejection and/or release E Fire control, or search- light directing G Recording and/or reproducing (graphic meteorological and sound) H Computing K Maintenance and/or test assemblies (including tools) M Navigational aids (including altimeters, beacons, compasses, racons, depth sounding, approach and landing) N Special, or combination of purposes Q Receiving, passive detecting R Detecting and/or range and bearing, search S Transmitting T Automatic flight or remote control W Identification and recognition X Surveillance (search detect, and multiple target tracking) and control (both fire control and air control) Y Facsimile or television Z


6-46

RADAR FUNDAMENTALS

Sample Problem: AN/FPS-35 A. B. C. D.

is categories as what type of radar? Airborne Countermeasure Radar Ship Bombing Radar Fixed Search Radar Ship Navigation Radar

Solution:

K. .COMMON RADAR TYPES. 1.

Search RADAR SEARCH RADAR is designed to continuously scan a volume of space to provide initial detection of all targets. Search radar is almost always used to detect and determine the position of new targets for later use by TRACK RADAR. a.

Surface-Search Radar A surface-search radar system has two primary functions: ª The detection and determination of accurate ranges and bearings of surface objects and low-flying aircraft and ª The maintenance of a 360- degree search pattern for all objects within line-of-sight distance from the radar antenna. Shipboard surface-search radar has the following design specifications 1. Transmitter frequency 5,450-5,825 MHz 2. Pulse width .25 or 1.3 microseconds 3. Pulse-repetition rate between 625 and 650 pulses per second 4. Peak power between 190 and 285 kW 5. Vertical beam width between 12 and 16 degrees 6. Horizontal beam width 1.5 degrees

b.

Air-Search Radar Air-search radar systems initially detect and determine the position, course, and speed of air targets in a relatively large area. The maximum range of air-search radar can exceed 300 miles, and the bearing coverage is a complete 360-degree circle

Self-Sufficient Guide to ECE by JASON AMPOLOQUIO c.

2.

L.

6-47

Height-Finding Search Radar The primary function of a height-finding radar (sometimes referred to as a three-coordinate or 3D radar) is that of computing accurate ranges, bearings, and altitudes of aircraft targets detected by air- search radars.

Track RADAR Track radar provides continuous range, bearing, and elevation data on one or more targets. a.

Carrier-Controlled Approach (CCA) and Ground-Controlled Approach (GCA) Both systems guide aircraft to safe landing under conditions approaching zero visibility. By means of radar, aircraft are detected and observed during the final approach and landing sequence.

b.

Airborne RADAR Airborne radar is designed especially to meet the strict space and weight limitations that are necessary for all airborne equipment. Even so, airborne radar sets develop the same peak power as shipboard and shore-based sets.

.COMMON RADAR DISPLAY. 1.

A scope The A-scope display presents only the range to the target and the relative strength of the echo. Such a display is normally used in weapons control radar systems. The bearing and elevation angles are presented as dial or digital readouts that correspond to the actual physical position of the antenna.

2.

Range-Height Indicator (RHI) The range-height indicator (RHI) scope is used with height-finding search radars to obtain altitude information. The RHI is a twodimensional presentation indicating target range and altitude.

3.

Plan Position Indicator (PPI) The PPI scope is by far the most used radar display. The PPI uses a radial sweep pivoting about the center of the presentation. This results in a map-like picture of the area covered by the radar beam.


6-48

RADAR FUNDAMENTALS

I

H

1.

Calculate the PW, and duty cycle of a radar with a peak power of 20 kW, PRR of 900 Hz, and the average power is 18 W. A. 10 μs, 0.9% B. 100 μs, 9% C. 0.1 μs, 0.09% D. 1 μs, 0.09%

2.

What is the received power for a radar with a 1-kW transmit power, 80 dBi of antenna gain, for the 1-m rectangular plate at 1 GHz at a range of 1,000 km? A. 6350 nW B. 0.635 nW C. 63.5 nW D. 635 nW

3.

Calculate the peak power of certain radar with an average power of 20 kW, pulse width of 20 μs and a PRF of 103 pulses/s A. 1 mW B. 1 MW C. 1 GW D. 1 kW

4.

If a radar system has a pulse width of 5 microseconds, the range resolution is A. 120 yards B. 620 yards C. 820 yards D. 420 yards

5.

With a CW transmit frequency of 5 GHz, calculate the Doppler frequency seen by a stationary radar when the target radial velocity is 100 kph. A. 729 Hz B. 297 Hz C. 279 Hz D. 927 Hz

6.

Determine the maximum allowable pulse repetition time for unambiguous reception for a radar to have a maximum range of 60 km A. 100 μs B. 800 μs C. 400 μs D. 600 μs

7.

A deep space RADAR facility is operating at 2.5 GHz with a peak pulse power of 25 MW, 64 m antenna diameter, 1.1 noise factor. If the receiver BW is only 5 kHz and the target cross section is 1m2, calculate the maximum range. A. 700,000 meters B. 132,700 meters C. 132,700,000 meters D. 132,000 meters

8.

A small airplane can fool a radar system into thinking it is much bigger by adding a radar reflector that increases the RCS. How large a flat plate should a jet fighter (RCS=6) add to its side so that it looks like a large jet (RCS=40) at 1 GHz? A. 0.7x0.7 m B. 5.7x5.7 m C. 4.7x4.7 m D. 2.7x2.7 m


6-49

An L-band radar operating at 1.25 GHz uses a peak pulse power of 3 MW, and must have a range of 100 nmi for objects whose RCS is 1 m2. If the minimum receivable power of the receiver is 200 fW, what is the smallest diameter the antenna reflector could have, assuming it to be a full paraboloid with k=0.65? A. 4.92 m B. 3.84 m C. 2.56 m D. 1.42 m

10. A Doppler radar at 15 GHz has a return signal with a frequency of 50 kHz higher than the transmitted signal. What is the component of the target velocity along a line joining the radar and the target assuming that the radar installation is stationary. a. 347 m/s B. 839 m/s c. 125 m/s D. 500 m/s 11. What is the distance in nmi to a target if it takes 123 μs for a radar pulse to travel from the radar antenna to the target, back to the antenna, and be displayed on the PPI scope? A. 100 mi B. 0.1 mi C. 10 mi D. 1000 mi 12. Calculate (a) the maximum unambiguous range, (b) duty cycle, (c) average power for a radar transmitter with a peak pulse power of 400 kW, PRF of 1500 Hz, and a PW of 0.8 μs. A. 10 km, 0.12%, 48 W B. 100 km, 0.2%, 48 W C. 10 km, 0.2%, 40 W D. 100 km, 0.12%, 480 W 13. An 8-GHz police radar measures a Doppler frequency of 1788 Hz, from a car approaching the stationary police station vehicle, in an 80 kph speed limit zone. By how many kph the car is overspeeding? A. 160 kph B. 40 kph C. 80 kph D. 120 kph 14. Calculate the lowest blindspeed of an MTI radar that operates at 10 GHz with a PRF of 3000 Hz. A. 162 kph B. 365 kph C. 812 kph D. 55 kph 15. A radar transmitter has a power of 10 kW and operates at a frequency of 9.5 GHz. Its signal reflects from a target 15 km away with a radar cross section of 10.2 m2. The gain of the antenna is 20 dBi. Calculate the received signal power. A. 10.1 fW B. 1.1 fW C. 1.1 nW D. 10.1 pW 16. A radar transmitter uses a magnetron to generate pulses with a power level of 1 MW. The pulses have a duration of 2μs and the pulse repetition period 2 ms. If the magnetron has an efficiency of 60%, calculate the average power dissipated in the magnetron A. 8320 W B. 3590 W C. 670 W D. 1729 W


6-50

RADAR FUNDAMENTALS

17. A radar has a pulse duration of 50 μs and a pulse repetition time of 5 ms. Calculate the maximum and minimum useful range A. 812 km, 8.12 km B. 750 km, 7.5 km C. 578 km, 57.8 km D. 136 km, 13.6 km 18. What is the minimum receivable power for a radar with a noise figure of 12 dB and 2.5 MHz operating bandwidth? B. 1.59 x 10-10 W A. 1.59 x 10-13 W C. 1.59 x 10-12 W D. 1.59 x 10-11 W 19. The AN/FPS-16 guided-missile tracking radar operates at 5 GHz, with a 1 MW peak power output. If the antenna diameter is 3.66 m, and the receiver has a bandwidth of 1.6 MHz and an 11 dB noise figure, what is its maximum detection range for a 1-m2 target? A. 123.4 km B. 456.7 km C. 345.6 km D. 678.9 km 20. Calculate the duty cycle if the pulse width is 20 μs a PRR of 500 pulses per second. A. 0.1% B. 10% C. 0.01% D. 1% 21. 32.8 mi is the maximum unambiguous range for a radar system with a PRT equal to _____μs. A. 400 B. 800 C. 200 D. 100 22. Calculate the minimum receivable signal in RADAR receiver which has an IF bandwidth of 1.5 MHz and a 9-dB noise figure. A. 183.5 fW B. 41.7 fW C. 0.269 fW D. 238.4 fW 23. What is the Doppler shift for a 500-MHz signal when the target is moving 45° toward the radar at 250 m/s. A. 849 Hz B. 589 Hz C. 123 Hz D. 234 Hz 24. What is the Doppler shift when the police radar unit operating at 2 GHz is aimed at a car traveling at 20 m/s at a curved part of the road where there is a 30° angle of return in the road? A. 965 Hz B. 96.5 Hz C. 23.1 Hz D. 231 Hz

Section 24 Laws & Ethics Section 25 Basic Signals & System

Miscellaneous Topics


7-1


Section

Laws

24

And Ethics


Law: Body of official rules and regulations, generally found in constitutions, legislation, judicial opinions, and the like, that is used to govern a society and to control the behavior of its members.

DEFINITION.

DEFINITION. Professional Electronics Engineer - a person who is qualified to hold himself/herself out as a duly registered/licensed Professional Electronics Engineer under this Act and to affix to his/her name the letters "PECE". DEFINITION. Electronics Engineer - a person who is qualified to hold himself/herself out as a duly registered/licensed Electronics Engineer under this Act and to affix to his/her name the letters "ECE". DEFINITION. Electronics Technician - a person who is qualified to hold himself/herself out as a duly registered/licensed Electronics Technician under this Act and to affix to his/her name the letters "ECT". Electronics and Communications Engineer - a person who is qualified to hold himself/herself out as a duly-registered/licensed Electronics and Communications Engineer under Republic Act No. 5734.

DEFINITION.

LAW STRUCTURE I.

.LAWS. A.

Telecoms 1.

Republic Acts 1.1.

Republic Act No. 3846 An act providing for the regulation of radio stations and radio communications in the Philippine islands, and for other purposes

Date of Approval: November 11, 1963 1.2.

Republic Act No. 5734 The Electronics and Communications Act of the Philippines

Date of Approval: June 21, 1969


LAWS AND ETHICS

7-2

1.3.

Republic Act No. 6849 An act providing for the installation, operation, and maintenance of public telephones in each and every municipality in the Philippines, appropriating funds therefore and for other purposes

Date of Approval: July 24, 1989 1.4.

Republic Act No. 7925 An act to promote and govern the development of Philippine telecommunications and the delivery of public telecommunications services

Date of Approval: July 25, 1995 1.5

Republic Act No. 9292 An act providing for a more responsive and comprehensive regulation for the registration, licensing and practice of professional electronics engineers, electronics engineers and electronics technicians, repealing republic act no. 5734, otherwise known as the "electronics and communications engineering act of the Philippines", and for other purposes

Date of Approval: April 17, 2004 II. .EXECUTIVE ORDERS. A.

Telecoms 1.

Executive Order No. 59 (E.O. 59) Prescribing the policy guidelines for compulsory interconnection of authorized public telecommunications carrier in order to create a universally accessible and fully integrated nationwide telecommunications network and thereby encourage greater private sector in telecommunications

Date of Approval: Feb 24, 1993 2.

Executive Order No. 109 (E.O. 109) Policy to improve the provision of Local Exchange Carrier service

Date of Approval: July 12, 1993

3.

Executive Order No. 125 (E.O. 125) Reorganizing the ministry of transportation and communications defining its powers and functions and for other purposes

Date of Approval: Jan 30, 1987

4.

Executive Order No. 196 (E.O. 196) Vesting the jurisdiction, control, and regulation over the Philippine Communications Satellite Corporation with the National Telecommunications Commission



7-3

Executive Order No. 467 (E.O. 467) Providing for the national policy on the operation and use of international satellite communications in the country

Date of Approval: March 17, 1998

6.

Executive Order No. 468 (E.O. 468) Providing for the creation of a National Council for the promotion of Electronic commerce in the country

Date of Approval: Feb 23, 1998

7.

Executive Order No. 469 (E.O. 469) Amending Executive Order No. 190 dated July 1994 approving and adopting the National Information Technology Plan 2000 and establishing the National Information Technology Council.

Date of Approval: Feb 23, 1998 B.

Broadcast B.1. General Broadcast Law 1. Executive Order No. 255 (E.O. 255) Requiring all radio stations with musical format programs to broadcast a minimum of four Original Pilipino Musical compositions in every clock-hour and for other purposes B.2. Related Cable Laws 1. Executive Order No. 205 (E.O. 205) Regulating the operation of Cable Antenna Television (CATV) systems in the Philippines, and for other purposes

Date of Approval: June 30, 1987 2.

Executive Order No. 436 (E.O. 436) Prescribing policy guidelines to govern the operations of Cable Television in the Philippines

Date of Approval: September 9, 1997

III. .PRESIDENTIAL OR ADMINISTRATIVE ISSUANCES. A.

Presidential Decree No. 36 Creating the mass media council and prescribing rules and regulations on the opening and operation of mass media

Date of Approval: November 02, 1972

B.

Presidential Decree No. 55 Penalizing unauthorized telephone installation



LAWS AND ETHICS

7-4

C.

Presidential Decree No. 576-A Regulating the ownership and operation of radio and television stations and for other purposes


D.

Presidential Decree No. 1986 Creating the Movie and Television Review and Classification Board

Date of Approval: October 5, 1985 E.

Presidential Decree No. 1987 An act creating the Video Regulatory Board

Date of Approval: October 5, 1985

IV. .DEPARTMENT ORDERS. 1.

2.

Department Order No. 4 Rules and regulations Philippines

governing

radio

training

schools

in

the

Date of Approval: January 3, 1972

Department Order No. 5 Rules and regulations governing commercial radio operators

Date of Approval: September 22 1948

3.

Department Order No. 6 Rules and regulations governing land mobile radio service and its operation


4.

Department Order No. 11 General rules and regulations governing the construction, installation, establishment or operation of radio stations and the possession or ownership, construction or manufacture, purchase, sale and transfer of transmitter or transceivers in the Philippines

Date of Approval: October 10, 1959 5.

Department Order No. 87 Rules and regulations governing the low power ship radiotelephone service in the Philippines


6.

Department Order No. 88 Rules and regulations requiring the service of duly registered electronics and communications engineer in the planning and designing, installation or construction governing commercial radio operators

Date of Approval: December 28, 1972


7-5

Department Order No. 13 Rules and regulations governing the amateur radio service in the Philippines as amended

Date of Approval: February 16, 1971

8.

Department Order No. 227 Rules and regulations governing land radio stations and radio communications in the Philippines

Date of Approval: October 14, 1933

9.

Department Order No. 287 Rules and regulations governing the citizen’s radio service


V. .IMPORTANT TERMINOLOGIES. 1.

Telecommunications Any process which enables a telecommunications entity to relay and receive voice, data, electronic messages, written or printed matter, fixed or moving pictures, words, music or visible or audible signals or any control signals of any design and for any purpose by wire, radio or other electromagnetic, spectral, optical or technological means.

2.

Public telecommunications entity Any person, firm, partnership or corporation, government or private, engaged in the provision of telecommunications services to the public for compensation.

3.

Broadcasting An undertaking the object of which is to transmit over-the-air commercial radio or television messages for reception of a broad audience in a geographic area.

4.

Franchise A privilege conferred upon a telecommunications entity by Congress, authorizing that entity to engage in a certain type of telecommunications service.

5.

Local exchange operator An entity providing transmission and switching of telecommunications services, primarily but not limited to voice-to-voice service, in a geographic area anywhere in the Philippines.

6.

Inter-exchange carrier An entity, sometimes referred to as carrier's carrier or national backbone network operator, authorized to install, own and operate facilities which connect local exchanges within the Philippines and to engage in the business of inter-exchange national long distance services.


LAWS AND ETHICS

7-6

7.

International carrier An entity primarily engaged in the business of providing transmission and switching of any telecommunications service between the Philippines and any other point of the world to which it has an existing correspondent or prospective interconnection agreements.

8.

Value-added service provider (VAS) An entity which relying on the transmission, switching and local distribution facilities of the local exchange and inter-exchange operators, and overseas carriers, offers enhanced services beyond those ordinarily provided for by such carriers.

9.

Public toll calling station A non-exclusive facility at which the public may, by the payment of appropriate fees, place as well as receive telephone calls and/or telegrams or other messages.

10. Mobile radio telephone system A wide area mobile, radio telephone system with its own switch, base stations and transmission facilities capable of providing high capacity mobile telecommunications by utilizing radio frequencies. 11. Interconnection The linkage, by wire, radio, satellite or other means, of two or more existing telecommunications carriers or operators with one another for the purpose of allowing or enabling the subscribers of one carrier or operator to access or reach the subscribers of the other carriers or operators.

7-7


REPUBLIC ACT 9292 H. No. 5224 S. No. 2683 REPUBLIC OF THE PHILIPPINES CONGRESS OF THE PHILIPPINES METRO MANILA TWELFTH CONGRESS THIRD REGULAR SESSION ⎯⎯⎯⎯⎯⎯ Begun and held in Metro Manila, on Monday, the twenty-eight day of July, two thousand three. [REPUBLIC ACT NO. 9292] AN ACT PROVIDING FOR A MORE RESPONSIVE AND COMPREHENSIVE REGULATION FOR THE REGISTRATION, LICENSING AND PRACTICE OF PROFESSIONAL ELECTRONICS ENGINEERS, ELECTRONICS ENGINEERS AND ELECTRONICS TECHNICIANS, REPEALING REPUBLIC ACT NO. 5734, OTHERWISE KNOWN AS THE "ELECTRONICS AND COMMUNICATIONS ENGINEERING ACT OF THE PHILIPPINES", AND FOR OTHER PURPOSES.

Be it enacted by the Senate and the House of Representatives of the Philippines in Congress assembled: ARTICLE I GENERAL PROVISIONS SECTION 1. Title. Law of 2004".

- This Act shall be known as the "Electronics Engineering

SEC. 2. Statement of Policy. - The State recognizes the importance of electronics engineering in nation-building and development. The State shall therefore develop and nurture competent, virtuous, productive and wellrounded Professional Electronics Engineers, Electronics Engineers and Electronics Technicians whose standards of practice and service shall be excellent, qualitative, world-class and globally competitive through inviolable, honest, effective and credible licensure examinations and through regulatory measures, programs and activities that foster their integrity, continuing professional education, development and growth. SEC. 3. Definition and Interpretation of Terms. following terms shall mean:

-

As used in this Act, the


7-8

LAWS AND ETHICS (a) Electronics - the science dealing with the development and application of devices and systems involving the flow of electrons or other carriers of electric charge, in a vacuum, in gaseous media, in plasma, in semiconductors, in solid-state and/or in similar devices, including, but not limited to, applications involving optical, electromagnetic and other energy forms when transduced or converted into electronic signals. (b) Professional Electronics Engineer - a person who is qualified to hold himself/herself out as a duly registered/licensed Professional Electronics Engineer under this Act and to affix to his/her name the letters "PECE". (c) Electronics Engineer - a person who is qualified to hold himself/herself out as a duly registered/licensed Electronics Engineer under this Act and to affix to his/her name the letters "ECE". (d) Electronics Technician - a person who is qualified to hold himself/herself out as a duly registered/licensed Electronics Technician under this Act and to affix to his/her name the letters "ECT". (e) Electronics and Communications Engineer - a person who is qualified to hold himself/herself out as a duly-registered/licensed Electronics and Communications Engineer under Republic Act No. 5734. (f) Computer – any of a variety of electronic devices that is capable of accepting data, programs and/or instructions, executing the programs and/or instructions to process the data and presenting the results. (g) Information and Communications Technology - the acquisition, production, transformation, storage and transmission/reception of data and information by electronic means in forms such as vocal, pictorial, textual, numeric or the like; also refers to the theoretical and practical applications and processes utilizing such data and information. (h) Communications - the process of sending and/or receiving information, data, signals and/or messages between two (2) or more points by radio, cable, optical wave guides or other devices and wired or wireless medium (i) Telecommunications - any transmission, emission or reception of voice, data, electronic messages, text, written or printed matter, fixed or moving pictures or images, words, music or visible or audible signals or sounds, or any information, intelligence and/or control signals of any design/format and for any purpose, by wire,

7-9

Self-Sufficient Guide to ECE by JASON AMPOLOQUIO radio, spectral, visual/optical/light, or electromagnetic and technological means.

other

electronic,

(j) Broadcast, Broadcasting - an undertaking the object of which is to transmit audio, video, text, images or other signals or messages for reception of a broad audience in a geographical area via wired or wireless means. (k) Industrial Plant - includes all manufacturing establishments and other business endeavors where electronic or electronicallycontrolled machinery or equipment are installed and/or are being used, sold, maintained, assembled, manufactured or operated. (l) Commercial Establishment - shall include but not be limited to office buildings, hotels, motels, hospitals, condominiums, stores, apartments, supermarkets, schools, studios, stadia, parking areas, memorial chapels/parks, watercraft and aircraft used for business or profit, and any other building/s or area/s for business purposes, where electronic or electronically-controlled machinery or equipment are installed and/or are being used, sold, maintained, assembled, manufactured or operated. (m) Consulting Services - as used in this Act, shall include services requiring adequate technical expertise, experience and professional capability in undertaking advisory and review, preinvestment or feasibility studies, design, planning, construction, supervision, management and related services, and other technical studies or special studies in the field of electronics engineering. (n) Accredited Professional Organization - the integrated and accredited national organization of Professional Electronics Engineers, Electronics Engineers and Electronics Technicians. SEC. 4. Categories of Practice. - The following shall be the engineering and technician categories covered by this Act: (a) Professional Electronics Engineer (PECE) (b) Electronics Engineer (ECE) (c) Electronics Technician (ECT) SEC. 5. Nature and Scope of Practice of Electronics Engineering and Electronics Technician Professions. (a) The scope and nature of practice of the Electronics Engineer shall embrace and consist of any work or activity relating to the application of engineering sciences and/or principles to the investigation, analysis, synthesis, planning, design, specification,


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LAWS AND ETHICS research and development, provision, procurement, marketing and sales, manufacture and production, construction and installation, tests/measurements/control, operation, repair, servicing, technical support and maintenance of electronic components, devices, products, apparatus, instruments, equipment, systems, networks, operations and processes in the fields of electronics, including communications and/or telecommunications, information and communication technology (ICT), computers and their networking and hardware/firmware/software development and applications, broadcast/broadcasting, cable and wireless television, consumer and industrial electronics, electro-optics/photonics/optoelectronics, electro-magnetics, avionics, aerospace, navigational and military applications, medical electronics, robotics, cybernetics, biometrics and all other related and convergent fields; it also includes the administration, management, supervision and regulatory aspects of such works and activities; similarly included are those teaching and training activities which develop the ability to use electronic engineering fundamentals and related advanced knowledge in electronics engineering, including lecturing and teaching of technical and professional subjects given in the electronics engineering and electronics technician curriculum and licensure examinations. (b) The scope and nature of practice of the Professional Electronics Engineer shall embrace and consist of all of the above plus the sole authority to provide consulting services as defined in this Act and to sign and seal electronics plans, drawings, permit applications, specifications, reports and other technical documents prepared by himself/herself and/or under his direct supervision. (c) The scope and nature of practice of the Electronics Technician profession shall embrace and consist of any non-engineering work or activity relating to the installation, construction, operation, control, tests and measurements, diagnosis, repair and maintenance, manufacture and production, sales and marketing of any electronic component/s, device/s, products, apparatus, instruments, equipment, system/s, network/s, operations and processes located on land, watercraft, aircraft, industrial plants or commercial establishments, including the teaching and training of technical and professional subjects given in the electronics technician curriculum and licensure examinations.


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ARTICLE II PROFESSIONAL REGULATORY BOARD OF ELECTRONICS ENGINEERING SEC. 6. Composition of the Board. - There is hereby created a Professional Regulatory Board of Electronics Engineering, hereinafter referred to as the Board, under the administrative control and supervision of the Professional Regulation Commission, hereinafter referred to as the Commission, composed of a chairman and two (2) members who shall be appointed by the President of the Philippines from the three (3) recommendees per position chosen and ranked by the Commission, which recommendees shall in turn be chosen from the five (5) nominees for each position submitted by the accredited professional organization, in accordance with rules and regulations presently in existence or that may be promulgated for such purpose. SEC. 7. Powers and Functions of the Board. - The Board is vested with the authority to: (a) Administer/Implement the provisions of this Act; (b) Administer oaths in connection with the administration of this Act; (c) Adopt an official seal of the Board; (d) Issue, suspend or revoke Certificates of Registration and accordingly the Professional Identification Cards of Professional Electronics Engineers, Electronics Engineers or Electronics Technicians, or otherwise suspend the holder thereof from the practice of his/her profession, for any justifiable cause and after due process; (e) Maintain a roster of Professional Electronics Engineers, Electronics Engineers and Electronics Technicians; (f) Issue, suspend and/or cancel special permits to foreign Professional Electronics Engineers, Electronics Engineers or Electronics Technicians in accordance with the provisions of this Act; (g) Prescribe, amend or revise the requirements for licensing of Professional Electronics Engineers, and prepare, adopt and issue the syllabi of the subjects for the licensure examination for Electronics Engineers and Electronics Technicians, and prepare the questions therefore, in strict conformance with the scope of the syllabi; (h) Adopt a program for the full computerization of the licensure examination;


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LAWS AND ETHICS (i) Grant registration without examination, subject to review and approval by the Commission, in accordance with the provisions of this Act; (j) Study, examine and recommend, in coordination with the Commission on Higher Education (CHED) and the Technical Education and Skills Development Authority (TESDA), and in consultation with other concerned government entities and the accredited professional organization, the essential requirements as to curricula and facilities of schools, colleges or universities, seeking permission to open courses or programs or already offering courses or programs in electronics engineering, electronics technician and related courses or programs and to see to it that these requirements, including employment of qualified faculty members, are properly complied with: Provided, That within three (3) years after the effectivity of this Act, the Board shall, in coordination with CHED, TESDA, and in consultation with other concerned government entities and the accredited professional organization, review and define/re-define the curricula for electronics engineering, electronics technician and/or allied courses or programs for the purpose of re-aligning, revising and/or consolidating the same and/or otherwise defining the minimum requirements by means of which graduates of related or allied courses or programs can qualify to take the Electronics Engineer and Electronics Technician licensure examinations; (k) Inspect educational institutions and based on their findings thereon, recommend to CHED and/or the TESDA and/or other government entities concerned with the granting of school permits or authorization, the opening, improvement/upgrading or closure of colleges or schools and universities offering electronics engineering and electronics technician courses or programs; (l) Adopt and administer a Code of Ethics and a Code of Technical Standards of Practice for Professional Electronics Engineers, Electronics Engineers and Electronics Technicians in the Philippines; (m) Promulgate rules and regulations on the scope of practice of Professional Electronics Engineers, Electronics Engineers and Electronics Technicians; (n) Promulgate a program for continuing professional education and/or development of Professional Electronics Engineers, Electronics Engineers and Electronics Technicians; (o) Prescribe the minimum manning and manpower requirements for Professional Electronics Engineers, Electronics Engineers and Electronics Technicians in industrial plants and commercial


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establishments for purposes of ensuring compliance with the provisions of this Act and all other ordinances, laws, rules and regulations that may be enacted hereinafter; (p) Formulate, prescribe and adopt such rules and regulations for electronics installations in industrial plants, commercial establishments and other buildings or structures covered by the National Building Code of the Philippines, in coordination with the Department of Public Works and Highways (DPWH), other concerned agencies, representatives of industry and the Accredited Professional Organization; (q) Study the conditions affecting the Professional Electronics Engineering, Electronics Engineering and Electronics Technician professions in the Philippines, and whenever necessary, exercise the powers conferred by this and other Acts, and adopt such measures as may be deemed proper for the enhancement and advancement of the professions and/or the maintenance of high professional, ethical and technical standards, and for this purpose, the Board may personally or through subordinate employees of the Commission or member/s of the Accredited Professional Organization, duly authorized by the Board and approved by the Commission, conduct ocular inspection or visit industrial plants and commercial establishments where Professional Electronics Engineers, Electronics Engineers and Electronics Technicians are employed for the purpose of determining compliance with the provisions of law relative thereto, in accordance with established policies promulgated by the Commission; (r) Hear and decide violations of this Act, its implementing rules and regulations, the Code of Ethics and the Code of Technical Standards of Practice for the profession, and for this purpose, issue subpoena ad testificandum and/or subpoena duces tecum to secure attendance of witnesses and the production of documents in connection with the charges presented to and/or any investigation pending before the Board; (s) Delegate the hearing or investigation of administrative cases filed before it to authorized officers of the Commission, except in cases where the issue involved strictly concerns the practice of the Professional Electronics Engineering, Electronics Engineering and Electronics Technician Professions, in which case the hearing shall be presided over by at least one (1) member of the Board assisted by a Legal or Hearing Officer of the Commission; (t) Promulgate resolutions, orders and/or decisions on such administrative cases: Provided, That such resolutions, orders and/or decisions shall be subject to appeal within fifteen (15) days from receipt thereof with the Commission, which may affirm or reverse the same, dismiss the case, deny the appeal or remand


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the case to the Board for further action or proceeding: Provided, further, That if after fifteen (15) days from the receipt of such decision no appeal is taken there from to the Commission, the same shall become final and immediately enforceable; (u) Submit an annual action plan and corresponding report at the beginning and close of each fiscal year on the activities, proceedings and accomplishments of the Board for the year, incorporating therein any recommendation to the Commission; and (v) Discharge such other powers and functions as the Board and the Commission may deem necessary for the practice of the profession and the upgrading, enhancement, development and growth of the Professional Electronics Engineer, Electronics Engineer and Electronics Technician professions in the Philippines. Except those in administrative cases, all resolutions embodying rules and regulations and other policies and measures issued and/or promulgated by the Board shall be subject to the review and approval by the Commission. SEC. 8. Qualifications of Board Members. - The chairman and members of the Board must possess the following qualifications at the time of their appointment: (a) Be a citizen and a resident of the Philippines for at least five (5) consecutive years prior to his/her appointment; (b) Be of good moral character and integrity; (c) Be a holder of a valid Certificate of Registration and a valid Professional Identification Card as a Professional Electronics Engineer, duly qualified to practice as a Professional Electronics Engineer in the Philippines; (d) Be a member of good standing of the Accredited Professional Organization; (e) Be in active practice of the electronics engineering profession for at least ten (10) years prior to his appointment, either in selfpractice, or employment in government service and/or in the private sector; (f) Must not have any pecuniary interest, directly or indirectly, in any school, academy, college, university or institution conferring an academic degree and/or certification/accreditation necessary for admission to the practice of Electronics Engineering and/or Electronics Technician or where review classes in preparation for the licensure examination are being offered or conducted nor shall


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he/she be a member of the faculty or of the administration thereof prior to taking his/her oath of office; and (g) Must not have been convicted of an offense involving moral turpitude. SEC. 9. Term of Office. - The members of the Board shall hold office for a term of three (3) years from date of appointment or until their successors shall have been appointed and qualified and may be re-appointed once for another term. Any vacancy occurring within the term of a member shall be filled for the unexpired portion of the term only: Provided, That the member appointed to serve the unexpired term may be re-appointed more than once for as long as his/her continuous tenure shall not exceed six (6) years. Each member of the Board shall take the proper oath prior to the assumption of office. SEC. 10. Compensation and Allowances of the Board. - The Chairman and members of the Board shall receive compensation and allowances comparable to that being received by the Chairman and members of existing regulatory boards under the Commission as provided for in the General Appropriations Act. SEC. 11. Removal of Board Members. - The President of the Philippines, upon recommendation of the Commission, may suspend or remove any member of the Board for neglect of duty, incompetence, manipulation or rigging of the licensure examination results, disclosure of secret information or the examination questions prior to the conduct of the said examination, or tampering of the grades therein, for unprofessional or unethical conduct, or for any final judgment or conviction of any criminal offense by the Courts, after having given the member concerned an opportunity to be heard and/or to defend himself/herself in a proper administrative investigation. SEC. 12. Custodian of Board Records, Secretariat and Support Services. -All records of the Board, including applications for examination, administrative cases involving Professional Electronics Engineers, Electronics Engineers and Electronics Technicians shall be kept by the Commission. The Commission shall designate the Secretary of the Board and shall provide the secretariat and other support services to implement the provisions of this Act.


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ARTICLE III EXAMINATION, REGISTRATION AND LICENSURE SEC. 13. Licensure Examination. - Except as otherwise specifically provided in this Act, all applicants seeking to be registered and licensed as Electronics Engineers and Electronics Technicians, shall undergo the required examinations to be given by the Board in such places and dates as the Commission may designate in accordance with the provisions of Republic Act No. 8981. SEC. 14. Qualifications for Examinations. - In order to be allowed to take the examination for Electronics Engineer or Electronics Technician, an applicant must, at the time of the filing of his/her application, establish to the satisfaction of the Board that: (a) He/She is a citizen of the Philippines or of a foreign country qualified to take the examination as provided for in Section 33 of this Act; (b) He/She is of good moral character and had not been convicted by a court of law of a criminal offense involving moral turpitude; (c) For the Electronics Engineering examinations, he/she is a holder of a degree of Bachelor of Science in Electronics and Communications Engineering or Electronics Engineering, or subject to compliance with minimum requirements to be prescribed by the Board, such equivalent and/or related engineering course or program from any school, institute, college, or university recognized by the Government or the State where it is established, after completing a resident collegiate course equivalent to that of a full baccalaureate degree; (d) For the Electronics Technician examinations: (1) is a graduate of an Associate, Technician, Trade or Vocational course in electronics or, subject to the evaluation of the Board, such equivalent and/or related formal or non-formal course or program from any school, college, university or training institution recognized by the Government or the State where it is established, after completing a resident course or program of not less than two (2) years, or (2) has completed at least the minimum third-year equivalent of a Bachelor of Science program in Electronics and Communications Engineering or Electronics Engineering according to CHED guidelines, or, subject to the evaluation of the Board such equivalent and/or related engineering course or program from any school, institute, college or university recognized by the Government or State where it is established;


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SEC. 15. Scope of Examination for Electronics Engineers and Electronics Technicians. - The examination for Electronics Engineers shall consist of written tests which shall cover subjects prescribed by the Board but including at least the following: Mathematics, Applied Sciences, Engineering Economics, Laws and Ethics, Electronics, Communications, Computers, and Information and Communications Technology. The examinations for Electronics Technician shall consist of written and/or practical tests covering subjects to be prescribed by the Board and shall cover subjects specific to the practice of Electronics Technicians. As urgent and important need arises so as to conform to technological and modern changes, the Board may re-cluster, rearrange, modify, add to or exclude any subject and prescribe the number of final examination/s per year after approval by the Commission. The PRC Board resolution thereon shall be officially published in the Official Gazette or major daily newspapers of general circulation and also circularized and disseminated to all colleges. SEC. 16. Ratings. To pass the licensure examination, a candidate for Electronics Engineer or Electronics Technician must obtain a passing rating of seventy percent (70%) in each subject given during the examination: Provided, however, That a candidate who obtains a passing rating in the majority of the subjects but obtains a rating in the other subject/s below seventy percent (70%) but not lower than sixty percent (60%), shall be allowed to take one removal examination on the subject/s where he/she failed to obtain the passing rating: Provided, finally, That should the examinee fail to obtain a passing rating in the removal examination, he/she shall be considered as having failed the entire licensure examination. SEC. 17. Release of the Results of Examination. - The Board and the Commission shall correct and rate the licensure examination papers and shall release the examination results within fifteen (15) days after the said examination. SEC. 18. Qualifications and Schedule of Registration for Professional Electronics Engineer. - For application for registration as a Professional Electronics Engineer, the following shall be required: (a) Valid Certificate of Registration and Professional Identification Card as Electronics Engineer; (b) Valid/current membership identification card from the Accredited Professional Organization; (c) Certified experience record of active self-practice and/or employment either in government service or in the private sector, in the format to be prescribed by the Board, indicating the inclusive dates, companies worked for, description of specific responsibilities, relevant accomplishments and name, position of immediate supervisors for a period of at least seven (7) years


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LAWS AND ETHICS (inclusive and/or aggregate), at least two (2) years of which are in responsible charge of significant engineering work, from the date applicant took his/her oath as an Electronics and Communications Engineer or Electronics Engineer; (d) Three (3) certifications signed by three (3) Professional Electronics Engineers attesting that the experience record submitted by the applicant is factual.

Applications for registration as Professional Electronics Engineer may be submitted anytime to the Commission. The Board shall then schedule an en banc oral interview of the applicant for the purpose of verifying, authenticating and assessing the submittals and establishing the competency of the applicant according to rules, regulations and competency standards to be formulated by the Board: Provided, That those who have been registered and licensed as Electronics and Communications Engineers under Republic Act No. 5734 for at least seven (7) years upon the effectivity of this Act need only to submit items (a), (b), and (c) above: Provided, further, That those who have been registered and licensed as Electronics and Communications Engineers under Republic Act No. 5734 for less than seven (7) years after the effectivity of this Act shall submit shall submit their certified experience records and certifications from three (3) Professional Electronics Engineers as in items (c) and (d) above, and submit to an en banc oral interview of the Board for competency assessment, upon passing of which he can be registered as a Professional Electronics Engineer. SEC. 19. Issuance of the Certificate of Registration and Professional Identification Card. - A Certificate of Registration shall be issued to examinees who pass the Electronics Engineer and Electronics Technician licensure examination, to Electronics Engineers who are registered as Professional Electronics Engineers and to Electronics Technicians who are registered without examination, subject to payment of fees prescribed by the Commission. The Certificate of Registration shall bear the signature of the Chairperson and Members of the Board, stamped with the official seal of the Commission and the Board, indicating that the person named therein is entitled to practice the profession with all the privileges appurtenant thereto, subject to compliance with all applicable requirements. The said certificate shall remain in full force and effect until withdrawn, suspended or revoked in accordance with this Act. A Professional Identification Card bearing the registration number, date of registration, duly signed by the Chairperson of the Commission, shall likewise be issued to every registrant who has paid the prescribed fee. This identification card will serve as evidence that the holder thereof is duly registered with the Commission. SEC. 20. Registration without Examination for Electronics Technicians. - Within five (5) years after the effectivity of this Act, the Board shall issue Certificates of Registration and Professional Identification Cards without examination to applicants for registration as Electronics Technicians who shall present evidence or other proof satisfactory to the Board that:


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(a) He/She is a graduate of at least a two-year Associate, Technician, Trade or Vocational course in Electronics as certified by the TESDA, or that he/she has completed at least the minimum thirdyear equivalent of a Bachelor's Degree of Science in Electronics and Communications Engineering or Electronics Engineering according to CHED guidelines, or, subject to the evaluation of the Board, an equivalent and/or related formal or non-formal course or program from any school, institute, college, university or training institution recognized by the Government or the State where it is established, and (b) He/She has rendered at least seven (7) years (inclusive or aggregate) of active self-practice and/or employment either in the Government or private sector, indicating therein his/her specific duties and responsibilities, relevant accomplishments, the complete names and addresses of clients and companies or persons worked for, as well as the names and positions of immediate superiors. The above submittals shall be accompanied by a certification from at least three (3) registered Professional Electronics Engineers vouching for the integrity, technical capability and good moral character of the applicant. SEC. 21. Non-issuance of a Certificate of Registration and/or Professional Identification Card for Certain Grounds. - The Board and/or the Commission shall not register and shall not issue a Certificate of Registration and Professional Identification Card to any person convicted by a court of competent jurisdiction of any crime involving moral turpitude, to any person of immoral or dishonorable conduct and to any person of unsound mind, furnishing the party concerned a written statement containing the reasons for such action, which statement shall be incorporated in the records of the Board. SEC. 22. Professional Oath. - All successful examinees and all those who have qualified for registration without examination shall be required to take a professional oath before any member of the Board or any person authorized by the Commission before he/she can be registered and issued a Certificate of Registration and Professional Identification Card, and as a prerequisite to practicing as a Professional Electronics Engineer, Electronics Engineer or Electronics Technician. SEC. 23. Revocation and Suspension of Certificate of Registration, Professional Identification Card and Cancellation of Special Permits. - The Board shall, upon proper notice and hearing, revoke or suspend the validity of a Certificate of Registration and accordingly the Professional Identification Card, or cancel a Special Permit granted under Section 26 herein, for any of the causes mentioned in the preceding sections, or for unprofessional or unethical conduct, malpractice, incompetence or any violation of this Act and its implementing rules and regulations, the Code of Ethics and the Code of


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Technical Standards of Practice, or where fraud, deceit, or false statement was found to have been employed in obtaining said Certificate of Registration, Professional Identification Card or Special Permit. SEC. 24. Reinstatement, Re-issuance or Replacement of Certificate of Registration and Professional Identification Card. - The Board may, two (2) years after the revocation of a Certificate of Registration and Professional Identification Card, upon application and for reasons deemed proper and sufficient, reinstate the validity of a revoked Certificate of Registration and Professional Identification Card, subject to compliance with the applicable requirements of the Commission, and the Board: Provided, That he/she did not commit any illegal practice of the profession or any violation of this Act, its rules, codes and policies during the time that his/her Certificate of Registration and Professional Identification Card was revoked. A new Certificate of Registration or Professional Identification Card to replace lost, destroyed, or mutilated certificate or registration card may be issued, subject to the rules promulgated by the Board and the Commission, upon payment of the required fees. SEC. 25. Roster of Professional Electronics Engineers, Electronics Engineers and Electronics Technicians. - The Board shall prepare and maintain a roster of the names, residence and/or office address of all registered Professional Electronics Engineers, Electronics Engineers and Electronics Technicians, which shall be updated annually in cooperation with the Accredited Professional Organization, indicating therein the status of the Certificate of Registration, Professional Identification Card and membership in the Accredited Professional Organization of the professional, whether valid, inactive due to death or other reasons, delinquent, suspended or revoked. The said roster shall be conspicuously posted within the premises of the Commission and the information therefrom made available to the public upon inquiry or request. SEC. 26. Exemptions from Examination and Registration. - No examination and registration shall be required for foreign Professional Electronics Engineers, Electronics Engineers or Electronics Technicians who are temporarily employed by the Philippine Government or by private firms in the Philippines in the following cases: (a) Where no qualified equivalent Filipino professional is available for the specific item of work to be rendered, as attested to by the Accredited Professional Organization; (b) Where the conditions of the scope and funding for the work or project are such that it stipulates the temporary employment of a foreign professional; (c) As defined in the General Agreement on Trade in Services, the ASEAN and APEC Engineer Registry programs and other similar international treaties, agreements and/or covenants to which the


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Philippine Government is a signatory and has ratified: Provided, however, That: (1) The said foreign professional is legally qualified to practice his/her profession in his/her own country in which the requirements for licensing and registration are not lower than those specified in this Act; (2) The work to be performed by said foreign professional shall be limited only to the particular work or project for which he/she was specifically contracted; (3) Prior to commencing the work, the foreign professional shall secure a Special Permit from the Board, which shall be subject to the approval of the Commission; Provided, That no working visa and/or permit shall be issued by concerned government agencies unless such Special Permit has been granted first; (4) The same foreign professional shall not engage in private practice on his/her own account; (5) For every foreign professional contracted for the work or project, at least two (2) corresponding Filipino professionals who are registered under this Act shall be employed as counterparts by the Philippine Government or the private firm utilizing the services of such foreign professional for at least the same duration of time as the foreigner's tenure of work; and (6) The Special Permit herein granted shall be valid only for a period of not more than six (6) months and renewable every six (6) months thereafter subject to the discretion of the Board and the approval of the Commission: Provided, That said permit shall cease to be valid if the foreigner terminates his/her employment in the work or project for which said permit was originally granted and thereafter engages in an occupation that requires another Special Permit or registration under this Act.


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ARTICLE IV PRACTICE OF PROFESSIONAL ELECTRONICS ENGINEERING, ELECTRONICS ENGINEERING AND ELECTRONICS TECHNICIAN SEC. 27. Practice of the Profession. - No person shall offer himself/herself in the Philippines as, or use the title "Professional Electronics Engineer", "Electronics Engineer" or "Electronics Technician", as defined in this Act, or use any word, letter figure, or sign whatsoever, tending to convey the impression that he/she is a Professional Electronics Engineer, Electronics Engineer or Electronics Technician, or advertise that he/she is qualified to perform the work of a Professional Electronics Engineer, Electronics Engineer or Electronics Technician, without holding a valid Certificate of Registration and a valid Professional Identification Card in accordance with this Act, except as provided under Section 26 hereof. SEC. 28. Prohibitions and Limitations on the Practice of Electronics Engineering and Electronics Technician Profession. - Unless otherwise prescribed by any supervening law, the practice of electronics engineering and electronics technician shall be a professional service, admission to which must be determined on the basis of the individual's personal qualifications. Hence, no firm, company, partnership, association or corporation may be registered or licensed as such for the practice of electronics engineering and electronics technician. However, persons properly registered and licensed as Professional Electronics Engineers, Electronics Engineers or Electronics Technicians may, among themselves or with any other allied professionals, form a partnership or association or corporation and collectively render services as such: Provided, That individual members of such partnerships or associations or corporations shall be responsible for their own respective acts. SEC. 29. Seal of the Professional Electronics Engineers. - All licensed Professional Electronics Engineers shall obtain and use a seal of a design prescribed by the Board bearing the registrant’s name, registration number and title. Plans, drawings, permit applications, specifications, reports and other technical documents prepared by and/or executed under the supervision of, and issued by the Professional Electronics Engineer shall be stamped on every sheet/page with said seal, indicating therein his/her current Professional Tax Receipt (PTR) number, date/place of payment and current membership number in the Accredited Professional Organization, when filed with government authorities or when used professionally. SEC. 30. Code of Ethics and Code of Technical Standards of Practice. - The Board shall adopt a Code of Ethics and the Code of Technical Standards of Practice for Electronics Engineers and Technicians, which shall be promulgated by the Accredited Professional Organization.


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ARTICLE V SUNDRY PROVISIONS SEC. 31. Continuing Professional Education (CPE) and/or Development Programs. – All registered Professional Electronics Engineers, Electronics Engineers, and Electronics Technicians, shall comply with pertinent rules and regulations already prescribed by and/or as may be prescribed and promulgated by the Commission and/or the Board, the Accredited Professional Organization and other government agencies, pursuant to this Act and other relevant laws, international treaties, agreements and/or covenants to which the Philippines is a signatory and has ratified, with respect to continuing professional education and/or development and/or other similar/related programs. SEC. 32. Integrated and Accredited Professional Organization. - There shall be one (1) integrated and Accredited Professional Organization of Professional Electronics Engineers, Electronics Engineers and Electronics Technicians in the country, which shall be registered with the Securities and Exchange Commission as a non-stock, non-profit corporation and recognized by the Board, the Commission and all government agencies as the one and only integrated and accredited national organization for the said professionals. Every Professional Electronics Engineer, Electronics Engineer and Electronics Technician, upon registration with the Commission as such, shall ipso facto become a member of this Accredited Professional Organization. Those who have been previously registered by the Board but are not members of this Accredited Professional Organization at the time of effectivity of this Act, shall be allowed to register as members of this organization within three (3) years after the effectivity of this Act. Membership in this Accredited Professional Organization shall not be a bar to membership in other associations of the electronics engineering and electronics technician professions. The Accredited Professional Organization shall implement the continuing professional education, accredit other organizations or entities to provide continuing professional education, and/or development program promulgated by the Board and/or the Commission, compliance with which shall be one of the requisites for the maintenance of membership in good standing of the professional in the Accredited Professional Organization. All members of good standing of this Accredited Professional Organization shall be issued an annual membership card indicating the membership number and validity period of the membership, which shall be affixed to all plans, specifications and any document signed by the member in the course of practice of his/her profession. Failure to maintain membership in good standing in the Accredited Professional Organization shall be a cause for listing of the individual as delinquent in the roster of professionals. SEC. 33. Foreign Reciprocity. - No foreigner shall be admitted for registration as Professional Electronics Engineer, Electronics Engineer or Electronics Technician with or without examination under this Act unless he/she proves in the manner as provided by the Board that, by specific provisions of law, the country, state or province of which he/she is a citizen, subject or national, or in


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accordance with international treaties, agreements and/or covenants to which their country, state or province is a signatory, admits Filipino citizens to practice as Professional Electronics Engineer, Electronics Engineer or Electronics Technician after an examination or registration process on terms of strict and absolute equality with the citizens, subjects or nationals of said country, including the unconditional recognition of professional licenses issued by the Board and/or the Commission and prerequisite degrees/diplomas issued by institutions of learning duly recognized by the government of the Philippines. Sec. 34. Positions in Government Requiring the Services of Registered and Licensed Professional Electronics Engineers, Electronics Engineers and Electronics Technicians. - Within three (3) years from the effectivity of this Act, all existing and proposed positions in the local and national government, whether career, permanent, temporary or contractual and primarily requiring the services of Professional Electronics Engineers, Electronics Engineers or Electronics Technicians shall accordingly be filled only by registered and licensed Professional Electronics Engineers, Electronics Engineers or Electronics Technicians. ARTICLE VI PENAL PROVISION AND ASSISTANCE OF LAW ENFORCEMENT AGENCIES SEC. 35. Penal Provision. - The following shall be punished by a fine of not less than One hundred thousand pesos (P100,000.00) nor more than One million pesos (P1,000,000.00), or by imprisonment of not less than six (6) months nor more than six (6) years, or both, in the discretion of the court: (a) Any person who shall give any false or fraudulent statement to the Board to obtain a Certificate of Registration and/or Professional Identification Card as Professional Electronics Engineer, Electronics Engineer or Electronics Technician; (b) Any person who shall present or use as his/her own a Certificate of Registration, Professional Identification Card, membership identification card in the Accredited Professional Organization and/or seal issued to another and any person who allows the use of his/her Certificate of Registration, Professional Identification Card, membership card in the Accredited Professional Organization and/or seal; (c) Any person who shall present or use a revoked or suspended Certificate of Registration as Professional Electronics Engineer, Electronics Engineer or Electronics Technician; (d) Any person who shall assume, use, or advertise as Professional Electronics Engineer, Electronics Engineer or Electronics Technician, or append to his/her name, any letter/s or words


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tending to convey the impression that he/she is a registered Professional Electronics Engineer, Electronics Engineer or Electronics Technician, when in fact he/she is not duly registered with the Board as such; (e) Any Professional Electronics Engineer, or any person on his/her behalf, who shall stamp or seal any document with his/her seal as such after his/her Certificate of Registration, Professional Identification Card and membership card in the Accredited Professional Organization has been revoked or suspended or after he/she has been suspended from practice or removed from the roster of Professional Electronics Engineer, Electronics Engineers or Electronics Technicians; (f) Any Professional Electronics Engineer who shall sign his/her name, affix his/her seal, or use any other method of signature on plans, technical descriptions or other documents prepared by or under the supervision of another Professional Electronics Engineer, unless the same is prepared in such manner as to clearly indicate the part of such work actually performed by the former; (g) Any person, except the Professional Electronics Engineer or Electronics Engineer in-charge, who shall sign for any electronics engineering work, or any function of electronics engineering practice, not actually performed by him/her; (h) Any person holding a Certificate of Registration and Professional Identification Card as Professional Electronics Engineer, Electronics Engineer or Electronics Technician who shall be involved in illegal wire-tapping, cloning, hacking, cracking, piracy and/or other forms of unauthorized and malicious electronic eavesdropping and/or the use of any electronic devices in violation of the privacy of another or in disregard of the privilege of private communications and/or safety to life, physical and/or intellectual property of others, or who shall maintain an unlicensed and/or unregistered communications system or device; and (i) Any person who shall violate any provision of this Act or any rules, regulations, the Code of Ethics and the Code of Technical Standards of Practice promulgated under this Act. SEC. 36. Assistance of Law Enforcement and Other Government Agencies. Any law enforcement agency shall, upon call or request of the Board and/or the Commission, render assistance in enforcing this Act including the Code of Ethics, Code of Technical Standards of Practice and the implementing rules and regulations and measures promulgated hereunder, by prosecuting violators thereof in accordance with law and the Rules of Court. Any department, instrumentality, office, bureau, institution or agency of the government including local governments, upon call or request from the


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Board and/or the Commission, shall render such assistance as it may require, cooperate and coordinate with it in carrying out, enforcing or implementing this Act, the codes, policies, measures, programs or activities of the Board and/or the Commission that it may undertake pursuant to the provisions of this Act. ARTICLE VII TRANSITORY PROVISIONS SEC. 37. Transitory Provision - Upon effectivity of this Act, the incumbent Board of Electronics and Communications Engineering shall complete all pending/unfinished works within a six (6)-month period, after which it shall cease to exist. The President of the Philippines shall before then appoint the Chairman and members of the first Board of Electronics Engineering in accordance with Sections 6 and 8 herein, who shall formulate and thereafter promulgate the rules and regulations for the implementation of this Act. SEC. 38. Vested Rights: Electronics and Communications Engineers when this Law is Passed. - Electronics and Communications Engineers holding a valid Certificate of Registration and Professional Identification Card at the time of effectivity of this Act shall be automatically registered and recognized as Electronics Engineers and shall be issued a new Certificate of Registration and Professional Identification Card as Electronics Engineers with the same license number as their original Electronics and Communications Engineer Certificate of Registration, subject to the payment of prescribed fees and other requirements of the Board and/or Commission. ARTICLE VIII FINAL PROVISION SEC. 39. Implementing Rules and Regulations. - Subject to the approval of the Commission, the Board, in coordination with the accredited professional organization, shall adopt and promulgate such rules, regulations, resolutions, the Code of Ethics and the Code of Technical Standards of Practice for Professional Electronics Engineers, Electronics Engineers and Electronics Technicians to carry out the provisions of this Act, which shall be published in the Official Gazette or a newspaper of general circulation and shall be effective fifteen (15) days after publication therein. SEC. 40. Appropriations. - The Chairperson of the Professional Regulation Commission shall include in the Commission's program the implementation of this Act, the funding of which shall be included in the annual General Appropriations Act. SEC. 41. Separability Clause. - If any provisions of this Act or any portion hereof is declared unconstitutional by any competent court, the other provisions hereof shall not be affected thereby.


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SEC. 42. Repealing Clause. - Republic Act No. 5734 is hereby repealed. All other laws, executive orders, rules and regulations or parts thereof in conflict with the provisions of this Act are hereby repealed or amended accordingly. SEC. 43. Effectivity. - This Act shall take effect fifteen (15) days following its full publication in the Official Gazette or any newspaper of general circulation. Approved. FRANKLIN M. DRILLON

President of the Senate

JOSE DE VENECIA, JR.

Speaker of the House of Representatives

This act which is a consolidation of House Bill No. 5224 and Senate Bill No. 2683 was finally passed by the House of Representatives and the Senate on February 2, 2004. OSCAR G. YABES

Secretary of the Senate

ROBERTO P. NAZARENO

Secretary General House of Representatives

Approved: April 17, 2004 GLORIA MACAPAGAL-ARROYO

President of the Philippines

Pledge of an Electronics & Communications Engineer I am an Electronics and Communications Engineer. In my profession, I take deep pride, but without vainglory; to it I owe solemn obligations that I am eager to fulfill. As an Electronics and Communications Engineer, I will participate in none but honest and legal enterprises. To him who has engaged my services, as employer or client, I will give the utmost of performance and fidelity. When needed, my skill and knowledge shall be given without reservation for the public good. From my special capacity springs the obligation to use it well in the service of humanity; and I accept the challenge that implies. Zealous of the high repute of my calling, I will strive to protect the interests and the good name of any engineer that I know to be deserving; but will not shrink, should duty dictate, from disclosing the truth regarding anyone who, by unscrupulous act, has shown himself unworthy of the profession.


LAWS AND ETHICS

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As others before me have vitalized and turned to practical account the principles of science and revelations of technology and have rendered usable to mankind nature’s vast resources of matter and energy so do I dictate myself to the analysis, synthesis and dissemination of engineering knowledge and practice, and especially to the instruction of younger members of any profession in all arts and traditions. To my colleagues I pledge in the same full measure I ask of them, integrity and fair dealing, tolerance and respect, and devotion to the standards and dignity of our profession, with the consciousness always, that our special expertise carries with it the obligation to serve humanity with complete dedication.

Engineering Code of Ethics A.

Fundamental Principles Engineers uphold and advance the integrity, honor, and dignity of the engineering profession by: ª Using their knowledge and skill for the enhancement of human welfare ª Being honest and impartial, and servicing with fidelity to the public, their employers, and clients ª Striving to increase the competence and prestige of the engineering profession ª Supporting the professional and technical societies of their disciplines

B.

Fundamental Canons 1. The engineer shall hold paramount safeguarding life, health and property and promoting the public welfare in the performance of their professional duties. ª

The engineer shall recognize that the lives, safety, health, welfare of the general public are dependent upon engineering judgments, decisions and practices incorporated into structures, machines, products, processes and devices.

ª

The engineer shall approve and seal only those design documents which in his considered opinion do not endanger the life, health, property and the public welfare in conformity with accepted engineering standards.

ª

The engineer shall not permit the use of his own, firms or associates’ name in business ventures with any person or firm which upon investigation he believes is engaging in fraudulent or dishonest business or professional practices.

ª

The engineer having knowledge of any alleged violation of the Code of Ethics shall be forthright and can did in cooperating with


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the Council in furnishing such information or assistance as may be required. 2.

3.

4.

The engineer shall perform services only in areas of his competence. ª

The engineer shall undertake to perform engineering only when qualified by education or experience in the areas of professional engineering involved.

ª

The engineer may accept an assignment requiring education or experience outside his areas will be performed by a Professional Engineer or otherwise qualified associates, consultants, or employees. He may then sign and seal the documents for the total project. The engineer shall not affix his seal to any such document not prepared under his supervisory control and review.

The engineer shall issue professional statements only in an objective and truthful manner. ª

The engineer shall be completely objective and truthful in all professional reports, statements, or testimony, and shall include all relevant and pertinent information.

ª

The engineer shall publicly express a professional opinion on technical subjects only when it is founded on technical subjects only when it is founded on adequate knowledge of the facts and competence in the subject matter.

ª

The engineer when acting as a representative of an individual or organization shall issue no statements, criticisms, or arguments on engineering matters unless he has prefaced those comments by explicitly identifying on whose behalf he is speaking. When the engineer is acting as a consultant his expression or professional opinion shall be prefaced by identifying his status as a consultant, without necessarily naming the client. The engineer shall reveal any personal interest he may have in the matter.

The engineer shall act in the professional matters for each employer or client as faithful agent or trustee, avoiding conflicts or interest. ª

The engineer shall disclose all known or potential conflicts of interest to his employer and/or client by promptly informing them of any business association, interest, or other circumstances which could influence his judgment or the quality of services.

ª

The engineer shall not accept compensation, financial or otherwise, from more than one party for services on the same project, unless the circumstances are fully disclosed and agreed to by all interested parties.


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5.

ª

The engineer shall not solicit or accept financial or other valuable consideration, directly or indirectly, from suppliers, contractors, their agents, or other parties in connection with his work for employees or clients.

ª

The engineer in public service as a member, advisor or employee of governmental body shall not participate in decisions on work which involves professional services solicited or provided by him or his organization.

ª

The engineer shall not solicit or accept a professional contract from a governmental body on which a principal or officer of his organization serves as a member, except upon public disclosure of all pertinent facts and circumstances and consent of the appropriate public authority.

ª

The engineer shall not reveal proprietary information obtained in a professional capacity without the prior consent of the client or employer, except as authorized or required by law.

The engineer shall avoid improper solicitation of professional employment. ª

The engineer shall not offer, give, solicit, or receive, either directly or indirectly, any commission, gift or other valuable consideration in order to secure work.

ª

The engineer shall not falsify or permit misrepresentation of his own, or associates’ academic or professional qualification. He shall not misrepresent his degree of responsibility in prior assignments. Brochures or other presentations incident to the solicitation of employment shall not misrepresent pertinent facts concerning employers, employees, associates, joint ventures, or past accomplishments.

ª

The engineer shall not submit any proposal for purpose of obtaining professional work in which he falsifies or misrepresents his capability of carrying out that proposal.


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Electronics and Communications Engineering Code of Ethics and Professional Conduct A.

Foreword Honestly, Justice, and Courtesy form a moral philosophy which, associated with mutual interest among men, constitutes the foundation of ethics. The engineer should recognize such a standard, not in passive observance, but as dynamic principles guiding his conduct and way of life. It is duty to practice his profession according to this Code of Ethics and Conduct. The keystone of professional conduct is integrity. Hence, it behooves the engineer to discharge his duties with fidelity to the public, his employers and clients, and with fairness and impartiality to all. It is his duty to interest himself in public welfare, and to be ready to apply his special knowledge for the benefit of mankind. He should uphold the honor and dignity of his profession and avoid association with any enterprise of questionable character. In his dealing with fellow engineers he should be fair and tolerant.

B.

Relations with the State (Section 1) 1. 2. 3. 4. 5.

6. 7. 8.

Each and every engineer shall recognize and respect the supreme authority of the state as expressed through its laws and implemented by its agencies. He shall recognize that the well-being of the public and the interests of the state are above the well being and interest of any individual. In the interest of justice, he shall aid the state, if and when the technology is needed for the prevention and/or prosecution of unjust, criminal, or unlawful acts. In the interest of good government, he shall in every way possible extend cooperation to the state in the accomplishment of its goals and objectives. In the interest of social efficiency, he shall extend assistance, guidance, and training to all subordinates under his jurisdiction in order to increase their skill and ability, knowledge, and experience for the purpose of eventually increasing their responsibilities. In the interest of the national economy and well-being, he shall always strive in the execution of his work for optimum efficiency, economy, and safety. In the interest of national security, the state shall be given primary consideration in all his inventions and/or devices in electronics and communications useful for national security and defense. In the event of any national emergency, he shall offer his technology, skill, ability, and experience to the services of the state, even if it will involve personal sacrifices.


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C.

Relations with the Public (Section 2) 1. 2.

3. 4.

5. 6.

7. 8.

D.

He shall interest himself in public welfare and be ready to apply his special knowledge for the benefit of mankind. He shall guard against conditions that are dangerous or threatening to life, limb, or property on work for which he is responsible, or if he is not responsible, he shall promptly call such conditions to the attention of those responsible so that the conditions can immediately and effectively be corrected. He shall have due regard for the safety of life and health of the public who may be affected by the work for which he is responsible. He shall endeavor to extend public knowledge of electronics and communications engineering and he shall strive to win or maintain the public confidence by discouraging the spread of untrue, unfair, and exaggerated statements regarding this engineering. As a witness before a court, commission, and/or tribunal, he shall express an opinion only when it is founded on adequate knowledge and honest conviction. He shall not issue on matters connected with public policy, statements, criticisms, or arguments which are inspired of paid for by private interests, unless he identifies on whose behalf he is making the statements. He shall refrain from expressing any public opinion on an engineering subject unless he is fully familiar and knowledgeable with all the facts relating to the subject. His integrity shall be unquestionable and he shall discharge his duties and responsibilities with fidelity to the public, his employers and clients, and with fairness and impartiality to all.

Relations with Clients, Employer, and Labor (Section 3) 1.

2.

3. 4. 5. 6.

He shall act in professional matters as a faithful agent or trustee, and treat as confidential all matters and information concerning the business affairs, technical processes, etc. of his clients and/or employers. He shall inform his client or employer of any financial interest on inventions, devices, equipment, or any other thing, before undertaking any engagement in which he may be called upon to decide on the use thereof. He shall not accept any other compensation, financial or otherwise, except from one interested party for a particular service or other services related therewith without the consent of all parties concerned. He shall exercise fairness and justice when dealing with contracts between his clients or employers and the contractors. He shall not accept any commissions or allowances, directly or indirectly, from contractors, suppliers, and all other connection with work for which he is responsible. He shall not be financially interested in the bid or bids of contractors, suppliers, and other interested parties participating in a competitive


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work or job on which he has been employed as engineer without full knowledge and consent of his clients or employers. 7. He shall promptly inform his clients or employers of any business in which he has any interest, business connection or affiliation which may compete with or affect the business of his clients or employers. 8. He shall not allow any decision in connection with work for which he has been employed or which he may be called upon to perform, to be affected by interests in any business. 9. He will present clearly the consequences to be expected from deviation proposed if his engineering or his judgment is overruled by nontechnical authority in case where he is responsible for the technical adequacy or engineering work. 10. He shall undertake only those engineering assignments for which he is qualified. He shall engage or advise his employer or client to engage in specialists and shall cooperate with them whenever his employer’s or client’s interests are served best by such an arrangement. E.

Relations with Engineers (Section 3) 1. 2.

3.

4. 5. 6. 7. 8.

He shall individually or collectively with others in the profession protect the profession from misunderstanding and/or misinterpretations. He shall not directly or indirectly injure the professional reputation, prospects, advancement, and/or practice of other engineers. However, if he has proof or personal knowledge that an engineer has been unethical and/or illegal in his practice, he shall inform in writing the proper authorities for appropriate action. He shall uphold the principle of appropriate and adequate compensation for those engaged in the engineering profession, including those in the subordinate capacities, in the interest of public service and maintenance of the standards of the profession. He shall not try to supplant another engineer in particular employment after becoming aware that definite steps have been taken toward the other’s employment. He shall not compete, by underbidding through reduction in his normal fess on the basis of charges for work, after having been informed of the charges submitted by another engineer. He shall be fair and tolerant in his dealings with fellow engineers and give credit to those whom credit is properly due. He shall uphold the honor and dignity of his profession and avoid association in responsibility for work with engineers who do not conform to ethical practices. He will exercise due restraint in criticizing another engineer’s work in public, recognizing the fact, that the engineering societies and the engineering press provide the proper forum for technical discussion and criticism.


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F.

Relation to the Profession (Section 5) 1.

2. 3.

He shall cooperate in extending the effectiveness of the engineering profession and endeavor to be well-informed of the latest development in the profession by sharing or exchanging information and experience with other engineers, other professionals and students; and by contributing to engineering publications and schools and by participating in the activities of engineering societies. He shall cooperate in upholding the integrity, dignity, and honor of the profession by avoiding all conducts and practices that will discrediting and injurious to the profession. He shall dignified and modest in explaining or discussing his work and/or merit and shall refrain from self-laudatory advertising or propaganda.

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I

H

1.

Which of the following is designated as the international distress, safety and calling frequency for radio telephony for station of the maritime mobile service when using frequencies in the authorized bands between 156 and 174 MHz? A. 168. 5 MHz B. 165.8 MHz C. 158.6 MHz D. 156.8 MHz

2.

This is a terminal where riser cable pairs are terminated to serve a portion or an entire floor of a building. A. Floor terminal distribution area B. Floor distribution terminal C. Riser terminal D. Raceway terminal

3.

Referred to as the device which diverts high transient voltage to the ground and away from the equipment thus protected. A. Arrester B. Alarm C. Anchor D. Alpeth

4.

The executive branch of government in charge of policy making in the telecommunication. A. Bureau of Telecommunications B. Department of Transportation and Communications C. Telecommunications Control Bureau D. National Telecommunications Communication

5.

A government regulation in telecommunication which provide policy to improve the provision of local exchange carrier service. A. E.O 546 B. E.O. 59 C. Act. 3846 D. E.O.109

6.

What is the maximum number of lines for any building other than a one or two story residential building to be required a service entrance facility under ECE building code? A. Not required B. Five lines C. Three lines D. Two lines The maximum power suggested by KBP on 919-1312 AM broadcast station in Metro Manila is A. 10 kW B. 5 kW C. 15 kW D. 20 kW

7.

8.

The potential difference between any exposed structure to ground in any electrical installation should not exceed ____ volts RMS. A. 45 B. 30 C. 10 D. 0


LAWS AND ETHICS

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9.

When is Electronics and Communications Engineer supervision required under DO 88? A. Standard AM broadcast with a carrier power of 5 kilowatts B. TV RF power booster with effective radiated power of 2 kilowatts C. TV translator with an RF carrier power of 0.5 watt D. FM broadcast station with a carrier power of 1000 watts

10. Which of the following penalty is provide under the existing telecommunication law, should an international carrier unable to comply with its obligation to provide local exchange un-served and under-served areas within three years from grant of authority? A. Given two years to comply B. Given one year to comply C. Financial penalty in accordance with existing schedules D. Cancellation of its authority 11. A law that specifically requires the services of a duly registered Electronics and Communications Engineer in the designing, installation and construction, operation and maintenance of radio stations. A. Act. 3846 B. Dept. Order 88 C. D.O. 1000 D. R.A. 5734 12. What is the basic qualification of an applicant for public carrier network before a Certificate of Public Convenience or a Provisional Authority is issued? A. Radio station license B. Franchise C. Business permit D. SEC document 13. Who issues an authority to install, operate, and maintain a cable television system or render a television service with a specified area in the country? A. National Telecommunications Communication B. Congress of the Philippines C. Board of communications D. Department or Transportation and Communications 14. An entity providing transmitting and switching of telecommunications services, primarily but not limited to voice, in a geographic area anywhere in the country A. Local exchange operator B. Public toil calling operator C. Value-added service operator D. Franchisee 15. When could cable TV operator lease or sub-lease its capacity? A. If lessor can pay higher cost B. If he can generate capacity for lease C. If he has capacity D. If lessor is within the franchise area 16. An entity, relying on the transmission, switching and local distribution facilities of local exchange and inter-exchange operators, and overseas carriers, offers enhanced services beyond those ordinarily provided by such carriers. A. Local exchange carrier B. Value-added service provider C. Inter-exchange carrier D. International carrier 17. Professional Regulation Commission was created under A. RA 223 B. PD 223 C. PD 323 D. PD 232

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18. What is the main principle used by ITU in determining the distribution of the orbit/spectrum resources? A. Efficient use and equitable access B. Depending on geographical boundary of a nation C. Equal distribution D. Depending on national sovereignty 19. When is coordination with the telephone company needed when an underground service entrance will be used as the most feasible and economical way? A. Expense of telephone company B. Expense of subscriber C. Decision of depth of conduit at interconnection point D. Length of a cable to be used by subscriber 20. An order signed by former President F. Ramos last March 1998 providing the policy in the operation and use of international satellite communications in the Philippines. A. E.O. 3846 B. E.O. 467 C. E.O. 456 D. E.O. 59 21. Which body in the present ITU structure took the place of CCITT? A. ITU-T B. ITU-D C. ITU-R D. RAG 22. One of the major components required under the global maritime distress and safety system. A. Provision of Morse code B. Provision of radio personnel C. Provision of radiotelegraph operator D. Provision of facsimile 23. Which conference in the Development Bureau report? A. WTSC C. RRB

ITU

structure B. D.

does

the

Telecommunication

WRC WTDC

24. One of the following is not a major components required on board ship under the global maritime distress and safety system. A. Radio operator telegraphy onboard B. Shore base facilities C. Radio personnel onboard D. On board radio facilities 25. Which of the following shall be complied by an international ship pursuant to the global maritime distress and safety system implemented last 1999? A. A facsimile B. A Morse code C. A radio personnel D. A radio telegraph operator


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LAWS AND ETHICS

26. Referred to as linkage by wire, radio, satellite or other means, of two or more telecommunications carrier or operators with one another for the purpose of allowing or enabling the subscriber of one carrier or operator to access or reach the subscriber of the other carrier or operator. A. Outside plant sharing B. Interconnection C. Toll patching D. Gateway 27. NTC cannot grant one of the following, an authority to operate a cable television system within the same franchise area covered by any provisional Authority or Certificate previously granted by the Commission. A. New entrant has more financial support B. Prior operator has not complied sufficiently with term and condition of the authorization C. Current service is grossly inadequate D. Issuance to new entrant will not cause ruinous competition 28. The institutionalization of the Continuing Professional Education (CPE) program of the various regulated professions under the supervision of the Professional Regulation Communication. A. E.O. No. 626 B. P.D. 381 C. E.O. No. 266 D. E.O. No. 662 29. Judgment on the case against an ECE shall become final and executory after A. 10 days B. 30 days C. 60 days D. 15 days 30. The basic law providing for the regulation of radio station, communication in the Philippines and other purposes. A. Act. No. 3846 B. D.O. No. 5 C. D.O. No. 11 D. D.O. No. 88 31. Which one is not the basic electrical protection measure in the Philippine Electronic Codes? A. Shielding B. Grounding and bonding C. Undergrounding D. Voltage/current limiting and interrupting 32. Designated year for the full implementation of the Global Maritime Distress and Safely System. A. 1999 B. 1992 C. 1996 D. 1998 33. Which part of the housing system in ECE code is a circular opening through the floor structure to allow the passage of a cable and wire? A. Insert B. Slot C. Sleeve D. Raceway 34. The latest government regulation in the telecommunication which provides policy for the provision of local exchange carrier service. A. E.O. 59 B. Act. 3948 C. E.O. 109 D. E.O. 546


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35. Any governmental office responsible in discharging the obligations undertaken in the convention of the ITU and the regulation. A. Administration B. Telecommunication office C. Country D. The Union 36. A privilege conferred upon a telecommunications entity by Congress, authorizing an entity to engage in a certain type of telecommunications service. A. Provisional Authority B. Franchise C. Certificate of Public Convenience and Necessity D. Authority to operate 37. Where does the secretary general in the organizational structure of present ITU report? A. WRC B. Council C. TDAB D. WTSC 38. Who is the principal administrator of Republic Act 7925? A. Congress B. NTC C. DOTC D. BOC 39. An act which was passed by Congress providing for the installation, operation and maintenance of a public telephone in each and every municipality in the country. A. R.A. No. 7925 B. R.A. No. 6849 C. R.A. No. 3846 D. R.A. No. 3396 40. Under NTC Memorandum Circular, what is the fine of any person, corporation or entity who will illegally purchase, sale, lease and/or retail of mobile phone acquired from (per unit/violation) any illegal source. A. PHP 5,000.00 B. PHP 25,000.00 C. PHP 15,000.00 D. PHP 35,000.00 41. Under NTC Memorandum Circular, what is the fine of any person, corporation or entity who will purchase, sale, lease and/or retail of mobile phones without NTC type (per unit/violation) approval/type acceptance labels/stickers. A. PHP 500.00 B. PHP 25,000.00 C. PHP 5,000.00 D. PHP 50,000.00 42. Under NTC Memorandum order what is the fine of any person, corporation or entity that will illegally import mobile phone, parts and accessories thereof (per unit). A. PHP 5,000.00 B. PHP 25,000.00 C. PHP 7,500.00 D. PHP 75,000.00 43. According to NTC M.C. 07-08-2004 the following are the fees and charges for each repair shop for Filing Fee, Permit Fee, and Inspection Fee respectively. A. PHP 180, PHP 1200.00/year, PHP 720.00/year B. PHP 720, PHP 180.00/year, PHP 1200.00/year C. PHP 720, PHP 1200.00/year, PHP 180.00/year D. PHP 1200, PHP 180.00/year, PHP 720.00/year


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LAWS AND ETHICS

44. Under the administrative sanction of NTC M.C. 07-08-2004 what is the fine for an MPSC (Mobile Phone Service Center) operating without valid NTC Permit. A. PHP 5,000.00 B. PHP 25,000.00 C. PHP 7,500.00 D. PHP 75,000.00 45. Under the administrative sanction of NTC M.C. 07-08-2004 what is the fine for an unauthorized servicing of 5 mobile phones? A. PHP 5,000.00 B. PHP 25,000.00 C. PHP 7,500.00 D. PHP 75,000.00 46. As per NTC M.C. 07-08-2004 an MPSC (Mobile Phone Service Center) shall properly identify its business and location by posting conspicuously at the entrance of its premises a signboard of at least ______ dimension which shall indicate clearly its business name, the type of services it offers, its full business address and the MPSC permit number. A. 50cm x 150cm B. 100cm x 50cm C. 50cm x 100cm D. 150cm x 100cm 47. Under section 1 of the proposed NTC memorandum circular on the rules and regulation on the allocation and assignment of 3G radio frequency band, what is the radio frequency band allocated for Band A and B to be used by 3G networks A. Band A : 1920 – 1935MHz/2110 – 2125MHz, Band B : 1935 – 1950MHz/2125 – 2140MHz B. Band A : 2010 – 2025MHz, Band B : 1935 – 1950MHz/2125 – 2140MHz C. Band A : 1950 – 1965MHz/2140 – 2155MHz, Band B : 1885 – 1900MHz/1965 – 1980MHz D. Band A : 1920 – 1935MHz/2110 – 2125MHz, Band B : 1950 – 1965MHz/2140 – 2155MHz 48. Under section 1 of the proposed NTC memorandum circular on the rules and regulation on the allocation and assignment of 3G radio frequency band, what is the radio frequency band allocated for Band C and D to be used by 3G networks A. Band C : 1920 – 1935MHz/2110 – 2125MHz, Band D : 1935 – 1950MHz/2125 – 2140MHz B. Band C : 1950 – 1965MHz/2140 – 2155MHz, Band D : 1885 – 1900MHz/1965 – 1980MHz C. Band C : 2010 – 2025MHz, Band D : 1935 – 1950MHz/2125 – 2140MHz D. Band C : 1920 – 1935MHz/2110 – 2125MHz, Band D : 1950 – 1965MHz/2140 – 2155MHz 49. Under section 7 of the proposed NTC memorandum circular on the rules and regulation on the allocation and assignment of 3G radio frequency band, what is the SUF (spectrum user fee) for the allocated 3G radio frequency band. A. PHP 100T for each band B. PHP 10M for each band C. PHP 100M for each band D. PHP 10B for each band


7-41

50. Under article VI entitled “Penal Provision and Assistance of Law Enforcement Agencies” of the new ECE law, a fine of not less than ______ nor more than ______, or by imprisonment of not less than six (6) months nor more than six (6) years, or both, in the discretion of the court shall be imposed to those who will violate all those listed under section 35 of RA 9292. A. PHP 10,000, PHP100,000 B. PHP 10,000, PHP 1,000,000 C. PHP 100,000, PHP 1,000,000 D. PHP 1,000,000, PHP 5,000,000


7-41


Section

Basic Signals

25

and System

DEFINITION.

Read it till it Hertz

Signal any quantity that conveys information.

DEFINITION. System is the abstraction of a process or objects that puts signal into some form of relationship. DEFINITION. Input signal: Exist independently of the system and its not affected by the system. A. .CLASSIFICATION OF SIGNALS. 1.

In terms of Source/s ª

Scalar Signal A signal generated by a single source.

ª

Vector Signal A signal generated by a multiple source sources, often called multi-channel signal.

A color video signal is an example of vector signal composed of three signals representing the three primary colors: red, green, and blue: ⎡R(x, y, t)⎤ ⎢ ⎥ ν ( x, y, t ) = ⎢ G(x, y, t)⎥ ⎢⎣ B(x, y, t) ⎥⎦ 2.

In terms of Independent Variable/s ª

One-dimensional 1-D A 1-D signal is a function of a single independent variable.

The speech signal is an example of a 1-D signal where the independent variable is time. ª

Two-dimensional 2-D A 2-D signal is a function of a two independent variable.

An image signal, such as a photograph, is an example of a 2-D signal where the two independent variables are two spatial variables. Loading ECE SUPERBook

Basic signals and system

7-42

ª

Multi-dimensional M-D A M-D signal is a function of more than one independent variable.

The black and white video signal can be considered an example of a 3-D signal where the three independent variables are two spatial variables and time. 3.

In terms of certainty by which the signal can be uniquely described ª

Deterministic signal A signal that can be uniquely determined by a well defined process such as mathematical expression or rule is called deterministic signal.

A good example of deterministic signal is the output of a sinusoidal oscillator. ª

Random signal A signal that is generated in random fashion and cannot be predicted ahead of time is called random signal.

Noise can be considered as an example of random signal. 4.

Time-dependent signals ª

Continuous-time signals A continuous-time signal is defined at every instant of time.

A continuous-time signal with continuous amplitude is usually called an analog signal. ª

Discrete-time signals A discrete-time signal is defined at discrete instant of time, and hence, it is a sequence of numbers.

A discrete-time signal with discrete-valued amplitudes represented by a finite number of digits is referred to as a digital signal. B. .CLASSIFICATION OF DISCRETE-TIME SEQUENCE. 1.

Classification Based on Length The discrete-time signal may be a finite or infinite length sequence. i.

For finite length sequence A finite length sequence is defined for a finite interval N1≤n≤N2, where -∞

ii.

For Infinite length sequence ª Right-sided Sequence Has zero-valued samples for n
ª

2.

Right-sided Sequence Has zero-valued samples for n>N2, x[n]=0

Classification Based on Symmetry i.

Conjugate Symmetric A sequence is said to be conjugate-symmetric if x[n]=x*[-n]

A real conjugate-symmetric sequence is called an odd sequence.


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7-44

ii.

Conjugate Anti-Symmetric A sequence is said to be conjugate-antisymmetric if x[n]=-x*[-n]

A real conjugate-antisymmetric sequence is called an even sequence. 3.

Classification Based on Periodicity A sequence x[n] satisfying x[n]=x[n+kN] for all n is called periodic sequence.

4.

Classification Based on Boundary A sequence is said to be bounded if each of its samples is of magnitude less than or equal to a finite positive number.

5.

Other Classification i.

Summable Sequence A sequence x[n] is said to be absolutely summable if ∞

x[n] =

∑ x[n] < ∞

n = −∞

ii.

Square-Summable Sequence A sequence x[n] is said to be square-summable if ∞

x[n] =

∑ x[n]

n = −∞

2

<∞

7-45

Self-Sufficient Guide to ECE by JASON AMPOLOQUIO iii. Energy and Power Signal ∞

E≡

∑

x[n]

2

N

1 2 x[n] 2N 1 + N→∞ n= −N

P ≡ lim N

n = −∞

∑

C. .BASIC SEQUENCE. 1.

Unit Sample Sequence The simplest and one of the most useful sequence, often called discrete-time impulse or unit impulse, denoted by δ[n], is defined by

⎧1, δ[n] = ⎨ ⎩ 0,

2.

for n = 0 for n ≠ 0

Unit Step Sequence It is denoted by μ[n] and is defined by

⎧1, μ[n] = ⎨ ⎩ 0,

for n ≥ 0 for n < 0



7-46

The unit sample and unit-step sequence are related ∞

μ[n] =

∑ δ[k]

δ[n] = μ[n] − μ[n − 1]

k = −∞

3.

Ramp Sequence It is denoted by r [n] and is defined by

⎧ n, r[n] = ⎨ ⎩ 0,

4.

for n ≥ 0 for n < 0

Sinusoidal Sequence

x[n] = A cos(ω0 + φ)

5.

Exponential Sequence where :

x[n] = Aα

n

A = A e jθ α = e j(σ + jω0 )

D. .BASIC OPERATION ON SEQUENCE. 1.

Modulation The product of two sequence x[n] and h[n] of length N yields a sequence of y[n], also of length N, as given by

y[n] = x[n] • h[n]


7-47

Addition The addition of two sequences x[n] and h[n] of length N yields a sequence of y[n], also of length N, as given by

y[n] = x[n] + h[n]

3.

Scalar Multiplication The multiplication of a sequence x[n] by a scalar A result in a sequence y[n] of length N, as given by

y[n] = A • x[n] In the case of analog signals, this operation, is usually called amplification if the magnitude of the multiplying constant, called gain, is greater than 1. If the magnitude of the multiplying constant is greater less than 1, the operation is called attenuation. 4.

Delay Operation The delay of a sequence x[n] of infinite length by a positive integer M results in a sequence y[n] of infinite length given by

y[n] = x[n − M]

5.

Time-Reversal Operation A time-reversal operation generates an output y[n], also of length N but reverse in time, as given by

y[n] = x[−n]

E. .CLASSIFICATION OF SYSTEM. A system is the abstraction of a process or object that puts a number of signals into some relationship. ª

Input Signal Exist independently of the system and are not affected by the system.


7-48


ª

Output Signal Bears information generated by the system, often in response to input signals.

Basic Classification: 1. Linearity System for which the superposition principle applies is called linear system. 2.

Time-Invariant A system that responds to a delayed input signal with a correspondingly delayed output signal is called time-invariant system. Linear Time-Invariant System System that is both linear and time-invariant is called a/an LTI system.

3.

Static or Memoryless A system is said to be static system if its output at any instant n depends only on the input sample at the same time.

4.

Dynamic or System with Memory If the output of a system at time n is completely determined by the input samples in the interval from n-N to N (N≥0), the system is said to have memory of duration N.

5.

Causality A system is said to be causal (non-anticipative) if the output of the system at any time n depends only on the present and past input but does not depend on future inputs.

6.

Stability A system is said to be stable if and only if, for every bounded input, the output is also bounded.

7.

Passivity A discrete-time system is said to be passive if, for every finite energy input sequence x[n], the output sequence y[n] has, at most, the same energy.


8-1

PRC Examinee’s Guide (Rev. March 15, 2004)

The data and instructions presented here are subject to changes depending upon PRC discretion. A. .WHAT TO DO AFTER FILING APPLICATION?. 1. 2. 3.

Secure a copy of the Program of Examination from the Application Division or Customer Service Center. (This is FREE) Your school/building assignment will be posted two or three days before the first day of the examination. Visit your school/building assignment for environment and transport familiarization.

B. .WHAT TO BRING IN EXAMINATION DAYS?. 1. 2. 3. 4. 5. 6. 7. 8.

Notice of Admission Application Stub PRC Official Receipt Two or more pencils (No.2) Ballpens with BLACK INK ONLY One (1) piece Metered-Stamp Window Envelope One (1) piece Long Brown Envelope One (1) piece Long Transparent (non-colored) Plastic Envelope (to keep above items).

C. .WHAT TO WEAR ON EXAMINATION DAYS?. 1. 2.

MALE – school uniform/white polo shirt or T-shirt (tucked-in) FEMALE - school uniform/white blouse or T-shirt

D. .GENERAL INSTRUCTION TO EXAMINEES?. 1. 2. 3. 4.

Report to the Test Center before 6:30 a.m. on the first day of examination to verify your room and seat numbers. Late examinees will not be admitted. Attend to your personal needs before the start of examination in every subject. No examinee will be allowed to go out of the examination room while the examination is in progress. Always put your answer sheet on top of the armchair while taking the examination.


prc examinee’s guide

8-2

5.

6.

E.

Stop answering the test questions at the end of the time allotted for the subject. Arrange your test papers as follows: (a) Notice of Admission (b) Answer Sheet (c) Test Questionnaire Do not leave the room until. (a) your answer sheet and test question set are received by the room watchers (b) you have signed, indicated the time and set (A or B) on the Examinees Record of Attendance, and (c) the lower portion of your Notice of Admission (Certification on the Receipt of Test Papers) is signed by the Room Watchers and returned to you.

.HANDLING OF EXAMINEE IDENTIFICATION SHEET/ANSWER SHEETS.

1.

2. 3. 4. 5.

Check if the Serial Number of Examinee Identification Sheet/ Answer Sheet are the same for all the sheets. If there is any discrepancy, return the set to your Room Watchers for replacement. The Serial Number is NOT the examination number. It has nothing to do with your examination. Check if there are defects or unnecessary marks on your Examinee Identification Sheet/Answer Sheets. Check if the number of Answer Sheets correspond to the number of examination subjects. Do not fold or mutilate, take extra care and keep clean your Examinee Identification Sheet/Answer Sheets. Before detaching an answer sheet, check of the brown envelope is yours and the set inside belongs to you.

F. .HOW TO MARK YOUR EXAMINEE IDENTIFICATION SHEET. 1. 2.

Use standard No.2 pencil only Do not use too much pressure

3.

Mark like this

4.

Mark the mark dark and straight

5.

Strictly no erasures allowed

not like these

9


8-3

G. .HANDLNG OF TEST QUESTION SETS. 1. 2. 3. 4. 5.

Indicate your seat number at the right top corner of page 1 Check if the number of pages of Test Question Set is complete and no misprint. If there is any problem, return the set to your Room Watchers for replacement. Mark A or B on the answer sheet to indicate the set of Test Question assigned to you. You can use your Test Question Set as your Scratch. Keep the Test Question Set stapled until the end of examination.

H. .PROHIBITED ACTS INSIDE THE EXAMINATION ROOM. 1. 2. 3. 4. 5.

6.

7.

Accepting or receiving anything, including food from any person while the examination is in progress. Giving money, food, or any favor and other consideration to the Room Watchers and other examination personnel. Loitering, talking, or discussing your answer inside the room or along the corridor while the examination is in progress. Putting any of the following markings on your answer sheets: name, seat number, unnecessary words or phrases, strokes, dots or any other marks not called for in the test questions. Taking out of the examination room test questions used or pages thereof, copying, an/or divulging or making known the nature or content of any examination question or answer to any individual or entity. Copying or referring to any solution, answer or work of another examinee or allowing anyone to copy or refer to your work, helping or asking help from any person or communication with anyone by means of words, signs, gestures, codes and other similar acts which may enable you to exchange, impart or acquire relevant information. Bringing inside the examination rooms the following: books, notes, review materials and other printed materials containing principles or excerpt thereof, coded data/information/formula which are relevant to or connected with the examination subject, PROGRAMMABLE CALCULATORS, CELLULAR PHONES, beeper, portable personal computers or other similar gadgets/devices. The act shall be ground for the cancellation of your examinations (PRC Resolution No. 463 dated November 27, 1996) MAINTAIN DISCIPLINE AT ALL TIMES, ANY MISCONDUCT OR IRREGULARITY ON YOUR PART OR ANY VIOLATION OF THE EXAMINATION RULES AND REGULATIONS AND ISNTRUCTIONS WILL BE SUFFICIENT CAUSE FOR THE CANCELLATION OF YOUR EXAMINATION PAPERS AND YOUR DEBARMENT FROM TAKING ANY FUTURE LICENSURE EXAMINATION.


prc examinee’s guide

8-4

PENALTIES FOR MULTIPLE MARKINGS & PLACING OF NAME IN THE SPACE PROVIDED FOR THE SUBJECT OF THE EXAMINATION

Resolution No. 2003-154

1.

Placing of Name – Cancellation suspension from taking the licensure Multiple Markings – Cancellation suspension from taking the licensure

2.

of the examination papers & examination for one (1) year. of the examination papers & examination for two (2) years.

CORRECTION, RATING AND RELEASE OF TEST RESULTS ARE FULLY COMPUTERIZED 1. 2. 3. 4.

Answer Sheets (with no names) are read by Optical Mark Reader (OMR). Answers are matched with the Answer Key and rated by computers. Test Results are FINAL when released via e-mail to leading newspapers. THERE IS NO RE-CHECKING OF TEST PAPERS.

LIST OF NON-PROGRAMMABLE CALCULATORS ALLOWED TO BE USED Examinees shall be allowed to bring in and use ONLY any of the following calculators that were identified as non-programmable (PRC Memorandum Circular No. 2004-06 dated March 4, 2004) CASIO STANDARD/DESK TOP SCIENTIFIC CALCULATORS CASIO CASIO CASIO CASIO CASIO CASIO CASIO CASIO CASIO CASIO CASIO CASIO CASIO CASIO CASIO

530 20V 470LA D-20M D-60M D-100V D-208H D-120TE DM-1600 DS-208H DS-881 DF100 DN 858A DS1800S EL310-A

CASIO CASIO CASIO CASIO CASIO CASIO CASIO CASIO CASIO CASIO CASIO CASIO CASIO CASIO CASIO

FX82C FX82LP FX85B FX85SA FX220 FX250HC FX350D FX350W FX570W HC100 HL821 HL-122L J-120TE JS-10LA JS-20LA

CASIO CASIO CASIO CASIO CASIO CASIO CASIO CASIO CASIO CASIO CASIO CASIO CASIO CASIO CASIO

JS-40LA JS-140 JS-SC LC-401B LC-1000T LC-403LD M-7LB MJ-100 MS 115 MS 373 MS-350 MS-470 MS-240T MS-808V MS-812V

CASIO MS-100TE CASIO MS 120TE CASIO MS 170LA CASIO MS-270LA CASIO MS-470LA CASIO MS-10V CASIO MS-10TE CASIO MS-80TE CASIO MS-8T CASIO MS-20/V CASIO MS-1002 CASIO 0-40M CASIO SL-320V CASIOSL-305TE CASIO ELECT. D-20M

8-5


Fx65 Fx75 Fx82 Fx95 Fx260 Fx401 Fx580 Fx824 Fx901 Fx991N Fx85S Fx82LB

Fx82MS Fx82SX Fx82TL Fx82 Super Fx95MS Fx100D Fx100S Fx100W Fx115D Fx115S Fx115W Fx122s

Fx300W Fx451M Fx500A Fx506G Fx506M Fx509D Fx509G Fx520G Fx546D Fx546L Fx570S Fx570W

Fx580D Fx825X Fx911W Fx991H Fx991S Fx991W Fx992S Fx992s Fx100MS Fx115MS Fx300SA Fx350TL

EL-233S EL-240S EL-242M EL-250S EL-310A EL-310A-GR EL-327S EL-331A

EL-376S EL-378S EL-501V EL-501V (BK) EL-506Ρ EL-506R EL-506V EL-506V (BK)

SHARP EL-509D EL-509G EL—509R EL-509V EL-510R EL-520G EL-520V EL-520V (BK)

EL-520VA EL-531GH EL-531RH EL-531VH EL-531VH (BK) EL-531VH (BL) EL-546L EL-546VA

CITIZEN CITIZEN TL-780 CITIZEN SLD-742N

CITIZEN STL-795 CITIZEN CITIZEN CITIZEN CITIZEN

SDC-8560 SDC-8610 SDC8620A SDC-8610A

OLYMPIA OLYMPIA OLYMPIA OLYMPIA OLYMPIA OLYMPIA

HL 88L HL 110 SD-81H SD-100H SD-100V

OLYMPIA SD-100HW

OLYMPIA SD-805M OLYMPIA SD-828 OLYMPIA SD-835

CANON CANON CANON CANON CANON CANON CANON CANON CANON CANON

F-604 F-720 LC 210HI LS-39H LS-154H LS-566H LS 120H LS 82Z LS-120V

Fx350HA Fx350HB Fx531GH Fx531LH Fx570AD Fx570MS Fx991MS Fx992VB Fx350TLG

EL-556G EL-771C EL-2125 EL-6053 EL-6750 EL-6810 EL-6850

AURORA AURORA AURORA AURORA AURORA

DT 393 DT 394 EB 964 HL 125

AURORA MS 270LA

AURORA SC 120 AURORA 2512

OTHER BRANDS CEBAR CD402 DAL 506X LIFELONG TL 30XS

KARCE-833 KARCE KC 250 KARCE 539 TEXAS TI 30

PORPO YH-105 PORPO YH-106

TAKSUN TS-217 TAKSUN TS-128B

TIME BIRD SJC-122

TAKSUN TS-128

Calculators whose brand and identifying mark do not appear in the list above provided shall be prohibited in the examinations.


9-1


Note: The answers to even-numbered questions are intentionally withheld but will be personally discuss by the author to ECE reviewees during review, refresher and coaching program for the reason that the author is anticipating ECE school teachers and co-reviewers will use some of the question as a safe and quick exam references since the answers are withheld and encourage readers to test their knowledge in the field of communications engineering.

SECTION 1 Basic Principles of Communications Engineering 1. 3. 5. 7. 9. 11. 13. 15. 17. 19. 21. 23. 25. 27.

D. D. A. A. B. C. B. D. A. A. C. B. B. D.

29. 31.

B. B.

33. 35. 37.

A. D. C.

39.

C.

41.

D.

43. 45. 47.

D. C. C.

24, 120 Quality Red, blue, and green Wideband ARPANET Andrei Marie Ampere Gigametric 500 nm 2.2% 20 74.82% Facsimile 60 atmospheric interference antenna a transmitter, a receiver, and a channel 2πf Multiplexing Telephony or telegraphy with ISB greater at low frequencies its amplitude, frequency, and phase angle purple noise Heinrich Hertz Guglielmo Marconi

49.

C.

51. 53.

C. B.

55. 57. 59. 61. 63. 65. 67. 69. 71. 73.

D. A. A. B. C. C. A. B. D. B.

75.

D.

77. 79. 81. 83.

A. D. D. A.

85. 87. 89. 91. 93. 95. 97. 99. 101. 103. 105. 107. 109. 111. 113.

C. D. C. A. C. B. A. B. D. B. B. D. D. C. D.

115. 117. 119.

A. D. B.

121. 123. 125. 127. 129. 131. 133. 135. 137. 139.

C. C. B. A. A. B. C. B. B. B.

Telegraphy; four-frequency duplex Myriametric Telephony; amplitudemodulated pulses Radio waves 0.3 to 3 Hz Longitudinal XXX Out Frequency -3dB bandwidth Telephony; SSBRC Infrared rays Telephony; pulse phase or position-modulated One-millionth of a meter Telephony; SSBSC Absolute bandwidth Multiplexing Telephony; pulsewidth modulated 3 to 30 GHz Telephony; SSBFC Narrowband SOS Break Necessary Bandwidth Michael Faraday 300 to 3000 GHz time 19th century Distortion 0.7 to 100 μm All of the above Attenuated distance a signal or bit has traveled low-pass EHF single-frequency; composite 12.4 peV to 124 peV Noise 7000 Ǻ time; frequency Ham radio bandpass Modulation frequency Phase Full duplex


Answers to odd-numbered questions

9-2

141. 143. 145. 147. 149.

D. C. D. B. B.

Facsimile -2 Reflected radio signals 300 to 3000 Hz P2 equals P1

SECTION 2 Amplitude Modulation 1.

B.

3. 5. 7. 9. 11.

C. C. B. A. B.

13. 15. 17. 19. 21. 23. 25. 27. 29.

B. A. D. B. D. B. C. C. D.

31. 33. 35. 37. 39. 41. 43. 45. 47. 49. 51. 53. 55. 57. 59. 61. 63. 65. 67. 69. 71. 73. 75.

A. D. B. C. B. A. C. A. B. A. D. D. B. A. A. D. C. C. A. D. B. D. B.

starts at 535 kHz and ends at 1605 kHz 102 watts 0.707 4 watts 13%, 50% one-fifth of the total signal power at 100% modulation 50 watts 75.76% Armstrong method 20 kHz 33.33% 952.4 mA 46.2%, 1.32 kW 0.68 v(t) = (Ec + Em sin(ωmt)) x sin(ωct) 1.56 to 2.08 W 15.78 A, 20.46% Coefficient of Modulation m<1, Vm1, Vm>Vc twice carrier amplitude and the modulation index

77. 79. 81. 83. 85. 87.

C. C. C. B. D. A.

89. 91. 93. 95. 97. 99. 101. 103. 105. 107. 109. 111. 113.

C. B. C. B. A. C. C. A. A. D. C. C. D.

115. 117.

B. C.

119. 121.

B. B.

123. 125. 127. 129.

B. C. A. C.

+sine, –cosine, +cosine 79.9% half the carrier power Envelope One sideband Vmin(20%)=25.3 V, Vmax(20%)= 37.9 V, Vmin(90%)=3.14 V, Vmax(90%)= 60.1 V Lattice modulator Analog multiplication Crystals Product detector Vsf=37.5 V Pin/dc=7.14 kW fo-fm Tuned circuit PSSB=167 W Phase inversion Resonant circuit Envelope detector Varying the gain of an amplifier Switches Vsb(20%)=2.24 V, Psb(20%)=0.1 W, Vsb(90%)=10.06 V, Psb(90%)=2.025 W Bandwidth Iload(20%)=0.451 A, Iload(90%)=0.53 A Low-level modulation The modulation index 7.78 dB 1.56 to 2.08 W

SECTION 3 Angle Modulation 1.

A.

3. 5. 7. 9.

B. C. B. D.

the carrier would advance and retard in phase 5,000 times each second 0.25 rad 10 kHz, 5 the threshold effect the instantaneous phase deviation is directly proportional to the amplitude of the modulation signal and unaffected by its

9-3


11. 13.

B. A.

15.

C.

17. 19. 21. 23. 25. 27.

C. A. D. D. C. A.

29. 31. 33. 35. 37.

B. C. B. C. A.

39. 41. 43. 45. 47. 49. 51. 53. 55. 57.

C. B. A. D. D. A. B. A. C. C.

59. 61. 63. 65. 67.

A. B. C. B. B.

69.

D.

71. 73. 75.

B. C. D.

77.

D.

frequency Decreases Carrier amplitude and frequency for de-emphasing high frequency component 90 kHz 144 kHz 5 rad Pre-emphasis 5 kHz/volt the instantaneous phase deviation is proportional to the integral of the modulating signal voltage 9 kHz 48.4 watts Track range 30 kHz changes would occur in the phase of the output frequency in respect to changes in the amplitude of the input voltage Acquire range 80% pre-emphasis 3 26 MHz 4.8 kHz 5 40 kHz 80 kHz remains constant even when modulation index varies 120 kHz 2.12 kHz 60 limiter increases with deviation and decreases with modulation frequency the power in the outer sidebands is negligible Bessel functions 120 kHz zero crossing of the modulating signal the time-constant of the filter circuits used

79. 81.

D. D.

83. 85.

D. C.

87. 89. 91. 93.

C. C. A. D.

95. 97. 99. 101. 103. 105. 107. 109. 111. 113. 115.

A. C. B. C. C. B. D. A. C. D. D.

all of the above starts at 88 MHz and ends at 108 MHz modulating signal the instantaneous frequency deviation is directly proportional to the amplitude of the modulation signal and inversely proportional to its frequency 160° percent modulation fc=100 MHz, Pt=40 kW mfm=2, fm=5 kHz, BW=30 kHz Sensitivity=3 μV 48 kHz Vsensitivity = 1 μV Angle Quadrature detector One shot multivibrator + 45 kHz Quadrature detector S/N=30 Crystal S/N=19.8:1

SECTION 4 Noise Analysis and dB Calculations 1. 3. 5. 7. 9. 11.

C. B. D. B. D. D.

13.

B.

15. 17. 19. 21. 23. 25. 27. 29. 31. 33. 35.

D. B. D. D. C. C. B. A. B. C. C.

31.6 W -90 dBm 23.85 mVrms 11.46 dB VN=8.51μV, S/N=1.4 dB S/N ratio is decreased by ½ Equivalent noise temperature 0.147 μVrms 10.44 dB 398 mA 5.658 μV Impulse noise 760 K 67.32 kHz -85 dBm and 1000 Hz dB 6,076.3 K -15



9-4

37. 39. 41. 43. 45. 47. 49. 51. 53. 55. 57. 59. 61. 63. 65.

B. C. D. D. B. C. C. C. C. A. C. B. B. A. D.

67. 69. 71. 73. 75. 77.

A. A. C. D. A. B.

79. 81. 83. 85. 87. 89. 91. 93. 95. 97. 99. 101. 103. 105. 107.

D. B. D. B. D. B. D. C. C. B. C. C. D. A. B.

3.45 dB 7.54 mW 21 dbrnCO -103.98 dBm Distortion -79 dBa 4.78 μV dBm0 Noise factor 10 dB 20 37.87 dB 19.8 dB 67 Intermodulation distortion 3.24 dB Interference 3.24 dB Partition noise 0 dBr R=5.984 kΩ, readout= 120 μV 69.44, 18.42 458 μV 5.01 μV Pn=5.52x10-15 W Resistance In the receiver front end All of these 15 to 160 MHz Impulse noise 174 K 30 MHz Quantizing level NPR Residual noise level Thermal agitation

SECTION 5 Transmitters and Receivers 1. 3. 5. 7. 9. 11.

A. B. B. C. B. A.

13. 15. 17.

D. C. B.

27.2064 MHz 17.9 W 1600 kHz 5 mV envelope detector the sensitivity and the selectivity 50 kHz, 0.05% 600 μV Local Oscillator

19.

B.

21. 23.

A. A.

25. 27. 29.

A. D. C.

31.

B.

33. 35. 37. 39. 41.

B. C. A. C. C.

43.

A.

45. 47. 49. 51. 53. 55. 57. 59. 61. 63. 65. 67. 69.

B. C. C. B. B. D. A. C. A. D. B. A. C.

71. 73. 75.

D. A. A.

77. 79. 81. 83.

B. C. D. C.

85. 87. 89.

D. A. B.

Varying the gain of an amplifier a tapped inductor to remove amplitude variations 108 dB 53 dB does not contain the input frequencies with two signals close in frequency, the ability to receive one and reject the other 150.0021 MHz 98.7 to 118.7 MHz 0.06 4 MHz keep the input to the detector at a constant amplitude the modulating amplifier 31.76 W 6 dB 1.83Ω squelch 50 watts 20 kHz 60 MHz 600 μV 3.75Vp 13 dB 10.005 MHz 46.52 dB enter the mixer, one below and one above the local oscillator by a difference equal to the IF it is cheaper 8 dBm cuts off an audio amplifier when the carrier is absent Phase discriminator 5.79 ppm 135 Hz the carrier frequency can be changed to any required value 4.02 ppm 1.58 mV 4 watts

9-5


Section 6 Acoustics Fundamentals 1. 3. 5. 7. 9. 11. 13. 15. 17. 19. 21. 23. 25. 27. 29. 31. 33. 35. 37. 39. 41. 43. 45.

D. C. D. D. B. A. B. D. A. A. C. B. B. C. C. D. A. C. B. D. A. C. D.

47. 49. 51. 53. 55.

B. B. A. D. A.

20 1.6 mW/ m2 89.1 dB the period is 0.5 s halving the wavelength twice as loud Flanking transmission Sound intensity 245 Hz 10 dB 75 dB 256 to 2048 Hz 71 dB Phonoscope 6 dB 115.56 dBSPL 35.3 137.7 lux 3 dB ¼ 140 dB 97.8 W/m2 111.6 dBPWL, 56.53 dBSPL 107. 55 dB dP/dA Pitch 2.2 dolby

Section 7 Television Fundamentals 1. 3. 5. 7. 9. 11.

C. B. A.

13. 15. 17. 19.

D.

74.1 IRE units 420 pixels 307 lines

B.

exact interlacing

A. A.

55.24 MHz P=45.75 MHz, S=41.25 MHz

B. A.

B.

B-Y

8% 151,761 pixels

a weak picture, a long warm up time, and

21. 23. 25. 27.

B.

29. 31.

B. A.

33. 35. 37. 39.

C. D. D.

B.

A. C.

C.

41. 43.

A.

45. 47. 49. 51. 53. 55. 57. 59. 61. 63. 65. 67. 69. 71. 73. 75. 77. 79.

A. B. C.

C.

C.

color imbalance Pincushion 2.23 m

saturation 67.25-MHz carrier frequency and 69.25MHz upper side frequency much higher

flicker, 60-Hz vertical, 30-Hz 1.51 in

400 pit depth symmetric envelope of amplitude variations 0.45

the brightness of the color no color footcandles monoscope 3.579545 MHz

B.

Gamma

B.

0.920455 MHz

D. B. A.

C. C.

C. C. C. C. C. B. B.

yellow, cyan, magenta decoding 8 and 16

I

426 picture elements

81.

B.

83. 85. 87. 89. 91. 93. 95. 97. 99. 101.

C. B. C.

30 4.5 MHz tint Aquadag Convergence 3.38 and 3.78 MHz the vertical scanning frequency is doubled from 30 to 60 Hz 31,500 for the vertical scanning frequency 426 Colorplexer zero

B. C.

960 μs 41%

B.

electrostatic

D. B.

347° 4 MHz IRE units 45.6%

C.

D.



9-6

103. 105. 107. 109. 111. 113. 115. 117. 119. 121. 123. 125. 127. 129. 131. 133. 135. 137. 139.

A. A.

blue 3x

B.

8%, 15.7%, 77.6%

D.

12.9 MHz blacker than black vestigial sideband AM

B. D. B. A. A. A. D. D. D. D. D. B. D. B. A.

Magnetic

Static 2 contrast FM

Fixed low beam current 15,750 Hz 2.2 centering white G-Y Trace part, retrace

Section 8 AM, FM, and TV Broadcasting Standards 1.

3. 5. 7. 9. 11. 13. 15. 17. 19. 21. 23. 25. 27. 29.

D.

B. C. D. C. A. B. A. B. B. B. B. C. D. A.

Loose connections in the oscillator, amplifier, or antenna circuits 80 kW 525 lines 525 lines MOPA class C 5 TVRO FM Television MF 1.3 MHz Surface loam soil 30 to 53 kHz Vestigial sideband

Section 10 Transmission Lines and Waveguides 1. 3. 5. 7. 9. 11. 13. 15. 17. 19. 21. 23. 25. 27. 29. 31. 33. 35. 37. 39. 41. 43. 45. 47. 49. 51. 53. 55. 57. 59. 61. 63. 65. 67. 69.

B. D. B. C. D. C. A. D. B. A.

71. 73. 75. 77. 79. 81.

C. A.

B. A. B.

83. 85.

D. A.

A.

C. D.

D. B. B. A. B. D. D.

C. B. C.

D. B. A. A.

D.

C. B. C. B. C. B. A.

D.

82.9 Ω, 195.4 Ω 151 W 168 m 2.15 112 Ω 417 ns, 5 V 43.1 W 9.1 W 40 - j30 Ω 225 nH/m increases 0.00521 dB/m 0.0493 dB/m dividing it by Z0 the dielectric constant Yttrium-Iron-Garnet -1, ∝ 6V 75∠-5° Ω 50∠-76° Ω the skin effect one 0.693 all of the above 54.38 Ω 0.569 Ω/m 0.296 rad/m would not reflect at all 6.4 pF 570 Ω 9.08 cm 1, 1 14.3 GHz 72.4 Ω 1.6x108 m/s, 5.51x108 m/s 0.448 m, 0.136 m 24.5 nH Tuned circuits 50 ± j0 Ω A capacitive impedance 1.60:1 (min), 2.50:1 (max) 0 ± j∞ Ω Noise figure increases by 10 dB

9-7


A.

89. 91.

D. D.

93. 95. 97. 99. 101. 103. 105. 107. 109.

B. A. C. B. B. D. B. B. D.

111.

D.

113.

A.

115. 117. 119.

A. B. A.

121. 123. 125.

A. D. C.

127. 129.

C. D.

jZ o tan(β A) Group velocity power out compared to reflected power back Infinity Propagation constant Standing waves High attenuation Reflectometer Velocity factor As pure inductor lesser Transmission System Engineering The velocity of the wave on the transmission line divided by the velocity of the light in a vacuum To reduce the possibility of internal arcing Velocity factor 2:1 transfer maximum power to the load Physical dimensions input resistive load at the resonant frequency ground TE

Section 11 Fiber Optics Communications 1. 3. 5.

C. A. B.

7. 9. 11. 13. 15. 17. 19.

C. B. D. B. C. B. C.

21. 23. 25.

A. A. A.

50% 11.88 degrees 6.54 x 109 photons/sec 2321 reflections/meter 62.5/125 μm 7 x 108 channels 6.33 x 1014 Hz 7.8 x 1011 photons/sec 70.38 degrees Maximum angle within the fiber acceptance cone Mode theory Macrobends 1.89 x 10 8 m/s

27. 29. 31. 33. 35. 37. 39.

B. B. B. A. B. C. C.

41. 43.

D. A.

45.

A.

47. 49. 51. 53.

B. A. D. B.

55. 57. 59. 61.

D. B. D. B.

63. 65. 67. 69. 71.

A. C. D. A. B.

73. 75. 77. 79. 81.

C. A. B. D. D.

83. 85. 87. 89.

B. A. C. B.

91. 93. 95. 97. 99. 101. 103. 105. 107. 109. 111.

D. C. D. D. C. B. A. C. C. B. A.

Intramodal Chromatic 3.68 dB 25 μm 0.351 0.5 mW Electromagnetic wave and particles of energy Established standards Electrical isolation and immunity to noise The wave magnitude varies perpendicular to the direction of wave motion 53 A/W, 0.38 nW 0 dBm, 22.2 dB 1.547 When the wavelength of the light is less than the cutoff wavelength 71.33 ns/km Coating 2.22 ns/km 0.1386 ns/km, 3.61 GHz-km 1.25 μW 5.175 million 0.5 mW 10 Launches the optical signal into the fiber 56.3°, 33.7° 66.7 ns/km 1.24 eV 74.9° Scattering, absorption, and dispersion 1 cm 3286 modes PIN diodes and APDs Impurities in the fiber material LED Reflected Law of Reflection Almost parallel 2,361 Scattered It is absorbed 700 ps 37.2 ps Snell's Law Core, cladding, and



9-8

113.

D.

115.

D.

117. 119. 121. 123. 125. 127.

C. B. A. B. B. C.

129.

B.

131. 133.

D. A.

135. 137. 139.

C. D. A.

141. 143. 145. 147. 149.

C. C. C. C. D.

151. 153. 155. 157.

C. C. D. A.

159.

C.

coating Reduces mechanical strength Single mode and multimode Single mode Core diameter and NA attenuation dB/km -23 dBm embedded; optical signals; multiplexing Dense wavelength division multiplexing; fiber-optic SONET in 1994, as part of the first Fast Ethernet standards scalability 90000 copper wire, coaxial cable, fiber, and wireless plesiochronous 51.84 Mbps 8 ISDN networks the time distortion of an optical signal that results from the time of flight differences of different components of that signal Dispersion light; silica spatial multiplexing the reduction of signal strength over the length of the light-carrying medium Fabric

Section 12 Telephone Networks and System 1. 3. 5. 7. 9. 11. 13. 15. 17. 19.

A. D. A. C. B. B. C. B. D. C.

82.13 Erlangs, 0.43% 20 mA to 80 mA 770 Hz, 1336 Hz 162 s 0.029 in 200 ohms 90 dBrn 65 dBrnC TLP 56 kbps 300 call-minute, 5 callhour

21. 23. 25.

A. B. D.

27. 29. 31. 33.

D. C. B. C.

35.

A.

37. 39. 41. 43. 45.

C. A. C. A. B.

47. 49. 51. 53.

D. C. D. B.

55. 57. 59. 61. 63. 65. 67. 69. 71. 73. 75. 77. 79. 81. 83. 85. 87. 89. 91. 93. 95. 97. 99. 101. 103.

B. D. C. C. B. C. A. D. B. C. B. A. B. C. B. D. C. B. C. D. C. D. D. B. B.

1 dB prevent oscillation occurs when the central office capacity is exceeded 90 volts, 20 Hz AC 0.045 414.9 CCS 26 gauge every 3000 ft with 66 mH inductors Public Switched Telephone Network 1225 D-channel 0.98966 3.533 Erlang reduce the attenuation of voice signals 6.1 sec occupancy 1.544 19 gauge every 6000 ft with 88 mH inductors 88 Erlangs 44.008 Mbps 2 24 calls will be block 0.8 dB B-channel 48 36 CCS cross bar Loop extender 1.544 lost DDD Terminal adapter MOSFET frame echo broadband ISDN intensity DTMF B-channel CCITT PABX 8 a wave which has been reflected or otherwise returned with sufficient magnitude and delay for it to perceptible in some manner as a wave distinct

9-9


105. 107.

A. C.

109.

B.

111. 113.

B. A.

115. 117. 119. 121. 123. 125. 127. 129. 131. 133. 135.

A. A. B. D. C. D. B. A. B. C. A.

137. 139. 141. 143. 145. 147. 149. 151. 153.

B. C. C. D. C. C. C. A. A.

155. 157. 159. 161. 163. 165.

C. C. B. D. C. C.

167. 169.

B. B.

171. 173. 175.

C. B. D.

from that directly transmitted terminal a telephone line connecting two central offices Two audio frequency tones TDM it is equal to the number of simultaneous calls originated during a specific hourly period Erlang B hunting green attenuator intertoll AWG #19 busy hour 300 to 3400 Hz 4 kHz first selector A circuit usually in the subscribers loop, between the telephone set and the local control office 852 and 1209 Hz 8 crosstalk occupancy fiftyfold 33 kbps T1/ E1 18,000 ft call setup, pre-image handshake, line testing sequence, image transmission 300 - 3400 Hz full duplex attenuation/gain distortion both B and C signal to noise ratio assigns time slots to each channel's packet sections assist in clock recovery a data link protocol that uses HDLC as a basis TA 2B + D encrypted division multiplexing

Section 13 Facsimile Transmission 1. 3.

A. A.

5.

A.

1653 pixels/sec IOC(CCITT)=400, IOC(IEEE)=1257 1610, 521.5

Section 14 Pulse Modulation 1. 3. 5. 7.

A. B. B. C.

9.

A.

11. 13. 15.

B. B. A.

17. 19.

D. D.

21.

C.

23. 25. 27. 29.

A. B. B. B.

31. 33.

C. C.

35. 37. 39.

B. B. C.

3.92-4.08 V 64 kb/s 69.5 dB synchronize the transmitter and receiver 66 kbps carry signaling 74 dB the strongest transmittable signal to the weakest discernible signal the μ Law They are the same thing preserve dynamic range while keeping bit-rate low 560 kbps the A Law 75.2% with a lower bit rate but the same quality 256, 80 Mbps too few samples per second 8-bit numbers C = 2B log2M C = B log2(1 + S/N)



9-10

Section 15 Digital Communications 1. 3.

B. C.

5. 7.

D. C.

9. 11. 13. 15. 17. 19. 21. 23. 25. 27. 29. 31. 33. 35. 37. 39. 41. 43. 45. 47. 49.

B. B. B. C. C. B. C. C. D. A. B. A. C. D. A. C. A. C. B. A. A.

51. 53. 55. 57. 59.

A. B. D. C. C.

61. 63. 65. 67. 69. 71. 73.

C. A. B. C. B. D. A.

75. 77. 79.

A. D. C.

P(x)=0.8, P(y)=0.2 31.895 kbps, 6.505 kbits will be corrupted per second 48.6 kbps 231.89 kHz, 6.65 bps/Hz 19.2 kbps 1000 bits 6 bits, 94 % 95.9%, 4.1% 3.32 5 MHz 10 kbps 9600 wpm 8-PSK 2.5x10-3 2.5x10-3, 25% 5 3, 5 intersymbol interference 300 bps, full-duplex, FSK bandwidth-delay Parallel 16 BnZS low-pass Number of bits used for quantization 64 the U.N. 1200 samples/s a training sequence number of possible states per symbol unipolar NRZ Manchester encoding bandpass 10 seconds noiseless all of the above Quadrature Amplitude Modulation Differential Manchester QAM 8

Section 16 Data Communications 1.

C.

3. 5.

C. D.

7.

D.

9.

C.

11. 13.

D. A.

15. 17.

B. A.

19. 21. 23. 25. 27. 29. 31. 33. 35. 37. 39. 41. 43. 45. 47. 49. 51. 53. 55. 57. 59. 61.

A. C. B. C. A. D. A. C. A. A. B. B. C. A. C. D. B. A. D. C. D. B.

63. 65.

C. B.

Reliability and Flow Control Network Layer Information Communication Entertainment age By transmitting extra data that may be used to detect and correct transmission errors Concerned with data structures and negotiation data transfer syntax. CAT 5 The clocking is derived from the data in synchronous transmission UDP Physical addressing, network topology, and media access. Application Demand priority A peer computer 10 Mbps 16 OSI user Physical Application Simplex Star FCC Reliability Security Point-to-point IMPs Forums point-to-point Topology WAN Transport Process-to-process delivery Transport Standards were needed to allow any two systems to communicate

9-11

Self-Sufficient Guide to ECE by JASON AMPOLOQUIO 67. 69. 71. 73. 75. 77. 79. 81. 83. 85. 87. 89. 91. 93.

D. D. C. C. C. D. A. B. B. C. B. D. C. B.

95. 97. 99. 101.

C. C. A. A.

103. 105. 107. 109. 111. 113. 115. 117. 119. 121.

C. C. B. D. A. B. B. D. D.

B.

123. 125. 127. 129. 131.

A.

133.

C.

135.

C.

137. 139. 141. 143. 145. 147. 149. 151. 153.

A. B.

D. B.

A.

C. A. B.

C. B.

C. D. A.

7 session logical 128 network none of the above TCP 6-byte Transport TCP/IP Semantics physical Data Link They use intermittent signals to transmit bits CSMA/CD Data Link-MAC synchronous Both ends can transmit at the same time Central processing Protocols User-server Bridge Multiplexer CSU/DSU Data link IEEE 802.7 Flow control VHS–quality video and audio 48 File delivery IP multicast multiplexed bridge ports supporting fullduplex transmission cyclic redundancy check (CRC) request that the bad character (and possibly, the rest of the message) be retransmitted physical Manchester secondary station only 48 bytes long HDLC 10 Mbps segmentation IEEE's 802.3 bit oriented, frame format

155. 157.

B.

159. 161. 163.

C. B.

165. 167. 169. 171.

B.

C.

A.

B.

D. D.

line of sight only reliable if a single error occurs in a character

1

transport interactive "real time" access to other users physical location of user dual attached station no one all of these

Section 17 Antenna Fundamentals 1. 3. 5. 7.

D. C.

0.48 26 dB

A.

40 dB, 1.75°

B.

Can be used for multiband operation

9. 11. 13. 15. 17. 19. 21. 23. 25. 27. 29. 31. 33. 35. 37. 39. 41.

C. B.

43.

A.

45. 47. 49. 51. 53.

D. A.

2.71 mV/m 2.6 dB

B. B.

Hertz Radiate harmonics

A. C. C. B. D. C. C. D. C. B. B. B. B. B. A.

A.

24 cm 6 cm

Dipole 1.5° More gain 2839.31 W Center of the antenna Omnidirectional 48.39 m Effective height mirror image principle horn antenna reciprocity Orthomode transducer aperiodic ¼ wavelength For receiving low and high band stations An element that receives its excitation from mutual coupling rather than from a transmission line

72.5 ft



9-12

23. 25.

A. D.

27.

B.

G=4x, φ=0.5

29.

C.

70λ

31. 33. 35. 37.

A. C. A. B.

11.9 m2

39.

B.

circular polarization 200 kph By adding an inductor in series

41. 43.

A. D.

55. 57.

B. A.

unchanged 2.8 m, 4 m respectively

59. 61. 63. 65. 67. 69. 71. 73. 75. 77. 79. 81. 83.

A. A.

Broadside array Major lobe radiation

B. A.

Radiation 1.76 dB

B. B. B.

6 dB horizontally polarized gain of the antenna

C. C. C.

D. B.

B.

85. 87. 89. 91. 93.

C. C.

-6 dB 5 dB

C.

Driven element

D.

95. 97. 99. 101. 103. 105.

A. C.

Reflector element is 5% longer At feed point Yagi

C. D. D.

Yagi-uda antenna quad antenna trailing wire antenna

C.

A.

13.98 dB

296 pW

Section 18 Radio-Wave Propagation 1. 3.

D. B.

5.

A.

7. 9.

D. B.

11. 13. 15. 17.

B. C. C. D.

19. 21.

D. C.

23.9 GW/m2 Because radiofrequency waves are below the sensitivity range of the human eye The angle of refraction is greater than the angle of incidence 328 sudden ionospheric (SID) 4.66 MHz (a) Vertical (b) Low 24.14 mi Space-wave reflections Ultraviolet radiation (a) Electric (b) earth

Seasonal variation Optimum working frequency Temperature inversions Maximum usable frequency Strong ground wave phase relationships Space Turbulence in the atmosphere When the density of the ionized layer is greatest below about 2 MHz All of the above

Section 19 Microwave Engineering 1. 3. 5.

B. D. B.

7. 9. 11. 13.

A. C. B. C.

15.

B.

17. 19. 21. 23.

A. A. D. B.

25. 27. 29. 31. 33. 35. 37. 39. 41. 43. 45. 47. 49.

A. C. C. B. A. C. B. B. C. B. C. B. C.

-134 dB 8493 km when distance exceeds line-of-sight 99.99% 28 dB -63.5 dBm Multichannel Multipoint Distribution System accumulation of noise is reduced 93. 52 dB, 14.2 mW 2 watts power Local Multipoint Distribution System 11.6 m 99.985% jitter 99.975% Grazing Path 16.4 meters noise level 0.521 -62 dBm 182 K -100 dBm 16 km -42.4 dBm

9-13


Section 20 Satellite Communications

Section 21 Cellular Communications System

1. 3. 5. 7. 9. 11. 13. 15. 17. 19. 21. 23. 25. 27. 29. 31.

B. C. A. D. D. A. C. A. A. C. C. D. A. C. D. D.

1. 3. 5. 7. 9. 11. 13. 15. 17. 19. 21. 23. 25. 27. 29. 31. 33. 35.

D. B. D. A. C. B. C. C. C. A. C. C. B. D. C. B. B. A.

33.

A.

37.

A.

35. 37. 39. 41. 43. 45. 47.

D. B. D. A. D. B. D.

49. 51. 53. 55.

A. D. D. C.

57.

D.

59. 61. 63. 65. 67. 69. 71. 73. 75.

C. B. C. A. C. C. A. A. D.

39. 41. 43. 45. 47. 49. 51. 53. 55. 57. 59. 61. 63. 65. 67. 69. 71. 73.

B. A. C. A. D. B. D. A. B. A. C. C. D. C. D. A. D. B.

75. 77. 79. 81. 83. 85.

D. D. D. D. C. A.

87. 89. 91.

A. A. B.

9.542 dB 71.5 dB 49 dB Centripetal force 77.96 dB 260 ms 600 ms 22,300 miles 253.33 ms 112.5 dB 7.6 km/s 24 satellites 290 K telemetry 6.35x10-16 W/m2 Gravitational pull of the earth and centripetal force of the revolving satellite Telemetry, Tracking, and Control 500-MHz, 12, 36-MHz DAMA Frequency Hopping line of apsides -165 dBW -208.6 dBW Equipped with two mixers Footprint 5 195.7 dB High directional antenna Frequency re-use technique Perigee 35,780 km footprint azimuth and elevation 103 watts orbital adjustments low-noise amplifier television receive only

Osumi, Alouette 1

15 x 25 mm TDD AMPS Adaptive power control 2 to 8 nodes IMT-MC Base station controller Blocked calls 3G WCDMA 270.833 kbps Clone 73 SIM -45.7 dBm 10 cm to 10 meters less than 600 mW increase the number of cells all frequencies are used in all cells 216 ft 416 E & 3333 E hard 1.76 minutes 3GPP 67.708 kHz 95 MHz 1 meter IS-95 666 250 mW CCITT LAI 300 Both A and B 2.4 GHz Special Interest Group the units is “handed off” to a closer cell 1981 Qualcomm 7 25 MHz 40,000 from the base to the mobile IMT-TC 36 s SAT



9-14

93. 95. 97. 99. 101. 103. 105. 107. 109. 111. 113. 115.

A. C. C. A. C. D. B. D. A. D. B. C.

117. 119. 121. 123.

C. A. C. D.

125.

C.

Basic Service Set AMPS 890-915 MHz FCCH GSM IMT-SC FM Bluetooth 100 m, 20 dBm IEEE 802.15.1 scatternet wireless USB, wireless Ethernet Intercell interference OFDM 3 the ratio of the distance between areas to the transmitter power of the areas frequency hopping

Section 22 Navigation System 1.

B.

3.

A.

5. 7. 9. 11. 13

A. A. D. A. D.

15.

B.

108.00 MHz to 117.95 MHz Its reception range is based on both the aircraft's altitude and the aircraft's line-of-sight to the VOR station Magnetic directions Variation Relative bearing 962 MHz to 1213 MHz Two antenna patterns are produced; one above the normal 2.5 degree ascent angle to the runway's surface at an audio modulated tone of 90 Hz and one below the normal 2.5 degree ascent angle to the runway's surface at an audio modulated tone of 150 Hz. Deviation

17. 19

B. A.

21. 23.

A. B.

25.

A.

27.

D.

29.

D.

31.

C.

33. 35. 37.

C. C. B.

39.

C.

41.

A.

43. 45. 47. 49.

A. B. B. A.

51.

D.

53.

C.

11.6 statute miles A measurable amount of time is required to send and receive a radio signal through the earth's atmosphere. radar beacon Pulse position modulation Differential phase shift keying A frequency modulated continuous wave An aircraft's ADF antennas can receive transmissions that are over the earth's horizon (sometimes several hundred miles away) since these signals will follow the curvature of the earth Quadrantal error is caused by the presence of the aircraft in the electromagnetic field of the NDB transmission 230o 3o If the strength of the 90 Hz audio signal is greater than the strength of the 150 Hz audio signal of the antenna patterns; the aircraft is to the left of the centerline of the runway. It is the line-of-sight distance between an aircraft and a selected ground-based navigation station. Transmit at 1090 MHz and receive at 1030 MHz Telemetry Gyrocompass Directions 180 degrees South, true bearing position of the VOR station bearing and distance indication

329 to 335 MHz

9-15

Self-Sufficient Guide to ECE by JASON AMPOLOQUIO 55. 57. 59.

B. B. B.

61.

C.

63. 65.

C. C.

67.

B.

69.

D.

VLF

108 to 112 MHz A horizontal deviation of an airplane from its optimum path of descent along the axis of the runway Echo time off the object to the source

10.

D.

11. 12. 13.

B. B. A.

Radio Detection and Ranging More affirmative at wider bandwidth Time delay

14.

A.

15. 16.

C. B.

17. 18.

B. A.

19.

C.

20. 21. 22.

B. A. C.

23. 24.

D. A.

25.

D.

26. 27.

B. A.

28. 29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. 41. 42. 43.

C. B. A. C. A. C. C. A. B. B. B. B. A. C. A. A.

44. 45.

A. B.

1227 to 1575 MHz

Section 23 Radar Fundamentals 1. 3. 5. 7. 9. 11. 13. 15. 17. 19. 21. 23.

D. B. D. C. B. C. B. A. B. C. A. B.

1 μs, 0.09% 1 MW 927 Hz 132,700,000 meters 3.84 m 10 mi 40 kph 10.1 fW 750 km, 7.5 km 345.6 km 400 589 Hz

Section 24 Laws and Ethics 1. 2. 3. 4.

D. C. A. B.

5. 6. 7. 8. 9.

D. C. D. A. A.

156.8 MHz Riser terminal Arrester Department of Transportation and Communications E.O.109 Three lines 20 kW 45 Standard AM

broadcast with a carrier power of 5 kilowatts Cancellation of its authority Dept. Order 88 Radio station license National Telecommunications Communication Local exchange operator If he has capacity Value-added service provider PD 223 Efficient use and equitable access Decision of depth of conduit at interconnection point E.O. 467 ITU-T Provision of radiotelegraph operator WTDC Radio operator telegraphy onboard A radio telegraph operator Interconnection New entrant has more financial support E.O. No. 266 30 days Act. No. 3846 Undergrounding 1999 Sleeve E.O. 109 Administration Franchise Council NTC R.A. No. 6849 PHP 5,000.00 PHP 5,000.00 PHP 5,000.00 PHP 180, PHP 1200.00/year, PHP 720.00/year PHP 5,000.00 PHP 25,000.00



9-16

46. 47.

C. A.

48.

B.

49. 50.

C. C.

50cm x 10 Band A : 1920 –1935MHz/ 2110 – 2125MHz, Band B : 1935 –1950MHz/ 2125 – 2140MHz Band C : 1950 – 1965MHz/ 2140 – 2155MHz, Band D : 1885 – 1900MHz/ 1965 – 1980MHz PHP 100M for each band PHP 100,000, PHP 1,000,000

--- End of Text ---

10-1


MEMORY ENHANCEMENT & EXAMINATION TIPS

Be careful for nothing; but in every thing by prayer and supplication with thanksgiving let your requests be made known unto God. Phi 4:6 I.

.MEMORY PREFERENCES. Each person will have his or her own set of ways of remembering any one thing. Each person will also find it easier to remember something difficult by using a different combination of strategies. Some people are outstanding at remembering things they hear aloud - so it makes sense for them to say out loud what they need to remember. Other people find they remember what they read if they sing it to themselves. Others remember things by associating them with places or people. This may feel unusual but use what works for you. For complex and important things, you increase you chances of remembering them if you: 1. 2. 3. 4.

use the strategies that work best for you use more than one strategy make a conscious effort to commit the information to memory do this on at least three occasions.

II. .USEFUL MEMORY STRATEGIES. 1. 2.

Putting the information to music. Try singing it. Using color. Color code information, or use different highlighter pens for different purposes. 3. Make visual aid materials that you always see everyday. 4. Walking to new places as you try to remember something. 5. Linking the information to other things that are relevant to you. 6. Working with the information: write it out, say it, draw it, color it in, discuss it, argue against it, summarize it, and organize it into the best order. 7. Numbering the different aspects. 8. Organizing the information into a smaller number of pieces, each with its own heading, name or label. 9. Finding out something else about it. 10. Remembering things in groups of three or five items. 11. Using several of these approaches for the same item.


10-2

M EM ORY ENHANCEM ENT and EXAM INATION TIPS

III. .IMPROVING YOUR MEMORY. Anyone 1. 2. 3.

can improve their memory by following a 3 step process: Paying attention Applying constructivist methods Making information easy to remember

A.

Paying attention Ä Take an active role in learning. Ä Memorization is sometimes needed but is not enough. Ä Review information and quiz yourself for true understanding.

B.

Constructivism Ä Constructivism deals with correlating new information with old information. Ä Thus, one constructs new understanding by fitting new information with prior understanding or experiences. Ä Rather than memorizing random facts, try to relate them to prior knowledge. Ä Think about new information and draw comparisons to other things you know. Ä Think about similar information learned earlier. Ä Draw analogies between old information and new information. Ä This allows you to see the big picture and not get swamped with new information.

C.

Make Information Memorable Repetition, Repetition, Repetition, Repetition, Repetition, Repetition, Repetition Read it, Write it, Say it, Explain it, Draw it, Ask questions about it..... Break down words by prefix or suffix. (Hydro relates to water) Use memory tricks: Mnemonics for lists: Examples: 1. ARMIDAS is for Antenna, RF amplifier, Mixer, IF amplifier, Detector, Audio, Speaker 2. PAA is for Phosphorous, Antimony, and Arsenic 3. BAG is for Boron, Aluminum, and Gallium Make silly rhymes or sayings to remember lists. Use silly analogies to remember examples. Humor is a powerful memory trigger; the dumber the better.


10-3

IV. .HOW WE LEARN BEST. We can improve the conditions for learning by being aware of some of the ways the brain works. Although we do not need to know a great deal about the brain, understanding some basics can help us to make the most of our minds. Some of the optimal conditions for learning are common sense and good for our general health. For example, the brain works well when: Ä it is rested - sleep affects our performance Ä it is hydrated-drinking water helps the electrical connections of the brain Ä it is unstressed- when it is stressed, it can focus only on 'escape', not on such matters as reading journals and writing assignments Ä it enjoys itself-it is important to look for any angle that can stimulate our interest in what we are learning Ä it has seen something several times -little and often works better than trying to understand something in one sitting Taking Better Notes Ä Go to review class. Note services are no substitute. Someone else's notes can contain fundamental errors, even with professional note takers. The physical act of writing helps some students remember information. Ä It's not "uncool" to sit up front. At the very least you'll hear everything clearly. Some rooms are big and some reviewers are very soft spoken. Ä Prime your mind before going to review class. Review the material from the textbook. This lets you become a little familiar with terms and images that may be covered in review class. Ä Also spend a little time trying to fit the upcoming material into the context of similar information you are already familiar with. This forces you to think about the upcoming material and see it in more familiar terms. It's hard to take accurate notes on a topic that you know absolutely nothing about.

V. .READING TIPS. A.

Reading for any subject Ä Ä

Ä Ä

Be selective. You are not expected to read books from cover to cover. Change strategy. You need to develop skills in changing from one kind of reading to another, depending on how useful the information is for your purposes. Use the index pages at the end of a book. Find the exact pages for what you need. Read from paper. Avoid reading for long periods from computer screens if using the internet: print out an electronic copy in a font that suits you.


10-4


Ä

Ä

B.

Reading for different purposes For all subjects, you will need to know how to change quickly from one kind of reading to another. Ä

Ä

Ä

Ä Ä

C.

Set targets. It is easy to lose focus when reading. Set yourself targets to complete a reading task, with clear objectives for what you want to achieve. Focus. Jot down a list of questions before you read and as you go along. This will improve your attention - and save you from getting side-tracked.

Browsing: looking over a text to see how it ‘feels’, whether it appears to be the right kind of book, what it contains that might be of use, getting a general feel of the contents. You often take in more information when browsing than you may think at the time. Checking: looking in the contents or index to see whether the book contains specific information that you know you want - or which looks useful. Focusing in: allowing yourself to read more closely when you spot something that looks more useful. It is also important to notice when the text is less useful, and to return to browsing. Fact-finding: looking for specific facts and data. Background. This is additional reading which gives you a sense of the bigger picture. Select texts that are general and which you find inviting or easy to read. Read these selectively and at your own pace. This is best undertaken in vacations if possible.

Reading for understanding The main purpose of reading is to understand - not to get through text at speed for the sake of it. Comprehension is increased if: Ä Ä Ä

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You are clear about what you are looking for. You discuss your reading with others. Each person is likely to make sense of different aspects, and you can pool your ideas. You read something that gives you a general overview first. For complex ideas, choose the easiest book first and work up to more complex texts. You keep active. Set yourself targets and jot down questions to answer. If the book is yours, underline key points, use highlighter pens selectively, write summaries in the margin. This prevents you from ‘drifting off’ or simply reading the same text over and over without taking it in. Read in short bursts of up to twenty minutes, then take a few minutes break before starting again. Make notes of key points as you go along. This can create natural breaks every few minutes in your reading that can help maintain attention. Change reading speed. Often, reading faster can help memory of what you are reading, so it makes more sense. Browse quickly and focus in more slowly only where needed.


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VI. .MULTIPLE CHOICE EXAM PREPARATIONS. Studying for a multiple choice exam requires a special method of preparation distinctly different from an essay exam. Multiple choice exams ask a student to recognize a correct answer among a set of options that include 3 or 4 wrong answers (called distractors ), rather than asking the student to produce a correct answer entirely from his/her own mind. To prepare for a multiple choice exam, consider the following steps: 1.

For Studying Tips Ä

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Find a quiet, well lit place to study. It seems obvious, but sometimes you have to hide from distractions. (i.e. friends in the dorm) Tell others you are working and ask them to leave you alone for a while. You can meet later to relax and have fun. Work at a desk or table. DO NOT lie down on a couch or bed to study. Take a break if you need it. Find the best time of day to study. Some students are most awake in the morning, others think best late at night. Learn the most optimal time of day for you to concentrate and study. Set up a study schedule that includes a few hours each day. Stick to it. Bribe yourself. Promise yourself that you'll do something fun IF you study for a period of time. Take frequent short breaks if you need it. Studies show people can concentrate for about 30 minutes before they lose focus. When you get to the point where your mind starts to wander, get up and take a walk. Get a drink and come back. Sometimes working with others can help you illustrate which material is understood and which isn't. If you can answer questions and explain concepts to others, odds are you know the information well. Otherwise you'll realize what need to study some more. Be cautious about studying with others. Sometimes certain members will actually provide a distracting force. Carefully choose with whom you will work. Cycle between working alone and working with others. Work alone to master concepts and information. Work with others to test your true understanding.

4. General Rules for taking Exams A.

Preparing for the exam Ä Multiple choice exams tend to focus on details, and you cannot retain many details effectively in short-term memory. If you learn a little bit each day and allow plenty of time for repeated reviews, you will build a much more reliable longterm memory.


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Ä Ä B.

Pay particular attention to fundamental terms and concepts that describe important events or features, or that tie related ideas and observations together. These are the items that most commonly appear on multiple choice exams. If your textbook highlights new vocabulary or key definitions, be sure that you understand them. Sometimes new words and concepts are collected at the end of a chapter. Check to be sure that you have not left any out by mistake. Do not simply memorize the book's definitions. Most instructors will rephrase things in their own words as they write exam questions, so you must be sure that you really know what the definitions mean. Practice on sample questions, if you have access to a study guide or old exams. Spaced practice is better than massed practice.

Taking the Test Ä Arrive early so you don't feel rushed trying to find a seat, it's an unnecessary distraction. Ä Scan the entire test first. Are all the pages there? Is there an answer sheet? Then you can budget your time as you go. Ä Don't guess too soon! You must select not only a correct answer, but the best answer. It is therefore important that you read all of the options and not stop when you come upon one that seems likely. Ä You must select not only a technically correct answer, but the most completely correct answer. Since "all of the above" and "none of the above" are very inclusive statements, these options, when used, tend to be correct more often than would be predicted by chance alone. Ä Be wary of the extra-long or "jargony option." These are frequently used as decoys. Ä Use your knowledge of common prefixes, suffixes, and word roots to make intelligent guesses about terminology that you don't know. Knowledge of the prefix "micro," for instance, would clue you that microwatts refer to a very low unit of power. Ä Utilize information and insights that you've acquired in working through the entire test to go back and answer earlier items that you weren't sure of. Ä If you are not certain of an answer, guess... but do so methodically. Eliminate ome choices you know are incorrect, then relate each alternative back to the stem of the question to see if it fits. Narrow down the choice to one or two alternatives and then compare them and identify how they differ. Finally, make an informed guess. Ä If you have absolutely no idea what the answer is, can't use any of the above techniques, and there is no scoring penalty for guessing, choose option C or B. Studies indicate that these


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10-7

are correct slightly more often than would be predicted by chance alone. When you get "all the above," "none of the above," or "a,b, not c" type questions, treat each alternative as a true-false question and relate it back to the question stem. Think the answer is wrong? Maybe you should change it? Studies indicate that when students change their answers they usually change them to the wrong answer. Therefore, if you were fairly certain you were correct the first time, leave the answer as it is. Finally, the best way to insure selection of the correct option is to know the right answer. A word to the test-wise is sufficient. Some of the worst problems occur when students enter a time warp and forget to check the clock, or when they spend too much time on one or two difficult items. To prevent this from happening, one trick you can use is to scribble the desired "finish time" time for each section right on the test booklet. That way, you'll be prompted to check the clock after completing each part of the exam. Try taking a few breaks during the exam by stopping for a moment, shutting your eyes, and taking some deep breaths. Periodically clearing your head in this way can help you stay fresh during the exam session. Remember, you get no points for being the first person to finish the exam, so don't feel like you have to race through all the items -- even two or three 30-second breaks can be very helpful. Don't speed through the items with the idea of going back to change answers you are unsure of. If you take time to think through each question, your initial answer will usually be the correct one. Although there are always exceptions to this rule, the best approach in most cases is to carefully answer each question the first time you go through the exam, and change only those answers that are clearly mistakes. Most importantly: Remain Calm! Panicking only leads to mistakes. If you studied well the answer is in your head somewhere. It may take a little concentration to retrieve the required information.


Major References: 1. R. Blake, Electronic Communication Systems, 2nd ed., Delmar, Singapore, 2002 2. A. Carlson, Communications System, McGraw-Hill, 1975 3. L. W. Couch, Digital and Analog Communication System, 6th ed., Prentice Hall, Upper Saddle River, 2001 4. R.L. Freeman, Telecommunication Transmission Handbook, 3rd ed., Wiley & Sons, Canada, 1991 5. R.L. Freeman, Fundamentals of Telecommunications, 3rd ed., Wiley & Sons, Canada, 1999 6. L.E. Frenzel, Communications Electronics, 2nd ed., McGraw-Hill, Singapore, 1994 7. B. Grob, Basic Television and Video Systems, 5th ed., McGraw Hill, Singapore, 1984 8. W.H. Hayt, Jr., Engineering Electromagnetics, 5th ed., McGraw-Hill, New York, 1988 9. W.C.Y. Lee, Mobile Communications Engineering, 2nd ed., McGraw Hill, New York, 1997 10. G. Kennedy, Electronic Communications System, 3rd ed., McGraw-Hill, New York, 1984 11. L.E. Kinsler, A.R. Frey, Fundamentals of Acoustics, 3rd ed., Wiley & Sons, Canada, 1982 12. J. Kraus, D. Fleisch, Electromagnetics with Applications, McGraw Hill, New York, 1999 13. G.M. Miller, Modern Electronic Communication, 4th ed., Prentice Hall, New Jersey, 1992 14. S.K. Mitra, Digital Signal Processing, “A Computer-Based Approach” McGraw Hill, Singapore, 1998 15. J.C. Palais, Fiber Optic Communications, 3rd ed., Prentice Hall, New Jersey, 1992 16. E. Pasahow, Electronics Pocket Reference, 3rd ed., McGraw-Hill, Singapore, 2000 17. J.G. Proakis, D.G. Manoakis, Digital Signal Processing, 3rd ed., Prentice Hall, Singapore, 2000 18. D. Roddy & J. Coolen, Electronic Communications, 4th ed., Prentice Hall, Singapore, 1995 19. A.S. Tanenbaum, Computer Networks, 3rd ed., Prentice Hall, New Jersey, 1996 20. R. Shrader, Electronic Communication, McGraw-Hill, New York,1995 21. W. Schweber, Electronic Communication Systems, 2nd ed., Prentice Hall, Singapore, 1997 22. W. Stallings, Wireless Communications and Networks, Prentice Hall, Upper Saddle River, N.J., 2002 23. W. Tomasi, Electronic Communications System, “Fundamental to Advanced”, Prentice Hall, Upper Saddle River, 2001 24. A. S. Tanembaum, Computer Networks, 3rd ed., Prentice Hall, Singapore, 1997 25. P. Young, Electronic Communications Techniques, Prentice Hall, New Jersey, 1999


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