Feedback Communications Systems boards Student Manual

k o o b k r o W s ' t n e d u t S n o i t a c i n u m m o C e u g o l a n A t i k i n k e T

S 1 0 0 3 5

Teknikit Analogue Ana logue Communications Student’s Workbook

53-001S

Feedback Feedback Instruments Ltd, Park Road, Crowborough, E. Sussex, TN6 2QR, UK. Telephone: +44 (0) 1892 653322, Fax: +44 (0) 1892 663719. email: [email protected] website: http://www.fbk.com Manual: 53-001S Ed03 072003

Printed in England by Fl Ltd, Crowborough

Feedback Part No. 1160–53001S

Teknikit Analogue Ana logue Communications Student’s Workbook

53-001S

Feedback Feedback Instruments Ltd, Park Road, Crowborough, E. Sussex, TN6 2QR, UK. Telephone: +44 (0) 1892 653322, Fax: +44 (0) 1892 663719. email: [email protected] website: http://www.fbk.com Manual: 53-001S Ed03 072003

Printed in England by Fl Ltd, Crowborough

Feedback Part No. 1160–53001S

Notes

ANALOGUE COMMUNICATIONS STUDENT’S WORKBOOK

Preface

THE HEALTH AND SAFETY AT WORK ACT 1974 We are required under the Health and Safety at Work Act 1974, to make available to users of this equipment certain information regarding its safe use.+ The equipment, when used in normal or prescribed applications within the parameters set for its mechanical and electrical performance, should not cause any danger or hazard to health or safety if normal engineering practices are observed and they are used in accordance with the instructions supplied. If, in specific cases, circumstances exist in which a potential hazard may be brought about by careless or improper use, these will be pointed out and the necessary precautions emphasised. While we provide the fullest possible user information relating to the proper use of this equipment, if there is any doubt whatsoever about any aspect, the user should contact the Product Safety Officer at Feedback Instruments Limited, Crowborough. This equipment should not be used by inexperienced users unless they are under supervision. We are required by European Directives to indicate on our equipment panels certain areas and warnings that require attention by the user. These have been indicated in the specified way by yellow labels with black printing, the meaning of any labels that may be fixed to the instrument are shown below:

CAUTION RISK OF DANGER

CAUTION RISK OF ELECTRIC SHOCK

CAUTION ELECTROSTATIC SENSITIVE DEVICE

Refer to accompanying documents

PRODUCT IMPROVEMENTS We maintain a policy of continuous product improvement by incorporating the latest developments and components into our equipment, even up to the time of dispatch. All major changes are incorporated into up-dated editions of our manuals and this manual was believed to be correct at the time of printing. However, some product changes which do not affect the instructional capability of the equipment, may not be included until it is necessary to incorporate other significant changes.

COMPONENT REPLACEMENT Where components are of a ‘Safety Critical’ nature, i.e. all components involved with the supply or carrying of voltages at supply potential or higher, these must be replaced with components of equal international safety approval in order to maintain full equipment safety. In order to maintain compliance with international directives, all replacement components should be identical to those originally supplied. Any component may be ordered direct from Feedback or its agents by quoting the following information:

1.

Equipment type

2.

Component value

3. Component reference 4. Equipment serial number Components can often be replaced by alternatives available locally, however we cannot therefore guarantee continued performance either to published specification or compliance with international standards.

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ANALOGUE COMMUNICATIONS ANALOGUE STUDENT’S WORKBOOK

Preface

DECLARATION CONCERNING ELECTROMAGNETIC COMPATIBILITY Should this equipment be used outside the classroom, laboratory study area or similar such place for which it is designed and sold then Feedback Instruments Ltd hereby states that conformity with the protection requirements of the European Community Electromagnetic Compatibility Directive (89/336/EEC) may be invalidated and could lead to prosecution. This equipment, when operated in accordance with the supplied documentation, does not cause electromagnetic disturbance outside its immediate electromagnetic environment.

COPYRIGHT NOTICE

© Feedback Instruments Limited All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without the prior permission of Feedback Instruments Limited.

ACKNOWLEDGEMENTS Feedback Instruments Ltd acknowledge all trademarks. IBM, IBM - PC are registered trademarks of International Business Machines. MICROSOFT, WINDOWS 2000, WINDOWS 98, WINDOWS 95, WINDOWS 3.1 are registered trademarks of of Microsoft Corporation.

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53-001S

ANALOGUE COMMUNICATIONS ANALOGUE COM MUNICATIONS STUDENT’S WORKBOOK

Contents

TABLE OF CONTENTS 1

Introduction

1-1

2

Assignments using the Amplitude Modulation Workboard

2-1

2.1

2-1

2.1.1

Objectives

2-1

2.1.2

Practicals

2-1

2.1.3

Workboard Wor kboard Required

2-1

2.1.4

Theory

2-2

2.1.5

Practical 1: A Simple Amplitude Modulator

2-5

2.1.6

Practical 2: Envelope Detectors

2-9

2.1.7

Practical 3: Product Detection

2.2

3

Amplitude Modulation with Full Carrier Assignment

Amplitude Modulation with No Carrier Assignment

2-12 2-15

2.2.1

Objectives

2-15

2.2.2

Practicals

2-15

2.2.3


2-15

2.2.4

Theory

2-16

2.2.5

Practical 1: Double Sideband Suppressed Carrier

2-19

2.2.6

Practical 2: Generation of Single Sideband Suppressed Carrier (SSB)

2-22

2.2.7

Practical 3: Demodulation of SSB

2-25

Assignments using the Frequency Modulation Workboard 3.1

Generation of Frequency Modulation Assignment

3-1 3-1

3.1.1

Objectives

3-1

3.1.2

Practicals

3-1

3.1.3


3-1

3.1.4

Theory

3-2

3.1.5

Practical 1: Concepts Of Frequency Modulation

3-5

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TOC-1


3.1.6

Practical 2: Generation of FM with a VCO

3.1.7

Practical 3: Spectrum of an FM signal with a large Modulation Index

3.2

Demodulation of Frequency Modulated Signals Assignment

3-9 3-11 3-13

3.2.1

Objectives

3-13

3.2.2

Practicals

3-13

3.2.3

Workboard Required

3-13

3.2.4

Theory

3-14

3.2.5

Practical 1: Quadrature Detector

3-16

3.2.6

Practical 2: Phase lock loop detector (PLL)

3-20

3.3

4

Contents

Limiters and the Effect of Noise on FM Demodulation Assignment

3-25

3.3.1

Objectives

3-25

3.3.2

Practicals

3-25

3.3.3

Workboard Required

3-25

3.3.4

Theory

3-26

3.3.5

Practical 1: A Quadrature Detector with Limiter

3-29

3.3.6

Practical 2: The Effect of Noise on a Quadrature Detector

3-32

3.3.7

Practical 3: PLL Detector with a Limiter

3-35

3.3.8

Practical 4: The Effect of Noise on a PLL Detector

3-37

Assignments using the Signal Sources Workboard 4.1

Wien Bridge Oscillator Assignment

4-1 4-1

4.1.1

Objectives

4-1

4.1.2

Practicals

4-1

4.1.3

Workboard Required

4-1

4.1.4

Theory

4-2

4.1.5

Practical 1: Basic Wien Bridge Oscillator

4-5

4.1.6

Practical 2: Amplitude Stabilisation

4-8

4.1.7

Practical 3: Changes from Standard

4.2

L-C Oscillator Assignment

4.2.1 TOC-2

Objectives

4-11 4-13 4-13 53-001S

ANALOGUE COMMUNICATIONS STUDENT’S WORKBOOK 4.2.2

Practicals

4-13

4.2.3

Workboard Required

4-13

4.2.4

Theory

4-14

4.2.5

Practical 1: Tuned-Collector Oscillator

4-17

4.2.6

Practical 2: Effect of Supply Variations

4-20

4.3

Crystal Oscillator Assignment

4-22

4.3.1

Objectives

4-22

4.3.2

Practicals

4-22

4.3.3

Workboard Required

4-22

4.3.4

Theory

4-23

4.3.5

Practical 1: Fundamental and Overtone Modes

4-27

4.4

5

Contents

Multivibrator Assignment

4-30

4.4.1

Objectives

4-30

4.4.2

Practicals

4-30

4.4.3

Workboard Required

4-30

4.4.4

Theory

4-31

4.4.5

Practical 1: Basic Circuit

4-35

4.4.6

Questions

4-38

4.4.7

Practical 2: Effect of Variable Supply

4-39

4.4.8

Practical 3: Mark/space Ratio Control

4-41

Assignments using the Tuned Circuits and Filters Workboard 5.1

Audio Low-Pass Filters Assignment

5-1 5-1

5.1.1

Objectives

5-1

5.1.2

Practicals

5-1

5.1.3

Workboard Required

5-1

5.1.4

Theory

5-2

5.1.5

Practical 1: Passive Low-Pass Filter

5-9

5.1.6

Practical 2: Passive Low-Pass Filter, Swept Frequency

5-12

5.1.7

Practical 3: Active Low-Pass Filter

5-14

5.1.8

Practical 4: Active Low-Pass Filter, Swept Frequency

5-17

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5.2

RF Selectivity Assignment

Contents

5-19

5.2.1

Objectives

5-19

5.2.2

Practicals

5-19

5.2.3

Workboard Required

5-19

5.2.4

Theory

5-20

5.2.5

Practical 1: L/C Tuned Circuit

5-24

5.2.6

Practical 2: L/C Tuned Circuit, Swept Frequency

5-26

5.2.7

Practical 3: L/C Tuned Circuit, Transient Response

5-28

5.2.8

Practical 4: Crystal Filter

5-31

5.2.9

Practical 5: Crystal Filter, Swept Frequency

5-33

5.3

RF Band-Pass Filters Assignment

5-35

5.3.1

Objectives

5-35

5.3.2

Practicals

5-35

5.3.3

Workboard Required

5-35

5.3.4

Theory

5-36

5.3.5

Practical 1: Coupled L/C Circuits

5-42

5.3.6

Practical 2: Coupled L/C Circuits, Swept Frequency

5-45

5.3.7

Practical 3: Ceramic Filter

5-48

5.3.8

Practical 4: Ceramic Filter, Swept Frequency

5-50

5.4

Tuned Amplifier with Gain Control Assignment

5-52

5.4.1

Objectives

5-52

5.4.2

Practicals

5-52

5.4.3

Workboard Required

5-52

5.4.4

Theory

5-53

5.4.5

Practical 1: Gain Control

5-55

5.4.6

Practical 2: Automatic Gain Control

5-58

5.4.7

Practical 3: Frequency Response with Automatic Gain Control

5-61

5.4.8

Practical 4: Decibel Gain

5-63

TOC-4

53-001S

Chapter 1 ANALOGUE COMMUNICATIONS STUDENT’S WORKBOOK

1

Introduction

Introduction

This manual provides computer-based assignments which make use of Analogue Workboards, the Discovery II environment and Analogue Telecommunications 53-921 software package to provide an understanding of the fundamental principles on which complex analogue communication systems are based. Assignments are divided into practicals whose objectives are clearly defined. Every practical is designed to contain as much circuit investigation, measurement and observation as possible. Explanatory text, diagrams and instrumentation are fully integrated. Details of hardware and software installation are given in manual 53-001-3 together with Discovery II environment and product operating instructions.

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1-1

Chapter 1 ANALOGUE COMMUNICATIONS STUDENT’S WORKBOOK

Introduction

Notes

1-2

53-001S


2

Chapter 2 Assignments using the Amplitude Modulation Workboard

Assignments using the Amplitude Modulation Workboard

2.1

Amplitude Modulation with Full Carrier Assignment

2.1.1 Objectives

On completion of this assignment you will be familiar with: •

Basic amplitude modulation and demodulation,

•

AM characteristics in the time domain,

•

AM characteristics in the frequency domain,

•

Envelope detectors,

•

Product detectors.

2.1.2 Practicals

Practical exercises are provided as follows: •

Practical 1: A simple amplitude modulator

•

Practical 2: Envelope detectors

•

Practical 3: Product detection

2.1.3 Workboard Required

Amplitude Modulation Workboard 53-130 which comprises the following blocks: •

Signal Generation

•

Modulation

•

Filters

•

Demodulation

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2-1



2.1.4 Theory

2.1.4.1 Modulation The equation of a sinusoidal voltage waveform is given by: v = Vmax.sin(ωt+Ø) where: •

v is the instantaneous voltage

•

Vmax is the maximum voltage amplitude

•

ω is the angular frequency

•

Ø is the phase

A steady voltage corresponding to the above equation conveys little information. To convey information the waveform must be made to vary so that the variations represent the information. This process is called modulation. Any of these may be varied to convey information. 2.1.4.2 Amplitude Modulation Amplitude modulation uses variations in amplitude (Vmax) to convey information. The wave whose amplitude is being varied is called the carrier wave. The signal doing the variation is called the modulating signal. For simplicity, suppose both carrier wave and modulating signal are sinusoidal; ie, vc = Vc sin ωc t (c denotes carrier) and vm = Vm sin ωm t (m denotes modulation) We want the modulating signal to vary the carrier amplitude, V c, so that: vc = (Vc + Vm sin ωmt).sin ωc t where (Vc + Vm sin ωm t) is the new, varying carrier amplitude.

2-2

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Expanding this equation gives: vc = Vc sin ωc t + Vm sin ωc t. sin ωm t which may be rewritten as: vc = Vc [sin ωc t + m sin ωc t. sin ωm t] where m = Vm /Vc and is called the modulation index. Now: sin ωc t.sin ωm t = (1/2) [cos(ωc - ωm) t - cos(ωc + ωm) t] so, from the previous equation: vc = Vc [sin ωc t + m sin ωc t. sin ωm t] we can express v c as: vc = Vc sin ωc t + (mVc /2) [cos(ωc - ωm) t] - (mVc /2) [cos(ωc + ωm) t] This expression for v c has three terms: 1. The original carrier waveform, at frequency ωc, containing no variations and thus carrying no information. 2. A component at frequency (ωc - ωm) whose amplitude is proportional to the modulation index. This is called the Lower Side Frequency. 3. A component at frequency (ωc + ωm) whose amplitude is proportional to the modulation index. This is called the Upper Side Frequency. It is the upper and lower side frequencies which carry the information. This is shown by the fact that only their terms include the modulation index m . Because of this, the amplitudes of the side frequencies vary in proportion to that of the modulation signal. 2.1.4.3 Sidebands If the modulating signal is a more complex waveform, for instance an audio voltage from a speech amplifier, there will be many side frequencies present in the total waveform. This gives rise to components 2 and 3 in the last equation being bands of frequencies, known as sidebands. Hence we have the upper sideband and the lower sideband, together with the carrier.

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2-3



2.1.4.4 Experimental Determination of the Modulation Index This is most easily done by measuring the maximum and minimum values which the instantaneous amplitude of the carrier reaches. Let us call these x and y . Taking our previous equation: vc = Vc [sin ωc t + m sin ωct. sin ωm t] and re-arranging it yet again, we can express v c as: vc = Vc sin ωc t [1 + m sin ωm t] so that the instantaneous amplitude of the carrier is: Vc [1 + m sin ωm t] Since sin wm t can vary between +1 and -1, x = Vc (1 + m) and y = V c (1 - m) To get the value of modulation index m from x and y, we eliminate V c between these equations by division, giving: y /x = (1 - m)/(1 + m). Solving for m gives: m = (x - y)/(x + y)

2-4

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2.1.5 Practical 1: A Simple Amplitude Modulator

This practical introduces the concept of Amplitude Modulation. You will meet the terms Carrier, Modulation and Modulated Signal and how they are related in both the time domain and frequency domain. A simple form of information transfer is Morse code, where the signal at the frequency selected for transmission is switched off and on in dots and dashes. The transmission frequency is selected by consideration of the transmission medium and, within reason, has nothing to do with the information it has to carry. For this reason it is called the carrier frequency. In order to carry any information some characteristic of the carrier must be changed, or modulated with that information; hence the term modulating signal. In the Morse code example where the carrier is switched off and on, it is the amplitude of the carrier that is carrying the information. This is a very crude form of Amplitude Modulation (or AM) because there are only two states: zero amplitude and full amplitude. In order to carry more complex information such as speech or television, the amplitude is varied linearly so that the instantaneous carrier amplitude is proportional to the amplitude of the modulation signal. Clearly, for a given carrier amplitude there are limits for the size of the modulating signal; the minimum must give zero carrier, the maximum gives twice the unmodulated carrier amplitude. If these limits are exceeded, the modulated signal cannot be recovered without distortion and the carrier is said to be overmodulated. When the modulating signal is varying the carrier from zero to twice its amplitude, the carrier is said to be fully, or 100%, modulated. Modulation depth is calculated from the formula: (x -y) / (x + y) where x is the maximum instantaneous carrier amplitude and y the minimum. The resulting fraction is often expressed as a percentage. If the fraction exceeds 1 (modulation depth greater than 100%), then the carrier is said to be over-modulated. 2.1.5.1 Amplitude Modulation in the Time Domain One of the easiest methods of examining a signal is by using the oscilloscope. This is in the time domain. The amplitude variation of the carrier in time with the modulation can be seen easily and the operation of a system analysed. 53-001S

2-5



In the practicals we use a sine wave as the modulation source as this makes the oscilloscope pattern clearest, in a practical system the modulating signal could be very complex. Also, to make the principles clearer, the modulating frequency is one ninth of the carrier so that individual carrier cycles show. Again, in a practical system, the modulating frequency would probably be lower. In order for the oscilloscope trace to be stable it must be triggered from the modulation source. 2.1.5.2 Amplitude Modulation in the Frequency Domain The use of a spectrum analyser for examining signals in the frequency domain is a very powerful tool. However, this facility is not always available, as a spectrum analyser costs much more than an oscilloscope. As an aid to understanding modulation principles it is invaluable. The spectrum analyser shows the modulated signal to have three components: 1. The Carrier, at the same frequency as the carrier source. 2. A lower side frequency at the carrier minus the modulation frequency. 3. An upper side frequency at the carrier plus the modulation frequency. The amplitude of the carrier is independent of the modulation, while the amplitude of the side frequencies depends entirely on modulation depth. In the practical the modulation source is a sine wave, containing only one frequency, and therefore is represented by a narrow line in the frequency spectrum. In practice, where the modulating signal is more complex, there would be a range, or band, of side frequencies above the carrier frequency and a band below it, the upper and lower sidebands. They extend either side of the carrier to an extent equal to the maximum modulating frequency.

2-6

53-001S



2.1.5.3 Procedure In this practical the hardware is configured as shown. You have available an oscilloscope and a spectrum analyser. Set the carrier level to maximum. Set modulation level to zero. Note the signals at all monitoring points. Now increase the modulation level and observe at monitor point 6.

Increase the modulation level until the carrier amplitude just reaches zero on negative modulation peaks. This is 100% modulation. Observe the signals at all the monitoring points both with the oscilloscope and the spectrum analyser at various modulation levels. Also, with a fixed modulation level try adjusting the carrier level . You will need to return to the practical and make some measurements in order to answer the questions.

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2.1.5.4 Questions 1. The 'envelope' of the modulated carrier wave is a curve joining its peaks. The positive envelope, joining the positive peaks, should follow the shape of the modulating signal in one polarity and the negative envelope, joining the negative peaks, in the opposite polarity. What happens to the positive and negative envelopes when over-modulation occurs?

2. How would you recognise over-modulation on the spectrum analyser display?

3. What is the amplitude of the two sidebands relative to the carrier expressed in dB for 50 percent modulation with a sine wave? (HINT: Use the oscilloscope to set the modulation level and the spectrum analyser to measure the sidebands)

4. See if you can use the theory mathematics to calculate the value in question three and compare it with the practical measurement.

2-8

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2.1.6 Practical 2: Envelope Detectors

In this practical you will investigate the demodulation of an AM signal with an envelope detector. The purpose of any detector or demodulator is to recover the original modulating signal with the minimum of distortion and interference. The simplest way of dealing with an AM signal is to use a simple half- wave rectifier circuit. If the signal were simply passed through a diode to a resistive load, the output would be a series of half-cycle pulses at carrier frequency. So the diode is followed by a filter, typically a capacitor and resistor in parallel. The capacitor is charged by the diode almost to the peak value of the carrier cycles and the output therefore follows the envelope of the amplitude modulation. Hence the term envelope detector. The time constant of the RC network is very important because if it is too short the output will contain a large component at carrier frequency. However, if it is too long it will filter out a significant amount of the required demodulated output. In this practical the output of the AM generator that you used in the Simple Amplitude Modulator practical is fed to an envelope detector. You can monitor the output and compare it with the original modulation source. The time constant of the filter following the detector can be adjusted. This filter is often called a post-detection filter. It also introduces a phase shift between the original signal and the output.

53-001S

2-9



2.1.6.1 Procedure Here the signal from the amplitude modulator from the previous practical is demodulated using an envelope detector. Confirm that the modulated signal is the same. Use the oscilloscope to monitor the detector output 16 and adjust the time constant . Note that a large carrier component is present if the time constant is too short.

Increase the time constant and note that the amplitude of the detected output decreases and becomes distorted as the filter cannot discharge in time to follow the required output. Use the spectrum analyser to observe the carrier component amplitude. Compare the original modulating signal with the detector output in both shape and phase at various time constants using the oscilloscope.

2-10

53-001S



2.1.6.2 Questions 1. Is the phase shift caused by the post detection filter a lead or lag?

2. Why do you think that the filter causes a phase shift?

3. How does the ratio of modulating frequency to carrier frequency affect the design of the detector and the post detection filter?

4. What problems could be caused if the range of modulating frequencies was quite large?

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2-11



2.1.7 Practical 3: Product Detection

In this practical you will investigate an alternative demodulator called a product detector. It has certain advantages over the simple envelope detector but at the expense of some complexity. It is not often used for AM but is the only type of detector that will demodulate the suppressed carrier amplitude modulations that are investigated in the next assignment. It is important to appreciate that a product detector will demodulate all forms of AM. 2.1.7.1 What is a Product Detector? If the AM signal is mixed with (ie, modulated by) a frequency equal to that of its carrier, the two sidebands are mixed down to the original modulating frequency and the carrier appears as a dc level. The mathematics of the process show that this will only happen if the mixing frequency is equal not only in frequency to that of the carrier, but also in phase; ie, the two signals are synchronous. This is why a product detector when used for AM is sometimes called a synchronous detector. For AM the effect is very similar to a full-wave rectifier rather than the half-wave of the envelope detector. The output still needs a post-detection filter to remove the residual ripple, but this time the ripple is at twice the carrier frequency and is therefore further away from the modulation and hence easier to remove. In general terms the product detector gives less distortion, partly because it uses both positive and negative peaks of the carrier. 2.1.7.2 Generating the Mixing Frequency This is produced by an oscillator which is usually referred to as a Beat Frequency Oscillator or BFO. This is because if it is not at the same frequency as the carrier the output of the product detector is a frequency equal to the difference between the two which is called a beat frequency. (You will be able to see this when you adjust the BFO for synchronism). As previously described, it is vital that the BFO be synchronised to the carrier. In practice this is achieved with a special recovery circuit but here for simplicity a sample of the carrier is fed directly to the BFO and when the free running frequency of the BFO is near to that of the carrier it locks into synchronism.

2-12

53-001S



2.1.7.3 Procedure Here is a product detector demodulating AM. The oscilloscope shows its input at monitor point 6, which is the output of the same modulator as before. Now monitor the BFO output with the oscilloscope and use the BFO frequency control to lock it to the carrier. This will be indicated by a stationary trace.

Use the oscilloscope to look at the output of the detector before the filter and note the frequency of the ripple compared with the carrier. Use the spectrum analyser to confirm this. Examine the output of the filter and compare it with the modulation source. Monitor the detector output before the filter with the oscilloscope, then unlock the BFO with the BFO frequency control and observe the result. Repeat whilst observing the filtered output.

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2-13



2.1.7.4 Questions 1. Are the design considerations for a post-detection filter different from those for the envelope detector?

2. Examine the filtered output, using the spectrum analyser at large size, with the BFO synchronised. The trace should show three points where the level is above the background ripple. What do they represent?

3. Again examine the filtered output, using the spectrum analyser at large size. Decrease the amplitude of the modulation signal as far as possible without the instrument trigger failing. Then vary the BFO control. How wide is the available range of beat frequency?

2-14

53-001S


2.2


Amplitude Modulation with No Carrier Assignment

2.2.1 Objectives


Amplitude modulation with suppressed carrier,

•

Double sideband suppressed carrier (DSB) modulation,

•

Single sideband suppressed carrier (SSB) modulation,

•

Balanced modulators,

•

Generating SSB with filters,

•

Demodulation methods.

2.2.2 Practicals

Practical exercises are provided: •

Practical1: Double sideband suppressed carrier

•

Practical 2: Generation of single sideband carrier (SSB)

•

Practical 3: Demodulation of SSB


Amplitude Modulation Workboard 53-130 which comprises the following blocks: •

Signal Generation

•

Modulation

•

Filters

•

Demodulation

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2-15



2.2.4 Theory

2.2.4.1 Double Sideband Suppressed Carrier In the theory for the Amplitude Modulation with Full Carrier assignment, Practical 1, it was established that if: vc = Vc sin ωct describes a carrier signal and, vm = Vm sin ωmt describes a modulating signal, then the normal AM signal is: vc = (Vc + Vm sin ωm t) .sin ωc t which may be rewritten as: vc = Vc sin ωc t + Vm sin ωc t . sin ωm t In DSB suppressed carrier modulation, the carrier term V c sin ωc t is suppressed, leaving just: Vm sin ωm t.sin ωc t = (Vm /2) [cos(ωc - ωm) t - cos(ωc + ωm) t] as the modulated signal. The two cosine terms represent the lower and upper sidebands respectively. In the case of SSB suppressed carrier modulation, one of these sidebands will also be suppressed. 2.2.4.2 Demodulating the DSB Signal In order to change the sideband frequencies back to the original modulating frequency, a locally-generated carrier frequency (from the BFO) is used to modulate the DSB signal. (Remember that modulation for the purpose of frequency changing is traditionally called mixing). Suppose that the BFO signal is: vo = Vo sin(ωo + ø)

2-16

53-001S



The modulation process will produce a signal proportional to: [Vo sin(ωo + ø)] (Vc /2) [cos(ωc - ωm) t - cos(ωc + ωm) t] or to: 2 sin(ωo + ø) [cos(ωc - ωm) t - cos(ωc + ωm) t] This can be divided into two terms: 2 sin(ωo + ø) cos(ωc - ωm) t ...

(1)

and: - 2 sin(ωo + ø) cos(ωc + ωm) t ... (2) but as 2 sin A cos B = sin(A + B) + sin (A - B) the first term, (1), becomes: sin(ωo + ø + ωc - ωm) t + sin(ωo + ø - ωc + ωm) t Since ωo is supposed to be equal to ωc, (ωo + ωc - ωm) will be a frequency roughly twice that of the carrier. This does not contribute to the desired signal. The rest of the expression, which does contribute, is: sin(wo + ø - ωc + ωm) t If ωo = ωc, then sin(ωo + ø - ωc + ωm) t becomes simply sin(ø + ωm) t, which is the original modulating frequency. Similarly the other term, (2), makes a contribution: - sin(ωo + ø - ωc - ωm) t which, for ωo = ωc , becomes: sin(- ø + ωm) t

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We now have two terms, or components of the output signal, each of the original modulating frequency. However, there is a problem when we combine them. The two terms are: sin(ø + ωm) t and sin(- ø + ωm) t If the phase ø is zero, the two terms become identical, so they combine to produce the signal: 2 sin ωm t i.e a signal at the original modulating frequency. Now suppose that the phase now changes through π /2 radians (90 degrees). The two sinusoids would now be π radians (180 degrees) apart in phase and would cancel each other out. We have assumed that ωo = ωc. If this were not true, the effect would be the same as if ø were continually changing, making the two terms alternately reinforce and cancel each other. This may be shown mathematically thus: sin(ø + ωm) t + sin(- ø + ωm) t = 2 sin ωm t cos ø Since cos 0 = 1, the strongest output is obtained for ø = 0. With ø = π /2, cos ø = 0, so no output is obtained.

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2.2.5 Practical 1: Double Sideband Suppressed Carrier

This practical introduces the idea of AM with suppressed carrier. After it you will understand the following ideas: •

Balanced modulators and carrier suppression

•

The BFO as a carrier insertion oscillator

In the assignment on basic amplitude modulation we saw that the modulated signal comprises a carrier and two sidebands. The carrier is of constant amplitude and only the sidebands vary in frequency and amplitude with the modulation. It is therefore clear that only the sidebands carry the modulating information while the carrier does nothing except, as we will see, help in the demodulation process. The transmission of the carrier takes a large proportion of the total transmitted power, so if the carrier were removed all the power could be used to transmit the sidebands which, after all, contain the information. If the modulation process is carried out by a balanced modulator, the output signal does not contain the carrier component because it is cancelled out by the balanced nature of the modulator. This signal is described as double sideband, suppressed carrier or DSB. 2.2.5.1 Carrier Unbalance If the modulator were perfectly balanced there would be no carrier in the output. In practice, due to circuit imperfections, some carrier is always present. The ratio of the actual carrier to that which would be there in a simple AM system is called the carrier suppression ratio and is an important parameter in such systems. Normally the ratio is expressed in dB to make the numbers manageable; 30 dB would be a typical figure. To calculate the carrier suppression ratio, you need to know what amplitude of carrier would have been present, if not suppressed. This is the carrier which would give 100% modulation by the maximum signal level for which the system is designed. Since 100% modulation produces side frequencies of half the carrier amplitude, the unsuppressed carrier amplitude may be taken as twice the allowable amplitude of either sideband.

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In this practical, a carrier and modulation source are connected to a balanced modulator to provide a DSB signal which can be examined with the oscilloscope and spectrum analyser. 2.2.5.2 Procedure Use the oscilloscope and spectrum analyser to examine the signals at monitor point 4 and monitor point 5. Set the carrier balance to mid-scale. Note that they are the same as for simple AM. Now examine at monitor point 6 and note the waveshape. Use the spectrum analyser to observe that there are two sidebands but no carrier.

Adjust the carrier balance ; note the effect on carrier amplitude. Adjust modulation level and carrier level and note the effects. Note that the output from the envelope detector is not the same as the modulating signal. Monitor at monitor point 13 and adjust the BFO frequency for a stable trace, so that the BFO is in phase with the original carrier. Observe that the product detector output is the same as the modulating signal. Unlock the BFO and observe the result.

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2.2.5.3 Questions 1. Why does AM have a low efficiency when the full carrier is transmitted?

2. How can you tell whether the modulator is balanced when using the oscilloscope?, and when using the spectrum analyser?

3. Measure the carrier suppression ratio for the system in Practical 1 when set for maximum modulation and minimum carrier amplitude.

4. Does the term ‘overmodulation’ have any meaning in a DSB system?

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2.2.6 Practical 2: Generation of Single Sideband Suppressed Carrier (SSB)

This practical introduces the concept of single sideband suppressed carrier or SSB operation. In the double sideband suppressed carrier practical it was demonstrated that it was possible to recover the modulating signal without the presence of a transmitted carrier. This is achieved by inserting a local carrier at the demodulator. That practical used DSB; ie, both sidebands were transmitted. It is obvious that as both sidebands are generated from the same carrier and modulation, they must contain the same information, and therefore the modulating frequency could be recovered from only one sideband. This saves further transmitter power. Another very important advantage is that the bandwidth is half that of simple AM or DSB. 2.2.6.1 Generating SSB The generator in the practical is a balanced modulator, producing DSB, followed by a bandpass filter for the required sideband. There are other methods but this filter method is the simplest to understand and is in very common usage in communication systems. It may be necessary for the bandpass filter to have a very good shape factor because, at normal carrier and audio frequencies, the upper and lower sidebands are quite close in frequency. Another consideration is that the sideband filter should offer significant attenuation to the carrier, so that the balanced modulator need not be so accurately balanced. In practice the balanced modulator might provide 30 db of carrier suppression and the filter a further 10db. The other sideband would normally be about 30 to 40 db down on the wanted one. In order to achieve this, the SSB filter has several poles and is, in most cases a ceramic filter or crystal filter. Various filters are commercially available with different specifications depending on the application. In the practical we use a high modulating frequency so you can see clearly the relationship between the various frequency components. This means that the filter specification can be relaxed and here a single tuned circuit is used. Separate filters are provided for upper and lower sidebands and the means is provided to monitor the output of both. You might be surprised that the output from the SSB filters is simply a sinusoidal signal but, since we use sinusoidal carrier and modulating frequencies, the sum or difference of the two must be a single frequency.

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When the modulation is a band of frequencies, the SSB output will also be a band of frequencies. 2.2.6.2 Upper or Lower Sideband? An obvious question is which sideband should be transmitted? The answer owes more to convention than theory! There is no reason why one sideband gives better results than the other, but general practice seems to favour the upper sideband. One convention is that with carrier frequencies below 10 MHz the lower sideband should be used, but this is not always the case. The result of this is that many pieces of communication equipment have to be able to deal with both. To begin the practical, please turn to the next page. 2.2.6.3 Procedure Use the spectrum analyser and oscilloscope to observe at monitor point 6. Note that the signal is DSB. Adjust the carrier balance as before. Monitor at monitor point 8, and at monitor point 9, and note that only one sideband is present. Note that the carrier suppression is less dependent on the carrier balance than before the filter.

Use the oscilloscope to observe that the SSB output is a sinusoidal signal. Use the spectrum analyser to note that the upper sideband frequency is the sum of the carrier and modulation frequencies and the lower sideband is the difference.

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2.2.6.4 Questions 1. Why is the balance of the modulator less important in a filter method SSB generator than for a DSB generator?

2. How is the width of the SSB filter related to the maximum and minimum modulating frequencies?

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2.2.7 Practical 3: Demodulation of SSB

This practical is about the demodulation of SSB. In the double sideband suppressed carrier practical we saw how DSB is demodulated using the BFO to reinsert the carrier. In the case of DSB the BFO must be in phase with the original carrier or the process will not work correctly. Since SSB is transmitted without a carrier it is not surprising that a similar method is employed. The main difference is that, for SSB, the BFO need not be in phase with the carrier. It does need to be at the same frequency but even a small error in the frequency results only in a small error in the frequency of the demodulated output. This means that in non-critical applications, such as speech, a small overall frequency error does not make the system useless. The effect on speech is to raise or lower the tone of the voice, which within limits does not reduce intelligibility. The fact that the BFO need not be locked, greatly simplifies the design of the receiver, and makes SSB one of the most powerful techniques for transmitting audio frequencies over radio links with its narrow bandwidth and efficient use of available transmitter power. In the practical you can use both upper and lower sidebands and see that with the BFO set correctly, near to the original carrier frequency, even though the two sidebands are at different frequencies the demodulated output is the same. You can also see that changing the BFO frequency causes the demodulated output to change in frequency by a similar amount.

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2.2.7.1 Procedure Monitor at monitor point 6, and observe the DSB signal. Move to monitor point 10 and note the upper sideband signal. Use the spectrum analyser to confirm that the frequency is that of the upper sideband. Change to lower sideband (by pressing the button) and repeat. Now monitor at monitor point 14 and compare the output with the modulation input. Use either the oscilloscope or analyser to set the BFO frequency to that of the carrier, by monitoring at monitor point 13. Note that both sidebands give the same output frequency.

Move the BFO frequency and observe the effect on the output using first one and then the other sideband.

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2.2.7.2 Questions 1. Why is SSB more efficient than either simple AM or DSB?

2. If the BFO frequency rises, what happens to the frequency: a) of the upper sideband? b) of the lower sideband?

3. Calculate the bandwidth of the transmitted signal when the modulation frequency band extends from 500 Hz to 50 kHz for simple AM, DSB and SSB.

4. If a SSB channel has no modulating signal, what is the modulated signal like?

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Notes

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3

Chapter 3 Assignments using the Frequency Modulation Workboard

Assignments using the Frequency Modulation Workboard

3.1

Generation of Frequency Modulation Assignment

3.1.1 Objectives


Frequency modulators,

•

Modulation index,

•

Bandwidth,

•

FM signals in the time domain,

•

FM signals in the frequency domain.

3.1.2 Practicals


Practical 1: Concepts of Modulation

•

Practical 2: Generation of FM with a VCO

•

Practical 3: Spectrum of an FM signal with a large Modulation Index


Frequency Modulation Workboard 53-140 which comprises the following blocks: •

Signal Generation,

•

Modulator,

•

Limiter,

•

Quadrative Demodulator,

•

VCO,

•

Phase Comparator.

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3.1.4 Theory

3.1.4.1 Modulation The equation of a sinusoidal voltage waveform is given by: v = Vmax.sin(ωt+Ø) where: •

v is the instantaneous voltage,

•

Vmax is the maximum voltage amplitude,

•

ω is the angular frequency,

•

Ø is the phase.

A steady voltage corresponding to the above equation conveys little information. To convey information the waveform must be made to vary so that the variations represent the information. This process is called modulation. Any of these may be varied to convey information. 3.1.4.2 Frequency Modulation Frequency modulation uses variations in frequency to convey information. We shall think in terms of the angular frequency ω. The wave whose frequency is being varied is called the carrier wave. The signal doing the variation is called the modulating signal. For simplicity, suppose both carrier wave and modulating signal are sinusoidal; ie:. vc = Vc sin ωc t (c denotes carrier) and vm = Vm cos ωm t (m denotes modulation)

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3.1.4.3 What is Frequency? If the frequency is varying, how do we define it? We can no longer count the number of cycles over a longish interval to count the cycles per second. Instead we define frequency as the rate of change of phase. This is consistent with the simple definition, because at a constant (angular) frequency ω radians/second the phase is changing at ω radians per second, which is ω /2π cycles per second. Since we can only define what the instantaneous frequency is by reference to the phase, we must look at the phase in order to arrive at an expression for the frequency-modulated signal. 3.1.4.4 Phase of the FM Signal For the unmodulated carrier v c = Vc sin ωc t, the phase is: s = ωc t We want the modulating signal to vary the carrier frequency, ωc, so that its frequency takes the form: ω = ωc + D cos ωm t

(where D denotes the peak value of the deviation) It is related to the amplitude of the modulating signal v m by the 'frequency slope' of the frequency modulator (VCO) say k radians/s per V. The peak value of v m produces deviation D, so: D = k Vm The total phase change undergone at time t is found by integrating the angular frequency. It is s = ! (ωc + D cos ωm t) dt = ωct + (D/ ωm) sin ωm t. (If you are not familiar with integration you will have to take this result on trust). So the FM signal can be expressed as: Vc sin [ωct + (D/ωm) sin ωm t]

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3.1.4.5 Modulation Index In the expression for the FM signal: Vc sin [ωc t + (D/ ωm) sin ωm t] the coefficient (D/ ωm) turns out to be quite important and is given the name modulation index. It is often represented by the Greek letter beta, β. So we may write the FM signal as: vc = Vc sin (ωct + β sin ωm) t where β is the modulation index (D/ ωm). In this expression, the factor sin (ωct + βsin ωm) t (let us call it F) is of the form sin (a + b) which can be expanded to sin a cos b + cos a sin b. Applying this expansion to F, we get: F = sin ωct cos(sin βωm) t + cos ωct sin (sin βωm) t 3.1.4.6 FM Sidebands These complicated functions can be expanded, using mathematics too elaborate to explain here, into a series of terms like this: F = J0(β) sin ωct+ J1(β) [ sin (ωc + ωm)t - sin (ωc - ωm)t ] + J2(β ) [ sin (ωc + 2ωm)t - sin (ωc - 2ωm)t ] + J3(β ) [ sin (ωc + 3ωm)t - sin (ωc - 3ωm)t ] + J4(β ) [ sin (ωc + 4ωm)t - sin (ωc - 4ωm)t ] + ... where J0(β ), J1(β), J2(β ) etc are constants whose values depend only on β . They are called Bessel Functions. There is an infinite series of these functions, and so an infinite number of FM sidebands. But in practice the values of the Bessel functions become very small as the series goes on. For example, when β = 2 J0(2) = 0.224 J1(2) = 0.577 J2(2) = 0.353 J3(2) = 0.129 3-4

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J4(2) = 0.034 J5(2) = 0.007 3.1.4.7 A Practical Approximate Rule Because the higher-order sidebands become very small, in practice the bandwidth of the FM signal may be restricted to a finite bandwidth. The practical rule that is used, often called Carson’s Rule, is to take the bandwidth required as: B = 2 ( Fd + Fm ) where B is the bandwidth, F d the deviation and F m is the bandwidth of the modulation, all in the same units. 3.1.5 Practical 1: Concepts Of Frequency Modulation

This practical introduces the idea of frequency modulation. Before you start it is necessary to appreciate some fundamental concepts. As in amplitude modulation, a carrier frequency is modulated by the information that is being sent. In AM it is the amplitude of the carrier that is varied in time with the modulation, in FM it is the frequency that is varied. The amplitude is constant as we will see. When no modulation is being applied the carrier is at its nominal frequency i.e the carrier frequency. The modulating signal causes the frequency to deviate, i.e. to move above and below its nominal value. With the greatest possible deviation, the minimum frequency could be near zero and, assuming the modulating signal to have no d.c component, the maximum frequency would then be about twice the carrier frequency. However, this would take a very large amount of frequency spectrum and the bandwidth would have no relationship to the modulating signal bandwidth. A set limit is normally made on the amount that the carrier can deviate from its nominal frequency and this is called the maximum deviation. Different systems use different values of maximum deviation, depending on a number of factors some of which are very complex

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3.1.5.1 Bandwidth of an FM Signal It is important that we can understand and estimate the bandwidth of the transmitted signal so that the transmission parameters can be chosen to fit into the available spectrum. Clearly the bandwidth must be at least equal to twice the deviation, as the carrier actually moves above and below its nominal frequency by that amount. But it also depends on how fast the frequency is being changed; ie, on the bandwidth of the modulating signal. The mathematical analysis of FM is quite involved and it shows that an FM signal has sidebands far above and below the maximum deviation. However the power in these sidebands decreases quickly as they become further away from the carrier and it can be shown that, for practical purposes, a good approximation to the bandwidth is given by: B = 2 ( Fd + Fm ) (where B is the bandwidth, Fd the deviation and F m is the bandwidth of the modulation.) This is sometimes called Carson’s Rule, and the bandwidth B can be viewed as containing the majority of the transmitted power, certainly sufficient for successful demodulation. 3.1.5.2 Modulation Index As we have seen, the bandwidth of an FM signal depends on both the deviation and the modulation bandwidth. It might be thought that, in order to keep the bandwidth as narrow as possible, all FM systems should be operated with a very small deviation. However, there are significant advantages to operating with a wide deviation. The main one is an apparent improvement in noise performance. So as you can see, a specific bandwidth can be the result of wide deviation with a low modulation bandwidth or a narrow deviation with a large modulation bandwidth. The ratio of deviation to modulation bandwidth is called the modulation index and is an important parameter in describing a FM system. Modulation index is given by: MI = Fd / Fm where MI is the modulation index. 3-6

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Modulation index is sometimes represented by the Greek letter beta ( β). In the practical a frequency modulator is formed from a voltage controlled oscillator. A voltage is applied to it from a control on the hardware board and the oscillator output can be examined on the oscilloscope and spectrum analyser. Using this configuration the fundamental concept of an oscillator frequency being changed by an external signal can be understood. 3.1.5.3 Procedure In this practical the hardware is configured as shown. You have available an oscilloscope and a spectrum analyser. Using this configuration you can see how the oscillator frequency can be controlled by an external signal. Set Carrier level to about half scale.

Use the oscilloscope to observe that when the manual frequency control is moved the frequency changes. Monitor at monitor point 16 to see the voltage applied to the oscillator and monitor point 4 to see the output. Monitoring monitor point 4 use the large oscilloscope calibration to measure the total frequency range of the oscillator. Use the spectrum analyser to confirm the frequency range measured on the oscilloscope.

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3.1.5.4 Questions 1. Is it easier to measure the frequency range on the oscilloscope or on the spectrum analyser?

2. Choose two voltages levels at the control input to the oscillator and measure the corresponding output frequencies. Hence calculate the 'frequency slope' of the oscillator in kilohertz per volt.

3. Can you see any amplitude variation over the frequency range? Should there be any?

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3.1.6 Practical 2: Generation of FM with a VCO

In this practical a sine wave signal is used to frequency-modulate a carrier so that you can investigate the appearance of such signals in both the time and frequency domains. You can adjust the amount of deviation and hence change the modulation index. Notice that the appearance of an FM signal on the spectrum analyser is similar to that of an AM signal when the modulation index is small. Try to reconcile the explanation of the bandwidth of an FM signal given in the previous background pages with the observations you make in this practical. Move to the next page to start the practical. 3.1.6.1 Procedure In this practical the variable voltage used to control the VCO frequency has been replaced by a sine wave oscillator. This sine wave now frequency-modulates the carrier. Set Carrier level to about half scale. Look at the signal at monitor point 4 with the oscilloscope. Turn the modulation level up and down and observe the effect.

Notice that the frequency is changing. Note where the output at monitor point 4 has a higher frequency. Change to monitor point 3 and observe how the instantaneous frequency depends on the instantaneous value of the modulating signal. Use the spectrum analyser to examine the sidebands of the signal. Adjust the modulation level and observe that, at low deviation, only the sidebands at Fc - Fm and Fc + Fm are present. At higher deviation, ie, a larger modulation index, higher-order sidebands appear.

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3.1.6.2 Questions 1. By looking at the spectrum of the modulated signal, can you estimate the frequency of the modulating signal? (Explain carefully how).

2. Would it be equally easy to estimate the bandwidth of the modulating signal from the spectrum if the modulating signal were complex, having many frequencies?

3. As the modulation level varies, how constant are: (a) the carrier-frequency component of the modulated signal? (b) the amplitude of the modulated signal?

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3.1.7 Practical 3: Spectrum of an FM signal with a large Modulation Index

This is a simple practical where the frequency modulator is connected to the spectrum analyser. The carrier frequency has been reduced to about 5 kHz so, since the maximum deviation is the same as in the other practicals of this Assignment, the modulation index is much greater. As we saw, the bandwidth is: B = 2 ( Fd+Fm ) where B is the bandwidth, Fd the deviation and F m is the bandwidth of the modulation. So if Fm is small compared with F d, ie, the modulation index is large, then: B = 2 Fd On the analyser the spectrum appears to be continuous but in reality it is made up of a large number of sidebands spaced at 5 kHz intervals from the carrier up to F d. This practical simply shows how when the modulation index is large the bandwidth is determined almost exclusively by the deviation. 3.1.7.1 Procedure In this practical the modulation frequency has been set to 5 kHz. This means that the modulation index can be very high. This enables you to see that under these conditions the bandwidth of an FM signal is almost equal to twice the deviation.

Set Carrier level to about half scale. Turn the 5 kHz level up and down and observe the bandwidth changing. Note that the bandwidth is almost proportional to the deviation.

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3.1.7.2 Questions 1. If the modulating frequency is 5 kHz and the deviation is 50 kHz, calculate the modulation index.

2. Calculate the signal bandwidth using Carson's rule.

3. If a bandpass filter were to be added at the input of an FM detector what factors determine the bandwidth required?

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3.2


Demodulation of Frequency Modulated Signals Assignment

3.2.1 Objectives


Quadrature detectors,

•

PLL detectors,

•

Noise.

3.2.2 Practicals



•


•


•




Signal Generation,

•

Modulator,

•

Limiter,

•


•

VCO,

•

Phase Comparator.

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3.2.4 Theory

3.2.4.1 Quadrature Detector

The quadrature detector splits the incoming FM signal into two paths. One path is connected directly to one input of a phase detector. The other path contains a simple network which shifts the phase of the signal in proportion to its frequency deviation. Let us consider the nature of a typical phase shifter ...

Regard this as a simple potential divider, with input at point 1, output at point 2. The upper arm has impedance jωL + R and the lower arm 1/(jωC) where w is the angular frequency. The circuit's transmission factor is e2 /e1 = [1/jwC] / [R + jωL + (1/jwC)] e2/e1 = 1 / [ 1 - ω2LC + jωCR ] 3-14

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It is convenient to express the transmission factor in terms of ωo, the resonant frequency, at which: ωoL = 1/(ωoC) , or ωo2 LC = 1 ,

and the quality factor Q, given by: Q = ωoL/R = 1/(ωoCR) Using these definitions, we can substitute 1/ ωo2 for LC and 1/Q ωo for CR. So the transmission factor: e2 /e1 = 1 / [ 1 - ω2LC + jωCR ] becomes: e2/e1 = 1 / [ 1 - ω2/ωo2 + jω/ωoQ ] The phase of this expression is: Ø = - arctan [ω / ωoQ ] / [1 - (ω / ωo)2] which, if we define y = ω / ωo can be written as: Ø = arctan [ y / Q (y2 - 1) ] Ø = arctan [ Q (y2 - 1) / y ] Ø = (π /2) - arctan [ Q (y2 - 1) / y ] Ø = (π /2) + arctan Q [ y - (1/y)] Replacing y by ω / ωo once more, this can be written as: Ø = (π /2) + arctan [Q (ω2- ωo2)/ ω ωo] Ø = (π /2) + arctan [Q (ω - ωo) (ω + ωo)/ ω ωo] Ø = (π /2) + arctan [Q dω ( 2ωo + dω) ]/[ (ωo + dω) ωo] where dω is defined as ω - ωo. Neglecting dω in comparison with ωo in:

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Ø = (π /2) + arctan [Q dω ( 2ωo + dω) ]/[ (ωo + dω) ωo] we get: Ø = (π /2) + arctan [Q dω ( 2ωo ) ]/[( ωo ) ωo] Ø = (π /2) + arctan [ 2 Q dω / ωo ] If dw is sufficiently small, the argument of the arctan is small, and therefore close to the arctan value in radians. So, still to a good approximation: Ø = (π/2) + [ 2 Q dω / ωo ] It must be noted however, that if Q has a high value, these approximations may become invalid, for quite moderate values of deviation (d ω). This indicates that for low distortion, the value of Q must be kept low. For good sensitivity, Q should be high. So here is an instance where a degree of compromise in the design is needed. Summing up, if the carrier is at its nominal frequency (dω = 0), the phase is just (π/2). The change in phase from (π/2) is proportional to the deviation (dω), provided that 2 Q dω/ωo is small. 3.2.5 Practical 1: Quadrature Detector

The purpose of an FM demodulator is to return the modulating signal to baseband. Since, in FM, the instantaneous frequency is proportional to the modulating signal, all that is needed is a circuit block which produces a voltage proportional to the input frequency. This is not quite as simple as it sounds. A very crude way of achieving this is to feed the signal through a filter with its cut-off near to the carrier frequency. The signal is then attenuated by an amount depending on its frequency. The filter output is now AM and can be demodulated by any AM detector. This type of detector uses the slope of the filter characteristic and therefore is called a Slope Detector. Slope detectors are not satisfactory in most applications as they are not very linear and do not reject noise well.

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3.2.5.1 Quadrature Detectors

Another way is to turn the FM to phase modulation and then use a phase detector. This is a very common technique and is the basis of the Quadrature Detector. In this practical you will see how a quadrature detector works. The incoming FM signal is split into two paths. One path is connected directly to one input of a phase detector. The other path contains a simple network which shifts the phase of the signal in proportion to its frequency deviation. The output of the phase shift network is connected to the other input of the phase detector. As the frequency varies, both detector inputs vary together in frequency, but one also shifts in phase relative to the other. It is this varying phase shift that produces the output from the detector. A minor complication is that most phase detectors produce their mean output for 90 degrees phase difference between the input signals. This is the required condition when the FM signal is at its centre frequency, so an additional constant 90 degree phase shift is added to one of the paths. When unmodulated, the two inputs to the phase detector are at 90 degrees apart, or in quadrature; hence the name of the detector. This constant phase shift is usually added by means of a simple inductor. The output of the phase detector still contains a large component at twice the carrier frequency and the detector is usually followed by a filter that passes the baseband but not the carrier. Quadrature detectors are used extensively in domestic FM radios and in a lot of communications equipment. In this practical, the same modulator is used as in the Generation of Frequency Modulation assignment. The modulation is a sine wave so that the signal can be followed through the circuit.

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3.2.5.2 Procedure This practical shows a quadrature detector working. Monitor at 9 and observe the FM signal at different settings of modulation level . Note the two signals at the inputs of the phase detector 9 and 11 . Set the modulation level to about half scale.

Observe the signal at the phase detector output 12 and then after the post-detection filter at 14 .

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3.2.5.3 Questions 1. Use the large oscilloscope to try and measure the phase shift between the two phase detector inputs when there is no modulation.

2. What frequencies must the output filter: (a) pass? (b) reject?

3. Would your answer to question 2(b) be altered if the phase comparator were imperfect in some way?

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3.2.6 Practical 2: Phase lock loop detector (PLL)

This practical introduces the phase locked loop (or PLL) demodulator. This type of detector offers some advantages over the quadrature detector when the signal to noise ratio is poor. Before trying to understand how a PLL can demodulate an FM signal it is necessary to understand what a PLL is. The concept is of an oscillator synchronised in phase to an external signal source using a feedback loop. As frequency is the same thing as rate of change of phase, once the phase of the local oscillator is synchronised to the external signal, the frequencies are automatically made identical.

A phase locked loop consists of three main blocks: 1. An oscillator, the frequency of which is controlled by an external voltage or current source. A voltage controlled oscillator or VCO is used in this assignment. 2. A phase detector, which compares the phase of the oscillator with that of the external signal. 3. A filter, which smoothes the output from the detector to provide the control signal to the VCO, adjusting its frequency so as to reduce the phase difference.

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3.2.6.1 Operation of a PLL

Imagine an incoming signal at a constant frequency within the range of the VCO. Its phase is compared with that of the VCO and a voltage produced that alters the VCO frequency. The phase of the VCO therefore starts changing relative to the incoming signal, until eventually the phases match. Once they are equal, the control signal goes to zero and the system settles into equilibrium. Any drift of the VCO will be corrected by the control voltage which again appears. The two signals are said to be phase locked. A filter is used in the control loop to keep the system stable and limit the maximum rate of change of oscillator frequency. This whole description is a very simplified view, and the parameters that set the filter characteristics are very complex. An important factor in the design is the time before the two signals become locked. Phase locked loops are used extensively in communications systems where it is necessary to produce a reference oscillator in phase with an incoming signal; also in special signal sources called frequency synthesizers.

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3.2.6.2 The PLL as an FM Detector Now that you have appreciated the concept of a PLL, how can it be applied to demodulate FM? Suppose that there was a PLL locked onto an incoming carrier which was unmodulated. The VCO would be at the same frequency as the carrier and the VCO control voltage would be constant. If the carrier were to change in frequency the VCO would follow the change by means of a change in control voltage. So the VCO control voltage varies with the carrier frequency, and if the carrier were frequency modulated the modulation would appear superimposed on the VCO control voltage. When a post-detection filter is added to the simple PLL to remove all the frequency components above the maximum modulating frequency we now have a PLL FM detector.

In this practical you will see a PLL detector demodulating the same FM signal as before. The PLL is used when the ability to demodulate in the presence of noise is important. The distortion produced by this type of detector is determined mainly by the linearity of the VCO but this is often less important in noisy applications. FM detection under noisy conditions is investigated in the next assignment.

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3.2.6.3 Procedure This practical shows a phase lock loop detector working. Monitor at 9 and observe the FM signal at different settings of modulation level . Examine the two signals at the input of the phase detector at 9 and the tracking VCO at 11 . Set carrier level to maximum.

Observe the signal at the phase detector output 12 and then after the post detection filter at 14 .

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3.2.6.4 Questions 1. What happens to the demodulated output when you reduce the control from maximum to half amplitude? Explain your answer in a sentence or two.

2. What happens to the demodulated output when you reduce the control to lower levels?

3. What is the special problem which occurs for very low signal levels?

4. Why do you think it happens and what do you think could be done about it?

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3.3


Limiters and the Effect of Noise on FM Demodulation Assignment

3.3.1 Objectives


Limiters,

•

Predetection noise,

•

Postdetection noise.

3.3.2 Practicals



•


•


•




Signal Generation,

•

Modulator,

•

Limiter,

•


•

VCO,

•

Phase Comparator.

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3.3.4 Theory

3.3.4.1 Frequency Modulation The equation of a sinusoidal voltage waveform is given by: v = Vmax.sin(ωt+ø) •

where v is the instantaneous voltage

•

Vmax is the maximum voltage amplitude

•

ω is the angular frequency

•

ø is the phase

A steady voltage corresponding to the above equation conveys little information. To convey information the waveform must be made to vary so that the variations represent the information. This process is called modulation. From the above equation, the basic parameters of such a waveform are: •

its amplitude, Vmax

•

its frequency, ω (or f)

•

its phase, ø

Any of these may be varied to convey information. Frequency modulation uses variations in frequency to convey information. We shall think in terms of the angular frequency w. The wave whose frequency is being varied is called the carrier wave. The signal doing the variation is called the modulating signal. For simplicity, suppose both carrier wave and modulating signal are sinusoidal; ie, vc = Vc sin ωc t (c denotes carrier) and vm = Vm cos ωm t (m denotes modulation)

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3.3.4.2 What is Frequency? If the frequency is varying, how do we define it? We can no longer count the number of cycles over a longish interval to count the cycles per second. Instead we define frequency as the rate of change of phase. This is consistent with the simple definition, because at a constant (angular) frequency w radians/second the phase is changing at ω radians per second, which is ω /2π cycles per second. Since we can only define what the instantaneous frequency is by reference to the phase, we must look at the phase in order to arrive at an expression for the frequency-modulated signal. 3.3.4.3 Phase of the FM Signal For the unmodulated carrier v c = Vc sin ωc t, the phase is s = ωc t We want the modulating signal to vary the carrier frequency, ωc, so that its frequency takes the form ω = ωc + D cos ωm t

(where D denotes the peak value of the deviation) It is related to the amplitude of the modulating signal v m by the 'frequency slope' of the frequency modulator (VCO) say k radians/s per V. The peak value of v m produces deviation D, so D = k Vm The total phase change undergone at time t is found by integrating the angular frequency. It is s = (ωc + D cos ωm t) dt = ωct + (D/ ωm) sin ωm t. (If you are not familiar with integration you will have to take this result on trust). So the FM signal can be expressed as: Vc sin [ωct + (D/ωm) sin ωm t] ,

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3.3.4.4 The Modulation Index In that expression for the FM signal: Vc sin [ωc t + (D/ ωm) sin ωm t] the coefficient (D/ ωm) turns out to be quite important and is given the name 'modulation index'. It is often represented by the Greek letter beta, . So we may write the FM signal as: vc = Vc sin (ωct + sin ωm) t where is the modulation index (D/ ωm) In this expression, the factor sin (ωct + sin ωm) t (let us call it F) is of the form sin (a + b) which can be expanded to sin a cos b + cos a sin b. Applying this expansion to F, we get F = sin ωct cos( sin ωm) t + cos ωct sin ( sin ωm) t 3.3.4.5 FM Sidebands These complicated functions can be expanded, using mathematics too elaborate to explain here, into a series of terms like this: F = J0( ) sin ωct+ J1( ) [ sin (ωc + ωm)t - sin (ωc - ωm)t ] + J2( ) [ sin (ωc + 2ωm)t - sin (ωc - 2ωm)t ] + J3( ) [ sin (ωc + 3ωm)t - sin (ωc - 3ωm)t ] + J4( ) [ sin (ωc + 4ωm)t - sin (ωc - 4ωm)t ] + ... where J0( ), J1( ), J2( ) etc are constants whose values depend only on . They are called Bessel Functions. There is an infinite series of these functions, and so an infinite number of FM sidebands. But in practice the values of the Bessel functions become very small as the series goes on. For example, when = 2 J0(2) J1(2) J2(2) J3(2) J4(2) J5(2) 3-28

= = = = = =

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3.3.4.6 A Practical Approximation Rule Because the higher-order sidebands become very small, in practice the bandwidth of the FM signal may be restricted to a finite bandwidth. The practical rule that is used, often called Carson's Rule, is to take the bandwidth required as: B = 2 ( Fd + Fm ) where B is the bandwidth, Fd the deviation and F m is the bandwidth of the modulation, all in the same units. 3.3.5 Practical 1: A Quadrature Detector with Limiter

In this practical you will see how a limiter works. In an FM system the information is carried by variations in carrier frequency. Since the variations in amplitude carry no information, they can be removed before the signal arrives at the detector. This is the function performed by a limiter. A limiter is simply a high gain amplifier that turns the usually sine-wave carrier of varying amplitude into a square wave of constant amplitude. The square wave still contains the frequency variations that contain the modulation. The addition of a limiter means that the FM detector has a constant amplitude signal to deal with which means that its output is only dependent on phase changes and not changes in amplitude. This can be shown in the practical by varying the carrier amplitude with no limiter in operation and finding that the output signal also varies in amplitude. When the limiter is placed in circuit the output no longer varies in this way. Of course, the limiter cannot produce a signal from nothing so as the input carrier amplitude to the limiter falls the noise content increases. Ultimately the signal becomes unrecognisable because of noise. The effect of noise on the detector performance is investigated in more detail in the next Practical.

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ANALOGUE COMMUNICATIONS ANALOGUE STUDENT’S STUDENT’S WORKBOOK


3.3.5.1 Procedure This practical shows the effect of a limiter on a quadrature detector. Start with the limiter out of circuit. Set modulation level to to maximum. Observe the signal at the detector output 14 while varying the carrier level .

Use the Limiter Button to to switch the limiter into circuit. Repeat your observations.

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3.3.5.2 Questions 1. In the absence of a limiter, does the demodulated output depend on the carrier amplitude?

2. Should it, ideally?

3. W ith the limiter in use, how does the demodulated output vary with carrier amplitude?

4. How does the waveform of the input to the phase comparator (point 9) differ with with and without the limiter?

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3.3.6 Practical 2: The Effect Effect of Noise on a Quadrature Detector

In the first Practical of this Assignment we saw how a limiter forces the input to the detector to be of constant amplitude. This has an important effect on how the detector behaves when the signal is noisy. 3.3.6.1 What is Noise? Noise is simply an unwanted signal which is mixed up with the required signal. In many cases it is not a specific frequency but is made up of random combinations of many frequencies. Such unwanted noise may be generated internally by circuit elements like amplifiers or come from the transmission medium such as cables or antennas. A very important characteristic of a communication system is how well it works in the presence of noise. noise. FM systems offer some advantage over AM systems in their noise performance. The theory behind this is quite complex and will not be dealt with here. 3.3.6.2 Signal/Noise Ratios One measure of the quality of the received signal applied to the detector is its Signal to Noise Ratio (SNR). (SNR). This is simply the ratio of signal power to noise power, usually expressed in decibels for convenience. A high SNR means that there is much more signal than noise. After passing through the detector the demodulated output also has noise on it, and therefore has a signal to noise ratio. These two ratios are often called predetection SNR (SNRi) and post detection SNR (SNRo). In an AM system these two values are approximately equal, but in an FM system the SNRo can be greater than SNRi. How much greater depends on many things, but especially on the modulation index. Another contrast between AM and FM is that in an AM channel, SNR i is proportional to SNRo but in FM it is not; instead, as SNRi is reduced below a certain level, called the threshold, SNRo drops very quickly. This means that FM systems tend to degenerate very quickly at low SNR. In general, the greater the modulation index (and therefore the bandwidth) the greater the improvement in SNR given by the detector. However, this is at the expense of a higher threshold. As is often the case, selecting the parameters to get the best performance out of the system is a compromise.

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An important point to note is that when noise is present it causes both amplitude and frequency variations, and to obtain the best performance on an FM system the amplitude noise should be removed with a limiter. In the practical, the FM generator is connected to a quadrature detector. A variable amount of noise is added to the signal fed to the detector generated by a special circuit that simulates the type of wide-band noise that would come from a telecommunications system. A limiter is also provided, which can be switched in and out of use. The purpose of the practical is to investigate the effect of noise on the demodulated output, and how this is affected by the limiter. 3.3.6.3 Procedure This practical shows a quadrature detector demodulating a noisy FM signal. Adjust Carrier level and Modulation level to examine each signal, leaving them both at maximum. Set the Noise level to minimum. The limiter should not be in use. Observe at 14 with the oscilloscope and note that the demodulator is working correctly. Increase the Noise level and observe that the output becomes noisy.

Now switch in the limiter with the Limiter Button and repeat the test. Note that the detector keeps working at lower SNR when the limiter is in use. Use the other monitoring points to see how the system is operating.

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3.3.6.4 Questions 1. What does the term SNR stand for and what does it mean?

2. Why do you think that it is more convenient to express it in decibels?

3. Why does the use of the limiter reduce the noise on the output?

4. With the limiter in circuit, observe that reducing the carrier level causes an increase in the noise on the output. Why is this? (Hint - a look at the limiter output may help).

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3.3.7 Practical 3: PLL Detector with a Limiter

In this practical you will see first how a Phase Locked Loop behaves at different signal amplitudes. Introducing a limiter will then show an improvement in performance. If the received signal is large enough, the PLL will lock the local oscillator phase to that of the received signal. Doubling the signal amplitude will not alter this situation, so will not affect the output signal. To this extent the PLL removes unwanted amplitude modulation of the received signal. If the signal is small and the deviation is large, the phase detector cannot give enough output to move the VCO over a large enough range to track the deviation. This can be shown in the practical by reducing the carrier amplitude with no limiter in operation. Failure to track over the whole range of deviation shows as a distortion of the output signal. For small enough signals, the PLL fails to lock altogether. The addition of a limiter means that the phase detector in the PLL has a constant amplitude signal to deal with. The gain of the phase lock control loop is therefore maintained as the signal level changes. 3.3.7.1 Procedure This practical shows the effect of a limiter on the performance of a PLL detector at various carrier levels. Start with the limiter out (this is the default). Set carrier level and 14 with the oscilloscope. Reduce the carrier level and observe the output. Note that the detector loses track of the signal below a certain level, causing distortion of the detector output. Note also that, when the modulation level is reduced, the carrier can be reduced further without distortion.

Now use the Limiter Button to switch in the limiter and repeat the tests. Note that the detector continues to work at much lower carrier levels. Use the other monitoring points to see how the system is operating.

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3.3.7.2 Questions 1. With the limiter switched out, why does a lower modulation level allow the carrier level to be reduced further before distortion occurs?

2. Does this also happen when the limiter is switched in?

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3.3.8 Practical 4: The Effect of Noise on a PLL Detector

Noise on the received signal causes both amplitude and phase changes. When a limiter is placed in the circuit, the amplitude changes are removed from the PLL input. The principal effect on the PLL is that as the input signal tends to zero amplitude, there remains an adequate amplitude of signal to drive the phase lock loop. This continues to track the phase of the noisy received signal effectively, and with minimum error caused by noise amplitude variations. Of course, the limiter cannot produce a signal from nothing, so as the carrier amplitude into the limiter falls, the noise from the limiter increases. This noise is faithfully detected by the PLL and degrades the output signal. This problem is due to fundamental noise problems and not due to any failing of the detector itself. 3.3.8.1 Procedure This practical has a PLL detector demodulating a noisy FM signal. Set the Carrier level and Modulation level to maximum. Set the Noise level to minimum. The limiter should not be in use. Observe at 14 that the demodulator is working correctly. Increase the Noise level and observe that the output becomes noisy. Decrease the signal/noise ratio (SNR) further by reducing the carrier level until the signal becomes unrecognisable.

Now switch in the limiter using the Limiter Button . Note that the detector keeps working at lower SNR when the limiter is in use. Use the other monitoring points to see how the system is operating.

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3.3.8.2 Questions 1. How does limiting the signal improve the signal to noise ratio?

2. How does the performance of the PLL demodulator compare with the phase comparator demodulator?

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4

Chapter 4 Assignments using the Signal Sources Workboard

Assignments using the Signal Sources Workboard

4.1

Wien Bridge Oscillator Assignment

4.1.1 Objectives


Conditions for oscillation of a Wien Bridge Oscillator,

•

The need for amplitude control of a Wien Bridge Oscillator.

4.1.2 Practicals


Practical 1: Basic Wien Bridge Oscillator

•

Practical 2: Amplitude Stabilisation

•

Practical 3: Changes from Standard


Signal Sources Workboard 53-110 which comprises the following blocks: •

Wein Bridge Oscillator,

•

L-C Pass Filter,

•

Crystal Oscillator,

•

Multivibrator.

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4.1.4 Theory

4.1.4.1 Wien Bridge Oscillator The theory of the Wien Bridge oscillator is usually presented for equal values of R and C in the series and parallel arms. Here we look at the more general case of arbitrary R and C values.

Here Z1 is the impedance of the combination of C1 and R1 and Z2 is the impedance of the combination of C2 and R2. It is easier to consider the reciprocal 1/B of the feedback fraction, since this is (Z1 + Z2)/Z2, which simplifies to: 1/B = 1 + Z1 / Z2 where Z1 = R1 + 1/jωC1 and Z2 = R2 / (1+jωC2R2) Routine manipulation produces the result: 1/B = [1 - ω2C1C2R1R2 + jω (C1R1 + C2R2 + C1R2)] / jωC1R2 1/B = [1 - ω2C1C2R1R2 + jω (C1R1 + C2R2 + C1R2)] / jωC1R2 which has zero phase shift when: 1 - ω2C1C2R1R2 = 0 which enables the frequency to be calculated. Now, 1/B = jω (C1R1 + C2R2 + C1R2) / jωC1R2 1/B = (C1R1} + C2R2 + C1R2) / C1R2 which is the required value of gain. 4-2

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It is not difficult to see that if all the CR terms have equal values, the required gain is 3. Every oscillator has the requirement that, following some initial disturbance, the behaviour is modified so as to increase that disturbance, until a useful level of signal has been built up. This happens at some particular frequency, so that a periodic disturbance is built up.

An oscillator typically comprises an amplifier with feedback around it, as shown. If a steady oscillation is to be maintained, then the input to the amplifier, must satisfy two conditions: it must have exactly the right amplitude to generate the output. its phase must be correct to generate the output. If we suppose that the amplifier has a positive gain A and no phase shift, then oscillations will be maintained if AB=1 where B is the gain of the feedback network. For sinusoidal oscillation, the condition AB=1 should be satisfied at the one required frequency, and at no other. This will then cause only the one, required, frequency to be produced. The Wien Bridge oscillator is so called because the circuit is based on a frequencyselective form of Wheatstone bridge known as the Wien bridge. The form of oscillator you will use is in fact a high-gain amplifier with a Wien bridge around it, but it is more usual to consider the circuit as being an amplifier of the requisite low value of gain, working with a feedback network equivalent to half of the Wien bridge.

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The usual Wien bridge oscillator has equal values of capacitance and of resistance arranged in the series/parallel arrangement shown. The feedback factor B may be expressed in terms of R and the reactance X= -1/ ωC of the capacitors, as follows: B = (R+jX)/[R+jX + jRX/(R+jX)] = (R+jX)2 /[(R+jX)2 + jRX)] The condition for oscillation is AB=1. Substituting B for the previously obtained expression: (R+jX)2 /[(R+jX)2 + jRX)] and cross-multiplying, the condition becomes: A(R+jX) = R+jX + jRX/(R+jX) A(R+jX)2 = (R+jX)2 + jRX (A-1) (R+jX)2 = jRX (A-1) (R2 -X2 +2jRX) = jRX (A-1) (R2 -X2) +(A - 3) jRX = 0 The condition: (A-1) (R2 -X2) +(A - 3) jRX = 0 becomes true if both the real and the imaginary parts of the left-hand side are zero; ie, if A=3 where A is the required gain of the amplifier, and R2 = X2, so that R = +/- X. Since X = -1/ ωC, the latter condition gives the frequency of oscillation thus: R = 1/ ωC so that ω = 1/CR

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4.1.5 Practical 1: Basic Wien Bridge Oscillator

In this practical an operational amplifier is provided with a simple negative feedback network which allows its gain to be adjusted within narrow limits. A second, positive, feedback network incorporates the Wien half-bridge, comprising series- and parallel-connected R and C The positive feedback has zero phase shift at just one frequency. It is shown in the Theory that both zero phase shift and a particular value of gain are needed to maintain oscillation. You will see that excessive gain causes oscillation to build up until limited by the available output swing from the amplifier. You will need to be careful when adjusting the gain control in order to see oscillations building up and dying down. Notice that the frequency meter may give high or erratic readings before falling to 0Hz as the amplitude is decreased. This is typical of frequency meter behaviour when the signal falls to a level it cannot distinguish from noise.

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4.1.5.1 Procedure You have available an oscilloscope and a frequency meter. The frequency meter is always connected to the oscillator output. R5 is adjustable to control the gain of the amplifier.

Start with the gain control at maximum. Look at the waveforms at both monitoring points, while varying the gain. Observe how a very slight change of gain can stop or start oscillations. You will need to return to the practical and make some measurements in order to answer the questions.

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4.1.5.2 Questions 1. What determines the amplitude of oscillator output when the gain slightly exceeds that required for oscillation?

2. Do you think that this oscillator could be designed to work satisfactorily, taking into account production tolerances and other component variations?

3. With the circuit adjusted for a nearly sinusoidal oscillation, what is the frequency : a) as measured by the frequency meter? b) measured another way?

4. Given that C 1 = C 2 = 1nF and that R 1 = R 2 = 15k,what does theory say the frequency should be? Compare with the answers to question 3.

5.

Why is the waveform at test point 2 not the same as at test point 3 (ignoring the amplitude difference)?

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4.1.6 Practical 2: Amplitude Stabilisation

In the Basic Wien Bridge Oscillator Practical you saw how if the gain is very slightly below the value required for oscillation, the oscillator stops working. On the other hand, with only slightly excessive gain, the oscillation builds up until the amplifier distorts. How do we ensure a stable sinusoidal signal? In practice, oscillations present in linear systems will either increase steadily or decrease steadily. An exact gain value cannot be achieved due to component tolerances and component variations with temperature and age. What we have to do is to find some kind of non-linearity which will allow the gain to vary with signal strength, but not distort the signal too much. This problem is common to all oscillators based on linear amplification. The solution adopted in this assignment is usable on such oscillators of many different kinds.

The diagram shows the negative feedback path modified by the inclusion of another path containing a pair of back-to-back diodes. These do not conduct significantly until the voltage across them is about 0.6 - 0.7 V. But once current has started to flow, the voltage drop will increase only slightly.

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Imagine that, with the diodes NOT conducting, the negative feedback sets the gain slightly higher than is needed for oscillation. Oscillations will start to build up. Once the amplitude has reached some 0.7 V the diodes will start to conduct, increasing the negative feedback. More negative feedback will be applied, tending to reduce the gain. There will be an amplitude for which the gain is just right to maintain oscillation. Any variation from this amplitude will change the gain, tending to keep the amplitude constant. 4.1.6.1 Procedure In this practical the hardware is as in the Basic Wien Bridge Oscillator Practical, except that an additional negative feedback path has been introduced, including the back-to-back diodes.

Start with the gain control at maximum. Look at the waveforms at both monitoring points, while varying the gain. Observe carefully any difference between the effect of the gain control in this Practical compared with the Basic Wien Bridge Oscillator Practical.

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4.1.6.2 Questions 1. How easy is it to control the amplitude, compared with Practical 1?

2. What limits the ratio over which (R4 + R5) can be varied, given that other components have their nominal values? Explain your answer.

3. What happens when (R4 + R5) goes outside the proper range of values?

4.

Set R5 in the middle of the range giving stable sinusoidal oscillation. Measure the signal amplitudes at the input and output of the amplifier. Calculate the amplifier gain and compare with the theoretical value.

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4.1.7 Practical 3: Changes from Standard

It is convenient, from a manufacturer's point of view, to have the same values R and C in the series and parallel-connected parts of the Wien half-bridge. This is not necessary for oscillation however. This Practical is provided simply to show that a Wien bridge oscillator can have differing R and C values, for which the standard theory does not apply. See the Wien Bridge Oscillator Theory. 4.1.7.1 Procedure In this practical the hardware is as in the Amplitude Stabilisation practical, except that the values of R2 and R4 have been increased. Set the gain control to give a steady sinusoidal oscillation, then, using the large oscilloscope, determine the gain of the amplifier from its '+' input terminal to the output.

You will find that the result is different from the value of 3 given in the Wien Bridge Oscillator Theory. This is not because the Theory is wrong, but because the circuit no longer conforms to the assumptions of the Wien Bridge Oscillator Theory.

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4.1.7.2 Question 1. What value of amplifier gain did you measure?

2. Was this a change from the value of 3 given for the standard basic theory?

3. Compare the gain and frequency with the theoretical values : gain = (C1R1 + C 2 R 2 + C 1 R 2 ) / C 1 R 2 and frequency = ω /2 (PI) , where 1 - ω 2 C 1 C 2 R 1 R 2 = 0.

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4.2


L-C Oscillator Assignment

There are several detailed kinds of oscillator which use a resonant circuit (inductance and capacitance together) to set the frequency of oscillation. This assignment deals with one particular circuit of this kind, which is a typical example. 4.2.1 Objectives


The operation of a tuned-collector oscillator,

•

The factors that determine its amplitude,

•

The stability of its frequency with supply variations.

4.2.2 Practicals


Practical 1: Tuned-Collector Oscillator

•

Practical 2: Effect of Supply Variations




•

L-C Pass Filter,

•

Crystal Oscillator,

•

Multivibrator.

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4.2.4 Theory

4.2.4.1 LC Oscillator

The diagram shows a simple transistor amplifier having the parallel tuned circuit, L,C as a load. The practical inductor has resistance, shown as r . The transistor will be considered as a simple device whose collector current is dependent only on the base-emitter voltage with a mutual conductance g. For small ac signals, ic = g v in This current flows in the load, whose impedance Z is given by: 1/Z = jωC + 1/(r + jωL) which can easily be rearranged to give: Z = (r + jωL) / (1 - w2LC + jωCr) so that the voltage across the load becomes: -g vin (r + jωL) / (1 - ω2LC + jωCr)

The load voltage, - g v in (r + jωL) / (1 - w2LC + jωCr), is applied to the impedance (r + jωL) to produce a current in L, iL = g vin / (1 - ω2LC + jωCr), 4-14

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whose phase is -90 degrees when ω2LC = 1.

A secondary coil is now coupled to L, with mutual inductance M. This secondary has induced in it a voltage jωMiL; ie: - jωM g vin / (1 - ω2LC + jωCr) which, when: ω2LC = 1,

becomes: - jωM g vin / jωCr = - (Mg/Cr) vin. Clearly, if Mg/Cr = -1, the secondary winding can supply the base voltage required to maintain oscillation.

The minus sign is important as it connected to the start of the winding L, then the base must be connected to the finish of shows the polarity with which the feedback winding must be connected. Conventionally winding polarities are identified as start and finish, with the winding sense taken to be the same for each winding. The value of M would be taken as positive if increasing positive current into the start of one winding produced a positive voltage at the start of the other winding. In this case, if the transistor's collector is the secondary winding. The 'o' symbols in the diagram conventionally show starts of windings.

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This fairly simple analysis has neglected the effect of base current flowing in the secondary winding. This is a reasonable approximation when the current gain of the transistor is large, and assuming also that stray capacitances do not provide significant additional current paths. There are numerous other factors in a complete analysis, of which a few instances are mentioned below. When the current in the secondary winding is significant, it becomes important to take into account the effect of the leakage inductance of the secondary winding. The transistor has various imperfections. The collector current is to some extent dependent on collector-emitter voltage. There is stray capacitance between collector and base, providing an additional feedback. There are delays due to the semiconduction process. One of the more important effects of these various imperfections is that the frequency is not exactly the resonant frequency of L,C, but is influenced by stray factors which can be variable. 4.2.4.2 Calculation of Stray Capacitance The resonant frequency of an L-C circuit is given by: ω2LC = 1

therefore: C = 1 / 4 π2 L f2 Suppose that we start with a nominal tuning capacitance C and stray capacitance C s, giving a total capacitance C + Cs and frequency f C. If the nominal tuning capacitance C is increased by a factor n, the new total capacitance will be nC + Cs and frequency f nC. So the square of the frequency will be altered in the ratio r given by: r = ( C + Cs ) / ( nC + Cs ) so C + Cs = r ( nC + Cs ) Collecting Cs terms on the left: Cs ( 1 - r ) = C ( nr - 1 ) so that the stray capacitance Cs = C ( nr - 1 ) / ( 1 - r ) , where r = ( fnC / fC )2. 4-16

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4.2.5 Practical 1: Tuned-Collector Oscillator

This practical introduces the LC-tuned oscillator, after it you will understand: 1. How feedback is taken to produce the oscillation. 2. How the presence of the tuned circuit sets the frequency and ensures that the oscillation will be at least approximately sinusoidal. 3. How stray capacitance affects the tuning.

The diagram shows a transistor amplifier having a parallel tuned circuit as a load. R1, C1 and C2 are present simply for biasing purposes. If an ac signal voltage were applied to the base of the transistor, an amplified version of the signal would appear at the collector. Maximum amplification would occur when the frequency of the signal equalled the resonant frequency of the tuned circuit. There is always some minute amount of noise current in any circuit, distributed over all frequencies. The noise at the resonant frequency of L and C builds up a voltage across L. The current in L lags that voltage by about 90 degrees.

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A secondary winding is coupled to L, and this has induced in it a voltage which leads the current by 90 degrees, so it is in phase with the voltage across LC. The amplifier inverts the signal at the resonant frequency - while the transistor base is going positive, the collector will go negative. So in order to build up oscillations, the secondary winding has to be reversed compared with L, the primary, as is shown by coloured 'o' symbols by the windings. Note that most of the discussion has been about what happens at the resonant frequency. At other frequencies, not only is the loop gain reduced, but its phase changes fairly rapidly with frequency. Since correct phase is necessary to maintain oscillation, no oscillations can be produced at frequencies away from resonance. 4.2.5.1 Procedure The hardware includes two switches. One, as shown, switches in a capacitor C' of equal capacitance to C. The other switch (not shown) reverses the two connections to the feedback (left-hand) winding.

Observe the effect of each of the switches. Use the spectrum analyser to measure the frequency of the signals.

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4.2.5.2 Questions 1. Is the polarity of the feedback important, and if so why?

2. Which components in the circuit do you think are the most important in determining the frequency of oscillation?

3. What two frequencies of oscillation can you measure?

4. What is the effect of doubling the tuning capacitance? Your answer should include a mention of a numerical factor or ratio, with explanation if possible.

5. Using your last answer, can you calculate the stray capacitance?

6. What is the value of inductance L?

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4.2.6 Practical 2: Effect of Supply Variations

For reliable operation, an oscillator usually has to be provided with a higher loop gain than is needed to start oscillation. Oscillations therefore build up in amplitude until something happens to reduce the effective gain, preventing further build up. Often, varying the supply voltage will greatly affect the oscillatory amplitude. This practical allows you to study the oscillator's behaviour over a range of supply voltage between about 5 and 10 V. The mechanism by which the amplitude is limited in this oscillator is the non-linearities due to the cutting-off of the transistor. Note : 1. How effective the amplitude control is. 2. How the presence of the L-C circuit prevents 4.2.6.1 Procedure The hardware is now configured as for the previous Tuned-Collector Oscillator Practical, (including the switched capacitor, not shown) except that the positive supply voltage is now variable between approximately +5 V and +10 V. Make sure that the feedback switch is set to allow oscillation.

Observe the effect of varying the supply voltage on the amplitude and frequency of the output. The changing dc conditions can be seen in terms of the dc voltage at the transistor's emitter.

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4.2.6.2 Questions 1. Some non-linearity in the circuit is needed to control the amplitude. Which Whi ch component provides this?

2. Name the non-linear behaviour which limits the amplitude.

3. W hat is the function of the two two diodes in the circuit?

4. Can you explain why why the amplitude decreases as the dc in the transistor increases?

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ANALOGUE COMMUNICATIONS ANALOGUE STUDENT’S WORKBOOK 4.3


Crystal Oscillator Assignment

4.3.1 Objectives


The behaviour of a quartz crystal,

•

The use of a quartz crystal for stabilising the frequency of an oscillator,

•

The fundamental and overtone modes of oscillation of a crystal oscillator.

4.3.2 Practicals


Practical 1: Fundamental and Overtone Modes




•

L-C Pass Filter,

•

Crystal Oscillator,

•

Multivibrator.

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4.3.4 Theory

4.3.4.1 Quartz Crystal Crystal oscillators are used when a very stable frequency of oscillation is desired. There are many circuits for a crystal oscillator. Many such oscillators are simple L-C oscillators with a crystal inserted into the circuit at some point. Alternatively, a crystal oscillator can be constructed using logic gates, again associated with a crystal in the circuit. The crystal effectively only conducts well at one particular frequency. So by putting the crystal in an oscillator circuit, one ensures that the oscillator can only oscillate at that one frequency (if at all). We will see why the crystal has this special frequency for conduction in a moment. A crystal for frequency control is usually of quartz, a substance which has two useful properties: 1. It has very little mechanical damping: it 'rings' very easily, like a bar of steel rather than a bar of lead. 2. It shows the piezo-electric effect. That is, when it is mechanically stressed it produces an electrical field; when it is electrically stressed it changes shape mechanically. Its mechanical 'ringing' (property number 1) produces (owing to property number 2) a resonant electrical effect. This is made available at terminals connected to electrodes on the crystal faces. 4.3.4.1.1 Equivalent Circuit of a Crystal Because the mechanical vibration or 'ringing' of the crystal has electrical effects, there appears to be an electrical circuit capable of 'ringing' between the terminals. It is called circuit'. the 'equivalent 'equivalent circuit'.

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Because the equivalent circuit can resonate or 'ring', it has inductance and capacitance in it, shown as L and C in the diagram. There is also some inevitable capacitance between the terminals, shown as C stray.

Component values in the equivalent circuit for a typical 1MHz crystal are shown above. Compare C with the typical capacitance of a tuning capacitor at 1MHz (a few hundred pF). L and C have high values of reactance (in the region of 63 megohms) so the slightest departure from resonant frequency causes the crystal's impedance to rise very rapidly. Inevitably there is some energy loss, or damping. This has an effect typically like that of placing one or two thousand ohms of resistance in series with L. 4.3.4.1.2 The Resonances

The important resonance is the series one, when jωL + (1/jωC) = 0 or ω2LC = 1

There is a parallel resonance also, which occurs when the ω2LC' = 1

where C' is the value of C and C stray in series. With the values in the diagram, C' will be about 0.1% different from C, so that the parallel and series resonant frequencies will differ by about 0.05%.

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4.3.4.2 Crystal Oscillator

Consider Fig 1, in which all dc-only components have been omitted. Imagine that the collector current has a component at the resonant frequency of L and C. Because of resonance, the current in L and the current in C greatly exceed the collector current. We can 'tap off' a small portion of the capacitor current, the portion still being as great as the collector current, and feed it into the emitter. To do this, some capacitance is removed from C and replaced by a capacitor connected to the emitter.

Fig 2 shows a form of oscillator using this principle, which is especially useful at the higher frequencies of tens of MHz upward. The fed-back emitter current is phase-shifted by the capacitor's reactance. To make the phase correct to maintain oscillation, some lagging phase shift must be added. This lag is provided by the transistor itself. Injection of emitter current causes holes to drift across the base barrier to the collector, but this takes time. The circuit in the Practical is slightly different in detail, in order to achieve similar functioning at two widely different frequencies.

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You can see that the difference from Fig 2 is just that the emitter feedback is derived in a slightly different way. Thevenin's Theorem states that any linear two-terminal network can be replaced by an equivalent circuit comprising a voltage source equal to the unloaded voltage at the terminals, and a series impedance such that the short-circuit currents of the original and equivalent circuits match.

Fig 4 shows how, in the basic circuit of Fig 1, splitting the capacitance C into two series capacitances, and feeding the emitter from their junction, is equivalent to Fig 2 with the feedback source voltage reduced.

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4.3.5 Practical 1: Fundamental and Overtone Modes

This Practical shows how the introduction of a suitable quartz crystal into an oscillator can stabilise its frequency. The particular form of oscillator chosen is just one example of many kinds of oscillator in which a crystal can be used. It is an L-C tuned oscillator, with variable tuning capacitance. Quartz is piezo-electric - mechanical vibration produces electrical effects and vice versa. Because a crystal is a block of material it can vibrate in several different modes at different frequencies. Ordinarily a crystal is designed for use in only one of these modes, with one specific frequency of oscillation. If the oscillator is so designed that it cannot oscillate at one of the crystal's resonant frequencies, it may still oscillate at another. This is shown in the Practical by providing an L-C tuning circuit which is tunable over a wide range. 4.3.5.1 Starting the Practical At the start of the Practical you must turn the tuning control fully counterclockwise (to its lowest frequency). This is because the crystal provided is designed to be used in fundamental mode, at 2MHz, and is not very good at starting oscillation at an overtone frequency. Because the crystal's performance at the overtone frequency is not defined, the oscillator has been designed to force oscillations even with poor crystal performance. This forceful (and not very typical) action can produce oscillations at various frequencies, not necessarily those of the crystal. But it should be clearly seen, with the aid of the Spectrum Analyser, that the oscillator does 'lock on' to two different crystal frequencies. A point to notice is the very erratic behaviour when the oscillator tuning is altered just far enough from the crystal's fundamental to stop the fundamental oscillation. This may be thought of as a kind of fight to set the frequency, between the crystal and the tuning circuit. A more normal oscillator design would simply refuse to oscillate at any frequency except the intended crystal frequency.

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4.3.5.2 Procedure The tuning control appears on the circuit diagram as C1 and C2. Start with the tuning control set fully counter-clockwise. Measure the signal frequency with the spectrum analyser and the large oscilloscope.

Alter the setting of the tuning control gradually and observe what happens. Note particularly any frequencies to which the circuit seems to 'lock', needing an appreciable movement of the tuning control before a change in frequency results. This is most easily seen using the spectrum analyser.

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4.3.5.3 Questions 1. What are the two frequencies at which the oscillator frequency becomes locked to the crystal?

2. What is the ratio between these two frequencies?

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ANALOGUE COMMUNICATIONS STUDENT’S WORKBOOK 4.4


Multivibrator Assignment

This assignment is about the simple multivibrator. This is a type of oscillator which uses the transistors as switches. It produces waveforms which are far from sinusoidal. 4.4.1 Objectives


Factors that determine the frequency of oscillation of a multivibrator,

•

Multivibrator waveforms and how they are affected by circuit design,

•

The effect of leakage in the transistors base-emitter diodes and how to prevent it,

•

Variation in the mark/space ratio of a multivibrator,

•

The spectrum of the multivibrator output signal.

4.4.2 Practicals


Practical 1: Basic Circuit

•

Practical 2: Effects of Variable Supply

•

Practical 3: Mark/Space Ratio Control




•

L-C Pass Filter,

•

Crystal Oscillator,

•

Multivibrator.

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4.4.4 Theory

4.4.4.1 The Simple Multivibrator The operation of a multivibrator is considered in the Basic Multivibrator Circuit Background, which should be read before continuing. As the Background explains, the multivibrator is characterised by quick transitions from one state to another, followed by 'relaxation' periods in which the charge on the capacitors changes comparatively slowly. To calculate the time taken over these changes, we have first to identify the paths taken by the charging and discharging currents. 4.4.4.1.1 Start of Discharge

The diagram shows the situation where the input terminal has just gone from near the supply potential down to zero volts. The capacitor previously had its left-hand terminal at Vs and its right-hand terminal at a small positive potential V f due to the forward voltage drop across the conducting base-emitter diode Before the input transition, the voltage across the capacitor was V s - Vf, or approximately Vs if Vf is neglected. Bringing the input terminal down to 0V must bring the base of the transistor down to - (Vs Vf). There is therefore a voltage across Rb equal to (2Vs - Vf). The resulting current (2Vs - Vf)/Rb starts to discharge the capacitor.

This current decays exponentially, with a time constant CR b.

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The current is therefore given by: i Rb = (2Vs - Vf) exp(-t/CRb) from which the voltage on the transistor base can be calculated as: Vb = Vs - i Rb Vb = Vs - (2Vs - Vf) exp(-t/CRb) This transistor ceases to be cut off when Vb has reached the value Vf; ie, at t given by: Vf = Vs - (2Vs - Vf) exp(-t/CRb) exp(t/CRb) = (2Vs - Vf) / (Vs - Vf) t = CRb ln [ (2Vs - Vf) / (Vs - Vf) ] It is not too easy to see the meaning of: t = CRb ln [ (2Vs - Vf) / (Vs - Vf) ] But since normally Vf is small compared with Vs, we can simplify to find an approximate value for t : t1 = CRb ln 2 = 0.69 CRb 4.4.4.1.2 Capacitor Charging

The other state that needs considering is when a transistor has just ceased conducting. At that time its collector is at approximately 0 V (with respect to the emitter). The circuit is effectively as shown here. The capacitor charges through the collector resistor and the other transistor's baseemitter diode. The latter is assumed to produce a constant small voltage drop V f. If we now measure time t' starting from this last transition, the current is given by: i Rc = (Vs - Vf) exp (-t'/CRc) 4-32

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But Ohm's Law leads us to: Vc = Vs - i Rc Vc = Vs - (Vs - Vf) exp (-t'/CRc) or neglecting Vf, Vc = Vs [1 - exp (-t'/CRc)] 4.4.4.1.3 A Check on the Collector Waveform We must check our previous assumption that the collector rose to nearly +V s. To do this we must substitute t' with the previously calculated time (the approximate value t1 will do): t1 = CRb ln 0.5 = 0.69 CRb. So at the end of this half-cycle, the collector voltage has risen to: Vc' = Vs [ 1 - exp (-0.69 CRb /CRc) ] Vc' = Vs [ 1 - exp (-0.69 Rb /Rc) ] Now, Rb is typically many times Rc, because the ratio can approach the current gain of the transistor. It is 68 in the multivibrator on the board for instance. Putting Rb /Rc = 68, we calculate: Vc' = Vs [ 1 - exp (-0.69 * 68) ] Vc' = Vs [ 1 - exp (-47) ] Vc' = Vs [ 1 - 0.0000000000000000000039 ] Even if we put Rb /Rc as small as 5, V c' comes very close to V s Vc' = Vs ( 1 - 0.0067) = 0.993 Vs

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4.4.4.2 The Effect of Diodes on Multivibrator Theory

In the The Simple Multivibrators Theory, the assumption was made that the base-emitter diode of each transistor had a substantial constant forward volt drop V(f) when conducting. If similarly we assume here that there is a constant voltage drop V df across the diodes added in series with each emitter, the theory previously presented can still be applied, with these adjustments: 1. Vs is replaced by V s - Vdf 2. Other voltages are unchanged if measured from the emitters. Alternatively, raise each of the values for Vb, Vc by Vdf. 4.4.4.3 Variable Mark/Space Ratios The multivibrator need not have equal time constants associated with its two transistors. If the time constants differ then correspondingly different times will be needed to discharge the two capacitors. It was shown in the previous Theory that the time to discharge one of the capacitors is: t = CRb ln [ (2Vs - Vf) / (Vs - Vf) ] which may for the present be written as t = k CR b, where k = ln [ (2Vs - Vf) / (Vs - Vf) ] In the Multivibrator with variable mark/space ratio Practical a control is introduced which enables the value of Rb to be increased in one half of the multivibrator and at the same time decreased in the other half. The period of the multivibrator is equal to the sum of the times for which one transistor is cut off and the other is cut off. It is therefore equal to: k C1Rb1 + k C2Rb2. 4-34

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In our particular case the values of C 1 and C2 are equal, say to C, so the period is T = kC (Rb1 + Rb2).

However, (Rb1 + Rb2) = R1 + R2 + Rvar, which is constant. So the period T is constant. The frequency, 1/T, is therefore also constant. 4.4.5 Practical 1: Basic Circuit

Consider a pair of amplifiers with feedback around them. If the product of their gains is positive and greater than one, any small disturbance in the circuit will be amplified, returning to the starting point greater than before. The process continues until something stops further amplification from occurring. If there are no delays in the loop caused by filters or tuned circuits this process will happen very quickly. What stops the process is usually limiting which takes place in one or both amplifiers.

Each amplifier stage in our simple multivibrator is an a.c. amplifier as shown above.

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If the signal source voltage goes positive, current flows through C and the emitter-base diode, charging C. This current dies away when the capacitor is charged, but there is also current flowing through R, causing the transistor to conduct heavily. The result is that the collector suddenly falls to a low potential.

Now consider what is happening in the second stage of our two-stage amplifier. The sudden drop in potential at the first stage's collector causes the base of the second stage transistor to go negative, cutting off the transistor. This transistor's collector therefore rises to a potential near the supply rail. Note that this is just the condition needed to operate the first stage in the way described previously. The transistor does not remain cut off indefinitely, however. The resistor R provides a current which alters the charge on the capacitor, raising the potential of the transistor base. Eventually the base goes positive with respect to the emitter and the transistor starts to conduct. This transistor now enters the state previously described for the first one, and it conducts heavily while the other one cuts off. The process repeats indefinitely, switching each transistor on in turn, and the circuit oscillates.

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4.4.5.1 Procedure The oscilloscope should be used to examine the waveforms at all four available test points. (They are NOT entirely symmetrical.)

The frequency meter is connected to the collector of the second stage's transistor at test point 11 . Make a note of the frequency value.

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4.4.6 Questions

1. Which part of which waveform(s) do you think is important in determining the frequency of oscillation?

2. What difference do you notice between the waveforms at the two collectors (apart from the phase difference). Can you think what causes it?

3. Does this difference have any significant effect on the frequency of oscillation?

4. Theoretically, the time for one half-period of oscillation is approximately 0.69 CR b . In this multivibrator C is 1nF and R b is 68k.Calculate the frequency of oscillation and compare it with the measured value.

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4.4.7 Practical 2: Effect of Variable Supply

In this Practical two new features are introduced: 1. The supply voltage is variable by a control on the board. 2. A pair of diodes is introduced. (The diodes were present in the first Practical of this assignment but not shown for simplicity.) They can be switched in and out of circuit. When the collector of one transistor drops from +V s to 0 V, the base of the other transistor is taken to about -Vs. The base-emitter diode is not designed to withstand more than about 5V reverse bias, so it partially breaks down in a similar way to a Zener diode. The purpose of the diodes is to prevent reverse current from flowing. This both protects the transistors and makes the behaviour of the multivibrator more predictable. This time the circuit is nominally symmetrical. This gives the output spectrum a particular character which should be noted. 4.4.7.1 Procedure In this practical the collector resistors are equal. Diodes in series with each emitter prevent breakdown of the base-emitter diodes of the transistors. The supply voltage can be varied. Start with it turned up to maximum.

Use the Diode button to switch the diodes in and out of the circuit. Observe that the frequency changes. Note any frequency changes caused by varying the supply voltage . Use the large-sized Spectrum Analyser to see which are the important harmonics of the collector waveforms.

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4.4.7.2 Questions 1. Which has the bigger effect on the frequency of oscillation - varying the supply voltage between 5 V and 10 V, or switching the diodes in and out of the emitter connections?

2. Is the behaviour of the base-emitter diodes like a Zener diode, with a sharp 'knee' in the characteristic, or more a general tendency to leak reverse current?

3. Which are the important harmonics in the spectrum of the collector waveforms?

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4.4.8 Practical 3: Mark/space Ratio Control

In this Practical a new feature is a variable mark/space ratio control. This is a potentiometer configured to appear as two resistors in series with the timing resistors R1, R2. The supply voltage is connected to the slider on the potentiometer so that while Rb1 is effectively increased, Rb2 is effectively decreased. As a result of this, altering the control makes one capacitor charge faster and the other slower. The charge time needed for each of the capacitors is changed, and the mark/space ratio is altered. It is shown in the Theory that changing the mark/space ratio in this way will not alter the period of the output signal. The term 'mark/space ratio' is a throwback to the early days of telegraphy, when a line was said to be in the mark condition when a Morse key was pressed and the space condition when the key was released. Mark and space are just the two possible states of a two-state waveform.

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4.4.8.1 Procedure The mark/space ratio can be adjusted by a control provided on the board. Do this while observing each of the monitor points in turn. Pay particular attention to the slopes of the waveforms at the transistor bases.

Examine the frequency as the mark/space ratio is varied. Observe the changes in the pattern of the frequency harmonics as the mark/space ratio is varied, using the large Spectrum Analyser.

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4.4.8.2 Questions 1. Explain with the aid of sketch waveforms what happens when the mark/space ratio control is adjusted, and why.

2. Does the frequency vary when the mark/space ratio is adjusted?

3. Would the frequency vary with mark/space ratio if the two capacitors were of different capacitance? Explain your answer briefly. (Hint - Calculate time constants for the two cases in which the variable control is set at the end of its travel.)

4. What is special about the spectrum of the collector waveform when the mark/space ratio is 1:1?

5. If the collector waveform were a true square wave, the fundamental, 3rd and 5th harmonics should be in the ratio 1:(1/3):(1/5). Compare these with the measured ratios.

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Notes

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5

Chapter 5 Assignments using the Tuned Circuits and Filters Workboard

Assignments using the Tuned Circuits and Filters Workboard

5.1

Audio Low-Pass Filters Assignment

5.1.1 Objectives


The frequency and transient response of a passive low-pass filter,

•

A filter's frequency response using a noise source and spectrum analyser,

•

The frequency and transient response of an active filter,

•

Matching considerations for passive and active filters.

5.1.2 Practicals


Practical 1: Passive Low-Pass Filter

•

Practical 2: Passive Low-Pass Filter, Swept Frequency

•

Practical 3: Active Low-Pass Filter

•

Practical 4: Active Low-Pass Filter, Swept Frequency


Tuned Circuits and Filters Workboard 53-120 which comprises the following blocks: •

Signal Generator,

•

Low Pass Filter,

•

Band Pass Filter,

•

Amplifier with AGC.

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5.1.4 Theory

5.1.4.1 Filters A filter is a circuit which passes signals of some frequencies freely, but attenuates at other frequencies. This assignment is about low-pass filters. 5.1.4.1.1 Low-pass Filters

An ideal low-pass filter passes signals of all frequencies below a certain value and does not pass signals of frequencies above that value. The frequency at which the signal starts to be attenuated is called the cut-off frequency. Since the cut-off is not quite sharp, the exact value must be based on some measure of how much the signal is reduced. This is typically 3 dB below the level in the pass band. To pass some frequencies and reject others, a filter must contain components that have impedances that vary with frequency. This means capacitors and/or inductors must be used, often together with resistors whose impedance (resistance) does not vary with frequency. If the filter does not have any active devices (transistors, IC's, etc.) in its circuit, it is called a Passive Filter. Filters can be classified as passive or active. The assignment includes one filter of each class. 'Active' implies that the circuit contains amplifying devices such as transistors or op-amps. 'Passive' means having no such devices. 5.1.4.1.2 Passive Filters Passive filters consist of networks of inductors and capacitors. They have the advantage of working without power supplies. Their disadvantages are mainly apparent at low frequencies, for which both inductors and capacitors need to be large and expensive. Inductors also tend to have appreciable resistance, making the action of the filter imperfect.

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5.1.4.1.3 Active Filters Passive filters consist of networks of inductors and capacitors. They have the advantage of working without power supplies. Their disadvantages are mainly apparent at low frequencies, for which both inductors and capacitors need to be large and expensive. Inductors also tend to have appreciable resistance, making the action of the filter imperfect. 5.1.4.2 Transmission Line Characteristic Impedance The concept of characteristic impedance first arose in the theory of transmission lines for telegraph purposes (they are now used for carrying RF and other signals).

Imagine an infinite transmission line. In practice this means a line so long that a signal takes so long to get to the far end that it virtually never comes back. Now suppose that we wish to send a signal of voltage v along the line. (v can be a function of time, like Vp sin wt). If that signal is applied at one end of the line, a current i will flow into the line. The characteristic impedance, often denoted by Z 0, is then defined as: Z0 = v / i.

Now let us look at the voltage and current at a point P, some way down the line. We find them to be v1 and i1 respectively. The line remaining is still infinite, so the ratio of voltage to current must again be Z 0, and we can see that Z 0 is also equal to v 1 / i1.

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This means that we can cut the line at P and connect an impedance Z 0 at point P without altering the behaviour of the first section of line at all. The line is now of finite length and has been terminated with Z 0. It is said to be matched to Z0, the Characteristic Impedance of the line. line. The important point to be made about matching the load to the characteristic impedance of the line is this : A properly matched load absorbs all the power which arrives at it, whereas a mismatched load causes 'spare' power to be reflected back towards the signal source. source. 5.1.4.2.1 Filter Matching

If a filter is inserted into a line, ideally it should have an impedance equal to the line's characteristic impedance at both pairs of terminals. If this condition is realised then the filter causes no reflections. In practice a resistive load is never a perfect match for a filter at all frequencies, so some reflection always takes place. 5.1.4.3 Passive Low-Pass Filters A filter is a circuit which passes signals in a selected range of frequencies freely (the passband'), but attenuates at all other frequencies (the 'stopband 'passband'), ' stopband'). ').

An ideal Low-Pass Filter lets through signals lower than a certain frequency without attenuating them, but does not let through any signal with a frequency above that value. The frequency above which no signals are passed is called the cut-off frequency. frequency.

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A real filter cannot have a response that changes instantly from zero attenuation to infinite attenuation at a certain frequency. In a practical low pass filter the response slopes off at higher frequencies, and the cut-off frequency is defined as being that at which the signal has been attenuated by 3 dB. To make the filter pass some frequencies and reject others, it must contain components whose impedances vary with frequency; ie, capacitors and inductors. If the filter does not have any active, amplifying devices in its circuit it is called a Passive Filter. Resistors cannot be used in a passive filter because they would introduce attenuation.

The theory of filters is too complex to explain here. But in simple terms, an elementary low-pass filter has inductance in series with the signal path and capacitance in parallel with it. As the diagram shows, a filter section can be arranged in a pi or a T form. Either form can be constructed from two identical half-sections. rad/s, for any of these filters is given by The cut-off frequency, ω rad/s, ω2LC = 1. 1.

The 'design 'design impedance', impedance', which approximates to the characteristic impedance at low frequencies, is Z where Z2 = L/C

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5.1.4.4 Active Low-Pass Filters

An active low-pass filter will often have several stages like this having the same circuit, though usually not identical component values. The gain G of the amplifier is normally set by a feedback network (not shown for simplicity). In the mathematics, 1/G occurs many times, so k is defined to be 1/G. The input to the amplifier carries a voltage k times the output voltage. So, considering the impedance of the right-hand capacitor: i = k jωC ... (1) Ohm's Law for the right-hand resistor gives: iR = ex - k eo ... (2) and the sum of the currents in the feedback capacitor and in the left-hand resistor is equal to i, so: i = (ei - ex)/R + (eo - ex) jωC ... (3) Combining equations (1) and (2), we get: k eo jωCR = ex - k eo ex = k eo (1 + jωCR) ... (4) Equating i from equations (1) and (3), then multiplying by R: keo jωCR = ei - ex + (eo - ex) jωCR keo jωCR = ei + eo jωCR - ex (1+ jωCR) Substituting for ex from (4): keo jωCR = ei + eo jωCR - keo (1+ jωCR)2

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which by routine manipulation, and replacing 1/k by G, becomes: eo [ - (ω2C2R2 + (3 - G) jωCR + 1] = G e i If ω = 0, and so at low frequencies, the gain is G. At high frequencies the gain approximates to 1/ ω2C2R2; ie, falling with increasing frequency. At frequencies given by ωCR approximately 1, the behaviour of the filter is very dependent on the j term, with G as a factor. This controls the damping of the response. A large value of G produces a large response in the ωCR=1 region. A small value produces a gradually drooping response. Usually several stages are used, with different damping factors, to give a response which is maintained well up to the cut-off frequency and then falls sharply. 5.1.4.4.1 Setting the Amplifier Gain G To keep the diagrams simple, the gain of the amplifier has been supposed to be G. In practice it is usual to employ operational amplifiers with very high gain, and with negative feedback which sets the gain.

Since the gain of the amplifier is very high, for any value of output voltage, the difference in potential between the two input terminals is negligibly small. So ein = eout R2 / (R1 + R2) The gain with this feedback is therefore G = (R1 + R2) / R2 = 1 + (R1 / R2) 5.1.4.5 Swept Frequency and Plotting In testing communications equipment, we very often need to discover the frequency response of some item. To do this an oscillator is used to provide the test signal, the frequency of which can be set to each of a range of different values in turn. For each frequency the output signal is measured. Usually the results of this are presented by plotting a graph of the response against frequency.

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5.1.4.5.1 Swept Frequency Facility Much tedious work can be saved if the oscillator frequency is varied automatically and the output response is simultaneously measured and plotted. A facility of this kind is provided in this equipment. A switch enables the oscillator to enter swept mode where the frequency is modulated slowly by a sweep oscillator. The end frequencies for the sweep can be set by controls on the board. 5.1.4.5.2 Using the Swept Frequency Facility In the Practical you will be told, or you can find out experimentally, sensible values for the start and end of the sweep. To set up the sweep oscillator with these values, set the Sweep switch to ON whilst viewing the oscilloscope. •

Press and hold down the button on the module marked Min. This will force the sweep oscillator to stay at its minimum value. The bottom frequency of the swept range can now be set using the usual Frequency control whilst observing the value of the frequency meter. Release the Min button when this has been done.

To set the top end of the swept frequency range: •

Press and hold down the button on the module marked Max. This will force the sweep oscillator to stay at its maximum value. The top frequency of the swept range can now be adjusted as an offset from the minimum frequency that has just been set. To do this, use the Range control on the module whilst observing the frequency meter. When this has been set, release the Max button.

Due to the fact that the maximum swept frequency is set as an offset from the minimum swept frequency, it is important to set the minimum and maximum in that order. When the range has been set, you can start the Plotter. The Sweep function is only enabled during those Practicals for which it required. For other Practicals the Sweep switch should always be set to OFF. Failure to do this will cause problems, usually failure of the oscilloscope to run or to synchronise. 5.1.4.5.3 How the Sweep Facility Works It is not necessary to read this section to be able to use the Plotter. When you select Plotter, the direct action of the Frequency control of the left-hand, signal source, portion of the module is disabled. The frequency is instead modulated over a preset range by a further, low-frequency, oscillator. 5-8

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The plotter takes readings and plots points for values of frequency and peak amplitude. When taking a reading, the signal frequency is frozen and the peak amplitude of the output signal is measured and stored to give the vertical deflection on the virtual plotter. At the same time the modulating oscillator's output is also stored to give the horizontal deflection. The plotted points will be scattered at different frequencies, so a graph will gradually build up as more and more points are plotted. 5.1.5 Practical 1: Passive Low-Pass Filter

This Practical uses a simple passive low-pass filter.

It comprises inductors and capacitors arranged as shown. It is designed to operate between a 600-ohm source and 600-ohm load. The impedance presented by the filter is not entirely resistive, so mismatching occurs near the cut-off frequency. There are two stages to the Practical. In the first, you will examine the filter's response to a sine wave at various frequencies. As you will see, this is quite markedly altered by whether the correct load is present or not.

The second stage is to examine the response to sudden changes in the input signal. A single such change is called a step function. This function is approximated by a square wave produced by the audio oscillator. 53-001S

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You will find that the effect of removing the high frequencies from the input signal is to cause rounding or smoothing of the output waveform. Unequal transmission of frequencies within the pass band gives rise to a distorted step function response. Since altering the matching also alters the frequency response, you will be able to see that a poor frequency response also worsens the transient response. 5.1.5.1 Procedure Switch Sweep OFF . Set the audio level to give a sine wave output of maximum deflection on the oscilloscope. Observe that this signal stays constant in amplitude as its frequency is changed. Examine and contrast this signal with that at the filter output, monitor point 13 .

Use the Load Button to remove the load from the filter. Repeat your previous observations. You should find quite noticeable changes. Switch the audio oscillator to give a square wave , and repeat the steps above.

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5.1.5.2 Questions 1. What happens to the output of the filter as the signal frequency rises with the load resister in circuit?

2. At what frequency does the output fall by 3 dB from its value at low frequencies?

3. If the oscillator's output is of constant amplitude, why does the input to the filter vary with frequency?

4. With square wave input, what is the effect of removing the higher frequencies?

5. With square wave input, what is the effect of having no load resistor on the shape of the waveforms?

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5.1.6 Practical 2: Passive Low-Pass Filter, Swept Frequency

Read the Swept frequency Background before continuing for information on how to use the sweep oscillator and plotter. This Practical is similar to the Passive Low-Pass Filter Practical, however the frequency response will be displayed as a graph by an automatic process. 5.1.6.1 Procedure Set the audio level for maximum sine wave output. Set the Sweep switch to ON . Set the minimum frequency to the lowest at which the frequency meter and oscilloscope still work. Set the maximum to the highest possible value.

Use the Load Button to remove the load from the filter. Repeat your previous observations. You should find quite noticeable changes.

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5.1.6.2 Questions 1. Describe the general shape of the frequency response of the complete filter.

2. Describe how the frequency response changes when the load is disconnected.

3. The effect of disconnecting the load can be described in terms of the signal being reflected back into the filter from the mismatched output terminals. Where else is matching important in this circuit?

4. If the filter were perfectly matched to the source and load resistances at the ends, the input to the filter would be half the source voltage. At what frequencies is this least true?

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5.1.7 Practical 3: Active Low-Pass Filter

This Practical uses an active low-pass filter. It comprises two filter stages, each of which has a circuit as shown although the second stage has different component values. You can see that the input impedance of this filter is high compared with the 600 ohms which is typical of a passive filter. The output impedance is low, because the output is generated by an operational amplifier. It is therefore far easier to match the filter to associated circuitry than a passive filter. As with the passive filter, tests will be performed both with a sinusoidal signal to check the frequency response, and with a slower square wave to check the transient response. In a filter of this kind the characteristics are almost entirely determined by the feedback networks around the operational amplifiers. The Practical allows you to change the value of one resistance in such a network, so that you can see the changes to the frequency and transient responses which result. Making that change should enable you to see once more that frequency and transient responses are related.

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5.1.7.1 Procedure Switch Sweep OFF . Set the audio level for maximum sine-wave output.

Observe that this signal stays constant in amplitude as its frequency is changed. Compare this with the signal at the filter input, monitor point 4 . Observe and contrast the signal at the filter output, monitor point 14 .

Set the oscillator to give a low-frequency square wave . Observe the output waveform. Use the Stage Button to adjust the second stage response. Then observe the effect on the output waveform. You may need to reduce the amplitude of the square wave to see the full effect.

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5.1.7.2 Question 1. How important is the matching of the signal source impedance to the active filter? Compare this with the situation when using the passive filter.

2. How does the frequency response of the active filter compare with the response of the passive one?

3. The adjustment to the second stage frequency response gave it a response more like the first stage. Why was this not a good thing?

4.

What is the effect of a sharply-peaked frequency response on the transient response of the filter?

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5.1.8 Practical 4: Active Low-Pass Filter, Swept Frequency

Read the Swept frequency Background before continuing for information on how to use the sweep oscillator and plotter. This Practical is similar to the Active Low-Pass Filter Practical. Indeed the principal object of this Practical is to show that the behaviour of any filter made of inductors and capacitors can be reproduced by a filter made of resistors and capacitors, together with active devices (amplifiers). Active filters are especially useful at low frequencies, because inductors having a good ratio of reactance to resistance at low frequencies are large, heavy and expensive. Conversely, passive filters are preferred at frequencies above about 50 kHz, because the phase shifts in amplifiers become difficult to control and reactors for high frequencies are comparatively inexpensive. 5.1.8.1 Procedure Set the audio level for maximum sine-wave output. Set the Sweep switch to ON . Adjust the minimum frequency to the lowest value at which the oscilloscope and frequency meter work. Set the sweep range to maximum. Then select the Plotter to plot responses measured at points 14 and 12 .

Use the Stage Button to alter the second stage of the filter. This will cause a peaky response, so you will need to reduce the audio oscillator signal by a factor of about 3, to avoid overloading the instruments. Repeat the previous observations.

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5.1.8.2 Questions 1. Describe the general shape of the frequency response of the complete filter.

2. The filter has two stages. Compare the response of the first stage only with that of the complete filter.

3. What kind of frequency response would you expect the second stage to have?

4. When the second stage has its component values altered, what happens?

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5.2


RF Selectivity Assignment

5.2.1 Objectives


The frequency response of a tuned circuit,

•

The transient response of a tuned circuit,

•

The effect of damping on the frequency response of a tuned circuit,

•

The frequency response of a crystal filter,

•

The effect of neutralisation on a crystal filter.

5.2.2 Practicals


Practical 1: L/C Tuned Circuit

•

Practical 2: L/C Tuned Circuit, Swept Frequency

•

Practical 3: L/C Tuned Circuit, Transient Response

•

Practical 4: Crystal Filter

•

Practical 5: Crystal Filter, Swept Frequency



Signal Generator,

•

Low Pass Filter,

•

Band Pass Filter,

•

Amplifier with AGC.

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5.2.4 Theory

5.2.4.1 Tuned Circuit The theory of the tuned circuit is most easily derived either for inductance L, resistance r and capacitance C all in series (where r is a usually a small resistance representing circuit losses), or for L, C and a usually high resistance R, all in parallel. We shall look at the parallel case, as it more nearly fits the Practicals.

For this circuit the parallel impedance Z is given by: 1 / Z = (1/jωL) + jωC + (1/R) If we write ω02LC = 1 then: 1 / Z = -j(ω02LC / jωL) + jω0C(ω / ω0) + (1/R) = jω0C y + (1/R) where y = (ω / ω0) - (ω0 / ω) and, if w is not too far from ω0, is a measure of the fractional mistuning, since: y = (ω2 - ω02) / ω ω0 = (ω - ω0) (1/ ω0 + 1/ ω) = 2 dω / ω0 approximately. The expression: 1/Z = jω0C y + (1/R) is clearly a minimum of 1/R when y = 0, so Z has its maximum value R when ω = ω0. For large mistunings, the term jω0C y becomes large, while (1/R) stays constant. So R does not greatly affect the response for large mistunings. But for zero mistuning, the larger R, the larger is Z. The selectivity is therefore highest for large R, implying small resistive losses. It is generally true that low losses give good selectivity, associated with a high impedance at resonance. This is true whether the losses are due to series resistance, parallel conductance or other effects. 5-20

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5.2.4.2 Crystal Filter A crystal for use in a filter is usually of quartz, a substance which has two useful properties: 1. It has very little mechanical damping - it 'rings' very easily, like a bar of steel rather than a bar of lead. 2. It shows the piezo-electric effect. That is, when it is mechanically stressed it produces an electrical field and when it is electrically stressed it changes shape mechanically. The crystal's mechanical 'ringing' (property number one) produces a resonant electrical effect (due to property number two). This is made available at terminals connected to electrodes on the crystal faces. A Crystal Filter behaves in the say way as a Tuned Filter, but has a much higher Q factor of the order of 10,000, compared with a Q of about 100 for a Tuned Filter. 5.2.4.2.1 Equivalent Circuit of a Crystal When the crystal vibrates or 'rings', it appears that there is an electrical circuit connected between its terminals that is ringing. This is due to the electrical effects of the crystal's vibration. The circuit is the 'equivalent circuit' of the crystal.

Because the equivalent circuit can resonate or 'ring', it must have inductance and capacitance in it, shown as L and C in the diagram. There is also some inevitable capacitance between the terminals, shown as C stray.

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The component values in the above diagram are for the equivalent circuit of a typical 1 MHz crystal. Compare C with the typical capacitance (a few hundred pF) of a tuning capacitor operating at 1 MHz. L and C have high values of reactance, in the region of 63 megohms, so the slightest departure from resonant frequency causes the crystal's impedance to rise very rapidly. Inevitably there is some energy loss, or damping. This has an effect similar to that of one or two thousand ohms of resistance placed in series with L. 5.2.4.2.2 The Resonances

The important resonance is the series one, when: jωL + (1/jωC) = 0 or: ω2LC = 1

There is a parallel resonance also, which occurs when the: ω2LC' = 1

where C' is the value of C and C stray in series. With the values in the diagram, C' will be about 0.1% different from C, so that the parallel and series resonant frequencies will differ by about 0.05%. 5.2.4.2.3 Neutralisation The stray capacitance has two effects. First it is responsible for the parallel resonance already mentioned, and secondly it causes a degradation of the response on the lowfrequency side of the series resonance peak. Neutralisation aims to remove or modify these effects.

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The diagram shows a neutralising circuit which is slightly easier to understand than the one in the Practical, although the principle is the same. Using Thevenin's Theorem, the circuit may be replaced by an equivalent circuit consisting of a voltage source, providing the open-circuit voltage at the two terminals, and an impedance equal to the impedance of the original circuit when all voltage sources are removed from it.

If the transformer provides equal and opposite voltages applied to C s and Cn, and if these capacitances are equal, they form a balanced bridge, providing no contribution to the output voltage. The voltage is therefore simply the original signal voltage. The impedance now is that of the crystal in parallel with 2C s. The circuit is therefore equivalent to the one above on the right.

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5.2.5 Practical 1: L/C Tuned Circuit

This Practical examines the frequency response of a tuned circuit. A sinusoidal oscillator is used to generate a single-frequency signal. You will be able to change the value of this frequency and observe the effect of the tuned circuit on the output. Initially the oscilloscope is connected to the signal source. Use it to check that the oscillator provides a constant amplitude as the frequency is varied. The next stage is to connect the oscilloscope to the output terminal and observe how the output signal amplitude varies with frequency. What you should expect to see is a large response when the oscillator frequency and the resonant frequency of the tuned circuit match one another. When the frequencies do not match the response will be much smaller. Both the oscillator frequency and the tuned circuit's resonant frequency can be varied. As you will see, the effect of altering either of the frequency controls is very similar; it is mainly the amount of mistuning which affects how much the signal is reduced. 5.2.5.1 Procedure Switch Sweep OFF .

Check that the RF oscillator delivers a signal of roughly constant amplitude, with variable frequency. Compare the way the signal at monitor point 10 is affected by changes in frequency. Observe also how the Capacitance control of the tuned circuit alters the frequency at which peak output is obtained.

The selectivity of the circuit is altered by applying damping. Click on the Damping Button to switch in the damping load resistor.

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5.2.5.2 Questions 1. Describe how the response of the tuned circuit varies with frequency. The output peaks at one frequency.

2. How is this response altered when the 'Frequency' control in the tuned circuit is varied? The peak frequency varies.

3. When damping is introduced the resonant frequency also shifts much more than theory would predict. Can you account for this? (Hint: The component which alters the tuning is NOT shown on the circuit diagram.) The stray capacity of the damping circuit.

4. When damping is introduced is there more reduction of the peak response, or of the mistuned response? Of the peak response.

5. Is the selectivity increased or decreased by damping? Decreased.

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5.2.6 Practical 2: L/C Tuned Circuit, Swept Frequency

In this Practical the process of observing the frequency response will be automated. Instead of controlling the signal frequency manually, you will only need to set up the resonant frequency of the tuned circuit and the range of frequencies to be swept. If you have not used the frequency sweeper and plotter before, you should read the Swept Frequency Background. You may need to take some care over the RF level setting. If the RF signal is too small, it will not drive the frequency meter properly. If it is too large, the oscilloscope and plotter will be over-driven. 5.2.6.1 Procedure Switch the Sweep OFF .

Set the RF oscillator to approximately 455 kHz. Tune the tuned circuit for maximum response, reducing the RF signal to avoid clipping on the oscilloscope. Switch the Sweep ON .

Set the minimum sweep frequency to 440 kHz and the maximum to 470 kHz, and finally select the Plotter.

Use the large plotter to see the frequency response in detail. To observe how the response is changed when damping is applied, switch in the damping load resistor by clicking on the Damping Button .

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5.2.6.2 Questions 1. Using the large plotter, what do you estimate to be the bandwidth of the circuit at a level 3 dB down from that at resonance?

2. What is the 3 dB bandwidth with added damping?

3. What values of Q apply, with and without the added damping?

4.

If C is the circuit capacitance before damping is added, what (in terms of C) is the stray capacitance associated with the damping resistor and its leads?

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5.2.7 Practical 3: L/C Tuned Circuit, Transient Response

In this Practical the driving signal will be set to a low frequency square wave. This will give the tuned circuit a periodic kick which should set it oscillating. This behaviour is called free oscillation. (The use of a sine-wave to drive the circuit at the resonant frequency is called forced oscillation ). Free oscillation is like the ringing of a bell after it has been struck by a hammer. There are two important characteristics of a free oscillation: 1. The frequency. 2. The rate at which it decays, or dies away. The frequency is exclusively determined by the resonant circuit; in this case by the inductance L and the capacitance C (including any stray capacitances in parallel with it). The rate of decay is expressed in various ways, such as the logarithmic decrement or the damping factor. It depends on the fraction of the energy stored in the reactances which is dissipated per cycle of oscillation. Anything in the circuit which causes energy loss contributes to the damping or decay of oscillations. In the Practical a resistor is provided which you can switch into circuit to see this effect.

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5.2.7.1 Procedure Switch the Sweep OFF . Switch the Audio Oscillator to square wave . Turn the main Frequency control fully counter-clockwise. Observe the step signal which is used to drive the tuned circuit, then look at monitor point 10 to see the tuned circuit 'ringing'. (The large oscilloscope will give a better display).

Increase the frequency of the square wave until the transient starts to be affected by the dying ripples of the previous one. Use the large oscilloscope to see in detail how the amplitude is affected by the phase of the free oscillation at the time the step arrives. Observe how the response is changed when damping is applied by switching in the damping load resistor by clicking on the Damping Button .

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5.2.7.2 Questions 1. How does the frequency of the 'ringing' or free oscillation compare with the frequency of peak response in Practical 2 with the same control settings?

2. What factors affect the frequency of the oscillation?

3. What is the principal effect of damping on the oscillation?

4. In a frequency multiplier circuit the signal which drives the resonant circuit is usually a series of sharp, narrow pulses. Can you suggest why this is more efficient than square wave drive?

5. In the frequency multiplier, why would the square wave be especially unsatisfactory when the multiplying factor is even?

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5.2.8 Practical 4: Crystal Filter

Crystal filters are used when a very narrow band of frequencies are required to be filtered; ie, for high selectivity. You will therefore find that great care is needed in tuning the signal source if you are to see the response of the crystal. 5.2.8.1 The Notch in the Response Stray capacitance in the circuit gives rise to a secondary resonance. You can observe this by looking at the output with the oscilloscope. When the frequency reaches this secondary resonance, the 'notch' will be seen as the complete disappearance of the output signal. 5.2.8.2 Neutralising Neutralising is an arrangement for cancelling this parallel resonance by using a similar small capacitance to provide a signal in anti-phase. This is switched in using the Conditions box. When it has been switched in, you can adjust the neutralising capacitor to see how the parallel resonant frequency is shifted. It can even shift to the other (lower) side of the series resonance. 5.2.8.3 Procedure Switch Sweep OFF . Check that the RF input signal is constant with frequency, then select monitor point 8 . Set the oscillator's fine frequency control to its mid-position and adjust the oscillator frequency to 455 kHz. Use the fine frequency control to explore the narrow band of frequencies around 455 kHz. Note particularly any frequency at which the output signal becomes zero, as well as the peak response frequency.

Click on the Neutralising Button to switch in the neutralising capacitor. Adjust its capacitance with the crystal capacitance control, noting the effects on the sharpness and the symmetry of the response peak. 53-001S

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5.2.8.4 Questions 1. Compare the selectivity of the crystal filter with that of the LC tuned circuit.

2. On which side of the peak response is the notch (zero response)?

3. With maximum neutralising capacitance switched in, what happens to the notch position?

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5.2.9 Practical 5: Crystal Filter, Swept Frequency

This Practical is the same as the Crystal Filter Practical, with the addition that you can plot the frequency response using the Swept Frequency plotter. Use the plotter to examine carefully the shape of the response curve with the neutralising capacitor in circuit and set: 1. To minimum capacitance. 2. To maximum capacitance. 3. To provide a symmetrical response curve. 5.2.9.1 Procedure Switch Sweep ON . Adjust the minimum frequency of the sweep to 450 kHz and the maximum frequency to 460 kHz. Select the Plotter. Note the asymmetric shape of the response curve.

Click on the Neutralising Button to switch in the neutralising capacitor. Adjust its capacitance using the crystal capacitance control, noting the effects on the sharpness and the symmetry of the response peak.

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5.2.9.2 Questions 1. Would a crystal filter be useful in a receiver for high-fidelity music reception? Explain your answer.

2. In what circumstances would you find it useful to be able to vary the tuning of the notch (ie, to alter the frequency of the parallel resonance)?

3.

If a receiver with a crystal filter is trying to receive a signal in the midst of a general spread of background noise, how should the neutralisation of the filter be adjusted? Explain why.

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5.3


RF Band-Pass Filters Assignment

5.3.1 Objectives


The effect of coupling on the combined response of two tuned circuits,

•

The response of a typical bandpass ceramic filter.

5.3.2 Practicals


Practical 1: Coupled L/C Circuits

•

Practical 2: Coupled L/C Circuits, Swept Frequency

•

Practical 3: Ceramic Filter

•

Practical 4: Ceramic Filter, Swept Frequency



Signal Generator,

•

Low Pass Filter,

•

Band Pass Filter,

•

Amplifier with AGC.

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5.3.4 Theory

5.3.4.1 Bandpass RF Circuits The RF Selectivity Assignment dealt with single resonators - an L/C circuit and a piezoelectric device, the crystal. Each of these tended to select a single frequency for transmission in preference to all others. If extreme selectivity is required - if one wants effectively to exclude all frequencies except one - then several L/C circuits may be used. Often several amplifier stages, each having a tuned circuit, are used together to increase the selectivity. What is often wanted is to transmit a band of frequencies, excluding signals outside the wanted band. In these cases a bandpass filter is required. This assignment is about two forms of RF bandpass filter. 5.3.4.1.1 Coupled LC Tuned Circuits

Several tuned circuits may be tuned to the same frequency and coupled together reactively in various ways. Some examples are shown above. If the coupling is loose (ie, if M or C1 are small, or if C2 is large), then the frequency response is very similar to that of uncoupled circuits such as those in separate amplifier stages. If the coupling is increased, however, more energy can flow back and forth between the two circuits. Each circuit then alters the resonance of the other. The result of this is that the frequency response 'spreads'. Initially, as the coupling increases the resonant peak simply broadens. With tighter coupling the peak splits into two separate peaks, either side of the uncoupled resonant frequency. 5.3.4.1.2 Ceramic Filters To get a wide pass-band with a sharp cut-off of frequencies outside the pass-band, a large number of reactive elements is necessary. Complex filters made of inductors and capacitors have limited performance because of losses in the inductors. Increasingly use is made of filters based on mechanical resonances, together with the piezo-electric effect.

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The quartz crystal is a single-resonance example. But for multiple coupled resonances the ceramic resonator is used. A plate of ceramic material has electrodes applied to it so that, like the quartz crystal, its mechanical behaviour can interact with an electric circuit. With a ceramic resonator quite complex resonances and couplings between resonances, can be set up in the ceramic plate by using electrodes of complex shapes. Fine tuning of these resonance effects is possible during manufacture by grinding the plate and/or electrodes. A cheap and highly effective bandpass filter can result. 5.3.4.2 Coupled Tuned Circuits

For this network of impedances: Zin = [Z1 (Z2 + Z3)] / (Z1 + Z2 + Z3) output = i Zin Z3 / (Z1 + Z3) = i (Z1 Z3) / (Z1 + 2 Z2 + Z3) so that if Z1 = Z3 = Z, say, then: output = i Z2 / 2(Z + Z2).

With two identical tuned circuits for Z1, Z3, we may write for each: Z = (jωL + R) / (1 - ω2 LC + jωCR) and if Z2 is a coupling capacitor, and multiplying by 2: 2 Z2 = 2 / jωCc.

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Substituting these values into the expression for the output, and using routine manipulation gives the result: output = numerator / denominator, where numerator = (jωL + R)2 jωCc and denominator = [1 - ω2LC + jωCR] [1 - ω2L(C + Cc) + jω (C + Cc)R]. The most obvious thing about this result is that the denominator has two factors, each similar to the denominator in the response of a single tuned circuit. What is more striking is that although the two tuned circuits are identical, the resonances are not - they are at different frequencies. This does not necessarily mean that there are two peaks in the frequency response curve. If the two resonant frequencies are close enough, the two peaks merge. The mathematics to show this is quite complicated, and is given in the Critical coupling Theory. 5.3.4.3 Critical Coupling We found in the Coupled tuned circuits Theory that for two coupled tuned circuits : output = numerator / denominator where numerator = (jωL + R)2 jωCc denominator = [1 - ω2LC' + jωC'R] [1 - ω2L(C' + Cc) + jω (C' + Cc)R] and the original C, discussed in the Coupled Tuned Circuit theory, has here been replaced by C'. It will be convenient to change the variables to make the expression simpler and more symmetrical. Let a new C in this critically coupled circuit be now set equal to C' - C c. This does not affect the numerator. The denominator can be written as denominator = [1 - ω2L(C - Cc) + jωR(C - Cc)] [1 - ω2L(C + Cc) + jωR(C + Cc)] Let us now define: ωo

... where ωo2LC = 1

y = (ωo / ω) - (ω / ωo) ... the fractional mistuning, which is approximately y = 2(ωo - ω) / ωo ... for small mistunings k = Cc / C ... coupling factor 5-38

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d = R / ωoL ... the reciprocal of Q, the quality factor Straightforward manipulation then leads to the relationship: den (ωo / ω)2 = [y + k(ω / ωo) + jd(1 - k)] [y - k(ω / ωo) + jd(1 + k)] which is of the form (A+B)(A-B) = A2-B2 , where: A = [ y + jd ] B = [ k(ω / ωo) - jdk ] = k [ (ω / ωo) - jd ] so it becomes: den (ωo/ω)2 = (y + jd)2 - k2[(ω/ωo) - jd]2 At this point some approximations have to be made to simplify this last result of: den (ωo / ω)2 = (y + jd)2 - k2[(ω / ωo) - jd]2 In any normal tuned circuit d will be small, and its square can be neglected in comparison with 1. We shall see shortly that k will be chosen to be of the same order of magnitude. If the mistuning is small, ω / ωo approximates to 1. If the mistuning is large then y2 becomes much larger than k 2, so again no great error will arise from setting ω / ω0 equal to 1. With these approximations, we have: den (ωo/ω)2 = (y + jd)2 - k2 ... (1) The principal factor affecting the response is the right-hand side of that equation: (y + jd)2 - k2 or y2 - k2 - d2 + 2jyd Its magnitude is the square root of: (y2 - k2 - d2)2 + 4y2d2 = y4 + k4 + d4 - 2k2y2 - 2d2y2 + 2k2d2 +4y2d2 = y4 + 2y2( d2 - k2 ) + k4 + d4 + 2k2d2 = y4 + 2y2( d2 - k2 ) + k4 + d4 + 2k2d2 To find maxima and minima, differentiate this expression with respect to y and equate to 0.

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Thus: 4y3 + 4y( d2 - k2 ) = 0. A first solution is y=0. There are two further solutions given by: y2 = k2 - d2 which are real if k > d; ie, if k > (1/Q). It can be shown by a further differentiation that the solution y=0 is a maximum if k < (1/Q). Otherwise it is a minimum and the further two solutions are maxima. The condition when k = 1/Q is called Critical Coupling. It will be useful next to find out how the maximum and minimum values vary as the coupling factor k is changed. 5.3.4.3.1 Maximum and Minimum Values Considering magnitudes, in the numerator (j ωL + R)2 jωCc the R will contribute very little, since (ωL/R)2 is approximately Q2, typically a very large number. Remembering that ωo2LC=1: InumeratorI = (ωL)2 ωkC / (ωo2LC) = ω3 L k / ω02 to a close approximation. Recalling equation (1): den (ωo / ω)2 = (y + jd)2 - k2 ... (1) then, for y2 = k2 - d2: den2 (ωo / ω)4 = (k2- d2)2 -2(k2 - d2)2 + (k2 + d2)2 which reduces to 4k2d2: So: den = 2 k d (ω/ωo)2

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The amplitude response at the peaks is therefore: num / den = ω3 L k / ωo2 / 2 k d (ω / ωo)2 = ω L / 2 d = Q ωL / 2 since Q = 1/k. Note that this expression is independent of k. for y=0, a similar process yields the result: ωLk / (d2 + k2)

Remembering that d is 1/Q, this can be expressed as: ωLQ (kQ) / [(1 + (kQ)2] ,

where kQ is the ratio of coupling to critical coupling. This expression reaches a maximum for kQ=1. The factor w appears in all these results. However, over the range say 440 to 470 kHz (as is relevant to the Practicals Capacitively coupled L/C circuits and Coupled L/C circuits with swept frequency, ω only varies by about 7%. 5.3.4.4 Ceramic Filters Ceramic filters in general are too complex to be analysed here. As explained in the Ceramic Filter Background, they are comprised of several coupled resonators. Their general theoretical background is therefore related: 1. via the piezo-electric effect to that of the crystal filter covered in the RF Selectivity Assignment. 2. as a system of coupled resonators, to the simpler case of two coupled L-C circuits covered in the Practicals Capacitively coupled L/C circuits and Coupled L/C circuits with swept frequency of this assignment. 3. as a system of reactors requiring a resistive termination for correct performance, to the treatment of the passive low-pass filter covered in the Audio Low-Pass Filters Assignment.

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ANALOGUE COMMUNICATIONS STUDENT’S WORKBOOK 5.3.5 Practical 1: Coupled L/C Circuits

The form of coupling used in this Practical is 'top-capacity coupling'. Controls are provided for you to adjust the tuning of one of the L/C circuits and the magnitude of the coupling capacitance. You can also adjust the frequency of the RF oscillator. Starting with the coupling capacitance at minimum, you will need to ensure that the frequencies of the two L/C circuits and the RF oscillator all match, in order to set the circuit up properly. Maximum signal output will then be achieved. Since only one of the L/C circuits is adjustable, you will have to adjust it and the RF oscillator together to match them to the other, fixed-tuned L/C circuit. IMPORTANT ! Once the circuit is correctly adjusted in this way, you must be careful not to alter the tuning of the L/C circuit until you have finished the Capacitively coupled L/C circuits Practical and started the Coupled L/C circuits with swept frequency Practical.

During the experiment you should: 1. Vary the RF oscillator frequency to explore the frequency response. 2. Do this again after altering the coupling between the two L/C circuits.

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5.3.5.1 Procedure Set the Sweep switch to OFF . Check that the RF oscillator delivers a signal of roughly constant amplitude with variable frequency. Set the coupling control fully counterclockwise and then change to monitor point 11. Adjust the oscillator's and the filter's frequency controls until a maximum response is attained at point 11. Note the frequency of this maximum.

DO NOT DISTURB THE FILTER'S FREQUENCY CONTROL AFTER THIS.

Set the coupling control fully counter-clockwise and explore the response of the circuit with variation of the oscillator frequency. Notice how two peak responses appear as the coupling is increased.

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5.3.5.2 Questions 1. Why were you asked to adjust both the RF oscillator and the first tuned circuit for maximum response?

2. Why was minimum coupling specified while doing this adjustment?

3. What was that frequency after you had done the adjustment?

4. With maximum coupling, what were the frequencies of the two peaks?

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5.3.6 Practical 2: Coupled L/C Circuits, Swept Frequency

This Practical should only be attempted after Practical: Capacitively coupled L/C circuits has been done. This is because it is important to get the two tuned circuits tuned to the same frequency. Practical: Capacitively coupled L/C circuits achieves this by getting : 1. The RF oscillator tuned to the fixed-tuned circuit. 2. The variably-tuned circuit tuned to the oscillator. Practical: Coupled L/C circuits with swept frequency will apply the swept-frequency technique to plot the response curves of the filter. With the two L/C circuits correctly tuned, the remaining variable is the coupling between them. This should initially be set to minimum. The resulting response will look like that of the single tuned circuit, but with steeper sides, showing more selectivity. As the coupling is increased, the peak will split into two peaks, giving a band of frequencies for which the signal is transmitted freely. The condition just before this split occurs is called 'critical coupling'. In practice the ideal response is flat-topped. A good approximation to this is got by using over-coupled and critically coupled circuit pairs together, so that the critically-coupled hump fills in the gap between two separated peaks.

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5.3.6.1 Procedure The frequency control in the bandpass filter should be set as it was in the Capacitively coupled L/C circuits Practical. If it is not, repeat practical before continuing. Set the coupling control fully counter-clockwise and switch the Sweep ON . Set the minimum sweep frequency to 445 kHz and the maximum to 465 kHz. Finally, select the Plotter.

You can superimpose responses for minimum and maximum coupling simply by changing the coupling quickly while the plotter is operating. If the two peaks are uneven, SLIGHT adjustment of the filter frequency control may help.

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5.3.6.2 Questions 1. Over-coupling pushes the sides of the response curve apart. Does it affect the slope of them much?

2. When the circuits are over-coupled, what is the principal effect of altering the resonant frequency of one of the tuned circuits?

3. From it can be found that the response at the centre frequency when over-coupled is down from the peak response by approximately 2 (kQ) / [1 + (kQ)2] where kQ is the ratio of coupling factor to critical coupling. What is the maximum value of kQ you can obtain with the equipment?

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5.3.7 Practical 3: Ceramic Filter

As was mentioned in the Coupled L/C circuits with swept frequency Practical, to get a good flat-topped response with steep sides requires several resonators, several pairs of L-C circuits being a possibility. This Practical is concerned with an alternative solution to the problem of providing several resonators. Just as the quartz crystal uses the piezo-electric effect to translate a mechanical resonance into an electrical one, so a device which has several coupled mechanical resonances can, through piezo-electricity, provide a system of several coupled electrical resonances. The combination of a piezo-electric ceramic plate with suitably shaped electrodes can provide this. The electrodes typically have several prongs or fingers, and the size and shape of each one determines one of the resonances in the system. Their closeness to each other determines the coupling. The tuning of the various resonances is done by grinding the electrodes or substrate ceramic during factory processing. Consequently, there is little you can do to adjust the circuit. You will find that, as for the low-pass passive filter Assignment, matching of the terminations is important. Now that you have seen the effect of damping on a resonant (LC) circuit, you should be able to appreciate that the matching load resistances do provide necessary damping on resonances which would otherwise give undesirable peaks in the response near cut-off. 5.3.7.1 Procedure Switch the Sweep OFF .

The RF oscillator delivers a signal of roughly constant amplitude, with variable frequency, as in the Capacitively coupled L/C circuits Practical. Compare the way the signal at monitor point 9 is affected by changes in frequency. Make a note of two frequencies between which the interesting part of the filter characteristic lies. 5-48

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5.3.7.2 Questions 1. What do you think is the most important difference between the ceramic filter and the L/C bandpass filter of Practicals 1 and 2?

2.

Over what range does the phase of the output signal vary between the two cut-off frequencies? A large phase change with frequency implies an appreciable delay (strictly, group delay) in the signal. This is typical of complex filters, having large numbers of reactive elements.

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5.3.8 Practical 4: Ceramic Filter, Swept Frequency

This Practical continues the Ceramic filter Practical by adding the automatic response plotter. 5.3.8.1 Procedure Switch the Sweep ON .

Set the minimum sweep frequency to 445 kHz and the maximum to 465 kHz. Select the Plotter. Use the large display to see the response in more detail.

The behaviour of the circuit is modified greatly when the correctly matching load or source impedance is not provided. Switch the load off by pressing the Load Button and observe its effect.

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5.3.8.2 Questions 1. Describe briefly the effects of removing the filter's load on: a) the shape of the frequency response. b) the max output amplitude.

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5.4


Tuned Amplifier with Gain Control Assignment

This assignment deals with the control of amplifier gain, both manual and automatic. 5.4.1 Objectives


The gain control characteristic of the amplifier stage (under manual gain control),

•

How automatic gain control can maintain a steady level of output signal,

•

Why automatic gain control can make difficulties during frequency-response testing.

5.4.2 Practicals


Practical 1: Gain Control

•

Practical 2: Automatic Gain Control

•

Practical 3: Frequency Response with Automatic Gain Control

•

Practical 4: Decibel Gain



Signal Generator,

•

Low Pass Filter,

•

Band Pass Filter,

•

Amplifier with AGC.

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5.4.4 Theory

5.4.4.1 Tuned Amplifiers Radio frequency (RF) amplifiers are usually tuned. That is, they include devices which select a particular frequency or band of frequencies. This is necessary for two main reasons: 1. The user is normally interested in only one at a time of the enormous number of radio signals which can be received. The others have to be rejected as unwanted interference. This aspect is dealt with in the RF Selectivity Assignment. 2. At high frequencies the reactance of stray capacitances becomes small. This makes it difficult and expensive to achieve large amplifier gains, unless inductive reactance is used to cancel out the effects of this capacitance. Where a fairly narrow band of frequencies is acceptable (and often is required), a parallel combination of inductance and capacitance resonates at a particular frequency to provide high impedance. 3. The resonant frequency must match the frequency of the signal to be received. Achieving this match is called 'tuning'. 5.4.4.1.1 Amplifier Gain Control The amplitude of signals, especially RF signals, can vary over an enormous range. The signal arriving at an aerial (antenna) terminal may well be a microvolt or so. But if a transmitter is near by, that signal may rise to several volts amplitude.

The signal required at the detector is typically in the order of 0.2 V to 10 V. To achieve this, one or more amplifiers process the aerial signal on its way to the detector. To achieve say 1 V at the detector, with input signals ranging from 1 microvolt to 1 V the gain has to be varied in the ratio one million to 1. The control of the gain of an amplifier is therefore very important. A typical RF amplifier in a radio or radar receiver is therefore tuned and has a very wide range of gain which can be controlled. This usually requires several amplifying stages to be gain-controlled; this assignment will look at just one such stage.

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5.4.4.2 Decibels The range of signal strengths in communications work is enormous. Signals may vary from a fraction of a microvolt to many volts, a ratio of many millions to one. In any given impedance the corresponding power ratio is trillions to one. In order to reduce the numbers used to manageable proportions, a logarithmic scale is convenient. This is because the common logarithm of say 1000000 is only 6 and the logarithm of 1/1000000 is -6. So the range of logarithms from -6 to +6 covers a ratio of ten to the power 12; ie, 1000000000000. 5.4.4.2.1 Decibel Power Ratio The bel is a unit of power ratio, simply defined so that the power gain, defined by the ratio: G = powerout / powerin may be expressed as log10 G bel. In practice the numbers arrived at tend to be rather small, so the unit generally used is the decibel (dB). The same power ratio G can therefore be expressed as 10 log10 G dB. 5.4.4.2.2 Decibel Voltage Ratio In many communication systems a signal travels along a system such as a line which has a characteristic impedance; any apparatus connected is matched to this same impedance. In these circumstances (constant impedance), the volt ratio between two points is simply the square root of the power ratio between them. It follows that if gv is the voltage ratio, then: G = (gv)2 and the power gain is: 10 log10 G dB = 10 log 10 (gv)2 dB = 20 log10 gv dB. If the resistive impedance at an input point is R i, the voltage is vi, at the output it is Ro, the voltage is v o and the power gain is G, then: G = ( vi2 / Ri ) / ( vo2 / Ro ) = ( v i / vo )2 / ( Ro / Ri )

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Electronic engineers on the other hand use circuits in which impedances vary widely. Voltage is easier to measure than power, so that, much of the time, they think more in terms of voltage gain than power gain. But the huge ratios of signal strength still have to be managed. As a result, the habit has grown up of speaking of voltage ratios expressed in terms of dB, but WITHOUT considering the impedance ratio. The 'voltage gain in dB' is taken as the value NOT including the impedance ratio. Thus: 'voltage gain in dB' = 20 log10 gv dB. It is important to understand when this convention is being used and when a true dB value based on power is required. 5.4.5 Practical 1: Gain Control

The gain of an RF amplifier is almost invariably controlled by adjusting a dc control voltage. This is because problems arise with any attempt to use a potentiometer or equivalent device in the manner of an audio gain control. Some of these problems are: 1. Ineffective control because stray capacitances provide leakage paths. 2. Unwanted alteration to the tuning. 3. Unwanted changes in selectivity. This assignment will examine some of the ideas associated with that of ' gain' and use a dc control voltage to adjust the gain in an amplifier. A manual control will enable you to adjust the dc level which controls the gain. Measurements with the oscilloscope will enable you to measure the gain for various dc settings. Output v. control voltage at constant input

You will need graph paper scaled as shown.

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5.4.5.1 Procedure Switch Sweep OFF .

Set the gain control of the tuned Amplifier fully clockwise. Adjust the output of the RF oscillator to a suitable level and then adjust its frequency for maximum response from monitor point 6 . Observe the oscilloscope and voltmeter while adjusting the gain control.

Use the large oscilloscope to measure the peak-to-peak output. Keeping the RF signal constant, set different values of dc voltage at about 0.1V intervals, using the gain control. Plot a graph of output signal against the dc voltage. Use both monitor points 6 and 7 to extend the range of measurements.

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5.4.5.2 Questions 1. What is the input signal voltage, measured at monitor point 4?

2. What is the maximum voltage gain from point 4 to point 6, assuming the one-fifth attenuation from point 6 to point 7 is correct? Express your answer: a) as a ratio b) in dB.

3. What is the minimum voltage gain from point 4 to point 6, expressed: a) as a ratio? b) in dB?

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5.4.6 Practical 2: Automatic Gain Control

To cope with signals which vary between the two input levels shown in the diagram would be extremely tiresome if the gain had to be adjusted manually each time a different signal were received. Also radio signals quite often vary in strength. This may be due to changes in propagation conditions, motion of the transmitter or receiver or other causes. Automatic gain control is intended to adjust the gain automatically so that the desired constant signal level is achieved at the detector input despite variations in the input signal.

In order to achieve this, a measure of the signal amplitude at the detector is required, in the form of a dc voltage. An envelope detector has an output containing dc proportional to the RF amplitude, together with ac corresponding to any amplitude modulation. The latter is removed by a filter. The dc voltage is compared against some reference value and, when it exceeds the reference value, the difference signal is used to decrease the gain. The reference voltage is often referred to as the ' AGC delay voltage'. The term 'delay' does not refer to a time delay. The early AGC systems simply fed back the dc component from the envelope signal detector to control the gain. This meant that even for undesirably weak signals, some gain reduction took place. 'Delaying' the start of the feedback as the signal level rose avoided this problem. In the circuit diagram of the Practical, the dc signals associated with AGC are distinctively coloured. You will see the envelope detector which generates the AGC signal is backbiassed by about 0.V. This is the delay voltage and the reference against which the signal level is compared. 5-58

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Output/Input showing effect of AGC.

You will need graph paper scaled as shown. 5.4.6.1 Procedure The sweep should be switched off . Set the RF level to maximum and the gain of the tuned Amplifier fully clockwise. Tune the RF oscillator for maximum response from monitor point 7 . Then observe how the output varies with changes of RF oscillator amplitude. Repeat using monitor point 6 .

Use button to switch in automatic gain control and repeat the observation. Use the large oscilloscope to measure peak-to-peak voltages. Set different amplitudes of RF signal, measured at monitor point 4 , and measure the corresponding output at monitor point 6 . Plot a graph of output signal against RF input voltage. 53-001S

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5.4.6.2 Questions 1. What causes lack of proportionality between output amplitude and input amplitude: a) with manual gain control? b) with automatic gain control?

2. How does the gain control voltage vary as the RF input is increased? Explain why: a) in relation to the required gain adjustment. b) in relation to how the control voltage is produced.

3.

Can you see a connection between the amplitude at which the output RF is automatically controlled and the 0.7 V marked on the circuit diagram?

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5.4.7 Practical 3: Frequency Response with Automatic Gain Control

Suppose that you are plotting the frequency response of an amplifier which includes AGC. A constant input voltage is applied at various frequencies and you hope to see the output vary in a way that shows the frequency response. What happens in fact? As the output tries to change, the action of the AGC system tries to prevent this variation. If it is a good AGC system it will virtually prevent any variation at all, once the output has reached a certain level. Consequently, when taking a frequency response, the AGC must be prevented from working. The Practical will show how important this is. 5.4.7.1 Procedure Switch the sweep ON . Set max RF oscillator amplitude. Set the gain control of the tuned Amplifier fully clockwise. Adjust the minimum frequency of the sweep to 445 kHz and the maximum to 465 kHz. Change to monitor point 6 and adjust the RF oscillator level so that a full height trace appears on the small oscilloscope at the peak frequency. Select Plot Response and observe the resulting humped curve showing the selectivity of the tuned circuit.

Use the Button to switch to Auto gain control. Observe the response now displayed.

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5.4.7.2 Questions 1. Why does the frequency response appear to become flat when AGC is switched in?

2. If the input signal were made small enough, would it be possible to plot the frequency response? (Explain your answer).

3. Suppose that the amplifier is operating with an input amplitude such that the AGC circuit holds a constant output amplitude. Does this mean that a signal on another frequency will not be relatively attenuated by the tuned circuit, as it would be with no AGC present? (Hint: Consider carefully whether the AGC control voltage acts on the active device in the amplifier or on the tuned circuit.)

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5.4.8 Practical 4: Decibel Gain

This Practical introduces the concept of using decibels to express the gain of an amplifier. In it you are asked to make some measurements and then translate the results into decibels. Read the theory section on how decibels work before attempting this practical and answering the questions. 5.4.8.1 Procedure Switch sweep OFF . Set the amplifier gain to maximum. Tune the RF oscillator for maximum output, adjusting the oscillator output to avoid overloading the oscilloscope. Change to the large oscilloscope and adjust the RF amplitude so that a full height trace appears.

Measure the voltages at points 4 and 6 . Answer questions 1 to 3.

Select Automatic Gain Control before answering question 4.

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