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In here you will find an assortment of Biology Paper 1 Multiple choice questions on Topic 6 - Human Health and Physiology. Please note that these are from the old IB Biology syllabus (First exam...
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QUESTIONS: 1. It is called the combining of signals s ignals from the same sources.Multiplexing 2.
It is called the combining of signals s ignals from multiple sources.Multiple access
3.
It demonstrates the theoretical limit to how much information can be delivered over a medium.Hartley’s Law
4.
It shows that time and bandwidth are equivalent.Hartley’s Law
5.
What does FDM mean? frequency-division multiplexing
6.
What does TDM mean? time-division multiplexing
7.
It is the most basic form of multiplexing and has been used since the first days of radio.
FDM/FDMA 8.
It is a technique by which the total bandwidth available in a communication medium is divided into a series of non-overlapping frequency sub-bands, each of which is used to carry a separate signal. Frequency-division multiplexing (FDM)
9.
It is where frequency division multiplexing is used as to allow multiple users to share a physical communications communications channel. Frequency-division multiple access (FDMA)
10.
It is used mainly for digital communication. communication. TDM
11.
It is a type of multiplexing in which two or more bit streams or signals are transferred apparently simultaneously as sub-channels in one communication channel, but are physically taking turns on the channel. Time-division multiplexing (TDM)
12.
It is a channel access method for shared medium networks. Time division multiple
access (TDMA) 13.
It is specially conditioned for use as T1 and T2 carriers. Twisted- pair lines
14.
It moves PCM samples from one time slot s lot to another in a TDM signal. time switch
15.
Each of the time switches has a separate bus at its output called highway
16.
System to allow multiple users to use the same frequency with separate PN codes and a spread spectrum modulation scheme. (Code Division Multiple Access (CDMA))
17.
Technique for increasing the bandwidth of a transmitted signal by combining it with a pseudorandom noise signal with a higher bit rate. (Direct Sequence Spread Spectrum)
18.
Form of spread spectrum communication in which RF carrier continually moves from one frequency to another according to a prearranged pseudo-random pattern. (Frequency
Hopping) 19. Improvement in interfere rejection due to spreading in a spread spectrum system. (Processing
Gain) 20. A transmitted series of ones and zeros that repeats after a set time and which appears random if the sequence is not known to the receiver. (Pseudo-Random Noise (PN) Sequence) 21. Variation in received signal strength s trength due to multipath propagation. pr opagation. (Rayleigh Fading)
22-23. Two important types of spread spectrum systems.
Frequency Hopping Direct Sequence 24. Extra bits used to spread the signal in a direct-sequence s pread spectrum system. (Chips) 25. Third form of multiple access. (CDMA)
26. Simpler of the two spread spectrum technique. (Frequency Hopping) 27. Is used to generate a carrier in the ordinary way. (Frequency Synthesizer) 28. Technique that spreads the signal over a broader spectrum of frequencies than is usual.
(Spread-Spectrum) 29. Form of spread spectrum communication that is commonly used with digital modulation schemes. (Direct Sequence) 30. Also called Spreading Gain. (Processing Gain)
PROBLEM SOLVING: 1. A voice transmission occupies a channel 40 kHz wide. Suppose a spread-spectrum system is used to increase its bandwidth to 20 MHz. If the signal has a total signal power of -100 dBm at the receiver input and the system noise temperature referred to the same point is 300 K. Calculate the signal-to-noise ratio for both system.
= -100.82 dBm A frequency-hopping spread-spectrum system hops to each of 150 frequencies every 10 seconds. How long does it spend on each frequency?
ANSWER: t=
= 0.0667 second per hop 3.
How many of each of the following signals would fit into a 2MHz spectrum allocation? a) Voice, with a maximum frequency of 4kHz, modulated using SSBSC AM N = 2MHz / 4kHz = 500 b) The same voice signal using DSB full-carrier AM N = 2MHz / 8kHz = 250 c) High- fidelity music with a maximum baseband frequency of 25kHz, using wideband FM with a maximum deviation of 75kHz. B = 2(δmax + fmmax ) = 2(75kHz + 25kHz) = 200kHz N = 2MHz / 200kHz = 10
