Blake: MCQ in Multiplexing and Multiple-Access Techniques

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(Last Updated On: March 27, 2020)

Blake: MCQ in Multiplexing and Multiple-Access Techniques

This is the Multiple Choice Questions in Chapter 13: Multiplexing and Multiple-Access Techniques from the book Electronic Communication Systems by Roy BlakeBlake: MCQ in Multiplexing and Multiple-Access Techniques. If you are looking for a reviewer in Communications Engineering this will definitely help. I can assure you that this will be a great help in reviewing the book in preparation for your Board Exam. Make sure to familiarize each and every questions to increase the chance of passing the ECE Board Exam.

See also: MCQ in Electronic Communication Systems by George Kennedy

Start Practice Exam Test Questions

Choose the letter of the best answer in each questions.

MULTIPLE CHOICE

1. TDMA stands for:

a. Time Domain Multiple Access

b. Time-Division Multiple Access

c. Tone Division Multiple Access

d. none of the above

View Answer:

Answer: Option B

Solution:

2. CDMA stands for:

a. Code-Division Multiple Access

b. Carrier Division Multiple Access

c. Compact Digital Multiplex Arrangement

d. none of the above

View Answer:

Answer: Option A

Solution:

3. TDMA is used instead of TDM when:

a. all the signals come from the same source

b. the signals come from different sources

c. TDM is used in RF communications

d. they mean the same thing

View Answer:

Answer: Option B

Solution:

4. When calculating the maximum number of users, a limiting factor in FDM is:

a. the type of media used

b. the length of the channel

c. the bandwidth of each signal

d. all of the above

View Answer:

Answer: Option C

Solution:

5. A DS-1 signal contains:

a. 12 channels

b. 24 channels

c. 32 channels

d. 64 channels

View Answer:

Answer: Option B

Solution:

6. The bit-rate of a DS-1 signal over a T-1 line is:

a. 64 kbps

b. 256 kbps

c. 1.536 Mbps

d. 1.544 Mbps

View Answer:

Answer: Option D

Solution:

7. Besides data bits, a DS-1 frame contains a:

a. timing bit

b. T-bit

c. signaling bit

d. framing bit

View Answer:

Answer: Option D

Solution:

8. In DS-1, a bit is “stolen” out of each channel:

a. every frame

b. every other frame

c. every sixth frame

d. every twelfth frame

View Answer:

Answer: Option C

Solution:

9. Moving signals from one line to another is called:

a. time switching

b. space switching

c. line switching

d. cross-point switching

View Answer:

Answer: Option B

Solution:

10. Moving PCM samples from one time-slot to another is called:

a. time switching

b. space switching

c. signal switching

d. crosspoint switching

View Answer:

Answer: Option A

Solution:

11. A digital space switch is a:

a. multiplexer

b. TDM switch

c. computerized Strowger switch

d. crosspoint switch

View Answer:

Answer: Option D

Solution:

12. Spread-spectrum can be done by using:

a. computer-controlled frequency reuse

b. frequency-hopping

c. direct-sequence method

d. all of the above

View Answer:

Answer: Option D

Solution:

13. The term “chip rate” is used in describing:

a. computer-controlled frequency reuse

b. frequency-hopping

c. direct-sequence method

d. all of the above

View Answer:

Answer: Option C

Solution:

14. For a given data rate, direct-sequence systems, compared to standard RF systems, use:

a. about the same bandwidth

b. much more bandwidth

c. much less bandwidth

d. approximately double the bandwidth

View Answer:

Answer: Option B

Solution:

15. “Processing gain” is another term for:

a. RF gain

b. computer speed

c. spreading gain

d. improved signal-to-noise ratio

View Answer:

Answer: Option C

Solution:

16. To calculate processing gain, divide the transmitted RF bandwidth by:

a. the digital data bit rate

b. bandwidth of original baseband

c. the S/N ratio

d. the chip size

View Answer:

Answer: Option B

Solution:

17. A receiver for frequency-hopping spread-spectrum would be:

a. a narrowband receiver

b. a wideband receiver

c. a direct-conversion receiver

d. a CDMA receiver

View Answer:

Answer: Option A

Solution:

18. A receiver for direct-sequence spread-spectrum would be:

a. a narrowband receiver

b. a wideband receiver

c. a direct-conversion receiver

d. a “chip-rate” receiver

View Answer:

Answer: Option B

Solution:

19. CDMA:

a. cannot be used with frequency-hopping spread-spectrum

b. cannot be used with direct-sequence spread-spectrum

c. cannot be used on an RF channel

d. allows many transmitters to use a band simultaneously

View Answer:

Answer: Option D

Solution:

20. For optimal performance, CDMA requires the use of:

a. orthogonal PN sequences

b. non-orthogonal PN sequences

c. true-random PN sequences

d. none of the above

View Answer:

Answer: Option A

Solution:

COMPLETION

1. Multiplexing allows many signals to ____________________ a channel.

View Answer:

Answer: share

Solution:

2. Three methods of multiple access are FDMA, TDMA, and ____________________.

View Answer:

Answer: CDMA

Solution:

3. In FDM, each signal uses part of the bandwidth ____________________ of the time.

View Answer:

Answer: all

Solution:

4. In TDM, each signal uses all of the bandwidth ____________________ of the time.

View Answer:

Answer: part

Solution:

5. Using CDMA on a radio channel, all signals can transmit ____________________ of the time.

View Answer:

Answer: all

Solution:

6. DS-1 is an example of ____________________-division multiplexing.

View Answer:

Answer: time

Solution:

7. The AM radio band is an example of ____________________-division multiplexing.

View Answer:

Answer: frequency

Solution:

8. A DS-1 frame contains one sample from each of ____________________ channels.

View Answer:

Answer: 24

Solution:

9. T1 uses the ____________________ line code.

View Answer:

Answer: AMI

Solution:

10. Each DS-1 frame contains a total of ____________________ bits.

View Answer:

Answer: 193

Solution:

11. A DS-1 frame is transmitted at a rate of ____________________ bits per second.

View Answer:

Answer: 1.544 Meg

Solution:

12. Each sample in a DS-1 frame contains ____________________ bits.

View Answer:

Answer: 8

Solution:

13. A group of twelve DS-1 frames is called a ____________________.

View Answer:

Answer: superframe

Solution:

14. Switching signals from one line to another is called ____________________ switching.

View Answer:

Answer: space

Solution:

15. Moving PCM samples from one time slot to another is called ____________________ switching.

View Answer:

Answer: time

Solution:

16. The deep fades caused by signal-cancellation due to reflection are called ____________________ fading.

View Answer:

Answer: Rayleigh

Solution:

17. A PN sequence is a ____________________-random noise sequence.

View Answer:

Answer: pseudo

Solution:

18. One method of spread-spectrum is frequency ____________________.

View Answer:

Answer: hopping

Solution:

19. It is ____________________ to jam a spread-spectrum signal.

View Answer:

Answer: difficult

Solution:

20. It is ____________________ to eavesdrop on a spread-spectrum signal.

View Answer:

Answer: difficult

Solution:

21. The extra bits added to the data in direct-sequence spread-spectrum are called ____________________.

View Answer:

Answer: chips

Solution:

22. A chipping-rate of at least ____________________ times the bit rate of the data is common.

View Answer:

Answer: ten

Solution:

23. The ‘C’ in CDMA stands for ____________________.

View Answer:

Answer: code

Solution:

24. In a frequency-hopping CDMA system, when no two transmitters use the same frequency at the same time the PN sequences are said to be ____________________.

View Answer:

Answer: orthogonal

Solution:

SHORT ANSWER

1. What does Hartley’s Law tell us about the relationship between time and bandwidth for digital transmission?

View Answer:

Answer: The more bandwidth, the less time it takes to send a given amount of information. So the more bandwidth available, the higher the possible bit rate.

Solution:

2. How many signals could fit into 1 MHz of bandwidth if each signal required 100 kHz of bandwidth and the separation between adjacent channels was 10 kHz?

View Answer:

Answer: 9

Solution:

3. Why is it difficult to jam a spread-spectrum signal?

View Answer:

Answer: Jamming requires an interference signal of sufficient power in the same part of the spectrum the information signal occupies. Because a spread-spectrum signal is, by definition, spread out over a very wide bandwidth, jamming can interfere with only a small fraction of the total signal.

Solution:

4. Why is it difficult to eavesdrop on a spread-spectrum signal?

View Answer:

Answer: In a spread-spectrum transmission, the signal power at any given frequency in its band is so low that it is virtually indistinguishable from noise. An eavesdropper would not know a signal was being sent. And without knowing the exact sequence being used, it is virtually impossible to “de-spread” the signal.

Solution:

5. Why is autocorrelation used to receive direct-sequence spread-spectrum signals?

View Answer:

Answer: Autocorrelation allows a signal to be “pulled out of” the noise even when the signal-to-noise ratio is less than one, as it is in spread-spectrum.

Solution:

6. What is meant by “orthogonal sequences” in CDMA?

View Answer:

Answer: During transmission, the PN sequences determine which parts of the available bandwidth the spread spectrum signal will occupy. Assume you have two PN sequences: PN1 and PN2. At some point in time, suppose PN1 would cause a transmission to occupy frequencies f11, f12, f13, and so forth. Now suppose PN2 would cause the transmission to occupy frequencies f21, f22, f23, and so forth. If the two sets of frequencies, (f11, f12, f13, …) and (f21, f22, f23, …), have no frequencies in common, then the two PN sequences are said to be orthogonal.

Solution:

Complete List of MCQ in Electronic Communication Systems by Blake

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