# Blake: MCQ in Multiplexing and Multiple-Access Techniques

(Last Updated On: January 3, 2020) This is the Multiple Choice Questions in Chapter 13: Multiplexing and Multiple-Access Techniques from the book Electronic Communication Systems by Roy Blake . 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.

#### 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

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

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

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

Solution:

5. A DS-1 signal contains:

a. 12 channels

b. 24 channels

c. 32 channels

d. 64 channels

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

Solution:

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

a. timing bit

b. T-bit

c. signaling bit

d. framing bit

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

Solution:

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

a. time switching

b. space switching

c. line switching

d. cross-point switching

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

Solution:

11. A digital space switch is a:

a. multiplexer

b. TDM switch

c. computerized Strowger switch

d. crosspoint switch

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

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

Solution:

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

b. much more bandwidth

c. much less bandwidth

d. approximately double the bandwidth

Solution:

15. “Processing gain” is another term for:

a. RF gain

b. computer speed

d. improved signal-to-noise ratio

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

Solution:

Solution:

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

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

Solution:

COMPLETION

1. Multiplexing allows many signals to ____________________ a channel.

Solution:

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

Solution:

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

Solution:

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

Solution:

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

Solution:

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

Solution:

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

Solution:

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

Solution:

9. T1 uses the ____________________ line code.

Solution:

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

Solution:

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

Solution:

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

Solution:

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

Solution:

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

Solution:

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

Solution:

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

Solution:

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

Solution:

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

Solution:

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

Solution:

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

Solution:

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

Solution:

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

Solution:

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

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 ____________________.

Solution:

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

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?

Solution:

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

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?

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:

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?

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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