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Digital communications: board exam questions with solutions

On the ECE exam under Electronic Systems and Technologies (EST). Try each question before you open its solution, then practise with fresh numbers below: 3 computed question types on this topic.

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Questions

1. What is the minimum (Nyquist) sampling rate for a signal band-limited to 4 kHz?

  1. 8 kHz
  2. 4 kHz
  3. 2 kHz
  4. 16 kHz
Show the answer and solution

Answer: A. 8 kHz

fs ≥ 2fmax = 2 × 4 kHz = 8 kHz.

2. A voice channel is sampled at 8 kHz and each sample is encoded with 8 bits. What is the PCM bit rate?

  1. 8 kbps
  2. 16 kbps
  3. 128 kbps
  4. 64 kbps
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Answer: D. 64 kbps

Rb = 8000 samples/s × 8 bits = 64 000 bit/s.

3. For a full-scale sine wave, what is the approximate signal-to-quantization-noise ratio of a 10-bit uniform quantizer?

  1. 60.2 dB
  2. 20.0 dB
  3. 62.0 dB
  4. 31.8 dB
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Answer: C. 62.0 dB

SQNR ≈ 6.02n + 1.76 = 6.02 × 10 + 1.76 = 62.0 dB.

4. A 3 kHz channel has a 30 dB signal-to-noise ratio. What is its Shannon capacity?

  1. 90.0 kbps
  2. 29.9 kbps
  3. 9.0 kbps
  4. 6.0 kbps
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Answer: B. 29.9 kbps

S/N = 10³; C = B log₂(1 + S/N) = 3000 × log₂(1001) = 29902 bit/s ≈ 29.9 kbps.

5. By Nyquist's formula, what is the maximum bit rate over a noiseless 4 kHz channel using 16 signal levels?

  1. 32 kbps
  2. 64 kbps
  3. 16 kbps
  4. 8 kbps
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Answer: A. 32 kbps

C = 2B log₂M = 2 × 4000 × 4 = 32 000 bit/s.

6. Which line code guarantees a transition in the middle of every bit, making it self-clocking?

  1. Unipolar NRZ
  2. Polar NRZ-L
  3. AMI with long runs of zeros
  4. Manchester
Show the answer and solution

Answer: D. Manchester

Manchester encodes each bit as a mid-bit transition (high-to-low or low-to-high), so the receiver can always recover the clock.

7. A signal band-limited to 15 kHz with a 10 V peak-to-peak range is sampled at the Nyquist rate and uniformly quantized with a step no larger than 10 mV. Using the fewest bits per sample, what is the bit rate?

  1. 150 kbps
  2. 270 kbps
  3. 300 kbps
  4. 330 kbps
Show the answer and solution

Answer: C. 300 kbps

Levels needed = 10/0.01 = 1000 → n = 10 bits (2¹⁰ = 1024). fs = 2 × 15 kHz = 30 kHz; Rb = 30k × 10 = 300 kbps.

8. A 16-QAM modem transmits 2.5 Mbaud. What is its bit rate?

  1. 2.5 Mbps
  2. 10 Mbps
  3. 40 Mbps
  4. 5 Mbps
Show the answer and solution

Answer: B. 10 Mbps

16-QAM carries log₂16 = 4 bits/symbol: 2.5 × 4 = 10 Mbps.

Worked examples with computed numbers

These numbers are fixed on this page; in practice they change every time.

9. What is the Shannon capacity of a 1 MHz channel with a 30 dB signal-to-noise ratio?

  1. 19.9 Mb/s
  2. 30 Mb/s
  3. 9.97 Mb/s
  4. 4.95 Mb/s
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Answer: C. 9.97 Mb/s

C = B log₂(1 + S/N) with S/N = 10^(30/10) = 1000: 9.97 Mb/s.

10. A 4 kHz signal is sampled at the Nyquist rate and coded at 16 bits per sample. What is the bit rate?

  1. 256 kb/s
  2. 64 kb/s
  3. 128 kb/s
  4. 524 Mb/s
Show the answer and solution

Answer: C. 128 kb/s

Nyquist rate fs = 2fm = 8 kHz; R = fs·n = 8 kHz × 16 = 128 kb/s.

11. What is the signal-to-quantization-noise ratio of an ideal 6-bit quantizer for a full-scale sine wave?

  1. 19.82 dB
  2. 37.88 dB
  3. 120 dB
  4. 36.12 dB
Show the answer and solution

Answer: B. 37.88 dB

For a full-scale sine, SQNR ≈ 6.02n + 1.76 dB = 37.88 dB.

Other topics in the same subjects

Communications fundamentals (dB, noise) · AM, FM and PM · Transmission lines · Antennas and wave propagation · Microwave, fiber optics and satellite · Data communications and networks · Broadcasting and acoustics

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