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AM, FM and PM: 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: 4 computed question types on this topic.

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Questions

1. An AM envelope has a maximum of 15 V and a minimum of 5 V. What is the modulation index?

  1. 0.33
  2. 0.67
  3. 0.50
  4. 2.00
Show the answer and solution

Answer: C. 0.50

m = (Vmax − Vmin)/(Vmax + Vmin) = 10/20 = 0.50.

2. A 100 W carrier is amplitude modulated to m = 0.8. What is the total transmitted power?

  1. 180 W
  2. 132 W
  3. 164 W
  4. 116 W
Show the answer and solution

Answer: B. 132 W

Pt = Pc(1 + m²/2) = 100(1 + 0.32) = 132 W.

3. A DSB-full-carrier AM transmitter is modulated by audio up to 5 kHz. What bandwidth does the signal occupy?

  1. 10 kHz
  2. 5 kHz
  3. 2.5 kHz
  4. 20 kHz
Show the answer and solution

Answer: A. 10 kHz

AM has upper and lower sidebands, so BW = 2fm(max) = 2 × 5 = 10 kHz.

4. Using Carson's rule, what is the approximate bandwidth of an FM signal with 75 kHz peak deviation and a 15 kHz maximum modulating frequency?

  1. 150 kHz
  2. 90 kHz
  3. 30 kHz
  4. 180 kHz
Show the answer and solution

Answer: D. 180 kHz

BW ≈ 2(Δf + fm) = 2(75 + 15) = 180 kHz.

5. What is the modulation index of an FM signal with 75 kHz peak deviation and a 15 kHz modulating tone?

  1. 0.2
  2. 1125
  3. 5
  4. 90
Show the answer and solution

Answer: C. 5

β = Δf/fm = 75/15 = 5.

6. Compared with full-carrier DSB AM at 100% modulation, what percentage of transmitted power is saved by sending only one sideband (SSB-SC)?

  1. 66.7%
  2. 83.3%
  3. 50.0%
  4. 16.7%
Show the answer and solution

Answer: B. 83.3%

At m = 1, Pt = 1.5Pc and each sideband = m²Pc/4 = 0.25Pc. Saving = (1.5 − 0.25)/1.5 = 83.3%. (66.7% is the saving from removing only the carrier.)

7. An AM transmitter's antenna current is 8.00 A unmodulated and 8.93 A when modulated by a sine wave. What is the modulation index?

  1. 0.70
  2. 0.12
  3. 0.49
  4. 0.50
Show the answer and solution

Answer: A. 0.70

It/Ic = √(1 + m²/2) → m = √(2[(8.93/8)² − 1]) = √(2 × 0.2460) = 0.70.

8. Why is pre-emphasis applied at an FM transmitter?

  1. To reduce the carrier power
  2. To widen the FM deviation at low audio frequencies
  3. To suppress one sideband
  4. To boost high audio frequencies, improving SNR after de-emphasis since FM noise rises with frequency
Show the answer and solution

Answer: D. To boost high audio frequencies, improving SNR after de-emphasis since FM noise rises with frequency

FM output noise increases with modulating frequency; boosting highs before transmission and cutting them (de-emphasis) at the receiver lowers high-frequency noise.

Worked examples with computed numbers

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

9. An AM transmitter has a 10 kW carrier modulated at 30%. What is the total transmitted power?

  1. 10.9 kW
  2. 10.5 kW
  3. 13 kW
  4. 11.5 kW
Show the answer and solution

Answer: B. 10.5 kW

Pt = Pc(1 + m²/2) = 10000(1 + 0.3²/2) = 10.5 kW.

10. An AM envelope swings between 103 V and 39 V peak. What is the modulation index?

  1. 0.689
  2. 0.621
  3. 0.379
  4. 0.451
Show the answer and solution

Answer: D. 0.451

m = (Vmax − Vmin)/(Vmax + Vmin) = 64/142 = 0.451.

11. An FM signal has a peak deviation of 5 kHz and a modulating frequency of 5 kHz. Estimate its bandwidth by Carson's rule.

  1. 20 kHz
  2. 30 kHz
  3. 40 kHz
  4. 10 kHz
Show the answer and solution

Answer: A. 20 kHz

Carson's rule: B ≈ 2(Δf + fm) = 2(5 kHz + 5 kHz) = 20 kHz.

12. What is the modulation index of FM with 15 kHz deviation and a 2 kHz tone?

  1. 0.133
  2. 3.75
  3. 7.5
  4. 15
Show the answer and solution

Answer: C. 7.5

β = Δf/fm = 15 kHz/2 kHz = 7.5.

Other topics in the same subjects

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

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