Analog Communications Questions and Answers – Power Relations in AM Waves

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This set of Analog Communications Questions and Answers for Campus interviews focuses on “Power Relations in AM Waves”.

1. Calculate power in each sideband, if power of carrier wave is 176W and there is 60% modulation in amplitude modulated signal?
a) 13.36W
b) 52W
c) 67W
d) 15.84W
View Answer

Answer: d
Explanation: Modulation index = 0.6 and Pc = 176W. Power in sidebands may be calculated as
Power in sidebands is 52W if power of carrier wave is 176W & 60% modulation in amplitude

2. For 100% modulation, power in each sideband is ________ of that of carrier.
a) 50%
b) 70%
c) 60%
d) 25%
View Answer

Answer: d
Explanation: Modulation index = 1. Power in sidebands may be calculated as
Power in each sideband is 25% for 100% modulation

3. Overmodulation results in ________________
a) Distortion
b) Weakens signal
c) Strengthens the signal
d) provides immunity to noise
View Answer

Answer: a
Explanation: When instantaneous level of modulating signal exceeds the value necessary to provide 100% modulation, the signal is said to over-modulated. In other words, when modulation index is greater than 1, it results in Overmodulation. Thus, Overmodulation results in distortion of the modulating signal.
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4. The maximum power efficiency of an AM modulator is?
a) 25%
b) 33%
c) 66%
d) 100%
View Answer

Answer: b
Explanation: Efficiency (ή) = m2 / (m2 + 2), m=Modulation Index
For maximum efficiency m = 1 so, ή = 1/(1+2) = 1/3
and ή% = (1/3)x100 = 33%.

5. Noise performance of a square law demodulator of AM signal is?
a) Better than that of synchronous detector
b) Weaker than that of synchronous detector
c) Better than that of envelope detector
d) Weaker than that of envelope detector
View Answer

Answer: a
Explanation: Process of recovering message signal from received modulated signal is called demodulation. It is exactly opposite to modulation. There are two most used AM demodulators: Square Law Demodulator and Envelope Demodulator. Noise performance of Square Law Demodulator is far better than that of Synchronous Detector.

6. For getting 100% modulation, carrier amplitude should ________
a) exceed signal amplitude
b) be equal to signal amplitude
c) be lesser than signal amplitude
d) be equal to 0
View Answer

Answer: b
Explanation: Modulation index is the amount of modulation present in a carrier wave. It is also described as the ration of the amplitude of message signal to that of carrier signal.
Modulation Index (m) = Vm/Vc, where Vm is maximum baseband or message signal amplitude and Vc is maximum carrier signal amplitude. So for m = 1, Vm should be equal to Vc.

7. For 100% modulation, total power is?
a) same as the power of unmodulated signal
b) twice as the power of unmodulated signal
c) four times as the power of unmodulated signal
d) one and half times as the power of unmodulated signal
View Answer

Answer: d
Explanation: Total power, Pt = Pc (1 + m22), where m is Modulated Signal, Pc is Power of Unmodulated Signal or Carrier Signal.
So, for m=1,
Pt = Pc (1 + 12/2) = 1.5 Pc.
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8. An AM signal is represented by x(t) = (30 + 2Sin(700πt)) Cos(2πt x 102t)V. What is the value of modulation index?
a) 0.7
b) 0.066
c) 0.341
d) 0.916
View Answer

Answer: b
Explanation: Given equation can be written as 30(1 + 0.066 Sin(700πt)).
Comparing it with general AM equation, s(t) = Ac(1 + mAm cos(wmt)) cos(wct),
Where, Ac = Amplitude of Carrier Signal, Am = Amplitude of Message Signal
m=Modulation Index
So modulation index(m) = 0.066.

9. An AM signal is represented by x(t) = (30 + 2Sin(700πt)) Cos(2πt x 102t)V. Carrier power of the wave is?
a) 555W
b) 675W
c) 450W
d) 310W
View Answer

Answer: c
Explanation: Carrier power of the wave is 450W if AM signal is (30 + 2Sin(700πt)) Cos(2πt x 102t)V
Hence Pc = 450W.
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10. An AM signal is represented by x(t) = (30 + 2Sin(700πt)) Cos(2πt x 102t)V. Find the total power of amplitude modulated wave?
a) 453W
b) 675W
c) 789W
d) 451W
View Answer

Answer: d
Explanation: Pt = Pc (1 + bm22) * pc
So, here, m = 0.066, Pc = 450W
Pt = (1+(0.0662/2))*450 = 451W.

11. An AM signal is represented by x(t) = (30 + 2Sin(700πt)) Cos(2πt x 102t)V. What is its sideband power?
a) 4W
b) 1W
c) 3W
d) 2W
View Answer

Answer: b
Explanation: Sideband power i = (m2* Pc)/2 = (Pt – Pc) i.e. 451 – 450 = 1W.

Sanfoundry Global Education & Learning Series – Analog Communications.

To practice all areas of Analog Communications for Campus Interviews, here is complete set of 1000+ Multiple Choice Questions and Answers.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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