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

Answer: d

Clarification: Modulation index = 0.6 and Pc = 176W. Power in sidebands may be calculated as

2. For 100% modulation, power in each sideband is ________ of that of carrier.

a) 50%

b) 70%

c) 60%

d) 25%

Answer: d

Clarification: Modulation index = 1. Power in sidebands may be calculated as

3. Overmodulation results in ________________

a) Distortion

b) Weakens signal

c) Strengthens the signal

d) provides immunity to noise

Answer: a

Clarification: 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.

4. The maximum power efficiency of an AM modulator is?

a) 25%

b) 33%

c) 66%

d) 100%

Answer: b

Clarification: Efficiency (ή) = m^{2} / (m^{2} + 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

Answer: a

Clarification: 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

Answer: b

Clarification: 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) = V_{m}/V_{c}, where V_{m} is maximum baseband or message signal amplitude and V_{c} is maximum carrier signal amplitude. So for m = 1, V_{m} should be equal to V_{c}.

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

Answer: d

Clarification: Total power, P_{t} = P_{c} (1 + ^{m2}⁄_{2}), where m is Modulated Signal, P_{c} is Power of Unmodulated Signal or Carrier Signal.

So, for m=1,

P_{t} = P_{c} (1 + 1^{2}/2) = 1.5 P_{c}.

8. An AM signal is represented by x(t) = (30 + 2Sin(700πt)) Cos(2πt x 10^{2}t)V. What is the value of modulation index?

a) 0.7

b) 0.066

c) 0.341

d) 0.916

Answer: b

Clarification: Given equation can be written as 30(1 + 0.066 Sin(700πt)).

Comparing it with general AM equation, s(t) = A_{c}(1 + mA_{m} cos(w_{m}t)) cos(w_{c}t),

Where, A_{c} = Amplitude of Carrier Signal, A_{m} = 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 10^{2}t)V. Carrier power of the wave is?

a) 555W

b) 675W

c) 450W

d) 310W

Answer: c

Clarification:

Hence P_{c} = 450W.