Analog Communications Multiple Choice Questions focuses on “Power Calculation”.

1. If peak voltage of a carrier wave is 10V, what is the peak voltage of modulating signal if modulation index is 50%?

a) 10V

b) 20V

c) 8V

d) 5V

Answer: d

Clarification: From the relation, Modulation Index (µ) = V_{m}/V_{c} = 50% = 0.5,

where V_{m} = Peak voltage of modulating signal,

V_{c} = Peak voltage of a carrier wave = 10V,

Therefore, V_{m} = 10 X 0.5 = 5V.

2. Maximum Amplitude of an amplitude modulated 10V and minimum amplitude is 5V. Find its modulation index?

a) 0.65

b) 0.9

c) 0.33

d) 1

Answer: c

Clarification: We know, Modulation Index(µ) = (V_{max}-V_{min})/(V_{max}+V_{min}),

Where, V_{max} = Maximum Amplitude of an amplitude modulated = 10V

V_{min} = Minimum amplitude of an amplitude modulated = 5V

Therefore, µ = (10-5)/(10+5) = 0.33.

3. 24 channels, each band limited to 3.4 KHz, are to be time division multiplexed. Find the bandwidth required for 128 quantization level? (Given that sampling frequency is 8 KHz)

a) 2436 KHz

b) 1002 KHz

c) 1536 KHz

d) 1337 KHz

Answer: c

Clarification: N = 24, f_{m} = 3.4 kHz

m = 128,

2^{n} = m = 128, n = 7

But f_{s} = 2f_{m}, where, f_{s} = sampling frequency

instead at 2f_{m} 2 x 3.4 kHz 6.8 KHz.

B.W. = N(n+1)X f_{s} = [24(7 + 1)] 8 kHz = 1536 KHz.

4. Sampling frequency of a signal is 6 KHz and is quantized using 7 bit quantizer. Find its bit rate?

a) 48kbPs

b) 64kbPs

c) 16kbPs

d) 8kbPs

Answer: a

Clarification: Bit rate refers to the rate at which data is processed or transferred. It is usually measured in seconds, ranging from bps for smaller values to kbps and mbps.

Bit rate is also known as bitrate or data rate.

Bit rate, R_{b} = ^{1}⁄_{Tb} where where n = number of bits and f_{s} = Sampling Frequency

T_{b} = 1/42, therefore Bit rate = 42 Kbps.

5. Calculate power in each sideband, if power of carrier wave is 96W and there is 40% modulation in amplitude modulated signal?

a) 11.84W

b) 6.84W

c) 3.84W

d) 15.84W

Answer: c

Clarification: Modulation index = 0.4 and Pc = 96W. Power in sidebands may be calculated as