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

Explanation: 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

Explanation: 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

Explanation: 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) 42kbPs

b) 64kbPs

c) 16kbPs

d) 8kbPs

Explanation: 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

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

6. For 50% modulation, power in each sideband is ________ of that of carrier.

a) 10%

b) 4.32%

c) 5%

d) 6.25%

Explanation: Modulation index = 0.5. Power in sidebands may be calculated as

7. For 100% modulation, total power is ________

a) 1.5P_{c}

b) 2P_{c}

c) 3.75P_{c}

d) 1.25P_{c}

Explanation: Total power, P

_{t}= P

_{c}(1 +

^{µ2}⁄

_{2}), where P

_{c}= Carrier Power

where µ = 1 (for 100% modulation),

so P

_{t}= P

_{c}(1 +(1

^{2}/2)). On solving it we get P

_{t}= 1.5P

_{c}.

8. If each element of signal occupies 70ms, what will its speed?

a) 11.23 bauds

b) 14.28 bauds

c) 17.39 bauds

d) 13.33 bauds

Explanation: The carrier signal is characterized by the number of signal intervals, or pulses, that are transmitted per second. Each pulse is called a baud. Bps stands for bits per second. Bps is a measure of how many bits can be transmitted during one pulse (one baud).

9. Power of carrier wave is 300W and modulation index is 0.75. Find its total power?

a) 465W

b) 384W

c) 323W

d) 502W

Explanation: Total power, P

_{t}= P

_{c}(1 +

^{µ2}⁄

_{2}), where P

_{c}= Carrier Power = 300W

where Modulation Index (µ) = 0.75,

10. If a wave is modulated by two waves. One of them has modulation index equal to 0.75 and other has 0.2, the total modulation index will be ________

a) 0.67

b) 0.58

c) 0.77

d) 0.35

Explanation: Given that m1 = 0.75 and m2 = 0.2. Total modulation index will be equal to By substituting values we have which is equal to 0.77.