Analog Communication Test Questions - Set 10

1)   The desired signal of maximum frequency wm centered at frequency w=0 may be recovered if

a. The sampled signal is passed through low pass filter
b. Filter has the cut off frequency wm
c. Both a and b
d. None of the above
Answer  Explanation  Related Ques

ANSWER: Both a and b

Explanation:
The original or desired signal may be recovered from the sampled signal by passing the signal through a low pass filter. The low pass filter should have the cut off frequency equal to wm so that it allows only low frequencies up to the cut off frequency to pass through. The other higher frequencies in the sampled signal are rejected by the low pass filter.


2)   A distorted signal of frequency fm is recovered from a sampled signal if the sampling frequency fs is

a. fs > 2fm
b. fs < 2fm
c. fs = 2fm
d. fs ≥ 2fm
Answer  Explanation  Related Ques

ANSWER: fs < 2fm

Explanation:
If the signal of frequency fm is sampled at the rate fs ≥ 2fm only then the spectrum of the sampled signal is obtained without overlapping. For fs < 2fm the sampled signal spectrum overlap each other, and therefore the signal cannot be recovered easily. For reconstruction of signal to be free from distortion, the important condition is
fs ≥ 2fm


3)   Calculate the minimum sampling rate to avoid aliasing when a continuous time signal is given by x(t) = 5 cos 400πt

a. 100 Hz
b. 200 Hz
c. 400 Hz
d. 250 Hz
Answer  Explanation  Related Ques

ANSWER: 400 Hz

Explanation:
In the given signal, the highest frequency is given by f = 400 π/ 2π
= 200 Hz

The minimum sampling rate required to avoid aliasing is given by Nyquist rate. The nyquist rate is = 2 * f
= 2 * 200
= 400 Hz.


4)   Calculate the Nyquist rate for sampling when a continuous time signal is given by
x(t) = 5 cos 100πt +10 cos 200πt - 15 cos 300πt


a. 300Hz
b. 600Hz
c. 150Hz
d. 200Hz
Answer  Explanation  Related Ques

ANSWER: 300Hz

Explanation:
For the given signal,
f1 = 100π/2π = 50Hz
f2 = 200π/2π = 100Hz
f3= 300π/2π = 150Hz

The highest frequency is 150Hz. Therefore fmax = 150Hz
Nyquist rate = 2 fmax
= 2 * 150
= 300Hz.


5)   A low pass filter is

a. Passes the frequencies lower than the specified cut off frequency
b. Rejects higher frequencies
c. Is used to recover signal from sampled signal
d. All of the above
Answer  Explanation  Related Ques

ANSWER: All of the above

Explanation:
The low pass filter should have the cut off frequency equal to wm so that it allows only lower frequencies up to the cut off frequency to pass through. The other higher frequencies in the sampled signal are rejected by the low pass filter. The original or desired signal may be recovered from the sampled signal by passing the signal through a low pass filter.


6)   The techniques used for sampling are

a. Instantaneous sampling
b. Natural sampling
c. Flat top sampling
d. All of the above
Answer  Explanation  Related Ques

ANSWER: All of the above

Explanation:
The techniques used for sampling are:

a) Instantaneous sampling
b) Natural sampling
c) Flat top sampling

The natural sampling and the flat top sampling techniques are used practically to sample a signal.


7)   The instantaneous sampling

a. Has a train of impulses
b. Has the pulse width approaching zero value
c. Has the negligible power content
d. All of the above
Answer  Explanation  Related Ques

ANSWER: All of the above

Explanation:
The instantaneous sampling is also called ideal sampling or impulse sampling. The instantaneous sampling has a train of impulses. The pulse width of the samples has almost zero value. Therefore it has negligible power content and thus may not be used for transmission purpose.


8)   The sampling technique having the minimum noise interference is

a. Instantaneous sampling
b. Natural sampling
c. Flat top sampling
d. All of the above
Answer  Explanation  Related Ques

ANSWER: Natural sampling

Explanation:
The natural sampling is the technique that has the minimum noise interference to the sampled signal. It is obtained by multiplying the input signal with the sampling function. It is a practical method used for sampling of signals. Chopping principle is used to sample the signal in natural sampling and it satisfies the Nyquist criteria for sampling of signals.


9)   Types of analog pulse modulation systems are

a. Pulse amplitude modulation
b. Pulse time modulation
c. Frequency modulation
d. Both a and b
Answer  Explanation  Related Ques

ANSWER: Both a and b

Explanation:
In pulse modulation systems, the carrier is a train of pulses rather than a continuous signal. The parameters of the pulses are varied according to the instantaneous value of the modulating signal. There are two types of pulse modulation systems:
- Pulse amplitude modulation
- Pulse time modulation


10)   In pulse amplitude modulation,

a. Amplitude of the pulse train is varied
b. Width of the pulse train is varied
c. Frequency of the pulse train is varied
d. None of the above
Answer  Explanation  Related Ques

ANSWER: Amplitude of the pulse train is varied

Explanation:
Pulse amplitude modulation is a type of analog pulse modulation system. In pulse modulation systems, the carrier is a train of pulses rather than a continuous signal. In PAM, the amplitude of the rectangular pulse train is varied according to the instantaneous value of the modulating signal. The pulses in PAM signal may be flat top type or natural type samples.


11)   Pulse time modulation (PTM) includes

a. Pulse width modulation
b. Pulse position modulation
c. Pulse amplitude modulation
d. Both a and b
Answer  Explanation  Related Ques

ANSWER: Both a and b

Explanation:
In pulse modulation systems, the carrier is a train of pulses rather than a continuous signal. The parameters of the pulses are varied according to the instantaneous value of the modulating signal. The carrier is a train of pulses rather than a continuous signal. In PTM, the timing of the pulses of the carrier is varied in accordance with modulating signal. PTM includes:
- Pulse width modulation
- Pulse position modulation


12)   Drawback of using PAM method is

a. Bandwidth is very large as compared to modulating signal
b. Varying amplitude of carrier varies the peak power required for transmission
c. Due to varying amplitude of carrier, it is difficult to remove noise at receiver
d. All of the above
Answer  Explanation  Related Ques

ANSWER: All of the above

Explanation:
In PAM, Bandwidth is very large as compared to modulating signal frequency. In PAM, the amplitude of the rectangular pulse train is varied according to the instantaneous value of the modulating signal. Due to this, the required power for transmission is also varied. Due to varying amplitude of carrier, the interference of noise is very high in PAM. So it is difficult to remove noise at receiver.


13)   In Pulse time modulation (PTM),

a. Amplitude of the carrier is constant
b. Position or width of the carrier varies with modulating signal
c. Pulse width modulation and pulse position modulation are the types of PTM
d. All of the above
Answer  Explanation  Related Ques

ANSWER: All of the above

Explanation:
In Pulse time modulation (PTM), amplitude of the carrier is kept constant and the Position or width of the carrier varies with the amplitude of the modulating signal at the time of sampling. Pulse width modulation and pulse position modulation are the types of Pulse Time Modulation.
As there is no variation in the amplitude of the carrier, the noise may be easily removed at the receiver.


14)   In different types of Pulse Width Modulation,

a. Leading edge of the pulse is kept constant
b. Tail edge of the pulse is kept constant
c. Centre of the pulse is kept constant
d. All of the above
Answer  Explanation  Related Ques

ANSWER: All of the above

Explanation:
There are types of Pulse Width Modulation. In one of the variations, leading edge of the pulse is kept constant and pulse width is measured with respect to leading edge. In second type, tail edge of the pulse is kept constant and pulse width is measured with respect to it. And the third type has a constant centre of the pulse and the pulse width changes on both the sides of the centre of the pulse.


15)   In pulse width modulation,

a. Synchronization is not required between transmitter and receiver
b. Amplitude of the carrier pulse is varied
c. Instantaneous power at the transmitter is constant
d. None of the above
Answer  Explanation  Related Ques

ANSWER: Synchronization is not required between transmitter and receiver

Explanation:
In pulse width modulation, the width of the carrier varies with the amplitude of the modulating signal at the time of sampling. Pulse width modulation is a type of Pulse Time Modulation. As there is no variation in the amplitude of the carrier, the noise may be easily removed at the receiver. It does not require synchronization between the transmitter and the receiver.


16)   In PWM signal reception, the Schmitt trigger circuit is used

a. To remove noise
b. To produce ramp signal
c. For synchronization
d. None of the above
Answer  Explanation  Related Ques

ANSWER: To remove noise

Explanation:
In pulse width modulation, the width of the carrier varies with the amplitude of the modulating signal at the time of sampling. In PWM signal reception, the received PWM signal is applied to the Schmitt trigger circuit. The Schmitt trigger circuit is used to remove noise in the PWM waveform. This output is the supplied further for detection of the original information.


17)   In Pulse Position Modulation, the drawbacks are

a. Synchronization is required between transmitter and receiver
b. Large bandwidth is required as compared to PAM
c. None of the above
d. Both a and b
Answer  Explanation  Related Ques

ANSWER: Both a and b

Explanation:
In Pulse Position Modulation, the position of the pulse of the carrier is varied with reference to the position of a reference pulse. The position is varied in accordance with the sampled modulating signal. In PPM, synchronization is required between the transmitter and the receiver. Large bandwidth is required in Pulse position Modulation as compared to the Pulse amplitude modulation.