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Sampling Theorem Quiz | Sampling Theorem Objective Type Questions and Answers

(21) Drawback of using PAM method is
[A] Due to varying amplitude of carrier, it is difficult to remove noise at receiver
[B] Bandwidth is very large as compared to modulating signal
[C] Varying amplitude of carrier varies the peak power required for transmission
[D] All of the above
Answer: All of the above
(22) In Pulse time modulation (PTM),
[A] Pulse width modulation and pulse position modulation are the types of PTM
[B] Amplitude of the carrier is constant
[C] Position or width of the carrier varies with modulating signal
[D] All of the above
Answer: All of the above

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(23) In different types of Pulse Width Modulation,
[A] Centre of the pulse is kept constant
[B] Leading edge of the pulse is kept constant
[C] Tail edge of the pulse is kept constant
[D] All of the above
Answer: All of the above
(24) In pulse width modulation,
[A] Amplitude of the carrier pulse is varied
[B] Instantaneous power at the transmitter is constant
[C] Synchronization is not required between transmitter and receiver
[D] None of the above
Answer: Synchronization is not required between transmitter and receiver
(25) In PWM signal reception, the Schmitt trigger circuit is used
[A] To produce ramp signal
[B] For synchronization
[C] To remove noise
[D] None of the above
Answer: To remove noise
(26) In Pulse Position Modulation, the drawbacks are
[A] Large bandwidth is required as compared to PAM
[B] Synchronization is required between transmitter and receiver
[C] None of the above
[D] Both a and b
Answer: Both a and b
(27) As per sampling theorem –
[A] Twice each cycle of its highest frequency
[B] Guard time should be as large as possible
[C] The signal should be sampled at least twice each cycle of its lowest frequency
[D] Nyquist rate should be as low as possible
Answer: Twice each cycle of its highest frequency
(28) In communication the sampling technique leads to –
[A] Higher speed of communication
[B] cheaper equipment
[C] Higher efficiency
[D] All of the above
Answer: Higher speed of communication
(29) Impulse modulation systems, The number of samples required to ensure no loss of information is given by –
[A] Parsevals theorem
[B] Fourier transmission
[C] Nyquist theorem
[D] None of the these
Answer: Nyquist theorem
(30) If the sampling time is less than the Nyquist interval :
[A] Simpler filters can be used to obtain the original signal
[B] Channel Capacity increases
[C] Bandwidth increases
[D] Guard time becomes less
Answer: Simpler filters can be used to obtain the original signal
31 The scaling of a sequence x[n] by a factor α is given by
[A] y[n] = α [x[n]]2
[B] y[n] = α x[n2]
[C] y[n] = α x[n]
[D] y[n] = x[n]x[-n]
Answer: y[n] = α x[n]
32 Causal systems are the systems in which
[A] The output of the system depends on the present and the past inputs
[B] The output of the system depends only on the present inputs
[C] The output of the system depends only on the past inputs
[D] The output of the system depends on the present input as well as the previous outputs
Answer: The output of the system depends on the present and the past inputs
33 Roll-off factor is
[A] The bandwidth occupied beyond the Nyquist Bandwidth of the filter
[B] The performance of the filter or device
[C] Aliasing effect
[D] None of the above
Answer: The bandwidth occupied beyond the Nyquist Bandwidth of the filter
34 A signal x[n] is anti symmetric or odd when
[A] x[-n] = x[n] • x[n]
[B] x[n] = -x[n]
[C] x[n] = [x[n]]2
[D] x[-n] = -x[n]
Answer: x[-n] = -x[n]
35 A signal of maximum frequency of 10 kHz is sampled at Nyquist rate. The time interval between two successive sample is –
[A] 50 µs
[B] 100 µs
[C] 1000 µs
[D] 5 µs
Answer: 50 µs
(16) Time reversal of a discrete time signal refers to
[A] y[n] = x[-n+k]
[B] y[n] = x[-n]
[C] y[n] = x[-n-k]
[D] y[n] = x[n-k]
Answer: y[n] = x[-n]
37 Time shifting of discrete time signal means
[A] y[n] = x[n-k]
[B] y[n] = x[-n-k]
[C] y[n] = -x[n-k]
[D] y[n] = x[n+k]
Answer: y[n] = x[n-k]

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