250+ TOP MCQs on Sampling and Answers

Signals & Systems Multiple Choice Questions on “Sampling”.

1. Find the Nyquist rate and Nyquist interval of sin(2πt).
A. 2 Hz, (frac{1}{2}) sec
B. (frac{1}{2}) Hz, (frac{1}{2}) sec
C. (frac{1}{2}) Hz, 2 sec
D. 2 Hz, 2 sec

Answer: A
Clarification: We know that sin⁡ ω₀ t ↔ jπ[δ(ω+ω₀) – δ(ω-ω₀)]
sin⁡ 2πt ↔ jπ[δ(ω+2π)-δ(ω-2π)]
Here ω_m = 2π
But ω_m = 2πf_m
∴ f_m = 1 Hz
Nyquist rate, F_s = 2f_m = 2 Hz
Nyquist interval, T = (frac{1}{2f_m} = frac{1}{2} )sec.

2. Find the Nyquist rate and Nyquist interval of sinc[t].
A. 1 Hz, 1 sec
B. 2 Hz, 2 sec
C. (frac{1}{2}) Hz, 2 sec
D. 2 Hz, (frac{1}{2})sec

Answer: A
Clarification: We know that sinc[t] ↔ G_2π(ω)
Here ω_m = 2π
2πf_m = π
∴ 2f_m = 1
Nyquist rate, F_s = 2f_m = 1 Hz
Nyquist interval, T = (frac{1}{2f_m}) = 1 sec.

3. Find the Nyquist rate and Nyquist interval of Asinc[t].
A. 2 Hz, 2 sec
B. 1 Hz, 1 sec
C. (frac{1}{2}) Hz, 1 sec
D. 1 Hz, (frac{1}{2}) sec

Answer: B
Clarification: Nyquist rate and Nyquist interval are independent of Amplitude (magnitude scaling). But time scaling will change the rate.
We know that sinc[t] ↔ G_2π(ω)
Here ω_m = 2π
2πf_m = π
∴ 2f_m = 1
Nyquist rate, F_s = 2f_m = 1 Hz
Nyquist interval, T = (frac{1}{2f_m}) = 1 sec.
∴F_s = 1 Hz, T = 1 sec.

4. Find the Nyquist rate and Nyquist interval of sinc[200t].
A. 200 Hz, (frac{1}{200}) sec
B. 200 Hz, 200 sec
C. (frac{1}{200}) Hz, 200 sec
D. 100 Hz, 100 sec

Answer: A
Clarification: Here ω_m=200π
2πf_m=200π
2f_m=200 Hz
Nyquist rate, F_s = 2f_m = 200 Hz
Nyquist interval, T = (frac{1}{2f_m} = frac{1}{200}) sec.

5. Which of the following is the process of ‘aliasing’?
A. Peaks overlapping
B. Phase overlapping
C. Amplitude overlapping
D. Spectral overlapping

Answer: D
Clarification: Aliasing is defined as the phenomenon in which a high frequency component in the frequency spectrum of the signal takes the identity of a lower frequency component in the spectrum of the sampled signal.
Aliasing can occur if either of the following condition exists:
• The signal is not band-limited to a finite range.
• The sampling rate is too low.

6. Find the Nyquist rate and Nyquist interval for the signal f(t)=(frac{sin⁡500πt}{πt}).
A. 500 Hz, 2 sec
B. 500 Hz, 2 msec
C. 2 Hz, 500 sec
D. 2 Hz, 500 msec

Answer: B
Clarification: Given f(t) = (frac{sin⁡500πt}{πt})
Frequency, ω_m = 500π
2πf_m = 500π
2f_m = 500 Hz
Nyquist rate, F_s = 2f_m = 500 Hz
Nyquist interval, T = (frac{1}{2f_m} = frac{1}{500}) = 2 msec.

7. Find the Nyquist rate and Nyquist interval for the signal f(t) = (Big[frac{sin⁡500πt}{πt}Big]^2).
A. 1000 Hz, 1 msec
B. 1 Hz, 1000 sec
C. 1000 Hz, 1 sec
D. 1000 Hz, 1000 sec

Answer: A
Clarification: Given f(t) = (Big[frac{sin⁡500πt}{πt}Big]^2 = frac{1-cos⁡1000πt}{(πt)^2})
Frequency, ω_m = 1000π
2πf_m = 1000π
2f_m = 1000 Hz
Nyquist rate, F_s = 2f_m = 1000 Hz
Nyquist interval, T = (frac{1}{2f_m} = frac{1}{1000}) = 1 msec.

8. Find the Nyquist rate and Nyquist interval for the signal f(t) = 1 + sinc300πt.
A. 300 Hz, 3 msec
B. 300 Hz, 3.3 msec
C. 30 Hz, 3 msec
D. 3 Hz, 3 msec

Answer: B
Clarification: Given f(t) = 1 + sinc300πt
Frequency, ω_m = 300π
2πf_m = 300π
2f_m = 300 Hz
Nyquist rate, F_s = 2f_m = 300 Hz
Nyquist interval, T = (frac{1}{2f_m} = frac{1}{300}) = 3.3 msec.

9. Find the Nyquist rate and Nyquist interval for the signal f(t) = rect(200t).
A. ∞ Hz, 0 sec
B. 0 Hz, ∞ sec
C. ∞ Hz, ∞ Hz
D. 0 Hz, 0 sec

Answer: A
Clarification: Given f(t) = rect(200t), which is a rectangular pulse signal having pulse width of 1/200 seconds. Since the signal is a finite duration signal, it is not band-limited. The signal spectrum consists of infinite frequencies.
Hence, Nyquist rate is infinity and Nyquist interval is zero.

10. The sampling frequency of a signal is F_s = 2000 samples per second. Find its Nyquist interval.
A. 0.5 sec
B. 5 msec
C. 5 sec
D. 0.5 msec

Answer: B
Clarification: Given F_s = 2000 samples per second
Nyquist interval, T = (frac{1}{F_s} = frac{1}{2000}) = 0.5 msec.

11. Determine the Nyquist rate of the signal x(t) = 1 + cos⁡ 2000πt + sin⁡ 4000πt.
A. 2000 Hz
B. 4000 Hz
C. 1 Hz
D. 6000 Hz

Answer: B
Clarification: Given x(t) = 1 + cos 2000πt + sin⁡ 4000πt
Highest frequency component in 1 is zero
Highest frequency component in cos⁡2000πt is ω_m1 = 2000π
Highest frequency component in sin⁡4000πt is ω_m2 = 4000π
So the maximum frequency component in x(t) is ω_m = 4000π [highest of 0, 2000π, 4000π]
∴ 2πf_m = 4000π
2f_m = 4000
Nyquist rate, F_s = 2f_m = 4000 Hz.

12. Find the Nyquist rate and Nyquist interval for the signal f(t) = -10 sin ⁡40πt cos ⁡300πt.
A. 40 Hz, 40 sec
B. 340 Hz, 340 sec
C. 300 Hz, 300 sec
D. 340 Hz, (frac{1}{340}) sec

Answer: D
Clarification: sin ⁡40πt has highest frequency ω_m1 = 40π
cos⁡300πt has highest frequency ω_m2 = 300π
As we know, multiplication in time domain is equivalent to convolution in frequency domain, the convoluted spectra will have highest frequency component ω_m = ω_m1 + ω_m2 = 40π + 300π
ω_m = 340π
2πf_m = 340π
2f_m = 340
Nyquist rate, F_s = 2f_m = 340 Hz
Nyquist interval, T = (frac{1}{F_s} =frac{1}{340}) sec.