# Signals & Systems Questions and Answers – Sampling

This set of Signals & Systems Multiple Choice Questions & Answers (MCQs) focuses 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

Explanation: We know that sin⁡ ω0 t ↔ jπ[δ(ω+ω0) – δ(ω-ω0)]
sin⁡ 2πt ↔ jπ[δ(ω+2π)-δ(ω-2π)]
Here ωm = 2π
But ωm = 2πfm
∴ fm = 1 Hz
Nyquist rate, Fs = 2fm = 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

Explanation: We know that sinc[t] ↔ G(ω)
Here ωm = 2π
2πfm = π
∴ 2fm = 1
Nyquist rate, Fs = 2fm = 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

Explanation: Nyquist rate and Nyquist interval are independent of Amplitude (magnitude scaling). But time scaling will change the rate.
We know that sinc[t] ↔ G(ω)
Here ωm = 2π
2πfm = π
∴ 2fm = 1
Nyquist rate, Fs = 2fm = 1 Hz
Nyquist interval, T = $$\frac{1}{2f_m}$$ = 1 sec.
∴Fs = 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

Explanation: Here ωm=200π
2πfm=200π
2fm=200 Hz
Nyquist rate, Fs = 2fm = 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

Explanation: 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.
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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

Explanation: Given f(t) = $$\frac{sin⁡500πt}{πt}$$
Frequency, ωm = 500π
2πfm = 500π
2fm = 500 Hz
Nyquist rate, Fs = 2fm = 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

Explanation: Given f(t) = $$\Big[\frac{sin⁡500πt}{πt}\Big]^2 = \frac{1-cos⁡1000πt}{(πt)^2}$$
Frequency, ωm = 1000π
2πfm = 1000π
2fm = 1000 Hz
Nyquist rate, Fs = 2fm = 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

Explanation: Given f(t) = 1 + sinc300πt
Frequency, ωm = 300π
2πfm = 300π
2fm = 300 Hz
Nyquist rate, Fs = 2fm = 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

Explanation: 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 Fs = 2000 samples per second. Find its Nyquist interval.
a) 0.5 sec
b) 5 msec
c) 5 sec
d) 0.5 msec

Explanation: Given Fs = 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

Explanation: Given x(t) = 1 + cos 2000πt + sin⁡ 4000πt
Highest frequency component in 1 is zero
Highest frequency component in cos⁡2000πt is ωm1 = 2000π
Highest frequency component in sin⁡4000πt is ωm2 = 4000π
So the maximum frequency component in x(t) is ωm = 4000π [highest of 0, 2000π, 4000π]
∴ 2πfm = 4000π
2fm = 4000
Nyquist rate, Fs = 2fm = 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

Explanation: sin ⁡40πt has highest frequency ωm1 = 40π
cos⁡300πt has highest frequency ωm2 = 300π
As we know, multiplication in time domain is equivalent to convolution in frequency domain, the convoluted spectra will have highest frequency component ωm = ωm1 + ωm2 = 40π + 300π
ωm = 340π
2πfm = 340π
2fm = 340
Nyquist rate, Fs = 2fm = 340 Hz
Nyquist interval, T = $$\frac{1}{F_s} =\frac{1}{340}$$ sec.

Sanfoundry Global Education & Learning Series – Signals & Systems.

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