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

View Answer

Explanation: We know that sin ω

_{0}t ↔ jπ[δ(ω+ω

_{0}) – δ(ω-ω

_{0})]

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

View Answer

Explanation: 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

View Answer

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

_{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

View Answer

Explanation: 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

View Answer

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.

6. Find the Nyquist rate and Nyquist interval for the signal f(t)=\(\frac{sin500πt}{πt}\).

a) 500 Hz, 2 sec

b) 500 Hz, 2 msec

c) 2 Hz, 500 sec

d) 2 Hz, 500 msec

View Answer

Explanation: Given f(t) = \(\frac{sin500πt}{πt}\)

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) = \(\Big[\frac{sin500πt}{πt}\Big]^2\).

a) 1000 Hz, 1 msec

b) 1 Hz, 1000 sec

c) 1000 Hz, 1 sec

d) 1000 Hz, 1000 sec

View Answer

Explanation: Given f(t) = \(\Big[\frac{sin500πt}{πt}\Big]^2 = \frac{1-cos1000πt}{(πt)^2}\)

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

View Answer

Explanation: 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

View Answer

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 F_{s} = 2000 samples per second. Find its Nyquist interval.

a) 0.5 sec

b) 5 msec

c) 5 sec

d) 0.5 msec

View Answer

Explanation: 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

View Answer

Explanation: Given x(t) = 1 + cos 2000πt + sin 4000πt

Highest frequency component in 1 is zero

Highest frequency component in cos2000πt is ω

_{m1}= 2000π

Highest frequency component in sin4000πt 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

View Answer

Explanation: sin 40πt has highest frequency ω

_{m1}= 40π

cos300πt 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.

**Sanfoundry Global Education & Learning Series – Signals & Systems.**

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