Updated Signals and Systems MCQs – Sampling Theorem MCQs ( Signals and Systems ) MCQs
Latest Signals and Systems MCQs
By practicing these MCQs of Sampling Theorem MCQs ( Signals and Systems ) MCQs – Latest Competitive MCQs , an individual for exams performs better than before. This post comprising of objective questions and answers related to “ Sampling Theorem MCQs ( Signals and Systems ) Mcqs “. As wise people believe “Perfect Practice make a Man Perfect”. It is therefore practice these mcqs of Signals and Systems to approach the success. Tab this page to check ” Sampling Theorem MCQs ( Signals and Systems )” for the preparation of competitive mcqs, FPSC mcqs, PPSC mcqs, SPSC mcqs, KPPSC mcqs, AJKPSC mcqs, BPSC mcqs, NTS mcqs, PTS mcqs, OTS mcqs, Atomic Energy mcqs, Pak Army mcqs, Pak Navy mcqs, CTS mcqs, ETEA mcqs and others.
Signals and Systems MCQs – Sampling Theorem MCQs ( Signals and Systems ) MCQs
The most occurred mcqs of Sampling Theorem MCQs ( Signals and Systems ) in past papers. Past papers of Sampling Theorem MCQs ( Signals and Systems ) Mcqs. Past papers of Sampling Theorem MCQs ( Signals and Systems ) Mcqs . Mcqs are the necessary part of any competitive / job related exams. The Mcqs having specific numbers in any written test. It is therefore everyone have to learn / remember the related Sampling Theorem MCQs ( Signals and Systems ) Mcqs. The Important series of Sampling Theorem MCQs ( Signals and Systems ) Mcqs are given below:
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
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 = 12fm=12sec.
2. Find the Nyquist rate and Nyquist interval of sinc[t].
a) 1 Hz, 1 sec
b) 2 Hz, 2 sec
c) 12 Hz, 2 sec
d) 2 Hz, 12sec
Answer: a
Explanation: We know that sinc[t] ↔ G2π(ω)
Here ωm = 2π
2πfm = π
∴ 2fm = 1
Nyquist rate, Fs = 2fm = 1 Hz
Nyquist interval, T = 12fm = 1 sec.
3. Find the Nyquist rate and Nyquist interval of Asinc[t].
a) 2 Hz, 2 sec
b) 1 Hz, 1 sec
c) 12 Hz, 1 sec
d) 1 Hz, 12 sec
Answer: b
Explanation: Nyquist rate and Nyquist interval are independent of Amplitude (magnitude scaling). But time scaling will change the rate.
We know that sinc[t] ↔ G2π(ω)
Here ωm = 2π
2πfm = π
∴ 2fm = 1
Nyquist rate, Fs = 2fm = 1 Hz
Nyquist interval, T = 12fm = 1 sec.
∴Fs = 1 Hz, T = 1 sec.
4. Find the Nyquist rate and Nyquist interval of sinc[200t].
a) 200 Hz, 1200 sec
b) 200 Hz, 200 sec
c) 1200 Hz, 200 sec
d) 100 Hz, 100 sec
Answer: a
Explanation: Here ωm=200π
2πfm=200π
2fm=200 Hz
Nyquist rate, Fs = 2fm = 200 Hz
Nyquist interval, T = 12fm=1200 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
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)=sin500πtπt.
a) 500 Hz, 2 sec
b) 500 Hz, 2 msec
c) 2 Hz, 500 sec
d) 2 Hz, 500 msec
Answer: b
Explanation: Given f(t) = sin500πtπt
Frequency, ωm = 500π
2πfm = 500π
2fm = 500 Hz
Nyquist rate, Fs = 2fm = 500 Hz
Nyquist interval, T = 12fm=1500 = 2 msec.
7. Find the Nyquist rate and Nyquist interval for the signal f(t) = [sin500πtπt]2.
a) 1000 Hz, 1 msec
b) 1 Hz, 1000 sec
c) 1000 Hz, 1 sec
d) 1000 Hz, 1000 sec
Answer: a
Explanation: Given f(t) = [sin500πtπt]2=1−cos1000πt(πt)2
Frequency, ωm = 1000π
2πfm = 1000π
2fm = 1000 Hz
Nyquist rate, Fs = 2fm = 1000 Hz
Nyquist interval, T = 12fm=11000 = 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
Explanation: Given f(t) = 1 + sinc300πt
Frequency, ωm = 300π
2πfm = 300π
2fm = 300 Hz
Nyquist rate, Fs = 2fm = 300 Hz
Nyquist interval, T = 12fm=1300 = 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
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
Answer: b
Explanation: Given Fs = 2000 samples per second
Nyquist interval, T = 1Fs=12000 = 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
Explanation: Given x(t) = 1 + cos 2000πt + sin 4000πt
Highest frequency component in 1 is zero
Highest frequency component in cos2000πt is ωm1 = 2000π
Highest frequency component in sin4000πt 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, 1340 sec
Answer: d
Explanation: sin 40πt has highest frequency ωm1 = 40π
cos300πt 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 = 1Fs=1340 sec.