250+ TOP MCQs on Discrete-Time Fourier Transform and Answers

Signals & Systems Multiple Choice Questions on “Discrete-Time Fourier Transform”.

1. Given a discrete time signal x[k] defined by x[k] = 1, for -2≤k≤2 and 0, for |k|>2. Then, y[k] = x[3k-2] is ______________
A. y[k] = 1, for k = 0, 1 and 0 otherwise
B. y[k] = 1, for k = 1 and -1 for k=-1
C. y[k] = 1, for k = 0, 1 and -1 otherwise
D. y[k] = 1, for k = 0, 1 and 0 otherwise
Answer: A
Clarification: y[k] = x [3k-2]
Now, y [0] = x [-2] = 1
Or, y [1] = x [1] = 1
Or, y [2] = x [4] = 0
∴y[k] = 1, for k = 0, 1 and 0 otherwise.

2. The time system which operates with a continuous time signal and produces a continuous time output signal is _________
A. CTF system
B. DTF System
C. Time invariant System
D. Time variant System
Answer: A
Clarification: DTF System operates with a discrete signal, on the other hand time invariant system is a system whose output does not depend explicitly on time. For continuous time system, the inputs as well as output both are CT signals.

3. A discrete time signal is given as X [n] = cos (frac{πn}{9}) + sin ((frac{πn}{7} + frac{1}{2})). The period of the signal X [n] is ______________
A. 126
B. 32
C. 252
D. Non-periodic
Answer: A
Clarification: Given that, N1 = 18, N2 = 14
We know that period of X [n] (say N) = LCM (N1, N2)
∴ Period of X [n] = LCM (18, 14) = 126.

4. What is the steady state value of The DT signal F (t), if it is known that F(s) = (frac{1}{(s+2)^2 (s+4)})?
A. (frac{1}{16})
B. Cannot be determined
C. 0
D. (frac{1}{8})
Answer: C
Clarification: The steady state value of the DT signal F(s) exists since all poles of the given Laplace transform have negative real part.
∴F (∞) = lims→0 s F(s)
= lims→0 (frac{s}{(s+2)^2 (s+4)})
= 0.

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5. F(t) and G(t) are the one-sided z-transforms of discrete time functions f(nt) and g(nt), the z-transform of ∑f(kt)g(nt-kt) is given by _____________
A. ∑f(nt)g(nt)z-n
B. ∑f(nt)g(nt)zn
C. ∑f(kt)g(nt-kt) z-n
D. ∑f(nt-kt)g(nt)z-n
Answer: A
Clarification: Given that F (t) and G (t) are the one-sided z-transforms.
Also, f (nt) and g (nt) are discrete time functions, which means that property of Linearity, time shifting and time scaling will be similar to that of continuous Fourier transform. Since, for a continuous Fourier transform, the value of ∑f(kt)g(nt-kt) is given by∑f(nt)g(nt)z-n.
∴ z-transform of ∑f(kt)g(nt-kt) is given by∑f(nt)g(nt)z-n.

6. A discrete time signal is as given below
X [n] = cos ((frac{n}{8})) cos ((frac{πn}{8}))
The period of the signal X [n] is _____________
A. 16 π
B. 16(π+1)
C. 8
D. Non-periodic
Answer: D
Clarification: We know that for X [n] = X1 [n] × X2 [n] to be periodic, both X1 [n] and X2 [n] should be periodic with finite periods.
Here X2 [n] = cos ((frac{πn}{8})), is periodic with fundamental period as 8/n
But X1 [n] = cos ((frac{n}{8})) is non periodic.
∴ X [n] is a non-periodic signal.

7. A Discrete signal is said to be even or symmetric if X(-n) is equal to __________
A. X(n)
B. 0
C. –X(n)
D. –X(-n)
Answer: A
Clarification: We know that any signal be it discrete or continuous is said to be even or symmetric when that signal f(x) = f (-x). Here given signal is X (n). It is a discrete time signal. So, the signal will be even symmetric if X (n) = X (-n).

8. The system described by the difference equation y(n) – 2y(n-1) + y(n-2) = X(n) – X(n-1) has y(n) = 0 and n<0. If x (n) = δ(n), then y (z) will be?
A. 2
B. 1
C. 0
D. -1

Answer: C
Clarification: Given equation = y (n) – 2y (n-1) + y (n-2) = X (n) – X (n-1) has y (n) = 0
For n = 0, y (0)2y (-1) + y (-2) = x (0) – x (-1)
∴ y(0) = x(0) – x(-1)
Or, y (n) = 0 for n<0
For n=1, y (1) = -2y (0) + y (-1) = x (1) – x (0)
Or, y (1) = x (1) – x (0) + 2x (0) – 2x (-1)
Or, y (1) = x (1) +x (0) – 2x (-1)
For n=2, y (2) = x (2) – x (1) + 2y (1) – y (0)
Or, y(2) = x(2) – x(1) + 2x(1) + 2x(0) – 4x(-1) – x(0) + x(-1)
∴y (2) = d (2) + d (1) + d (0) – 3d (-1).

9. A discrete time signal is as given below
X [n] = cos ((frac{πn}{2})) – sin ((frac{πn}{8})) + 3 cos ((frac{πn}{4} + frac{π}{3}))
The period of the signal X [n] is _____________
A. 16
B. 4
C. 2
D. Non-periodic
Answer: A
Clarification: Given that, N1 = 4, N2 = 16, N3 = 8
We know that period of X [n] (say N) = LCM (N1, N2, N3)
∴ Period of X [n] = LCM (4, 16, 8) = 16.

10. The Nyquist frequency for the signal x (t) = 3 cos 50πt + 10 sin 300πt – cos 100t is ___________
A. 50 Hz
B. 100 Hz
C. 200 Hz
D. 300 Hz
Answer: D
Clarification: We know that Nyquist frequency is twice the maximum frequency, i.e. fs = 2 fm.
The maximum frequency present in the signal is ωm = 300 π or fm = 150 Hz. Therefore the Nyquist frequency fs = 2 fm = 300 Hz.

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