Signals & Systems Questions and Answers – Discrete-Time Fourier Transform

«
»

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

Explanation: 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

Explanation: 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

Explanation: 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) $$\frac{1}{8}$$

Explanation: 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.

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

Explanation: 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

Explanation: 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)

Explanation: 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

Explanation: 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

Explanation: 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

Explanation: 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.

Sanfoundry Global Education & Learning Series – Signals & Systems.

To practice all areas of Signals & Systems, here is complete set of 1000+ Multiple Choice Questions and Answers.