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

View Answer

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

View Answer

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

View Answer

Explanation: Given that, N

_{1}= 18, N

_{2}= 14

We know that period of X [n] (say N) = LCM (N

_{1}, N

_{2})

∴ 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) \(\frac{1}{8}\)

View Answer

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 (∞) = lim

_{s→0}s F(s)

= lim

_{s→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)z^{n}

c) ∑f(kt)g(nt-kt) z^{-n}

d) ∑f(nt-kt)g(nt)z^{-n}

View Answer

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

View Answer

Explanation: We know that for X [n] = X

_{1}[n] × X

_{2}[n] to be periodic, both X

_{1}[n] and X

_{2}[n] should be periodic with finite periods.

Here X

_{2}[n] = cos (\(\frac{πn}{8}\)), is periodic with fundamental period as 8/n

But X

_{1}[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)

View Answer

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

View Answer

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

View Answer

Explanation: Given that, N

_{1}= 4, N

_{2}= 16, N

_{3}= 8

We know that period of X [n] (say N) = LCM (N

_{1}, N

_{2}, N

_{3})

∴ 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

View Answer

Explanation: We know that Nyquist frequency is twice the maximum frequency, i.e. f

_{s}= 2 f

_{m}.

The maximum frequency present in the signal is ω

_{m}= 300 π or f

_{m}= 150 Hz. Therefore the Nyquist frequency f

_{s}= 2 f

_{m}= 300 Hz.

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

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