Discrete Time Signal Analysis MCQ Quiz – Objective Question with Answer for Discrete Time Signal Analysis

1. What is the Fourier series representation of a signal x(n) whose period is N?

A. $$\sum_{k=0}^{N+1}c_k e^{j2πkn/N}$$

B. $$\sum_{k=0}^{N-1}c_k e^{j2πkn/N}$$

C. $$\sum_{k=0}^Nc_k e^{j2πkn/N}$$

D. $$\sum_{k=0}^{N-1}c_k e^{-j2πkn/N}$$

Here, the frequency F0 of a continuous-time signal is divided into 2π/N intervals.
So, the Fourier series representation of a discrete-time signal with period N is given as

x(n)=$$\sum_{k=0}^{N-1}c_k e^{j2πkn/N}$$

where ck is the Fourier series coefficient

2. What is the expression for Fourier series coefficient ck in terms of the discrete signal x(n)?

A. $$\frac{1}{N} \sum_{n=0}^{N-1}x(n)e^{j2πkn/N}$$

B. $$N\sum_{n=0}^{N-1}x(n)e^{-j2πkn/N}$$

C. $$\frac{1}{N} \sum_{n=0}^{N+1}x(n)e^{-j2πkn/N}$$

D. $$\frac{1}{N} \sum_{n=0}^{N-1}x(n)e^{-j2πkn/N}$$

We know that, the Fourier series representation of a discrete signal x(n) is given as

x(n)=$$\sum_{n=0}^{N-1}c_k e^{j2πkn/N}$$

Now multiply both sides by the exponential e-j2πln/N and summing the product from n=0 to n=N-1. Thus,

$$\sum_{n=0}^{N-1} x(n)e^{-j2πln/N}=\sum_{n=0}^{N-1}\sum_{k=0}^{N-1}c_k e^{j2π(k-l)n/N}$$

If we perform summation over n first in the right hand side of above equation, we get

$$\sum_{n=0}^{N-1} e^{-j2πkn/N}$$ = N, for k-l=0,±N,±2N…
= 0, otherwise

Therefore, the right hand side reduces to Nck
So, we obtain ck=$$\frac{1}{N} \sum_{n=0}^{N-1}x(n)e^{-j2πkn/N}$$

3. Which of the following represents the phase associated with the frequency component of discrete-time Fourier series(DTFS)?

A. ej2πkn/N
B. e-j2πkn/N
C. ej2πknN
D. none of the mentioned

We know that,

x(n)=$$\sum_{k=0}^{N-1}c_k e^{j2πkn/N}$$

In the above equation, ck represents the amplitude and ej2πkn/N represents the phase associated with the frequency component of DTFS.

4. The Fourier series for the signal x(n)=cos√2πn exists.

A. True
B. False

For ω0=√2π, we have f0=1/√2. Since f0 is not a rational number, the signal is not periodic. Consequently, this signal cannot be expanded in a Fourier series.

5. What are the Fourier series coefficients for the signal x(n)=cosπn/3?

A. c1=c2=c3=c4=0,c1=c5=1/2
B. c0=c1=c2=c3=c4=c5=0
C. c0=c1=c2=c3=c4=c5=1/2
D. none of the mentioned

In this case, f0=1/6 and hence x(n) is periodic with fundamental period N=6.
Given signal is x(n)=cosπn/3=cos2πn/6

=$$\frac{1}{2} e^{j2πn/6}+\frac{1}{2} e^{-j2πn/6}$$

We know that -2π/6=2π-2π/6=10π/6=5(2π/6)

∴ x(n)=$$\frac{1}{2} e^{j2πn/6}+\frac{1}{2} e^{j2π(5)n/6}$$

Compare the above equation with

x(n)=$$\sum_{k=0}^{N-1}c_k e^{j2πkn/N}$$

So, we get c1 = c2 = c3 = c4 = 0 and c1 = c5 = 1/2.

6. What is the Fourier series representation of a signal x(n) whose period is N?

A. $$\sum_{k=0}^{\infty}|c_k|^2$$

B. $$\sum_{k=-\infty}^{\infty}|c_k|$$

C. $$\sum_{k=-\infty}^0|c_k|^2$$

D. $$\sum_{k=-\infty}^{\infty}|c_k|^2$$

The average power of a periodic signal x(t) is given as

$$\frac{1}{T_p}\int_{t_0}^{t_0+T_p}|x(t)|^2 dt$$

=$$\frac{1}{T_p}\int_{t_0}^{t_0+T_p} x(t).x^* (t) dt$$

=$$\frac{1}{T_p}\int_{t_0}^{t_0+T_p}x(t).\sum_{k=-∞}^∞ c_k^* e^{-j2πkF_0 t} dt$$

By interchanging the positions of integral and summation and by applying the integration, we get

=$$\sum_{k=-∞}^∞|c_k |^2$$

7. What is the average power of the discrete time periodic signal x(n) with period N?

A. $$\frac{1}{N} \sum_{n=0}^{N}|x(n)|$$

B. $$\frac{1}{N} \sum_{n=0}^{N-1}|x(n)|$$

C. $$\frac{1}{N} \sum_{n=0}^{N}|x(n)|^2$$

D. $$\frac{1}{N} \sum_{n=0}^{N-1}|x(n)|^2$$

Let us consider a discrete-time periodic signal x(n) with period N.
The average power of that signal is given as

Px=$$\frac{1}{N} \sum_{n=0}^{N-1}|x(n)|^2$$

8. What is the equation for the average power of discrete-time periodic signal x(n) with period N in terms of Fourier series coefficient ck?

A. $$\sum_{k=0}^{N-1}|c_k|$$

B. $$\sum_{k=0}^{N-1}|c_k|^2$$

C. $$\sum_{k=0}^N|c_k|^2$$

D. $$\sum_{k=0}^N|c_k|$$

We know that

Px=$$\frac{1}{N} \sum_{n=0}^{N-1}|x(n)|^2$$

=$$\frac{1}{N} \sum_{n=0}^{N-1}x(n).x^*(n)$$

=$$\frac{1}{N} \sum_{n=0}^{N-1}x(n) \sum_{k=0}^{N-1}c_k * e^{-j2πkn/N}$$

=$$\sum_{k=0}^{N-1}c_k * \frac{1}{N} \sum_{n=0}^{N-1}x(n)e^{-j2πkn/N}$$

=$$\sum_{k=0}^{N-1}|c_k |^2$$

9. What is the Fourier transform X(ω) of a finite energy discrete time signal x(n)?

A. $$\sum_{n=-∞}^∞x(n)e^{-jωn}$$

B. $$\sum_{n=0}^∞x(n)e^{-jωn}$$

C. $$\sum_{n=0}^{N-1}x(n)e^{-jωn}$$

D. None of the mentioned

If we consider a signal x(n) which is discrete in nature and has finite energy, then the Fourier transform of that signal is given as
X(ω)=$$\sum_{n=-∞}^∞x(n)e^{-jωn}$$

10. What is the period of the Fourier transform X(ω) of the signal x(n)?

A. π
B. 1
C. Non-periodic
D. 2π

Let X(ω) be the Fourier transform of a discrete time signal x(n) which is given as

X(ω)=$$\sum_{n=-∞}^∞x(n)e^{-jωn}$$

Now X(ω+2πk)=$$\sum_{n=-∞}^∞ x(n)e^{-j(ω+2πk)n}$$

=$$\sum_{n=-∞}^∞ x(n)e^{-jωn}e^{-j2πkn}$$

=$$\sum_{n=-∞}^∞ x(n)e^{-jωn}= X(ω)$$

So, the Fourier transform of a discrete-time finite energy signal is periodic with a period 2π.

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