# Digital Signal Processing Questions and Answers – Frequency Analysis of Discrete Time Signal – 1

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This set of Digital Signal Processing Multiple Choice Questions & Answers (MCQs) focuses on “Frequency Analysis of Discrete Time Signals-1”.

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

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

Explanation: 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

Explanation: 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.
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4. The Fourier series for the signal x(n)=cos√2πn exists.
a) True
b) False

Explanation: 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

Explanation: 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)
Therefore, 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) $$\sum_{k=-\infty}^{\infty}|c_k|^2$$

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

Explanation: 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 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) $$\sum_{k=0}^N|c_k|$$

Explanation: 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

Explanation: 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

Explanation: 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 period 2π.

11. What is the synthesis equation of the discrete time signal x(n), whose Fourier transform is X(ω)?
a) $$2π\int_0^2π X(ω) e^jωn dω$$
b) $$\frac{1}{π} \int_0^{2π} X(ω) e^jωn dω$$
c) $$\frac{1}{2π} \int_0^{2π} X(ω) e^jωn dω$$
d) None of the mentioned

Explanation: We know that the Fourier transform of the discrete time signal x(n) is
X(ω)=$$\sum_{n=-∞}^∞ x(n)e^{-jωn}$$
By calculating the inverse Fourier transform of the above equation, we get
x(n)=$$\frac{1}{2π} \int_0^{2π} X(ω) e^{jωn} dω$$
The above equation is known as synthesis equation or inverse transform equation.

12. What is the value of discrete time signal x(n) at n=0 whose Fourier transform is represented as below?
a) ωc
b) -ωc
c) ωc
d) none of the mentioned

Explanation: We know that, x(n)=$$\frac{1}{2\pi} \int_{-\pi}^{\pi}X(\omega)e^{j\omega n} dω$$
=$$\frac{1}{2\pi} \int_{-ω_c}^{ω_c}1.e^{j\omega n} dω$$
At n=0,
x(n)=x(0)=$$\int_{-ω_c}^{ω_c}1 dω=\frac{1}{2\pi}(2 ω_c)=\frac{ω_c}{\pi_ω}$$
Therefore, the value of the signal x(n) at n=0 is ωc/π.

13. What is the value of discrete time signal x(n) at n≠0 whose Fourier transform is represented as below?
a) $$\frac{ω_c}{\pi}.\frac{sin ω_c.n}{ω_c.n}$$
b) $$\frac{-ω_c}{\pi}.\frac{sin ω_c.n}{ω_c.n}$$
c) $$ω_c.\pi \frac{sin ω_c.n}{ω_c.n}$$
d) None of the mentioned

Explanation: We know that, x(n)=$$\frac{1}{2\pi} \int_{-\pi}^{\pi}X(\omega)e^{j\omega n} dω$$
=$$\frac{1}{2\pi} \int_{-ω_c}^{ω_c}1.e^{j\omega n} dω=\frac{sin ω_c.n}{ω_c.n}$$ =$$\frac{ω_c}{\pi}.\frac{sin ω_c.n}{ω_c.n}$$

14. The oscillatory behavior of the approximation of XN(ω) to the function X(ω) at a point of discontinuity of X(ω) is known as Gibbs phenomenon.
a) True
b) False

Explanation: We note that there is a significant oscillatory overshoot at ω=ωc, independent of the value of N. As N increases, the oscillations become more rapid, but the size of the ripple remains the same. One can show that as N→∞, the oscillations converge to the point of the discontinuity at ω=ωc. The oscillatory behavior of the approximation of XN(ω) to the function X(ω) at a point of discontinuity of X(ω) is known as Gibbs phenomenon.

15. What is the energy of a discrete time signal in terms of X(ω)?
a) $$2π\int_{-π}^π |X(ω)|^2 dω$$
b) $$\frac{1}{2π} \int_{-π}^π |X(ω)|^2 dω$$
c) $$\frac{1}{2π} \int_0^π |X(ω)|^2 dω$$
d) None of the mentioned

Explanation: We know that, Ex=$$\sum_{n=-∞}^∞ |x(n)|^2$$
=$$\sum_{n=-∞}^∞ x(n).x^*(n)$$
=$$\sum_{n=-∞}^∞ x(n)\frac{1}{2π} \int_{-π}^π X^*(ω) e^{-jωn} dω$$
=$$\frac{1}{2π} \int_{-π}^π|X(ω)|^2 dω$$

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