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

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

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

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.

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.

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.

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

=\(\sum_{k=-∞}^∞|c_k |^2\)

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

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

We know that

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

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

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

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

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

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)=\(\sum_{n=-∞}^∞x(n)e^{-jωn}\)

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.