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

Answer: B

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

Answer: D

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,

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

Answer: A

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

Answer: B

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

Answer: A

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

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

Answer: A

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

Answer: D

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.