Digital Signal Processing Multiple Choice Questions 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})
Answer: B
Clarification: 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. (Nsum_{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
Clarification: 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
Answer: A
Clarification: 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
Clarification: 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
Clarification: 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)
Answer: B
Clarification: 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 )
Answer: D
Clarification: 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|)
Answer: B
Clarification: 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
Answer: A
Clarification: 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
Clarification: 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. (2πint_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
Answer: C
Clarification: 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.π
B. -ωc/π
C. ωc/π
D. none of the mentioned
Answer: C
Clarification: We know that, x(n)=(frac{1}{2pi} int_{-pi}^{pi}X(omegA.e^{jomega n} dω)
=(frac{1}{2pi} int_{-ω_c}^{ω_c}1.e^{jomega n} dω)
At n=0,
x(n)=x(0)=(int_{-ω_c}^{ω_c}1 dω=frac{1}{2pi}(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
Answer: A
Clarification: We know that, x(n)=(frac{1}{2pi} int_{-pi}^{pi}X(omegA.e^{jomega n} dω)
=(frac{1}{2pi} int_{-ω_c}^{ω_c}1.e^{jomega 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
Answer: A
Clarification: 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. (2πint_{-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
Answer: B
Clarification: 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)