Digital Signal Processing Questions and Answers – Properties of DFT

This set of Digital Signal Processing Multiple Choice Questions & Answers (MCQs) focuses on “Properties of DFT”.

1. If x(n) and X(k) are an N-point DFT pair, then x(n+N)=x(n).
a) True
b) False
View Answer

Answer: a
Explanation: We know that the expression for an DFT is given as
X(k)=\(\sum_{n=0}^{N-1} x(n)e^{-j2πkn/N}\)
Now take x(n)=x(n+N)=>X1(k)=\(\sum_{n=0}^{N-1} x(n+N)e^{-j2πkn/N}\)
Let n+N=l=>X1(k)=\(\sum_{l=N}^0 x(l)e^{-j2πkl/N}\)=X(k)
Therefore, we got x(n)=x(n+N)

2. If x(n) and X(k) are an N-point DFT pair, then X(k+N)=?
a) X(-k)
b) -X(k)
c) X(k)
d) None of the mentioned
View Answer

Answer: c
Explanation: We know that
x(n)=\(\frac{1}{N}\sum_{k=0}^{N-1} x(k)e^{j2πkn/N}\)
Let X(k)=X(k+N)
=>x1(n)=\(\frac{1}{N} \sum_{k=0}^{N-1}X(k+N)e^{j2πkn/N}\)=x(n)
Therefore, we have X(k)=X(k+N)

3. If X1(k) and X2(k) are the N-point DFTs of X1(n) and x2(n) respectively, then what is the N-point DFT of x(n)=ax1(n)+bx2(n)?
a) X1(ak)+X2(bk)
b) aX1(k)+bX2(k)
c) eakX1(k)+ebkX2(k)
d) None of the mentioned
View Answer

Answer: b
Explanation: We know that, the DFT of a signal x(n) is given by the expression
X(k)=\(\sum_{n=0}^{N-1}x(n)e^{-j2πkn/N}\)
Given x(n)=ax1(n)+bx2(n)
=>X(k)= \(\sum_{n=0}^{N-1}(ax_1(n)+bx_2(n))e^{-j2πkn/N}\)
=\(a\sum_{n=0}^{N-1} x_1(n)e^{-j2πkn/N}+b\sum_{n=0}^{N-1}x_2(n)e^{-j2πkn/N}\)
=>X(k)=aX1(k)+bX2(k).
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4. If x(n) is a complex valued sequence given by x(n)=xR(n)+jxI(n), then what is the DFT of xR(n)?
a) \(\sum_{n=0}^N x_R (n) cos⁡\frac{2πkn}{N}+x_I (n) sin⁡\frac{2πkn}{N}\)
b) \(\sum_{n=0}^N x_R (n) cos⁡\frac{2πkn}{N}-x_I (n) sin⁡\frac{2πkn}{N}\)
c) \(\sum_{n=0}^{N-1} x_R (n) cos⁡\frac{2πkn}{N}-x_I (n) sin⁡\frac{2πkn}{N}\)
d) \(\sum_{n=0}^{N-1} x_R (n) cos⁡\frac{2πkn}{N}+x_I (n) sin⁡\frac{2πkn}{N}\)
View Answer

Answer: d
Explanation: Given x(n)=xR(n)+jxI(n)=>xR(n)=1/2(x(n)+x*(n))
Substitute the above equation in the DFT expression
Thus we get, XR(k)=\(\sum_{n=0}^{N-1} x_R (n) cos⁡\frac{2πkn}{N}+x_I (n) sin⁡\frac{2πkn}{N}\)

5. If x(n) is a real sequence and X(k) is its N-point DFT, then which of the following is true?
a) X(N-k)=X(-k)
b) X(N-k)=X*(k)
c) X(-k)=X*(k)
d) All of the mentioned
View Answer

Answer: d
Explanation: We know that
X(k)=\(\sum_{n=0}^{N-1} x(n)e^{-j2πkn/N}\)
Now X(N-k)=\(\sum_{n=0}^{N-1} x(n)e^{-j2π(N-k)n/N}\)=X*(k)=X(-k)
Therefore,
X(N-k)=X*(k)=X(-k)
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6. If x(n) is real and even, then what is the DFT of x(n)?
a) \(\sum_{n=0}^{N-1} x(n) sin⁡\frac{2πkn}{N}\)
b) \(\sum_{n=0}^{N-1} x(n) cos⁡\frac{2πkn}{N}\)
c) -j\(\sum_{n=0}^{N-1} x(n) sin⁡\frac{2πkn}{N}\)
d) None of the mentioned
View Answer

Answer: b
Explanation: Given x(n) is real and even, that is x(n)=x(N-n)
We know that XI(k)=0. Hence the DFT reduces to
X(k)=\(\sum_{n=0}^{N-1} x(n) cos⁡\frac{2πkn}{N}\) ;0 ≤ k ≤ N-1

7. If x(n) is real and odd, then what is the IDFT of the given sequence?
a) \(j \frac{1}{N} \sum_{k=0}^{N-1} x(k) sin⁡\frac{2πkn}{N}\)
b) \(\frac{1}{N} \sum_{k=0}^{N-1} x(k) cos⁡\frac{2πkn}{N}\)
c) \(-j \frac{1}{N} \sum_{k=0}^{N-1} x(k) sin⁡\frac{2πkn}{N}\)
d) None of the mentioned
View Answer

Answer: a
Explanation: If x(n) is real and odd, that is x(n)=-x(N-n), then XR(k)=0. Hence X(k) is purely imaginary and odd. Since XR(k) reduces to zero, the IDFT reduces to
\(x(n)=j \frac{1}{N} \sum_{k=0}^{N-1} x(k) sin⁡\frac{2πkn}{N}\)
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8. If X1(n), x2(n) and x3(m) are three sequences each of length N whose DFTs are given as X1(k), X2(k) and X3(k) respectively and X3(k)=X1(k).X2(k), then what is the expression for x3(m)?
a) \(\sum_{n=0}^{N-1}x_1 (n) x_2 (m+n)\)
b) \(\sum_{n=0}^{N-1}x_1 (n) x_2 (m-n)\)
c) \(\sum_{n=0}^{N-1}x_1 (n) x_2 (m-n)_N \)
d) \(\sum_{n=0}^{N-1}x_1 (n) x_2 (m+n)_N \)
View Answer

Answer: c
Explanation: If X1(n), x2(n) and x3(m) are three sequences each of length N whose DFTs are given as X1(k), x2(k) and X3(k) respectively and X3(k)=X1(k).X2(k), then according to the multiplication property of DFT we have x3(m) is the circular convolution of X1(n) and x2(n).
That is x3(m) = \(\sum_{n=0}^{N-1}x_1 (n) x_2 (m-n)_N \).

9. What is the circular convolution of the sequences X1(n)={2,1,2,1} and x2(n)={1,2,3,4}?
a) {14,14,16,16}
b) {16,16,14,14}
c) {2,3,6,4}
d) {14,16,14,16}
View Answer

Answer: d
Explanation: We know that the circular convolution of two sequences is given by the expression
x(m)= \(\sum_{n=0}^{N-1}x_1 (n) x_2 (m-n)_N\)
For m=0, x2((-n))4={1,4,3,2}
For m=1, x2((1-n))4={2,1,4,3}
For m=2, x2((2-n))4={3,2,1,4}
For m=3, x2((3-n))4={4,3,2,1}
Now we get x(m)={14,16,14,16}.
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10. What is the circular convolution of the sequences X1(n)={2,1,2,1} and x2(n)={1,2,3,4}, find using the DFT and IDFT concepts?
a) {16,16,14,14}
b) {14,16,14,16}
c) {14,14,16,16}
d) None of the mentioned
View Answer

Answer: b
Explanation: Given X1(n)={2,1,2,1}=>X1(k)=[6,0,2,0]
Given x2(n)={1,2,3,4}=>X2(k)=[10,-2+j2,-2,-2-j2]
when we multiply both DFTs we obtain the product
X(k)=X1(k).X2(k)=[60,0,-4,0]
By applying the IDFT to the above sequence, we get
x(n)={14,16,14,16}.

11. If X(k) is the N-point DFT of a sequence x(n), then circular time shift property is that N-point DFT of x((n-l))N is X(k)e-j2πkl/N.
a) True
b) False
View Answer

Answer: a
Explanation: According to the circular time shift property of a sequence, If X(k) is the N-point DFT of a sequence x(n), then the N-pint DFT of x((n-l))N is X(k)e-j2πkl/N.

12. If X(k) is the N-point DFT of a sequence x(n), then what is the DFT of x*(n)?
a) X(N-k)
b) X*(k)
c) X*(N-k)
d) None of the mentioned
View Answer

Answer: c
Explanation: According to the complex conjugate property of DFT, we have if X(k) is the N-point DFT of a sequence x(n), then what is the DFT of x*(n) is X*(N-k).

Sanfoundry Global Education & Learning Series – Digital Signal Processing.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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