250+ TOP MCQs on Properties of DFT & Answers

Digital Signal Processing Multiple Choice Questions 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
Answer: A
Clarification: 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
Answer: C
Clarification: 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. e^akX₁(k)+e^bkX₂(k)
D. None of the mentioned
Answer: B
Clarification: 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})
=(asum_{n=0}^{N-1} x_1(n)e^{-j2πkn/N}+bsum_{n=0}^{N-1}x_2(n)e^{-j2πkn/N})
=>X(k)=aX₁(k)+bX₂(k).

4. If x(n) is a complex valued sequence given by x(n)=x_R(n)+jx_I(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})
Answer: D
Clarification: Given x(n)=x_R(n)+jx_I(n)=>x_R(n)=1/2(x(n)+x*(n))
Substitute the above equation in the DFT expression
Thus we get, X_R(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

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Answer: D
Clarification: 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)

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
Answer: B
Clarification: Given x(n) is real and even, that is x(n)=x(N-n)
We know that X_I(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
Answer: A
Clarification: If x(n) is real and odd, that is x(n)=-x(N-n), then X_R(k)=0. Hence X(k) is purely imaginary and odd. Since X_R(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})

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 )
Answer: C
Clarification: 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 ).

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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}
Answer: D
Clarification: 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}.

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
Answer: B
Clarification: 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
Answer: A
Clarification: 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
Answer: C
Clarification: 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).

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