15. 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
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)
16. 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
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
17. 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
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
18. 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
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) = \(\sum_{n=0}^{N-1}x_1 (n) x_2 (m-n)_N \).
19. 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
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}.
20. 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
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}.
21. If X(k) is the N-point DFT of a sequence x(n), then the circular time-shift property is that the N-point DFT of x((n-l))N is X(k)e-j2πkl/N.
A. True
B. False
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.
22. 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
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).
