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

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Explanation: We know that the expression for an DFT is given as

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

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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^{ak}X1(k)+e^{bk}X2(k)

d) None of the mentioned

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Explanation: We know that, the DFT of a signal x(n) is given by the expression

=>X(k)= aX1(k)+bX2(k).

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)?

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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,

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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6. If x(n) is real and even, then what is the DFT of x(n)?

d) None of the mentioned

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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

7. If x(n) is real and odd, then what is the IDFT of the given sequence?

d) None of the mentioned

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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

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)?

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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).

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}

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Explanation: We know that the circular convolution of two sequences is given by the expression

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}

a) {16,16,14,14}

b) {14,16,14,16}

c) {14,14,16,16}

d) None of the mentioned

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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

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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

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**Sanfoundry Global Education & Learning Series – Digital Signal Processing.**

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