# DFT Properties MCQ Quiz – Objective Question with Answer for DFT Properties MCQ

1. If x(n) and X(k) are an N-point DFT pair, then x(n+N)=x(n).

A. True
B. False

We know that the expression for an DFT is given as

X(k)=$$\sum_{n=0}^{N-1} x(n)e^{-j2πkn/N}$$

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

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

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).

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. $$\sum_{n=0}^{N-1} x_R (n) cos⁡\frac{2πkn}{N}+x_I (n) sin⁡\frac{2πkn}{N}$$

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

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

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

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

$$x(n)=j \frac{1}{N} \sum_{k=0}^{N-1} x(k) sin⁡\frac{2πkn}{N}$$

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. $$\sum_{n=0}^{N-1}x_1 (n) x_2 (m+n)_N$$

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}

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