## Digital Signal Processing Questions and Answers Part-19

1. What is the DFT of the four point sequence x(n)={0,1,2,3}?
a) {6,-2+2j-2,-2-2j}
b) {6,-2-2j,2,-2+2j}
c) {6,-2-2j,-2,-2+2j}
d) {6,-2+2j,-2,-2-2j}

Explanation: Given x(n)={0,1,2,3}
We know that the 4-point DFT of the above given sequence is given by the expression
X(k)=$$\sum_{n=0}^{N-1}x(n)e^{-j2πkn/N}$$
In this case N=4
=>X(0)=6,X(1)=-2+2j,X(2)=-2,X(3)=-2-2j.

2. If W4100=Wx200, then what is the value of x?
a) 2
b) 4
c) 8
d) 16

Explanation: We know that according to the periodicity and symmetry property,
100/4=200/x=>x=8.

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

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)

4. 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

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)

5. 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

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).

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

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

7. 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

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)

8. 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

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

9. 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

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

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

That is x3(m) = $$\sum_{n=0}^{N-1}x_1 (n) x_2 (m-n)_N$$.