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 W_{4}^{100}=W_{x}^{200}, 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)=>X

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

Let n+N=l=>X

^{1}(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)

=>x

^{1}(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 X_{1}(k) and X_{2}(k) are the N-point DFTs of X_{1}(n) and x_{2}(n) respectively, then what is the N-point DFT of x(n)=ax_{1}(n)+bx_{2}(n)?

a) X_{1}(ak)+X_{2}(bk)

b) aX_{1}(k)+bX_{2}(k)

c) e^{ak}X_{1}(k)+e^{bk}X_{2}(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)=ax

_{1}(n)+bx

_{2}(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)=aX

_{1}(k)+bX

_{2}(k).

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

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

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 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

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 X

_{R}(k)=0. Hence X(k) is purely imaginary and odd. Since X

_{R}(k) reduces to zero, the IDFT reduces to

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

10. If X_{1}(n), x_{2}(n) and x_{3}(m) are three sequences each of length N whose DFTs are given as X_{1}(k), X_{2}(k) and X3(k) respectively and X3(k)=X_{1}(k).X_{2}(k), then what is the expression for x_{3}(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) \(\sum_{n=0}^{N-1}x_1 (n) x_2 (m+n)_N \)

Explanation: If X

_{1}(n), x

_{2}(n) and x

_{3}(m) are three sequences each of length N whose DFTs are given as X

_{1}(k), x

_{2}(k) and X

_{3}(k) respectively and X

_{3}(k)=X

_{1}(k).X

_{2}(k), then according to the multiplication property of DFT we have x

_{3}(m) is the circular convolution of X

_{1}(n) and x

_{2}(n).

That is x

_{3}(m) = \(\sum_{n=0}^{N-1}x_1 (n) x_2 (m-n)_N \).