# FFT Algorithms Application MCQ Quiz – Objective Question with Answer for FFT Algorithms Applications MCQ

1. FFT algorithm is designed to perform complex operations.

A. True
B. False

The FFT algorithm is designed to perform complex multiplications and additions, even though the input data may be real-valued. The basic reason for this is that the phase factors are complex and hence, after the first stage of the algorithm, all variables are basically complex-valued.

2. If x1(n) and x2(n) are two real valued sequences of length N, and let x(n) be a complex valued sequence defined as x(n)=x1(n)+jx2(n), 0≤n≤N-1, then what is the value of x1(n)?

A. $$\frac{x(n)-x^* (n)}{2}$$

B. $$\frac{x(n)+x^* (n)}{2}$$

C. $$\frac{x(n)-x^* (n)}{2j}$$

D. $$\frac{x(n)+x^* (n)}{2j}$$

Given x(n)=x1(n)+jx2(n)

=>x*(n)= x1(n)-jx2(n)

Upon adding the above two equations, we get

x1(n)=$$\frac{x(n)+x*(n)}{2}$$.

3. If x1(n) and x2(n) are two real valued sequences of length N, and let x(n) be a complex valued sequence defined as x(n)=x1(n)+jx2(n), 0≤ n≤ N-1, then what is the value of x2(n)?

A. $$\frac{x(n)-x*(n)}{2}$$

B. $$\frac{x(n)+x*(n)}{2}$$

C. $$\frac{x(n)+x*(n)}{2j}$$

D. $$\frac{x(n)-x*(n)}{2j}$$

Given x(n)=x1(n)+jx2(n)

=>x*(n) = x1(n)-jx2(n)

Upon subtracting the above two equations, we get

x2(n)=$$\frac{x(n)-x*(n)}{2j}$$.

4. If X(k) is the DFT of x(n) which is defined as x(n)=x1(n)+jx2(n), 0≤ n≤ N-1, then what is the DFT of x1(n)?

A. $$\frac{1}{2} [X*(k)+X*(N-k)]$$

B. $$\frac{1}{2} [X*(k)-X*(N-k)]$$

C. $$\frac{1}{2j} [X*(k)-X*(N-k)]$$

D. $$\frac{1}{2j} [X*(k)+X*(N-k)]$$

We know that if

x(n)=x1(n)+jx2(n) then

x1(n)=$$\frac{x(n)+x*(n)}{2}$$

On applying DFT on both sides of the above equation, we get

X1(k)=$$\frac{1}{2} {DFT[x(n)]+DFT[x*(n)]}$$

We know that if X(k) is the DFT of x(n), the DFT[x*(n)]=X*(N-k)

=>X1(k)=$$\frac{1}{2} [X*(k)+X*(N-k)]$$.

5. If X(k) is the DFT of x(n) which is defined as x(n)=x1(n)+jx2(n), 0≤ n≤ N-1, then what is the DFT of x1(n)?

A. $$\frac{1}{2} [X*(k)+X*(N-k)]$$

B. $$\frac{1}{2} [X*(k)-X*(N-k)]$$

C. $$\frac{1}{2j} [X*(k)-X*(N-k)]$$

D. $$\frac{1}{2j} [X*(k)+X*(N-k)]$$

We know that if x(n)=x1(n)+jx2(n) then

x2(n)=$$\frac{x(n)-x^* (n)}{2j}$$.

On applying DFT on both sides of the above equation, we get

X2(k)=$$\frac{1}{2j} {DFT[x(n)]-DFT[x*(n)]}$$

We know that if X(k) is the DFT of x(n), the DFT[x*(n)]=X*(N-k)

=>X2(k)=$$\frac{1}{2j} [X*(k)-X*(N-k)]$$.

6. If g(n) is a real valued sequence of 2N points and x1(n)=g(2n) and x2(n)=g(2n+1), then what is the value of G(k), k=0,1,2…N-1?

A. X1(k)-W2kNX2(k)
B. X1(k)+W2kNX2(k)
C. X1(k)+W2kX2(k)
D. X1(k)-W2kX2(k)

Given g(n) is a real valued 2N point sequence. The 2N point sequence is divided into two N point sequences x1(n) and x2(n)

Let x(n)=x1(n)+jx2(n)

=> X1(k)=$$\frac{1}{2} [X*(k)+X*(N-k)]$$ and X2(k)=$$\frac{1}{2j} [X*(k)-X*(N-k)]$$

We know that g(n)=x1(n)+x2(n)

=>G(k)=X1(k)+W2kNX2(k), k=0,1,2…N-1.

7. If g(n) is a real valued sequence of 2N points and x1(n)=g(2n) and x2(n)=g(2n+1), then what is the value of G(k), k=N,N-1,…2N-1?

A. X1(k)-W2kX2(k)
B. X1(k)+W2kNX2(k)
C. X1(k)+W2kX2(k)
D. X1(k)-W2kNX2(k)

Given g(n) is a real valued 2N point sequence. The 2N point sequence is divided into two N point sequences x1(n) and x2(n)

Let x(n)=x1(n)+jx2(n)

=> X1(k)=$$\frac{1}{2} [X*(k)+X*(N-k)]$$ and X2(k)=$$\frac{1}{2j} [X*(k)-X*(N-k)]$$

We know that g(n)=x1(n)+x2(n)

=>G(k)=X1(k)-W2kNX2(k), k=N,N-1,…2N-1.

8. Decimation-in frequency FFT algorithm is used to compute H(k).

A. True
B. False

The N-point DFT of h(n), which is padded by L-1 zeros, is denoted as H(k). This computation is performed once via the FFT and the resulting N complex numbers are stored.

To be specific we assume that the decimation-in-frequency FFT algorithm is used to compute H(k). This yields H(k) in the bit-reversed order, which is the way it is stored in the memory.

9. How many complex multiplications are needed to be performed for each FFT algorithm?

A. (N/2)logN
B. Nlog2N
C. (N/2)log2N
D. None of the mentioned

The decimation of the data sequence should be repeated again and again until the resulting sequences are reduced to one-point sequences. For N=2v, this decimation can be performed v=log2N times. Thus the total number of complex multiplications is reduced to (N/2)log2N.

10. How many complex additions are required to be performed in linear filtering of a sequence using FFT algorithm?

A. (N/2)logN
B. 2Nlog2N
C. (N/2)log2N
D. Nlog2N