Test: FFT Algorithms Applications - Electronics and Communication Engineering (ECE) MCQ

Explanation: 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.

Explanation: Given x(n)=x1(n)+jx2(n)
=>x*(n)= x1(n)-jx2(n)
Upon subtracting the above two equations, we get x2(n)= (x(n)-x*(n))/2j.

Explanation: We know that if x(n)=x1(n)+jx2(n) then x1(n)= (x(n)+x*(n))/2
On applying DFT on both sides of the above equation, we get
X1(k)= 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)= 1/2 [X*(k)+X*(N-k)].

Explanation: 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)= 1/2 [X*(k)+X*(N-k)] and X2(k)= 1/2j [X*(k)-X*(N-k)] We know that g(n)= x1(n)+x2(n)
=>G(k)= X1(k)+W₂^kNX2(k), k=0,1,2…N-1.

Explanation: 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)= 1/2 [X*(k)+X*(N-k)] and X2(k)= 1/2j [X*(k)-X*(N-k)] We know that g(n)= x1(n)+x2(n)
=>G(k)= X1(k)-W₂^kNX2(k), k= N,N-1,…2N-1.

Explanation: 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 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.

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

Explanation: The number of additions to be performed in FFT are Nlog₂N. But in linear filtering of a sequence, we calculate DFT which requires Nlog₂N complex additions and IDFT requires Nlog₂N complex additions. So, the total number of complex additions to be performed in linear filtering of a sequence using FFT algorithm is 2Nlog₂N.

Explanation: In the overlap add method, the N-point data block consists of L new data points and additional M-1 zeros and the number of complex multiplications required in FFT algorithm are (N/2)log₂N. So, the number of complex multiplications per output data point is [Nlog₂2N]/L.