Digital Signal Processing

Applications of FFT Algorithms

Question 1
Marks : +2 | -2
Pass Ratio : 100%
How many complex multiplications are need to be performed for each FFT algorithm?
(N/2)logN
Nlog2N
(N/2)log2N
None of the mentioned
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=2v, this decimation can be performed v=log2N times. Thus the total number of complex multiplications is reduced to (N/2)log2N.
Question 2
Marks : +2 | -2
Pass Ratio : 100%
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)?
\\(\\frac{1}{2} [X*(k)+X*(N-k)]\\)
\\(\\frac{1}{2} [X*(k)-X*(N-k)]\\)
\\(\\frac{1}{2j} [X*(k)-X*(N-k)]\\)
\\(\\frac{1}{2j} [X*(k)+X*(N-k)]\\)
Explanation:
We know that if x(n)=x1(n)+jx2(n) then x2(n)=\\(\\frac{x(n)-x^* (n)}{2j}\\).
Question 3
Marks : +2 | -2
Pass Ratio : 100%
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?
X1(k)-W2kNX2(k)
X1(k)+W2kNX2(k)
X1(k)+W2kX2(k)
X1(k)-W2kX2(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)
Question 4
Marks : +2 | -2
Pass Ratio : 100%
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)?
\\(\\frac{x(n)-x^* (n)}{2}\\)
\\(\\frac{x(n)+x^* (n)}{2}\\)
\\(\\frac{x(n)-x^* (n)}{2j}\\)
\\(\\frac{x(n)+x^* (n)}{2j}\\)
Explanation:
Given x(n)=x1(n)+jx2(n)
Question 5
Marks : +2 | -2
Pass Ratio : 100%
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?
X1(k)-W2kX2(k)
X1(k)+W2kNX2(k)
X1(k)+W2kX2(k)
X1(k)-W2kNX2(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)
Question 6
Marks : +2 | -2
Pass Ratio : 100%
How many complex additions are required to be performed in linear filtering of a sequence using FFT algorithm?
(N/2)logN
2Nlog2N
(N/2)log2N
Nlog2N
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.
Question 7
Marks : +2 | -2
Pass Ratio : 100%
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)?
\\(\\frac{1}{2} [X*(k)+X*(N-k)]\\)
\\(\\frac{1}{2} [X*(k)-X*(N-k)]\\)
\\(\\frac{1}{2j} [X*(k)-X*(N-k)]\\)
\\(\\frac{1}{2j} [X*(k)+X*(N-k)]\\)
Explanation:
We know that if x(n)=x1(n)+jx2(n) then x1(n)=\\(\\frac{x(n)+x*(n)}{2}\\)
Question 8
Marks : +2 | -2
Pass Ratio : 100%
FFT algorithm is designed to perform complex operations.
True
False
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.
Question 9
Marks : +2 | -2
Pass Ratio : 100%
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)?
\\(\\frac{x(n)-x*(n)}{2}\\)
\\(\\frac{x(n)+x*(n)}{2}\\)
\\(\\frac{x(n)+x*(n)}{2j}\\)
\\(\\frac{x(n)-x*(n)}{2j}\\)
Explanation:
Given x(n)=x1(n)+jx2(n)
Question 10
Marks : +2 | -2
Pass Ratio : 100%
Decimation-in frequency FFT algorithm is used to compute H(k).
True
False
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.