Digital Signal Processing

Efficient Computation of DFT FFT Algorithms

Question 1
Marks : +2 | -2
Pass Ratio : 100%
The following butterfly diagram is used in the computation of __________
Decimation-in-time FFT
Decimation-in-frequency FFT
All of the mentioned
None of the mentioned
Explanation:
The above given diagram is the basic butterfly computation in the decimation-in-frequency FFT algorithm.
Question 2
Marks : +2 | -2
Pass Ratio : 100%
If N=LM, then what is the value of WNmqL?
WMmq
WLmq
WNmq
None of the mentioned
Explanation:
We know that if N=LM, then WNmqL = WN/Lmq = WMmq.
Question 3
Marks : +2 | -2
Pass Ratio : 100%
If we split the N point data sequence into two N/2 point data sequences f1(n) and f2(n) corresponding to the even numbered and odd numbered samples of x(n) and F1(k) and F2(k) are the N/2 point DFTs of f1(k) and f2(k) respectively, then what is the N/2 point DFT X(k) of x(n)?
F1(k)+F2(k)
F1(k)-WNk F2(k)
F1(k)+WNk F2(k)
None of the mentioned
Explanation:
From the question, it is given that
Question 4
Marks : +2 | -2
Pass Ratio : 100%
How many complex multiplications are required to compute X(k)?
N(N+1)
N(N-1)/2
N2/2
N(N+1)/2
Explanation:
We observe that the direct computation of F1(k) requires (N/2)2 complex multiplications. The same applies to the computation of F2(k). Furthermore, there are N/2 additional complex multiplications required to compute WNk. Hence it requires N(N+1)/2 complex multiplications to compute X(k).
Question 5
Marks : +2 | -2
Pass Ratio : 100%
The total number of complex multiplications required to compute N point DFT by radix-2 FFT is?
(N/2)log2N
Nlog2N
(N/2)logN
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 6
Marks : +2 | -2
Pass Ratio : 100%
Which of the following is true regarding the number of computations required to compute an N-point DFT?
N2 complex multiplications and N(N-1) complex additions
N2 complex additions and N(N-1) complex multiplications
N2 complex multiplications and N(N+1) complex additions
N2 complex additions and N(N+1) complex multiplications
Explanation:
The formula for calculating N point DFT is given as
Question 7
Marks : +2 | -2
Pass Ratio : 100%
If we split the N point data sequence into two N/2 point data sequences f1(n) and f2(n) corresponding to the even numbered and odd numbered samples of x(n), then such an FFT algorithm is known as decimation-in-time algorithm.
True
False
Explanation:
Let us consider the computation of the N=2v point DFT by the divide and conquer approach. We select M=N/2 and L=2. This selection results in a split of N point data sequence into two N/2 point data sequences f1(n) and f2(n) corresponding to the even numbered and odd numbered samples of x(n), respectively, that is
Question 8
Marks : +2 | -2
Pass Ratio : 100%
Which of the following is true regarding the number of computations required to compute DFT at any one value of ‘k’?
4N-2 real multiplications and 4N real additions
4N real multiplications and 4N-4 real additions
4N-2 real multiplications and 4N+2 real additions
4N real multiplications and 4N-2 real additions
Explanation:
The formula for calculating N point DFT is given as
Question 9
Marks : +2 | -2
Pass Ratio : 100%
Divide-and-conquer approach is based on the decomposition of an N-point DFT into successively smaller DFTs. This basic approach leads to FFT algorithms.
True
False
Explanation:
The development of computationally efficient algorithms for the DFT is made possible if we adopt a divide-and-conquer approach. This approach is based on the decomposition of an N-point DFT into successively smaller DFTs. This basic approach leads to a family of computationally efficient algorithms known collectively as FFT algorithms.
Question 10
Marks : +2 | -2
Pass Ratio : 100%
How many complex multiplications are performed in computing the N-point DFT of a sequence using divide-and-conquer method if N=LM?
N(L+M+2)
N(L+M-2)
N(L+M-1)
N(L+M+1)
Explanation:
The expression for N point DFT is given as