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-time FFT algorithm.
Question 2
Marks : +2 | -2
Pass Ratio : 100%
If X(k) is the N/2 point DFT of the sequence x(n), then what is the value of X(k+N/2)?
F1(k)+F2(k)
F1(k)-WNk F2(k)
F1(k)+WNk F2(k)
None of the mentioned
Explanation:
We know that, X(k) = F1(k)+WNk F2(k)
Question 3
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
Question 4
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 5
Marks : +2 | -2
Pass Ratio : 100%
The total number of complex additions 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 additions is reduced to Nlog2N.
Question 6
Marks : +2 | -2
Pass Ratio : 100%
What is the real part of the N point DFT XR(k) of a complex valued sequence x(n)?
\\(\\sum_{n=0}^{N-1} [x_R (n) cos⁡\\frac{2Ï€kn}{N} – x_I (n) sin⁡\\frac{2Ï€kn}{N}]\\)
\\(\\sum_{n=0}^{N-1} [x_R (n) sin⁡\\frac{2πkn}{N} + x_I (n) cos⁡\\frac{2πkn}{N}]\\)
\\(\\sum_{n=0}^{N-1} [x_R (n) cos⁡\\frac{2πkn}{N} + x_I (n) sin⁡\\frac{2πkn}{N}]\\)
None of the mentioned
Explanation:
For a complex valued sequence x(n) of N points, the DFT may be expressed as
Question 7
Marks : +2 | -2
Pass Ratio : 100%
WNk+N/2=?
WNk
-WNk
WN-k
None of the mentioned
Explanation:
According to the symmetry property, we get WNk+N/2=-WNk.
Question 8
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 9
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 10
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.