Digital Signal Processing

Efficient Computation of DFT FFT Algorithms

Question 1
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 2
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 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), 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 4
Marks : +2 | -2
Pass Ratio : 100%
If the arrangement is of the form in which the first row consists of the first M elements of x(n), the second row consists of the next M elements of x(n), and so on, then which of the following mapping represents the above arrangement?
n=l+mL
n=Ml+m
n=ML+l
none of the mentioned
Explanation:
If we consider the mapping n=Ml+m, then it leads to an arrangement in which the first row consists of the first M elements of x(n), the second row consists of the next M elements of x(n), and so on.
Question 5
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 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%
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 8
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 9
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 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.