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

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

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## 10 Questions MCQ Test Signals and Systems - Test: FFT Algorithms Applications

Test: FFT Algorithms Applications for Electronics and Communication Engineering (ECE) 2024 is part of Signals and Systems preparation. The Test: FFT Algorithms Applications questions and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus.The Test: FFT Algorithms Applications MCQs are made for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: FFT Algorithms Applications below.
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Test: FFT Algorithms Applications - Question 1

### FFT algorithm is designed to perform complex operations.

Detailed Solution for Test: FFT Algorithms Applications - Question 1

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.

Test: FFT Algorithms Applications - Question 2

### 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)?

Detailed Solution for Test: FFT Algorithms Applications - Question 2

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.

Test: FFT Algorithms Applications - Question 3

### 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)?

Detailed Solution for Test: FFT Algorithms Applications - Question 3

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)].

Test: FFT Algorithms Applications - Question 4

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)?

Detailed Solution for Test: FFT Algorithms Applications - Question 4

Test: FFT Algorithms Applications - Question 5

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?

Detailed Solution for Test: FFT Algorithms Applications - Question 5

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)+W2kNX2(k), k=0,1,2…N-1.

Test: FFT Algorithms Applications - Question 6

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?

Detailed Solution for Test: FFT Algorithms Applications - Question 6

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)-W2kNX2(k), k= N,N-1,…2N-1.

Test: FFT Algorithms Applications - Question 7

Decimation-in frequency FFT algorithm is used to compute H(k).

Detailed Solution for Test: FFT Algorithms Applications - Question 7

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.

Test: FFT Algorithms Applications - Question 8

How many complex multiplications are need to be performed for each FFT algorithm?

Detailed Solution for Test: FFT Algorithms Applications - Question 8

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.

Test: FFT Algorithms Applications - Question 9

How many complex additions are required to be performed in linear filtering of a sequence using FFT algorithm?

Detailed Solution for Test: FFT Algorithms Applications - Question 9

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.

Test: FFT Algorithms Applications - Question 10

How many complex multiplication are required per output data point?

Detailed Solution for Test: FFT Algorithms Applications - Question 10

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.

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