1. Which is the correct order of the following steps to be done in one of the algorithm of divide and conquer method?
i) Store the signal column wise
ii) Compute the M-point DFT of each row
iii) Multiply the resulting array by the phase factors WNlq.
iv) Compute the L-point DFT of each column.
v) Read the result array row wise.
a) i-ii-iv-iii-v
b) i-iii-ii-iv-v
c) i-ii-iii-iv-v
d) i-iv-iii-ii-v
Explanation: According to one of the algorithm describing the divide and conquer method, if we store the signal in column wise, then compute the M-point DFT of each row and multiply the resulting array by the phase factors WNlq and then compute the L-point DFT of each column and read the result row wise
2. If we store the signal row wise then the result must be read column wise.
a) True
b) False
Explanation: According to the second algorithm of divide and conquer approach, if the input signal is stored in row wise, then the result must be read column wise.
3. If we store the signal row wise and compute the L point DFT at each column, the resulting array must be multiplied by which of the following factors?
a) WNlq
b) WNpq
c) WNlq
d) WNpm
Explanation: According to the second algorithm of divide and conquer approach, if the input signal is stored in row wise, then we calculate the L point DFT of each column and we multiply the resulting array by the factor WNpm
4. FFT algorithm is designed to perform complex operations.
a) True
b) 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.
5.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)?
a) \(\frac{x(n)-x^* (n)}{2}\)
b) \(\frac{x(n)+x^* (n)}{2}\)
c) \(\frac{x(n)-x^* (n)}{2j}\)
d) \(\frac{x(n)+x^* (n)}{2j}\)
Explanation: Given x(n)=x1(n)+jx2(n)
=>x*(n)= x1(n)-jx2(n)
Upon adding the above two equations, we get x1(n)=\(\frac{x(n)+x*(n)}{2}\).
6. 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)?
a) \(\frac{x(n)-x*(n)}{2}\)
b) \(\frac{x(n)+x*(n)}{2}\)
c) \(\frac{x(n)+x*(n)}{2j}\)
d) \(\frac{x(n)-x*(n)}{2j}\)
Explanation: Given x(n)=x1(n)+jx2(n)
=>x*(n) = x1(n)-jx2(n)
Upon subtracting the above two equations, we get x2(n)=\(\frac{x(n)-x*(n)}{2j}\).
7. 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)?
a) \(\frac{1}{2} [X*(k)+X*(N-k)]\)
b) \(\frac{1}{2} [X*(k)-X*(N-k)]\)
c) \(\frac{1}{2j} [X*(k)-X*(N-k)]\)
d) \(\frac{1}{2j} [X*(k)+X*(N-k)]\)
Explanation: We know that if x(n)=x1(n)+jx2(n) then x1(n)=\(\frac{x(n)+x*(n)}{2}\)
On applying DFT on both sides of the above equation, we get
X1(k)=\(\frac{1}{2} {DFT[x(n)]+DFT[x*(n)]}\)
We know that if X(k) is the DFT of x(n), the DFT[x*(n)]=X*(N-k)
=>X1(k)=\(\frac{1}{2} [X*(k)+X*(N-k)]\).
8. 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)?
a) \(\frac{1}{2} [X*(k)+X*(N-k)]\)
b) \(\frac{1}{2} [X*(k)-X*(N-k)]\)
c) \(\frac{1}{2j} [X*(k)-X*(N-k)]\)
d) \(\frac{1}{2j} [X*(k)+X*(N-k)]\)
Explanation: We know that if x(n)=x1(n)+jx2(n) then x2(n)=\(\frac{x(n)-x^* (n)}{2j}\).
On applying DFT on both sides of the above equation, we get
X2(k)=\(\frac{1}{2j} {DFT[x(n)]-DFT[x*(n)]}\)
We know that if X(k) is the DFT of x(n), the DFT[x*(n)]=X*(N-k)
=>X2(k)=\(\frac{1}{2j} [X*(k)-X*(N-k)]\).
9. 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?
a) X1(k)-W2kNX2(k)
b) X1(k)+W2kNX2(k)
c) X1(k)+W2kX2(k)
d) 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)
Let x(n)=x1(n)+jx2(n)
=> X1(k)=\(\frac{1}{2} [X*(k)+X*(N-k)]\) and X2(k)=\(\frac{1}{2j} [X*(k)-X*(N-k)]\)
We know that g(n)=x1(n)+x2(n)
=>G(k)=X1(k)+W2kNX2(k), k=0,1,2…N-1.
10. 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?
a) X1(k)-W2kX2(k)
b) X1(k)+W2kNX2(k)
c) X1(k)+W2kX2(k)
d) 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)
Let x(n)=x1(n)+jx2(n)
=> X1(k)=\(\frac{1}{2} [X*(k)+X*(N-k)]\) and X2(k)=\(\frac{1}{2j} [X*(k)-X*(N-k)]\)
We know that g(n)=x1(n)+x2(n)
=>G(k)=X1(k)-W2kNX2(k), k=N,N-1,…2N-1