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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)(1/2) [X*(k)+X*(N-k)].
- b)(1/2) [X*(k)-X*(N-k)].
- c)(1/2j) [X*(k)-X*(N-k)].
- d)(1/2j) [X*(k)+X*(N-k)].
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
If X(k) is the DFT of x(n) which is defined as x(n)=x1(n)+jx2(n), 0&le...
Explanation: We know that if x(n)=x1(n)+jx2(n) then x2(n)= (x(n)-x^* (n))/2j.
On applying DFT on both sides of the above equation, we get
X2(k)= (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)= (1/2j) [X*(k)-X*(N-k)].
On applying DFT on both sides of the above equation, we get
X2(k)= (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)= (1/2j) [X*(k)-X*(N-k)].
Most Upvoted Answer
If X(k) is the DFT of x(n) which is defined as x(n)=x1(n)+jx2(n), 0&le...
Explanation:
- The given signal x(n) can be represented as a sum of two signals x1(n) and x2(n), where x1(n) and x2(n) are real and imaginary parts of x(n) respectively.
- Given x(n) = x1(n) + jx2(n), the DFT of x(n) is X(k).
- By linearity property of DFT, the DFT of x(n) can be written as the sum of DFTs of x1(n) and x2(n) i.e., X(k) = X1(k) + jX2(k).
- We need to find the DFT of x1(n) i.e., X1(k).
- From the DFT property of conjugate symmetry for real signals, X1(k) = X1*(N-k).
- Therefore, the DFT of x1(n) can be written as (1/2j) [X*(k) - X*(N-k)].
- Hence, the correct answer is option 'C' i.e., (1/2j) [X*(k) - X*(N-k)].
- The given signal x(n) can be represented as a sum of two signals x1(n) and x2(n), where x1(n) and x2(n) are real and imaginary parts of x(n) respectively.
- Given x(n) = x1(n) + jx2(n), the DFT of x(n) is X(k).
- By linearity property of DFT, the DFT of x(n) can be written as the sum of DFTs of x1(n) and x2(n) i.e., X(k) = X1(k) + jX2(k).
- We need to find the DFT of x1(n) i.e., X1(k).
- From the DFT property of conjugate symmetry for real signals, X1(k) = X1*(N-k).
- Therefore, the DFT of x1(n) can be written as (1/2j) [X*(k) - X*(N-k)].
- Hence, the correct answer is option 'C' i.e., (1/2j) [X*(k) - X*(N-k)].
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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)(1/2) [X*(k)+X*(N-k)].b)(1/2) [X*(k)-X*(N-k)].c)(1/2j) [X*(k)-X*(N-k)].d)(1/2j) [X*(k)+X*(N-k)].Correct answer is option 'C'. Can you explain this answer?
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