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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 'A'. Can you explain this answer?
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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:
Given:
x(n) = x1(n) + jx2(n), 0 ≤ n ≤ N-1
X(k) = DFT of x(n)
DFT of x1(n):
- Let X1(k) be the DFT of x1(n)
- Using linearity property of DFT, we have X1(k) = DFT{x1(n)}
Using given relation:
x(n) = x1(n) + jx2(n)
- Taking DFT on both sides, we get X(k) = X1(k) + jX2(k)
Applying conjugate symmetry property of DFT:
- X*(k) = DFT{x*(-n)} = DFT{x(n)}
Using conjugate symmetry:
X*(k) = DFT{x(n)} = X(k)
- X*(k) = X(k) (due to conjugate symmetry property)
Substitute X(k) = X1(k) + jX2(k) in X*(k):
X*(k) = X1*(k) + jX2*(k)
Given X*(k) = X(k):
X(k) = X1*(k) + jX2*(k)
- Substitute X(k) = X1(k) + jX2(k):
X1(k) + jX2(k) = X1*(k) + jX2*(k)
- Separate real and imaginary parts:
X1(k) = X1*(k) and X2(k) = -X2*(k)
Using X1(k) = X1*(k), we can write:
X1(k) = 1/2 [X*(k) + X*(-k)]
- As X(k) = X1(k) + jX2(k) and X*(-k) = X1*(-k) + jX2*(-k):
X1(k) = 1/2 [X*(k) + X*(N-k)] (Option A)
Therefore, the DFT of x1(n) is given by 1/2 [X*(k) + X*(N-k)].
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Community Answer
If X(k) is the DFT of x(n) which is defined as x(n)=x1(n)+jx2(n), 0&le...
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)].
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)].
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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 'A'. Can you explain this answer?
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