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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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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 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)].
Most Upvoted Answer
If X(k) is the DFT of x(n) which is defined as x(n)=x1(n)+jx2(n), 0&le...
If x(n) = x1(n) + jx2(n), where x1(n) and x2(n) are real-valued sequences, then the DFT of x(n), X(k), can be defined as:
X(k) = Σ [x1(n) + jx2(n)] * exp(-j2πnk/N)
where Σ denotes the summation from n = 0 to N-1, and N is the length of the sequences x1(n) and x2(n).
Using the linearity property of the DFT, we can split the DFT into two parts:
X(k) = Σ x1(n) * exp(-j2πnk/N) + jΣ x2(n) * exp(-j2πnk/N)
The first term corresponds to the DFT of x1(n), denoted as X1(k), and the second term corresponds to the DFT of x2(n), denoted as X2(k). Therefore, we can express X(k) as:
X(k) = X1(k) + jX2(k)
In other words, the DFT of a complex-valued sequence can be obtained by calculating the DFTs of its real and imaginary parts separately.
X(k) = Σ [x1(n) + jx2(n)] * exp(-j2πnk/N)
where Σ denotes the summation from n = 0 to N-1, and N is the length of the sequences x1(n) and x2(n).
Using the linearity property of the DFT, we can split the DFT into two parts:
X(k) = Σ x1(n) * exp(-j2πnk/N) + jΣ x2(n) * exp(-j2πnk/N)
The first term corresponds to the DFT of x1(n), denoted as X1(k), and the second term corresponds to the DFT of x2(n), denoted as X2(k). Therefore, we can express X(k) as:
X(k) = X1(k) + jX2(k)
In other words, the DFT of a complex-valued sequence can be obtained by calculating the DFTs of its real and imaginary parts separately.
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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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