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If we split the N point data sequence into two N/2 point data sequences f1(n) and f2(n) corresponding to the even numbered and odd numbered samples of x(n) and F1(k) and F2(k) are the N/2 point DFTs of f1(k) and f2(k) respectively, then what is the N/2 point DFT X(k) of x(n)?
- a)F1(k)+F2(k)
- b)F1(k)- WNk F2(k)
- c)F1(k)+WNkNk F2(k)
- d)None of the mentioned
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
If we split the N point data sequence into two N/2 point data sequence...
Explanation: From the question, it is given that
f1(n)=x(2n)
f2(n)=x(2n+1) ,n=0,1,2…N/2-1
f1(n)=x(2n)
f2(n)=x(2n+1) ,n=0,1,2…N/2-1
Most Upvoted Answer
If we split the N point data sequence into two N/2 point data sequence...
Explanation:
To understand why the correct answer is option 'C', let's break down the given information and analyze it step by step.
Given:
- We have an N-point data sequence x(n).
- We split this sequence into two N/2 point data sequences, f1(n) and f2(n).
- f1(n) consists of the even-numbered samples of x(n), and f2(n) consists of the odd-numbered samples of x(n).
- We also have the N/2 point DFTs of f1(n) and f2(n), denoted as F1(k) and F2(k) respectively.
To Find:
We need to determine the N/2 point DFT X(k) of x(n).
Solution:
When we split the N-point data sequence x(n) into two N/2 point data sequences, we can represent x(n) as the sum of f1(n) and f2(n). Mathematically, this can be written as:
x(n) = f1(n) + f2(n)
Now, we know that the Discrete Fourier Transform (DFT) is a linear transform. This means that if we have two sequences a(n) and b(n) and their respective DFTs A(k) and B(k), then the DFT of their sum can be written as:
DFT(a(n) + b(n)) = A(k) + B(k)
Applying this property to our problem, we can write the DFT of x(n) as:
X(k) = DFT(f1(n) + f2(n))
Since we know the DFTs of f1(n) and f2(n) as F1(k) and F2(k) respectively, we can substitute these values into the equation above:
X(k) = F1(k) + F2(k)
This matches the expression given in option 'C': F1(k) + WN^k * F2(k), where WN^k represents the twiddle factor.
Therefore, the correct answer is option 'C': F1(k) + WN^k * F2(k).
Summary:
The N/2 point DFT of x(n) is given by F1(k) + WN^k * F2(k), where F1(k) and F2(k) are the N/2 point DFTs of the even and odd numbered samples of x(n) respectively.
To understand why the correct answer is option 'C', let's break down the given information and analyze it step by step.
Given:
- We have an N-point data sequence x(n).
- We split this sequence into two N/2 point data sequences, f1(n) and f2(n).
- f1(n) consists of the even-numbered samples of x(n), and f2(n) consists of the odd-numbered samples of x(n).
- We also have the N/2 point DFTs of f1(n) and f2(n), denoted as F1(k) and F2(k) respectively.
To Find:
We need to determine the N/2 point DFT X(k) of x(n).
Solution:
When we split the N-point data sequence x(n) into two N/2 point data sequences, we can represent x(n) as the sum of f1(n) and f2(n). Mathematically, this can be written as:
x(n) = f1(n) + f2(n)
Now, we know that the Discrete Fourier Transform (DFT) is a linear transform. This means that if we have two sequences a(n) and b(n) and their respective DFTs A(k) and B(k), then the DFT of their sum can be written as:
DFT(a(n) + b(n)) = A(k) + B(k)
Applying this property to our problem, we can write the DFT of x(n) as:
X(k) = DFT(f1(n) + f2(n))
Since we know the DFTs of f1(n) and f2(n) as F1(k) and F2(k) respectively, we can substitute these values into the equation above:
X(k) = F1(k) + F2(k)
This matches the expression given in option 'C': F1(k) + WN^k * F2(k), where WN^k represents the twiddle factor.
Therefore, the correct answer is option 'C': F1(k) + WN^k * F2(k).
Summary:
The N/2 point DFT of x(n) is given by F1(k) + WN^k * F2(k), where F1(k) and F2(k) are the N/2 point DFTs of the even and odd numbered samples of x(n) respectively.
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If we split the N point data sequence into two N/2 point data sequences f1(n) and f2(n) corresponding to the even numbered and odd numbered samples of x(n) and F1(k) and F2(k) are the N/2 point DFTs of f1(k) and f2(k) respectively, then what is the N/2 point DFT X(k) of x(n)?a)F1(k)+F2(k)b)F1(k)- WNkF2(k)c)F1(k)+WNkNkF2(k)d)None of the mentionedCorrect answer is option 'C'. Can you explain this answer?
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