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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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Most Upvoted Answer
If we split the N point data sequence into two N/2 point data sequence...
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
X(k) = F1(k)+WNkNk F2(k)
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Community Answer
If we split the N point data sequence into two N/2 point data sequence...
Introduction:
In this question, we are given a data sequence x(n) of length N and we need to split it into two sequences f1(n) and f2(n) corresponding to the even and odd numbered samples of x(n). We also have N/2 point DFTs of f1(n) and f2(n) denoted as F1(k) and F2(k) respectively. We need to find the N/2 point DFT X(k) of x(n).
Explanation:
To solve this problem, let's first understand the concept of splitting a data sequence and its DFT.
When we split a data sequence x(n) into even and odd numbered samples, we get two sequences f1(n) and f2(n). The even numbered samples are given by f1(n) = x(2n) and the odd numbered samples are given by f2(n) = x(2n+1).
The N/2 point DFT of f1(n) is denoted as F1(k), and it is defined as:
F1(k) = Σ[n=0 to N/2-1] f1(n) * exp(-j2πkn/(N/2))
Similarly, the N/2 point DFT of f2(n) is denoted as F2(k), and it is defined as:
F2(k) = Σ[n=0 to N/2-1] f2(n) * exp(-j2πkn/(N/2))
Now, let's find the N/2 point DFT X(k) of x(n) using F1(k) and F2(k).
Solution:
To find the N/2 point DFT X(k) of x(n), we can write it in terms of f1(n) and f2(n) as follows:
X(k) = Σ[n=0 to N-1] x(n) * exp(-j2πkn/N)
= Σ[n=0 to N/2-1] x(2n) * exp(-j2πk(2n)/N) + Σ[n=0 to N/2-1] x(2n+1) * exp(-j2πk(2n+1)/N)
= Σ[n=0 to N/2-1] f1(n) * exp(-j2πkn/(N/2)) + Σ[n=0 to N/2-1] f2(n) * exp(-j2πk(2n+1)/N)
Now, let's compare this expression with the definition of F1(k) and F2(k):
X(k) = F1(k) + Σ[n=0 to N/2-1] f2(n) * exp(-j2πk(2n+1)/N)
We can rewrite the second term on the right-hand side using the identity exp(-j2πk(2n+1)/N) = exp(-j2πkn/N) * exp(-j2πk/N), where exp(-j2πk/N) is denoted as WNk:
X(k) = F1(k) + Σ[n=0 to N/2-1] f2(n) * exp(-j2πkn/N) * exp
In this question, we are given a data sequence x(n) of length N and we need to split it into two sequences f1(n) and f2(n) corresponding to the even and odd numbered samples of x(n). We also have N/2 point DFTs of f1(n) and f2(n) denoted as F1(k) and F2(k) respectively. We need to find the N/2 point DFT X(k) of x(n).
Explanation:
To solve this problem, let's first understand the concept of splitting a data sequence and its DFT.
When we split a data sequence x(n) into even and odd numbered samples, we get two sequences f1(n) and f2(n). The even numbered samples are given by f1(n) = x(2n) and the odd numbered samples are given by f2(n) = x(2n+1).
The N/2 point DFT of f1(n) is denoted as F1(k), and it is defined as:
F1(k) = Σ[n=0 to N/2-1] f1(n) * exp(-j2πkn/(N/2))
Similarly, the N/2 point DFT of f2(n) is denoted as F2(k), and it is defined as:
F2(k) = Σ[n=0 to N/2-1] f2(n) * exp(-j2πkn/(N/2))
Now, let's find the N/2 point DFT X(k) of x(n) using F1(k) and F2(k).
Solution:
To find the N/2 point DFT X(k) of x(n), we can write it in terms of f1(n) and f2(n) as follows:
X(k) = Σ[n=0 to N-1] x(n) * exp(-j2πkn/N)
= Σ[n=0 to N/2-1] x(2n) * exp(-j2πk(2n)/N) + Σ[n=0 to N/2-1] x(2n+1) * exp(-j2πk(2n+1)/N)
= Σ[n=0 to N/2-1] f1(n) * exp(-j2πkn/(N/2)) + Σ[n=0 to N/2-1] f2(n) * exp(-j2πk(2n+1)/N)
Now, let's compare this expression with the definition of F1(k) and F2(k):
X(k) = F1(k) + Σ[n=0 to N/2-1] f2(n) * exp(-j2πk(2n+1)/N)
We can rewrite the second term on the right-hand side using the identity exp(-j2πk(2n+1)/N) = exp(-j2πkn/N) * exp(-j2πk/N), where exp(-j2πk/N) is denoted as WNk:
X(k) = F1(k) + Σ[n=0 to N/2-1] f2(n) * exp(-j2πkn/N) * exp
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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?
Question Description
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? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about 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? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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?.
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? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about 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? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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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