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If g(n) is a real valued sequence of 2N points and x1(n)=g(2n) and x2(n)=g(2n+1), then what is the value of G(k), k=N,N-1,…2N-1?
- a)X1(k)-W2kX2(k)
- b)X1(k)+W2kNX2(k)
- c)X1(k)+W2kX2(k)
- d)X1(k)-W2kNX2(k)
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
If g(n) is a real valued sequence of 2N points and x1(n)=g(2n) and x2(...
Explanation: Given g(n) is a real valued 2N point sequence. The 2N point sequence is divided into two N point sequences x1(n) and x2(n)
Let x(n)= x1(n)+jx2(n)
=> X1(k)= 1/2 [X*(k)+X*(N-k)] and X2(k)= 1/2j [X*(k)-X*(N-k)] We know that g(n)= x1(n)+x2(n)
=>G(k)= X1(k)-W2kNX2(k), k= N,N-1,…2N-1.
Let x(n)= x1(n)+jx2(n)
=> X1(k)= 1/2 [X*(k)+X*(N-k)] and X2(k)= 1/2j [X*(k)-X*(N-k)] We know that g(n)= x1(n)+x2(n)
=>G(k)= X1(k)-W2kNX2(k), k= N,N-1,…2N-1.
Most Upvoted Answer
If g(n) is a real valued sequence of 2N points and x1(n)=g(2n) and x2(...
0,-1,-2,...,-N+1?
To find G(k), we need to take the DFT of x1(n) and x2(n) and use the formula:
G(k) = X1(k) + e^(-2πik/N) X2(k)
where X1(k) and X2(k) are the DFTs of x1(n) and x2(n), respectively.
Since x1(n) contains the even-indexed points of g(n) and x2(n) contains the odd-indexed points of g(n), we can write:
x1(n) = {g(0), g(2), ..., g(2N-2)}
x2(n) = {g(1), g(3), ..., g(2N-1)}
Taking the DFT of x1(n) and x2(n), we get:
X1(k) = Σ[n=0 to 2N-2] g(2n) e^(-2πikn/(2N))
X2(k) = Σ[n=0 to 2N-2] g(2n+1) e^(-2πikn/(2N))
Substituting these values in the formula for G(k), we get:
G(k) = Σ[n=0 to 2N-2] g(2n) e^(-2πikn/(2N)) + e^(-2πik/N) Σ[n=0 to 2N-2] g(2n+1) e^(-2πikn/(2N))
To simplify this expression, we can write it as:
G(k) = Σ[n=0 to N-1] g(2n) e^(-2πikn/N) + e^(-2πik/N) Σ[n=0 to N-1] g(2n+1) e^(-2πikn/N)
Using the fact that g(n) is a real-valued sequence, we have:
g(2N-2-n) = g(2n)
g(2N-1-n) = -g(2n+1)
Substituting these values in the expression for G(k), we get:
G(k) = Σ[n=0 to N-1] g(2n) e^(-2πikn/N) + e^(-2πik/N) Σ[n=0 to N-1] (-1)^n g(2n+1) e^(-2πikn/N)
Now, let's consider the cases k=N and k=N-1 separately:
For k=N, we have:
G(N) = Σ[n=0 to N-1] g(2n) e^(-2πiNn/N) + e^(-2πi/N) Σ[n=0 to N-1] (-1)^n g(2n+1) e^(-2πiNn/N)
Since e^(-2πiNn/N) = e^(-2πin) = 1 for all n, we can simplify this expression to:
G(N) = Σ[n=0 to N-1] g(2n) + e^(-2πi/N) Σ[n=0 to N-
To find G(k), we need to take the DFT of x1(n) and x2(n) and use the formula:
G(k) = X1(k) + e^(-2πik/N) X2(k)
where X1(k) and X2(k) are the DFTs of x1(n) and x2(n), respectively.
Since x1(n) contains the even-indexed points of g(n) and x2(n) contains the odd-indexed points of g(n), we can write:
x1(n) = {g(0), g(2), ..., g(2N-2)}
x2(n) = {g(1), g(3), ..., g(2N-1)}
Taking the DFT of x1(n) and x2(n), we get:
X1(k) = Σ[n=0 to 2N-2] g(2n) e^(-2πikn/(2N))
X2(k) = Σ[n=0 to 2N-2] g(2n+1) e^(-2πikn/(2N))
Substituting these values in the formula for G(k), we get:
G(k) = Σ[n=0 to 2N-2] g(2n) e^(-2πikn/(2N)) + e^(-2πik/N) Σ[n=0 to 2N-2] g(2n+1) e^(-2πikn/(2N))
To simplify this expression, we can write it as:
G(k) = Σ[n=0 to N-1] g(2n) e^(-2πikn/N) + e^(-2πik/N) Σ[n=0 to N-1] g(2n+1) e^(-2πikn/N)
Using the fact that g(n) is a real-valued sequence, we have:
g(2N-2-n) = g(2n)
g(2N-1-n) = -g(2n+1)
Substituting these values in the expression for G(k), we get:
G(k) = Σ[n=0 to N-1] g(2n) e^(-2πikn/N) + e^(-2πik/N) Σ[n=0 to N-1] (-1)^n g(2n+1) e^(-2πikn/N)
Now, let's consider the cases k=N and k=N-1 separately:
For k=N, we have:
G(N) = Σ[n=0 to N-1] g(2n) e^(-2πiNn/N) + e^(-2πi/N) Σ[n=0 to N-1] (-1)^n g(2n+1) e^(-2πiNn/N)
Since e^(-2πiNn/N) = e^(-2πin) = 1 for all n, we can simplify this expression to:
G(N) = Σ[n=0 to N-1] g(2n) + e^(-2πi/N) Σ[n=0 to N-
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If g(n) is a real valued sequence of 2N points and x1(n)=g(2n) and x2(n)=g(2n+1), then what is the value of G(k), k=N,N-1,…2N-1?a)X1(k)-W2kX2(k)b)X1(k)+W2kNX2(k)c)X1(k)+W2kX2(k)d)X1(k)-W2kNX2(k)Correct answer is option 'D'. Can you explain this answer?
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If g(n) is a real valued sequence of 2N points and x1(n)=g(2n) and x2(n)=g(2n+1), then what is the value of G(k), k=N,N-1,…2N-1?a)X1(k)-W2kX2(k)b)X1(k)+W2kNX2(k)c)X1(k)+W2kX2(k)d)X1(k)-W2kNX2(k)Correct answer is option 'D'. 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 g(n) is a real valued sequence of 2N points and x1(n)=g(2n) and x2(n)=g(2n+1), then what is the value of G(k), k=N,N-1,…2N-1?a)X1(k)-W2kX2(k)b)X1(k)+W2kNX2(k)c)X1(k)+W2kX2(k)d)X1(k)-W2kNX2(k)Correct answer is option 'D'. 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 g(n) is a real valued sequence of 2N points and x1(n)=g(2n) and x2(n)=g(2n+1), then what is the value of G(k), k=N,N-1,…2N-1?a)X1(k)-W2kX2(k)b)X1(k)+W2kNX2(k)c)X1(k)+W2kX2(k)d)X1(k)-W2kNX2(k)Correct answer is option 'D'. Can you explain this answer?.
If g(n) is a real valued sequence of 2N points and x1(n)=g(2n) and x2(n)=g(2n+1), then what is the value of G(k), k=N,N-1,…2N-1?a)X1(k)-W2kX2(k)b)X1(k)+W2kNX2(k)c)X1(k)+W2kX2(k)d)X1(k)-W2kNX2(k)Correct answer is option 'D'. 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 g(n) is a real valued sequence of 2N points and x1(n)=g(2n) and x2(n)=g(2n+1), then what is the value of G(k), k=N,N-1,…2N-1?a)X1(k)-W2kX2(k)b)X1(k)+W2kNX2(k)c)X1(k)+W2kX2(k)d)X1(k)-W2kNX2(k)Correct answer is option 'D'. 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 g(n) is a real valued sequence of 2N points and x1(n)=g(2n) and x2(n)=g(2n+1), then what is the value of G(k), k=N,N-1,…2N-1?a)X1(k)-W2kX2(k)b)X1(k)+W2kNX2(k)c)X1(k)+W2kX2(k)d)X1(k)-W2kNX2(k)Correct answer is option 'D'. Can you explain this answer?.
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