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What is the sequence y(n) that results from the use of four point DFTs if the impulse response is h(n)={1,2,3} and the input sequence x(n)={1,2,2,1}?
- a){9,9,7,11}
- b){1,4,9,11,8,3}
- c){7,9,7,11}
- d){9,7,9,11}
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
What is the sequence y(n) that results from the use of four point DFTs...
Explanation: The four point DFT of h(n) is H(k)=1+2e-jkπ/2+3 e-jkπ (k=0,1,2,3)
Hence H(0)=6, H(1)=-2-j2, H(3)=2, H(4)=-2+j2
The four point DFT of x(n) is X(k)= 1+2e-jkπ/2+2 e-jkπ+3e-3jkπ/2(k=0,1,2,3)
Hence X(0)=6, X(1)=-1-j, X(2)=0, X(3)=-1+j
The product of these two four point DFTs is
Y ̂(0)=36, Y ̂(1)=j4, Y ̂(2)=0, Y ̂(3)=-j4
The four point IDFT yields y ̂(n)={9,7,9,11}
We can verify as follows
We know that from the previous question y(n)={1,4,9,11,8,3}
y ̂(0)=y(0)+y(4)=9
y ̂(1)=y(1)+y(5)=7
y ̂(2)=y(2)=9
y ̂(3)=y(3)=11.
Hence H(0)=6, H(1)=-2-j2, H(3)=2, H(4)=-2+j2
The four point DFT of x(n) is X(k)= 1+2e-jkπ/2+2 e-jkπ+3e-3jkπ/2(k=0,1,2,3)
Hence X(0)=6, X(1)=-1-j, X(2)=0, X(3)=-1+j
The product of these two four point DFTs is
Y ̂(0)=36, Y ̂(1)=j4, Y ̂(2)=0, Y ̂(3)=-j4
The four point IDFT yields y ̂(n)={9,7,9,11}
We can verify as follows
We know that from the previous question y(n)={1,4,9,11,8,3}
y ̂(0)=y(0)+y(4)=9
y ̂(1)=y(1)+y(5)=7
y ̂(2)=y(2)=9
y ̂(3)=y(3)=11.
Most Upvoted Answer
What is the sequence y(n) that results from the use of four point DFTs...
Explanation:
- Given impulse response, h(n) = {1, 2, 3} and input sequence, x(n) = {1, 2, 2, 1}
- To find the sequence y(n) using 4-point DFTs, we need to calculate the DFT of h(n) and x(n) separately using the formula:
- X(k) = Σx(n)exp(-j2πnk/N) where N is the size of DFT and k = 0, 1, 2, ..., N-1
- H(k) = Σh(n)exp(-j2πnk/N) where N is the size of DFT and k = 0, 1, 2, ..., N-1
- Using the above formula, we can calculate the DFT of x(n) and h(n) as follows:
- X(0) = 1 + 2 + 2 + 1 = 6
- X(1) = 1 + 2exp(-jπ/2) - 2 - 2exp(-jπ/2) = -2j
- X(2) = 1 - 2 + 2 - 1 = 0
- X(3) = 1 - 2exp(-jπ/2) - 2 + 2exp(-jπ/2) = 2j
- H(0) = 1 + 2 + 3 = 6
- H(1) = 1 + 2exp(-jπ/2) - 3 - 2exp(-jπ/2) = -1-j
- H(2) = 1 - 2 + 3 = 2
- H(3) = 1 - 2exp(-jπ/2) - 3 + 2exp(-jπ/2) = -1+j
- Now, we can calculate the DFT of the output sequence y(n) using the formula:
- Y(k) = X(k)H(k) where k = 0, 1, 2, ..., N-1
- Using the above formula, we can calculate the DFT of y(n) as follows:
- Y(0) = X(0)H(0) = 6*6 = 36
- Y(1) = X(1)H(1) = (-2j)*(-1-j) = -2+2j
- Y(2) = X(2)H(2) = 0*2 = 0
- Y(3) = X(3)H(3) = (2j)*(-1+j) = -2-2
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What is the sequence y(n) that results from the use of four point DFTs if the impulse response is h(n)={1,2,3} and the input sequence x(n)={1,2,2,1}?a){9,9,7,11}b){1,4,9,11,8,3}c){7,9,7,11}d){9,7,9,11}Correct answer is option 'D'. Can you explain this answer?
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