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Find circular convolution and linear using circular convolution for the following sequences x1(n) = {1, 2, 3, 4} and x2(n) = {1, 2, 1, 2}. Using Time Domain formula method.

Mumbai University > Computer Engineering > Sem 7 > Digital Signal Processing

Marks: 10 Marks

Year: Dec 2015

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answer is right but you have given wrong headings. first one is linear using circular and second one is circular convolution.

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Circular convolution using circular convolution:

$x_1$(n) = {1, 2, 3, 4}

and $x_2$ (n) = {1, 2, 1, 2}

L=4, M=4

Length of y(n) = L+M-1=4+4-1=7

$\therefore, x_1$(n) = {1, 2, 3, 4, 0, 0, 0}

& $x_2$(n) = {1, 2, 1, 2, 0, 0, 0}

For y(0),

enter image description here

$\therefore$, y(0)= 1×1=1

For y(1),

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$\therefore$, y(1)= 2×1+1×2=4

For y(2),

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$\therefore$ , y(2)= 1×1+2×2+3×1=8

For y(3),

enter image description here

y(3)=1×2+2×1+3×2+4×1=14

For y(4),

enter image description here

$\therefore$, y(4)= 4×2+3×1+2×2=15

For y(5),

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$\therefore$, y(5) = 4×1+3×2=10

For y(6),

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$\therefore$, y(6) = 4×2=8

$\therefore$ ,y(n) = {1, 4, 8, 14, 15, 10, 8}

enter image description here

Result: y(n) = {2, 4, 8, 14, 15, 10, 8}

Linear using circular convolution:

For y(0),

enter image description here

$\therefore$ , y(0)= 1+4+3+8=16

For y(1),

enter image description here

$\therefore$ , y(1)= 2+2+6+4=14

For y(2),

enter image description here

$\therefore$, y(2)= 1+4+3+8=16

For y(3),

enter image description here

$\therefore$, y(3)= 2+2+6+4=14

y(n) = {16, 14, 16, 14}

enter image description here

Result: y(n) = {14, 16, 14, 16}

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