ECE 101 lab_solution2

ECE 101 lab_solution2 - ECE 101 Linear Systems, Winter 2008...

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ECE 101 – Linear Systems, Winter 2008 Lab #2 Solutions January 24, 2008 (send questions/comments to psiegel@ucsd.edu) Problem 1 . The plots for each pair of x[n] and h[n] are shown below: When a signal is convolved with a scaled, shifted unit impulse, the result is the scaled and shifted version (by the same amount) of the input signal. Problem 2 . The plots for this problem are shown below:
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-2 0 2 4 6 0 20 40 x(k)h(-k) -2 0 2 4 6 0 10 20 x(k)h(1-k) -2 0 2 4 6 0 10 20 x(k)h(2-k) -2 0 2 4 6 -10 0 10 x(k)h(3-k) -2 0 2 4 6 -10 0 10 x(k)h(4-k) -2 0 2 4 6 -5 0 5 x(k)h(5-k) -2 0 2 4 6 0 1 2 x(k)h(6-k) 0 2 4 6 0 20 40 y -2 0 2 4 6 8 0 20 40 conv(x,h) The convolution of the two signals results in conv(x,h) = [30 33 31 6 7 1 2], which matches the plot of y which was formed by summing the values of each x[k]h[n-k] plot at time n. Problem 3 Using the relationship y[n] = x[n] + 2x[n-2], we have: -2 -1 0 1 2 3 4 5 0 1 2 3 4 5 6 7 8 9 10 n y(n) y = x(n) + 2x(n-2) Using the result from Problem 1, h[n] = δ [n] + 2 δ [n-2], which in MATLAB can be written as h = [1 0 2]; The convolution of x and h is shown below, verifying that h[n] is the correct impulse response:
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-2 -1 0 1 2 3 4 5 6 7
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This note was uploaded on 03/16/2009 for the course ECE 101 taught by Professor Siegel during the Fall '08 term at UCSD.

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ECE 101 lab_solution2 - ECE 101 Linear Systems, Winter 2008...

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