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ee20-hw03-f10sol

# ee20-hw03-f10sol - tered on the active point for the dirac...

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HW 3.2 (a) In this part, we recognize that the output will simply be the convolution of two boxes of magnitude 1 and width 1. We know to use convolution because even if we did somehow ﬁnd the frequency response of our system, our input signal is not periodic. As we discussed in lecture, the convolution of two equally wide boxes will give us the triangular ﬁgure shown below. (b) For this part, we are convolving of two boxes of magnitude 1, but dif- fering widths. This will give us the saturated triangular ﬁgure shown below. (c) Since convolution with the dirac delta just makes two copies of x cen-

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Unformatted text preview: tered on the active point for the dirac delta, we get by inspection the following simple ﬁgure. Note the maximum magnitude is when the two copies are added (so it is 2, not 1). (d) In this case, we simply have an inﬁnite number of copies of our box, centered on the active point for each delta. 1-3-2-1 1 2 3 4 5 6 0.2 0.4 0.6 0.8 1 t y(t) Figure 1: y ( n ) 2-3-2-1 1 2 3 4 5 6 0.2 0.4 0.6 0.8 1 t y(t) Figure 2: y ( n ) 3-1-0.5 0.5 1 1.5 2 0.5 1 1.5 2 t y(t) Figure 3: y ( n ) 4-10-8-6-4-2 2 4 6 8 10 0.5 1 t y(t) Figure 4: y ( n ) 5...
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ee20-hw03-f10sol - tered on the active point for the dirac...

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