Hw02_Sol_Su09

Hw02_Sol_Su09 - ECE220 Homework #2 Solutions Solution to...

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Unformatted text preview: ECE220 Homework #2 Solutions Solution to Problem 2.37 s 3 ( t ) = p 1 (2 t ) + p 1 (0 . 5 t ) + p 2 (2 t + 2) where the subscript in p ( t ) defines the width of the centered pulse. Lets consider the first term. Using the formula 2.13 on page 60, we have: p 1 (2 t ) = 1 ,- 1 2 2 t < 1 2 , otherwise Similarly p 1 (0 . 5 t ) = 1 ,- 1 2 . 5 t < 1 2 , otherwise And finally p 2 (2 t + 2) = 1 ,- 1 2 t + 2 < 1 , otherwise The following MATLAB code sketches the function s 3 ( t ): clear close all t = -4:0.001:4; f1 = (2*t >= -1/2) & (2*t < 1/2); f2 = (0.5*t >= -1/2) & (0.5*t < 1/2); f3 = (2*t+2 >= -1) & (2*t+2 < 1); s3 = f1+f2+f3; % Plot the function plot(t,s3, linewidth,2); xlabel(t -->) ylabel(s_3(t) -->) title(ECE220 Problem 2.37. The function s_3(t)) grid on axis([-4 4 -1 3]) print -deps sol2_37.eps Figure 1 shows the sketch for s 3 ( t ) 1-4-3-2-1 1 2 3 4-1-0.5 0.5 1 1.5 2 2.5 3 t --> s 3 (t) --> ECE220 Problem 2.37. The function s 3 (t) Figure 1: Sketches for Problem 2.37. Solution to Problem 2.41 Similarly, we can detail the properties of the four distinct pulses observed in Fig. 2.21 (left to right direction) in p 2 ( t ), as follows: amplitude -2, shifted by -7, scale factor 1/2 amplitude 2, shifted by 0.5, scale factor 2 amplitude 1, shifted by 2.5, scale factor 2 amplitude 3, shifted by 6.5, scale factor 2/5amplitude 3, shifted by 6....
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Hw02_Sol_Su09 - ECE220 Homework #2 Solutions Solution to...

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