hw1_sol

# Then solve for homega as before the answer will be the

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Unformatted text preview: \omega). As before, the answer will be the same. r-?,rJtd = =Z = &-rr ( g + '1 04; )/z This did not get used in this derivation. usa This solution is correct; but it is not clear how it is obtained. My approach to solving this problem would be to do a direct computation of the response to an impulse as shown on the right. =-++ 4<w <o Kril<-rf -fiev/<cl x_0[n] x[n] b Page 10 This derivation shows some of the intermediate steps needed to obtain the final result shown below. y_0[n] y[n] 4., Again, at this stage of the course, this problem should be solved by setting x[n] = e^{j \omega n}, and y[n] = H (\omega) x[n]. The answer will be the same. b. C. fu"*, f"* a. F ll"t"t), \tu,t) - yTfi = 7frA4 q l+{^"0 =) / [Xcrrl - e+\v,.z,) :7H,Lt^))--W=fu ,T i,-t a)"u <trwrltu in frt z Tzs-;tr HC"r) 9",UIAW I t- - zg-Iw I+ b. 2 l+ z e-I,, -zotd zsjw 1I* b,w tl"!r@ flip w &* gw l,es",bt. _L...
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## This note was uploaded on 01/16/2014 for the course ECE 438 taught by Professor Staff during the Fall '08 term at Purdue.

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