Then solve for homega as before the answer will be the

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

This note was uploaded on 01/16/2014 for the course ECE 438 taught by Professor Staff during the Fall '08 term at Purdue.

Ask a homework question - tutors are online