Transformations-Digital Signal Processing-Assignment Solution, Exercises of Digital Signal Processing

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229 7.1. Using the partial fraction technique, we see Sta 05 + 0.5 (stay'+h statjb sta-jd Now we can use the Laplace transform pair HAs) = etu(g} a sta to get al (entotame 4 .-(a-sit helt) = 5 ( ots ag ) ut). {a) Therefore, Ayfn} = Ae(nT) = ; [erteraner + ert] ufr] 0.8 . A(z) = Toes b> ee? {b) Since helrydr a FE) 5.46) oo 5 we get _ sta _ A Az AS Ss) = yaa jbstengb) ~ s }s4aajb stay where a 05 "aap Oa AL Thongh the system hg{zi} is related by step invariance to h,(t), the signal sa[n] is related to s(t) by impulse invariance. Therefore, we know the poles of the partial fraction expansion of S.(s) above must transform as z, = e**7, and we can find L Ar AS S22) = pa + Term + ese aire Now, since the relationship between the step response and the impulse response is sala] = 7 Aalk} = xs ha{kjufn — k] = hain] + u[n] kano oe — F(z) Sale} = 1-27! We can finally solve for H2(z) Fh{z) = $2(2(1- 27) l-zt . 1-27 -oT = At Ameer tAipoy {z| >e* where A; and A are as given above.