Question

kindaly first FIND THE LIMItS AND RANGEs THRough FOR thIS QUestion BY graphical methods AND BY HAND ON

NO OVERLaping , partial overlAPing ,COMPLETE oVerLAPING AND THEN AGAIN ZERO OR NO OVER LAPING .

Task 01: Suppose that a LTI system is described by impulse response h(t)=efu(t). Compute the response of the system (by both methods) to the input signal -1<t< 0.5 10.6 X(t)={0.3 0 0.5<t<3 t<-land t > 3

Answer 1

ht) = é et utt) x Gt) = 10.6 of Kt Los 0-3 of t23 o else AnG 2 XCH 0.6 t O oos 3 Since, ych) = xe6+)*nC - 2 x(e) h(4-7) ar Steps :- find hilt-?) where x-axis is of & not t haft) sht 2) 2 the Now, wer slide the function convolution integral For [ta-1 h (t-t) & fing No overlapping There i yGt) = foode = 0 be +

For Het <0.5 There is partial overlapping you = 1 606) (4-3) de Flo HE 3 0.5 -(t-C) -T) 7+ 0.6 e -(t+1) 0.6 le -(t+1) yH = 0.6 ( 1 -e (6+1)] for 105¢ t < 3 is partial overlapping $ de Again theiße y (+) = 1086)2+(8-3) X(T) & (t-T) -1 0.5 & 3 -(1-2) at dr t 0.5 de → y(t) = {o-3ett-egt 05 +0.3 [6 (4-1) +0.3 461) t + f (03e ycey = (org (18+4)** 0.6 l (t = 0.5) efe+t) ] e(t-os)] 0.3 (e48-0-5) 10-62-(+1) et (0.3eos - o.Get] +0.3 0.3 ta •178 e-t ly(t) = 0.3 + 0 2738 é © Feb complete overlatting yet) = 1 xce) e (tt) de yet) = los obeft de +13 0.3 0 (4-7 de o.se * {egos +03e & fel? 0.6 e-t ( 605 ) + 0.30-t ( e² - etois) t3 There is 3 +5.531e-t 0.7685e-t 6.3 e-t yet

Hence, o 036(1-8 (4+)) 0.3 +0.2378et 6.se ; t<-| gtst<0.5 ; 05<t<3 y (t) = t23