Signal Processing and Linear Systems-B.P.Lathi copy

# 4 1 fzt et h t fzt 1 1 h tfzt r dt

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Unformatted text preview: (t) is causal, I (r) = 0 for r < O. T herefore, I (r) = 0 for r < 0, a s illustrated in Fig. 2.5a. Similarly, if h(t) is causal, h (t-r) = 0 for t - r < 0; t hat is, for r > t, a s depicted in Fig. 2.5a. Therefore, t he p roduct f (r)h(t - r) = 0 e verywhere except over t he n onshaded interval 0:::; r :::; t s hown in Fig. 2.5a ( assuming t ~ 0). O bserve t hat if t is negative, I (r)h(t - r) = 0 for all r a s shown in Fig. 2.5b. Therefore, Eq. (2.37) reduces t o (2.38) T he lower limit of integration in Eq. (2.38) is t aken as 0 - to avoid t he difficulty in integration t hat c an arise if I (t) c ontains a n i mpulse a t t he origin. In subsequent discussion, t he lower limit will b e shown as 0 w ith t he u nderstanding t hat it means 0 -. T his r esult shows t hat if I (t) a nd h (t) a re b oth c ausal, t he r esponse y (t) is also causal. Because of t he c onvolution's c ommutative p roperty [Eq. (2.31)], we c an also express Eq. (2.38) as [assuming causal I (t) a nd h(t)] Zero-State Response and Causality T he ( zero-state) response y (t) o f a n L TIC s ystem is = I (t) * h(t) = i~ I (r)h(t - =0 * y(t) * h (t) = ( I) 1- 1 ",0 1 (1:) = 0 (2.36) 6. The Width Property: I f t he d urations ( widths) of l 1(t) a nd /2(t) a re T l a nd T2 r espectively, t hen t he d uration ( width) of l1(t) * /2(t) is T l + T2 (Fig. 2.4). T he p roof o f t his p roperty follows readily from t he g raphical considerations discussed l ater i n Sec. 2.4-2. T his r ule m ay superficially a ppear t o b e v iolated in some s pecial cases discussed later. 1- ; ;;;;;,,;mHh. (a) Actually t his r esult h as been derived earlier in Eq. (2.28). 12 123 ~ (2.39) 0 As in Eq. (2.38), t his r esult assumes t hat b oth t he i nput a nd t he s ystem are causal. • E xample 2 .4 For an LTIC system with the unit impulse response h(t) response y(t) for the input = e - 2t u(t), determine the (2.40) Here both f (t) and h(t) are causal (Fig. 2.6). Hence, we need only to perform the convolution's integration over the range (0, t) [see Eq. (2.38)]. The system response is therefore given by y(t) = i t f (r)h(t - r) dr t ~0 Because f(t) = e -tu(t) and h(t) = e - 2t u(t) f (r) = e-Tu(r) and h(t - r) = e - 2(t-T)u(t - r) Remember t hat t he integration is performed with respect to r (not t ), and the region of integration is 0 :::; r :::; t. In other words, r lies between 0 and t. Therefore, if t ~ 0, then r ~ 0 and t - r ~ 0, so t hat u(r) = 1 and u(t - r) = 1; consequently y(t) = i t e - Te - 2 (t-T) dr t ~0 124 2 Time-Domain Analysis of Continuous-Time Systems I (t) 125 2.4 System Response t o E xternal Input: T he Zero-State Response TABLE 2.1: Convolution Table h ( t) / 1(t) * h (t) = h (t) * / 1(t) f (t - T) 1 . At ~u(t) - ), t u(t) n! e)'t ),n+l u (t) - n! t n - n 2: ) ,i+1(n _ j j )! u (t) j =o m in! tm+n+lu(t) ( m+n+l)! e A,t - eA,t + ( ),l - )'2)te A,t u (t) ( ),l - ),2)2 E xercise E 2.5 For a n L TIC system with t he impulse response h(t) = 6 e- t u(t), d etermine t he s ystem response t o t he i nput; ( a) 2u(t) a nd ( b) 3 e- 3t u(t). £:, Answer; (a) 12(1 - e-t)u(t) ( b) 9 (e- t - e- 3t )u(t) 11 'V E xercise E 2....
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## This note was uploaded on 04/14/2013 for the course ENG 350 taught by Professor Bayliss during the Spring '13 term at Northwestern.

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