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

2 8 convolution of f t and ht in example 26 t2o e

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: ) * I (t) r ather t han I (t) * g (t). T hus c (t) = g(t) * I (t) = l :g(r)/(t-r)dr t- 134 2 T ime-Domain A nalysis o f C ontinuous-Time S ystems 2.4 S ystem R esponse t o E xternal I nput: T he Z ero-State R esponse F irst, we d etermine t he e xpressions for t he s egments o f f (t) a nd g(t) u sed in finding c(t). A ccording t o Figs. 2 .10a a nd 2.lOb, these segments c an b e expressed as Therefore f (t) = 1 (a) = ( Jo'H g (r)f(t 1 1H ( b) g (r) = ~r and Fig. 2.10c shows g (r) a nd f (-r), w hereas Fig. 2.10d shows g(r) a nd f (t - r), w hich is f ( - r) s hifted by t. B ecause t he edges o f f (-r) a re a t r = - 1 a nd 1, t he edges of f (t-r) a re a t - 1 + t a nd 1 + t. T he two functions overlap over t he interval (0, 1 + t) ( shaded interval), so t hat c(t) = g et) g(t) = ~t a nd f (t - r) = 1 I (t) 1 35 -I 0 tg (~) I(-~) l(t-~)rl _ _- ' -1" t " 1 "(e) r) d r ~rdr -I =~(t+1)2 t- 3 - 1::;t::;1 0 (d) 1:- 1:- 1 " t" 2 (2.46a) T his s ituation, d epicted i n Fig. 2.10d, is valid only for - 1 ::; t ::; 1. For t > 1 b ut < 2, t he 8 ituation is as i llustrated i n Fig. 2.lOe. T he t wo functions overlap o nly over t he r ange - 1 + t t o 1 + t ( shaded i nterval). N ote t hat t he e xpressions for g (r) a nd f (t - r) d o n ot c hange; only t he r ange o f i ntegration c hanges. Therefore (e) 1:- 1H c(t) = 1 ~rdr l(t-~) - 1+' 2" t " 4 (2.46b) ( f) Also n ote t hat t he e xpressions in Eqs. (2.46a) a nd (2.46b) b oth a pply a t t = 1, t he t ransition p oint b etween t heir r espective ranges. We c an r eadily verify t hat b oth e xpressions yield a value of 2 /3 a t t = 1, so t hat c(1) = 2 /3. T he c ontinuity o f c(t) a t t ransition p oints i ndicates a high p robability o f a r ight answer. t F or t :::: 2 b ut < 4 t he s ituation is as shown in Fig. 2.1Of. T he f unctions g (r) a nd f (t - r) o verlap over t he i nterval from - 1 + t t o 3 ( shaded interval), s o t hat 1 3 c(t) = - 1H l +t 0 t < -1 I(t-~) ~- 1(t-1:) t" 4 ~r dr (2.46c) Again, b oth Eqs. (2.46b) a nd (2.46c) apply a t t he t ransition p oint t = 2. We c an r eadily verify t hat c(2) = 4 /3 w hen either o f t hese expressions is used. For t :::: 4, f (t - r) h as been shifted so far t o t he r ight t hat i t no longer overlaps w ith g (r) as d epicted i n Fig. 2. 109. Consequently c(t) = 0 - 1+t l +t o l +t ( g) ( h) c ( t) (2.46d) We now t urn o ur a ttention t o n egative values o f t. We have a lready d etermined c(t) u p t o t = - 1. F or t < - 1 t here is no overlap between t he t wo functions, a s i llustrated in Fig. 2.10h, so t hat (2.46e) c(t) = 0 t < -1 F igure 2.lOi shows c(t) p lotted a ccording t o Eqs. (2.46a) t hrough (2.46e). • t Even if c( t) is continuous at the transition, the answer could be wrong in the unlikely event of two or more errors canceling out their effects. Our discussion assumes t hat there are no impulses in f (t - r) and g(r) after the transition which were not present before. 2 ( i) 3 4 t- F ig. 2 .10 C onvolution of f (t) a nd g(t) i n E xample 2.8. 136 2 T ime-Domain Analysis of Continuous-Time Systems r ~*r= o t...
View Full Document

Ask a homework question - tutors are online