Unformatted text preview: ) * I (t) r ather t han I (t) * g (t). T hus
c (t) = g(t) * I (t)
= l :g(r)/(tr)dr t 134 2 T imeDomain A nalysis o f C ontinuousTime S ystems 2.4 S ystem R esponse t o E xternal I nput: T he Z eroState 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 (tr)
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(t1:) 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 imeDomain Analysis of ContinuousTime Systems r ~*r=
o t...
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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.
 Spring '13
 Bayliss
 Signal Processing, The Land

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