This preview shows page 1. Sign up to view the full content.
Unformatted text preview: t he m ost i mportant signal in t he s tudy of LTI systems.
Interestingly t he forced response for a n e xponential i nput signal t urns o ut t o b e
very simple. , From Table 2.2 we see t hat t he forced response £ t h ' put e ( t h as
lor e m
t he form {3e(t. We now show t hat (3 = Q «()/ P «()·t To determine t he c onstant (3,
we s ubstitute Y4>(t) = /3e(t i n t he s ystem equation [Eq. (2.52)] t o o btain 2 d {2
= dt 2 (32t + {3lt + (3o ) = 2{32 D f(t) 143 Classical Solution of Differential Equations '2.5 T hus, for t he i nput f (t) = e (tu(t), t he forced response is given by = K I + K2 + 1 3 =  KI  2K2 The solution of these two simultaneous equations is K I
yet) = 4 e t  3e 2t +t +1 Y4>(t) = H«()e(t +1 t >O (2.55) where = 4 and K2 =  3. Therefore t2': 0 • Comments o n Initial Conditions
I n t he classical method, t he i nitial conditions are required a t t = 0+. T he
reason is t hat because a t t = 0 , only t he zeroinput component exists, a nd t he
i nitial conditions a t t = 0  can be applied to t he zeroinput component only. In t he
classical m ethod, t he z eroinput a nd z erostate components cannot be separated.
Consequently, t he initial conditions must be applied to t he t otal response, which
begins a t t = 0 +. P«()
H«() = Q«() (2.56) T his is a n i nteresting a nd significant result. I t s tates t hat for a n e xponential input e (t t he forced response Y4>(t) is t he s ame exponential multiplied by H (O =
P (O/Q«(). T he t otal s ystem response y (t) t o a n e xponential input e (t is t hen
given by
n y (t) = L :Kje AJt + H (Oe(t j =l T his r esult is valid only if ( is n ot a c haracteristic r oot o f t he s ystem. (2.57) 144 2 T imeDomain A nalysis o f C ontinuousTime S ystems w here t he a rbitrary c onstants K 1 , K 2, . .. , K n a re d etermined f rom a uxiliary c onditions.
Recall t hat t he e xponential s ignal i ncludes a l arge v ariety o f s ignals, s uch a s a
c onstant
= 0), a s inusoid ( ( = ± jw), a nd a n e xponentially g rowing o r d ecaying
sinusoid
= (7" ± j w). L et u s consider t he f orced response for s ome o f t hese c ases. 2.5 For this case P (C) ( a) For input f (t) = l Oe 3t , Y 4>(t)=lOH(3)e T he Constant Input f (t) = C
B ecause C = C eat, t he c onstant i nput is a special case o f t he e xponential i nput
C e,t w ith ( = O. T he forced response t o t his i nput is t hen g iven b y y</>(t) = C H«()e,t w ith 3' ( =  3, a nd [
3 3'
15 3'
= 10 ( 3)2+3(3)+2 e
= e yet) = K le t (2.58) + 3( + 2 + K 2e 2t  15e 3• t t >O H ere ( = j w a nd y</>(t) = H (jw)e jwt (2.59) T he initial conditions are y(O+) = 2 a nd y(O+) = 3. S etting t = 0 in t he above equations
and t hen s ubstituting the initial conditions yields Kl T he Sinusoidal Input f (t) = cos wot
W e know t hat t he f orced r esponse for t he i nput e±jwt is H (±jw)e±jwt. S ince
c os w t = (e jwt + e  jwt )/2, t he f orced r esponse t o cos w t is y</>(t) = ~ [ H(jw...
View
Full
Document
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

Click to edit the document details