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

# E xample 29 solve t he differential equation d2 3d

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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&gt;(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&gt;(t) = H«()e(t +1 t &gt;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 zero-input component exists, a nd t he i nitial conditions a t t = 0 - can be applied to t he zero-input component only. In t he classical m ethod, t he z ero-input a nd z ero-state 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&gt;(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 ime-Domain A nalysis o f C ontinuous-Time 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&quot; ± 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&gt;(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&lt;/&gt;(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 &gt;O H ere ( = j w a nd y&lt;/&gt;(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&lt;/&gt;(t) = ~ [ H(jw...
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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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