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Unformatted text preview: )e jwt + K2  15 = 2 Solution of these equations yields K l + H (  jw ) ejwtj ( b) For input f (t) =5 = H (jw) = I H(jw)lejLH(jw) [wt + 8 + L HUw)] t >O To find {3, we s ubstitute Y4>(t) in t he system equation t o o btain (2.61) (D2 + 3 D + 2) Y4>(t) = D f(t) or
B ut
D [{3te  2tJ = {3(1  2t)e  2t D2 [{3te 2t ] = 4{3(t _ l )e 2t Solve t he differential equation + 3 D + 2) yet) = D f(t) if t he i nitial conditions are y(O+) = 2 a nd y(O+) = 3 a nd the input is
( b) 5 >0 (2.60) E xample 2 .10 (D2 t Y4>(t) = {3te 2t T his r esult c an b e g eneralized for t he i nput f (t) = c os (wt + 8). T he f orced r esponse
i n t his c ase is IHUw)1 cos ( c) e  2 • D e 2t = _ 2e 2t Consequently {3(4t  4 + 3  6t + 2 t)e 2• = _ 2e 2t ( d) lOcos (3t + 30°). According t o Example 2.9, t he n atural response for this case is 3 5eo., ( = 0, a nd y</>(t) = R e { IH ( jw )lej[wt+LH(jW)]} y</>(t) = + 45 = Therefore T he complete solution is K le t + K 2e 2t . Using t he initial conditions, we determine K l
and K2 as in p art ( a).
( c) Here ( =  2, which is also a characteristic root of t he system. Hence, (see P air
2, Table 2.2, or t he comment a t t he b ottom of t he t able) so t hat + L H(jw)]  Kl  2K2 Y4>(t) = 5 H(0) = 0 B ut = IH(jw)1 cos [wt a nd =  8 and K2 = 25. yet) =  8e' + 2 5e 2•  15e 3 • B ecause t he t wo t erms o n t he r ighthand s ide a re c onjugates, ( a) l Oe 3 • >0 a nd T he Exponential Input e jwt • t> 0 T he t otal solution (the sum of t he forced and t he n atural response) is ( =0 = C H(O) ( H«() = Q«() = ( 2 «(
«( 145 C lassical S olution o f D ifferential E quations or
Therefore, {3 = 2 so t hat 146 2 T imeDomain A nalysis o f C ontinuousTime Systems 2.6 147 S ystem S tability Therefore
y",(t) = 2 te 2t
t
T~e complete solution is K le + K 2e 2t + 2 te 2t . Using the initial conditions, we determme K l and K2 as in part (a).
(d) For t he i nput f (t) = 1 0cos(3t+300), t he forced response [see Eq. (2.61)J is y(t) = _ lOe t + 2 5e 2t _ 1 5e 3t which agrees with the solution found previously in Eq. 2.51b. o • C omputer E xample C 2.S
Solve t he differential equation y",(t) = 10IH(j3)1 cos [3t + 30° + LH(j3)J (D 2 + 3D where + 2)y(t) = f (t) for t he input f (t) = 5t + 3.
H ( '3) = P (j3) _
j3
_
j3
J
Q (j3)  (j3)2 + 3(j3) + 2  7 + j 9 27  j2l = "'""130 = 0.263e  J37 gO Therefore
IH(j3)1 = 0.263, L H(j3) =  37.9° a nd
y",(t) = 10(0.263) cos (3t + 30°  37.9°) = 2.63 cos (3t  7.9°) The complete solution is K le t + K2e 2t + 2.63cos ( 3t7.9°). Using the initial conditions,
we determine K 1 and K2 as in p art (a). •
E xample 2 .11
Using t he classical method, find the loop current y(t) in the R LC circuit of Fig. 2.1,
Example 2.2 if t he i nput voltage f (t) = l Oe 3t and the initial conditions are y (O) = 0
and v c(O) = 5.
The zeroinput and zerostate responses for this problem are found in Examples 2.2
a nd 2.5, respectively. The natural and f...
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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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