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

# 55 where 4 and k2 3 therefore t2 0 comments o

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Unformatted text preview: )e jwt + K2 - 15 = 2 Solution of these equations yields K l + H ( - jw ) e-jwtj ( 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 ight-hand 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 ime-Domain A nalysis o f C ontinuous-Time 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 ( 3t-7.9°). 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 zero-input and zero-state 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.

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