《电路》（英文版）15-7 The complete response

§15-7 The compl lete response When initial energy is present in a circuit, the Laplace transform method may be used to obtain the complete response The first is the more fundamental. for it involves writing the differential equations for the network and then taking the laplace transform of those equations. Letu,0)=10V 24 ()50 cos 2tu(t)I 30U2 1 F andu2(07)=25V, F findo

§15-7 The complete response The first is the more fundamental, for it involves writing the differential equations for the network and then taking the Laplace transform of those equations. (0 ) 1 − Let  =10V and =25V, find .  2 (0 ) 2 −  When initial energy is present in a circuit, the Laplace transform method may be used to obtain the complete response. 1 50cos 2tu(t)V − +  2 − + − + F 48 1 F 24 1 30 20 24 2 1

20 LetD(0)=10V 24 andu,(0 )=25V 50c0s2(t) 30U2 24 findu F Solution: at the node 1: Dr +D1=0or2U2=2U1+U1(1) 24 48 at the node 2: U2-50 cos 2tu(t) 20 24 3024020 —+—D orU1=D2+32-60c0s2t()(2) Identifying U2 as the desired response, we eliminate D, and D1 by taking the derivative of (2), remembering that du(t)/dt=d(t) U2=U2+302+2×60sim2m()-606()(3)

Solution: at the node 1: 0 2 2 (1) 48 1 24 ' 2 1 1 ' 1 1 2       + = = + − or 0 24 1 20 24 30 50cos2 ( ) ' 2 2 2 1 2 + + = − + −   t u t    3 60cos2 ( ) (2) 2 ' 1 2 or  =  +  − t u t at the node 2: Identifying as the desired response, we eliminate '  2 1 and 1 by taking the derivative of (2), remembering that du(t)/ dt =  (t). 3 2 60sin2 ( ) 60 ( ) (3) ' 2 '' 2 ' 1  = +  +  t u t −  t (0 ) 1 − Let  =10V and =25V, find .  2 (0 ) 2 −  1 50cos 2tu(t)V − +  2 − + − + F 48 1 F 24 1 30 20 24 2 1

2 2U1+U U1=U2+32-60c0s2t(t)(2) U1=U2+3U2+2×60sin2(t)-606()(3) substituting() and 3)into(1), 2=22+32-60c00+u2+32+2×60s0m(0-606( 0rU2+5U2+4U2=120c0s2t(t)-120sin2m(t)+605(1) We now take the Laplace transform 120s-120×2 s22(s)-sU2(0)-U2(0)+5y2(s)-5U2(0)+42() +60 s2+4 20s-120×2 0r(S2+5s+4)V2(s)=sU2(0)+U2(0)+5U2(0)+ +60

substituting (2) and (3) into (1), 2 2 2 = + 3 2 − 60cos 2 ( ) + ' 2   tu t 3 2 60sin2 ( ) 60 ( ) ' 2 '' 2  +  +  t u t −  t 5 4 120cos2 ( ) 120sin2 ( ) 60 ( ) 2 ' 2 '' 2 or  +  +  = t u t − t u t +  t We now take the Laplace transform, 60 4 120 120 2 ( ) (0 ) (0 ) 5 ( ) 5 (0 ) 4 ( ) 2 2 2 2 ' 2 2 2 2 + + −  − − + − + = − − − s s s V s s  sV s  V s 60 4 120 120 2 ( 5 4) ( ) (0 ) (0 ) 5 (0 ) 2 2 ' 2 2 2 2 + + −  + + = + + + − − − s s or s s V s s   2 2 (1) '  2 = 1 +1 3 60cos 2 ( ) (2) 2 ' 1 2  =  +  − t u t 3 2 60sin2 ( ) 60 ( ) (3) ' 2 '' 2 ' 1  = +  +  t u t −  t

120s-120×2 (2+5s+4)V2(s)=sU2(0)+u2(0)+5U2(0)+ s2+4 LetU2(0)=25 120s-120×2 (s2+5s+4)V2()=25s+125+U2(0)+ +60 And need a value forv2(0 This we may obtain from(1)and (2) by evaluating each term at t=0, Actually, we need use only (2)in this problem: U1=U2+3U2-60c0s2u(t)(2) (0)=U2(0)+32(0)-0 amdU2(0)=-65 120s-120×2 (s2+5s+4)2(s)=25s+125-65+ +60 s2+4 120s-120×2 =25s+120+ +4 25s+120+(120s-240)/(s2+4 (s+1)(s+4)

60 4 120 120 2 ( 5 4) ( ) (0 ) (0 ) 5 (0 ) 2 2 ' 2 2 2 2 + + −  + + = + + + − − − s s s s V s s   2 (0 ) = 25 − Let  60 4 120 120 2 ( 5 4) ( ) 25 125 (0 ) 2 ' 2 2 2 + + −  + + = + + + − s s s s V s s  And need a value for . This we may obtain from (1) and (2) by evaluating each term at t=0 - , Actually, we need use only (2) in this problem: (0 ) ' 2 −  (0 ) (0 ) 3 2 (0 ) 0 ' 1 = 2 + − − − −    (0 ) 65 ' 2 = − − and  60 4 120 120 2 ( 5 4) ( ) 25 125 65 2 2 2 + + −  + + = + − + s s s s V s s 4 120 120 2 25 120 2 + −  = + + s s s ( 1)( 4) 25 120 (120 240)/( 4) ( ) 2 2 + + + + − + = s s s s s V s 3 60cos 2 ( ) (2) 2 ' 1 2  =  +  − tu t

2(s) 25s+120+(120s-240)/(2+4) (s+1)(S+4) 23/316/312s+24 s+1s+4s2+4 23/3,16/3.6-j6,6+j6 十 s+1s+4s2+4s2+4 U2()=[ 23 16 e-+0e4+12√2cos(2t-45°) 3 3

( 1)( 4) 25 120 (120 240)/( 4) ( ) 2 2 + + + + − + = s s s s s V s 4 6 6 4 6 6 4 16/ 3 1 23/ 3 2 2 + + + + − + + + + = s j s j s s t  e e t u t V t t 12 2 cos(2 45 ) ( ) 3 16 3 23 ( ) 4 2   = + + − − − 4 12 24 4 16 3 1 23 3 2 + + + + + + = s s s / s /

The second technique requires each initial capacitor voltage or inductor current to be replaced by an equivalent dc source. The frequency-domain equivalent of an inductor L i(t i(0) Li(o SL i(0 u(t=L di(t) dt V(S=sLI(S)-Li(o)I(s) V(s),i(0

The second technique requires each initial capacitor voltage or inductor current to be replaced by an equivalent dc source. The frequency-domain equivalent of an inductor L: dt di t t L ( ) ( ) = ( ) ( ) (0 ) − V s = sLI s − Li s i sL V s I s ( ) (0 ) ( ) − = + i( t ) L ( ) − +  t (0 ) − i ( a ) + − I( s ) sL ( ) − + V s ( b ) (0 ) − Li I( s ) ( ) sL − + V s ( c ) s i(0 ) −

The frequency-domain equivalent of a capacitor C: 一+ I(s) 1/SO U(0) C(0 U(0 1/sC (b) do(t) v(s) I(s),D(0) i(t)=C 十 1(S)=SCV(s)-C0(0) dt

The frequency-domain equivalent of a capacitor C: sC s I s V s ( ) (0 ) ( ) − = +  dt d t i t C ( ) ( )  = ( ) ( ) (0 ) − I s = sCV s − C i( t ) C ( ) − +  t (0 ) −  ( a ) − + I( s ) ( ) − + V s ( b ) s (0 ) −  1 / sC I( s ) ( ) − + V s ( c ) (0 ) − C 1 / sC

0 20 24 24 24 ①52m0 Os 48 30 F 30 02T24F s2+4 tvi(s) 48 Solution: V2(50s 10 5 s V2( (S) s-+4 十 0 20 24+48/S30 24/s 25s+120+(120-240)/(2+4)23/3,16/3,12s+24 (s+1)(s+4) s+1s+4s2+4 23 16 u2(t=e+e+12 2t+12sin 2tu(t)v or U,(t)= 23-t16。-4t1+122cs(2t-45u() e+—e

0 24 / 25 ( ) 30 ( ) 24 48 / 10 ( ) 20 4 50 ( ) 2 2 2 2 2 = − + + + − + + − s s V s V s s s V s s s V s 4 12 24 4 16/ 3 1 23/ 3 2 + + + + + + = s s s s Solution: t e e t t u t V t t 12cos2 12sin2 ] ( ) 3 16 3 23 ( ) [ 4 2 = + + + − −  ( 1)( 4) 25 120 (120 240)/( 4) ( ) 2 2 + + + + − + = s s s s s V s − + 4 50 2 s + s s 48 30 20 2 1 − + ( ) − + V s 1 s 24 s 10 s 25 − + ( ) − + V s 2 24 1 50cos 2tu(t)V − +  2 − + − + F 48 1 F 24 1 30 20 24 2 1 or t e e t u t V t t 12 2 cos(2 45 ] ( ) 3 16 3 23 ( ) [ 4 2 = + + −  − − 

Example 3: Find i(t)and u(t), fort>0 0.5F i(t) Solution: Vc(s)-4/s+8/3s 十 IF IF 2+2/s 4V c(s)-4/3s,Vc(s) 十 1/s 1/s 22/s8/3s I(s) 0.8-2/15 Vc(s)=+ 十 ss+1.25 s(s) ①4/3s U2(t)=(4/5-2/15e-)n(t U2(01)=4/5-2/15=2/3 Vc(s)4(2/15)s2 5s+1.25s=+ U(0)=0 1/s 36s+125 i(t)=(t)+e12u(t)4 电荷守恒1F2/3+1F2/3=1F4/3

Example 3: Find i(t) and (t), fort  0. c 0 1/ ( ) 1/ ( ) 4 / 3 2 2 / ( ) 4 / 8 / 3 + = − + + − + s Vc s s Vc s s s Vc s s s 1 25 0 8 2 15 s . / s . Vc(s) + − = + c (0 ) = 4 / 5 − 2 /15 = 2 / 3V +  [ (0 ) = 0] −  c s Vc s I s 1/ ( ) ( ) = i t t e u t A t ( ) 6 1 ( ) 3 2 ( ) −1.25 =  + Solution: t e u t V t c ( ) (4 / 5 2 / 15 ) ( ) −1.2 5  = − − + VC (s) 2 I(s) + − 8 / 3s 1/ s − + 4 / 3s − + 4 / s 2/s 1/ s 2 i(t) − + (t)  C − + 4V 1F 1F 0.5F 电荷守恒 1F2/3+ 1F2/3 =1F4/3 1.25 1 6 1 3 2 + = + 1.25 s (2/15) 5 4 + = − s s

DP: 2 (a)Draw the frequency domain circuit that is valid for the circuit shown in fig. (b)Find i(t), vc(t) andu(t).for 0. 40 l(-t)4 UL(OS2H nF TUC( 10

DP: 2 (a)Draw the frequency domain circuit that is valid for the circuit shown in Fig. (b)Find i(t), (t) and .for t>0.  C (t)  L u(−t)A − + (t) C  i( t ) 10 t = 0 F 32 1 40 20 + − 30 V 2H − + (t)  L