习题三答案uc(0_)=4 V1<0时,uc(0,)=4Vt>0时,uc(o) = 5 vRo =2.502电容左边的戴维南等效电阻t=lsuc(t)=(5-e-")V, 1>0得三.(1)%云o21-1F(2)1%1I1止2+%0211-1F(3)1%1F%oII2元-1F2三.uc(t)=10 V,1<0T=4suc(0,)=13Vt>0 时,uc(0)=10 V1uc(t) =10+3e7e(t)V得-=[660-号00]a2ducic(t) = 2dt2L得.四.i,(t)=2.5AA,ui(t)=0,1<012i(0)=7At>0 时,iz(o)=2.5Ai () =[2.5+4.5e-2r e 0] A得0=2 %-[98 0-18e2* 0]得dt
习题三答案 一. t 0 时， uC (0− ) = 4 V t 0 时， uC (0+ ) = 4 V uC()= 5 V 电容左边的戴维南等效电阻 R0 = 2.5 = 1s 得 ( ) (5 e ) V t C u t − = − ，t 0 二. （1） O 1 −1 2 t s u V （2） O 1 −1 2 2 t s u V （3） O 1 −1 2 2 t s u V 三. u t C ( ) = 10V ， t 0 = 4 s t 0 时， uC (0+ ) = 13V uC () = 10V 得 u t t C t ()= + e () V − 10 3 1 4 ε 得 i t u t t t C C t () d d = = ()− e () A − 2 6 3 2 1 δ 4 ε 四. i t L ( ) = 2.5A ，u t L ( ) = 0，t 0 = 1 2 s t 0 时， i L (0+ ) = 7A iL () = 2.5A 得 iL t t t ()= . + . e () A − 2 5 4 5 2 ε 得 u t i t t t L L t () d d = = ()− e () V − 2 9 18 δ 2 ε
五.解法一us(0)=[2e(t-1)- 2e(t-2)t = lsu(0)=[e-(-1) e(t-1)-e-(-2) (t-2)V解法T = lsIs<t<2s时,u(1)=IV得u(0)=e-(-1)V, 1s<t<2s(0)=[1-e-(-)]/A,ls<t<2s<2s时, u(2.)=-i,(2.)=-(1-e-)v得u (0)=-(1-e-")e-(-2)V, t>2s或u ()=-0·632e-(-2)V, t>2s六.0.1s952+2420.125s)Ur(s)740.4.s20.1s4×0.125sU(s)=$2 +242 4 +0.125s(s+ 32)(s2 +242)K,K22 +j242S+32K, =0.256Kz=0.12Z53.130[o0.256e-32t +0.24 cos(24t+53.10)(t) V0.4s-j24，求得u取实部）(或将七.25×10-3s$2 +252=10340 uc(s)s40103103sZ(s) =103S+2540+s25sUc(s) =(s+ 25)(s2 +252)125tcos(25t-45)s(t)Vuc:-e22八
五. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0 632e V 2s 1 e e V 2s 2s 2 2 1 e V 1 e A 1s 2s e V 1s 2s 1s 2s 1 1V 1s e 1 e 2 V 1s 2 1 2 2 V 2 1 2 1 1 1 1 2 S = − = − − = − = − − = − = = = = − − − = = − − − − − − − − − + + − − − − + − − − − u t t u t t t u i i t t u t t t u u t t t u t t t t t L t L t t t 或 ， 得 ， 时， ， 得 ， 时， 解法 ε ε 解法一 ε ε 六. + _ 4 0.125s U s L ( ) 01 24 2 2 . s s + U s s s s s s s s ( ) . . . . ( )( ) = + + = + + 01 24 4 0125 4 0125 0 4 32 24 2 2 2 2 2 = + + + K s K s 1 2 2 2 32 j24 K1 = 0.256 K2 = 0.1253.13 u t t t = + + − 0 256 0 24 24 531 32 . e . cos( . ) ( ) V （或将 I s = − 0 4 24 . j ，求得 u 取实部） 七. + _ 40 103 s 25 10 25 3 2 2 + − s s U s C ( ) Z s s s s ( ) = + = + 4010 40 10 10 25 3 3 3 U s s s s C ( ) ( )( ) = + + 25 25 25 2 2 u t t C t = − + − 1 − 2 1 2 25 45 25 e cos( ) ( ) V 八
100I(s)10s0.1s52 +106100sI(s) =(s+103(s2+10°)100-50e-10,cos(103t-45°) |(t)mAV2九.t(s)+12.50.50.55J0.5U(s)s2+-22(s+25)12.5U(s)=+2+2s2 +50s+25s$2+12.5s4(s+25)Ii(s)=$(s +50s+25)43.960.0404I(s) =$+49.495s+0.5051s, =(43.96e0505s-0.04e-495)()A十.0.5s0.5+250=Uc(s)J30sUc(s)=Uc'(s)+Uc"(s)3037.57.5Uc(s)=ss+10s+506.256.25Uc"(s)=$+10$+50303125125Uc(s)=s$+10$+50uc =(30-31.25e-101 +1.25e-50 )(0)V+1
+ _ 100 0.1s 10 10 2 6 s s + I(s) I s s s s ( ) ( )( ) = + + 100 10 10 3 2 6 i t t t = − + − − 50 100 2 10 45 10 3 3 e cos( ) ( ) mA 九. + _ + _ 0.5 0.5 0.5s I s L ( ) 2 s 12.5 s U(s) U s s s s s s s s ( ) . . ( ) = + + + + = + + + 2 2 12 5 2 12 5 2 2 2 25 50 25 2 I s s s s s L ( ) ( ) ( ) = + + + 4 25 50 25 2 I s s s s L ( ) . . . . = − + − + 4 396 05051 0 0404 49 495 i t L t t = − − − − ( . e . e ) ( )A . . 4 396 004 0 505 49 5 十. + _ _ + 30 0.5s 0.5 1 s 250 s U s C ( ) U s U s U s C C C ( ) = '( ) + ' '( ) U s s s s C '( ) . . = − + + + 30 37 5 10 7 5 50 U s s s C ' '( ) . . = + − + 6 25 10 6 25 50 U s s s s C ( ) . . = − + + + 30 3125 10 125 50 u t C t t = − + − − (30 31.25e 1.25e ) ( ) V 10 50 十一
15Ic(s)AC75euci(0_)=5Vuc2(0_)=5V玉%1-1+=)U(s) =-(++14+7453+3+%174+7541./4s74s2.2s2+9s+5U(s)=2s(s+ 1)(s + 4)41Icr(s)=(9 s+19s+44e-")e(t)Ae~icr=(-+99十二.uc(0_)=2Vi(0_)=1A复频域模型如图C2=-1-2(s+2+-)Ic(s)-(s+2)×2回路方程sS5+2Ic(s) =得s2 +2s+111-$+1 (s+1)3ic =(e- +te-')A，>0十三.ii(0_)=i2(0_)=1Auc(0_) =1V复频域模型如图
_ _ _ + + + 5 5 3 U (s) 5 s 5 s 6 s 75 4s 15 4 1 s I s C1 ( ) uC1 (0− ) = 5V uC2 (0− ) = 5V ( ) ( ) 1 5 1 3 6 1 15 4 75 4 1 5 1 5 3 6 5 15 4 75 4 + + + + + = + + + + s s U s s s s s s U s s s s s s ( ) ( )( ) = + + + + 2 2 9 5 1 4 2 I s s s C1 4 9 1 1 1 9 1 4 ( ) = (− ) + + + i t C t t 1 4 4 9 1 9 = − + − − ( e e )( ) A 十二. uC (0− ) = 2 V iL (0− ) = 1A 复频域模型如图 + _ _ + 1 2 1s 1 s 2 s 2 s I s C ( ) 回路方程 (s ) ( ) ( ) s I s s s s + 2 + C − + = − − 1 2 2 1 2 得 I s s s s C ( ) = + + + 2 2 1 2 = + + + 1 1 1 1 2 s (s ) i e t C t t = + − − ( e )A ， t 0 十三. i i L1 (0− ) = L2 (0− ) = 1A uC (0− ) = 1V 复频域模型如图
52sIs11D2DI(s)3sQ!O1F+!!2sSI(s) =12s+1s5+3s得0.211-X-1$+0.5Ss+15--1i =(1+t+0.2e~5*-e2)At>0十四.21=-10Ai(0_)=-30×A2 + 4uc(0_)=40-u(0_)=80Vu,(0)-SLSscD Li(0_)U(s)2030-30,i(0.)+Cuc(0_)+ssU(s)=-11+ sCsL1/=-4005+2×103+S+8×103u(t) = -40(e-2x101 +e-8x10%)8(t)V十五、1、D2、A3、B
_ + + + + _ _ _ 5 2 1 1 2s s 1 1 s 1 3s 1 s I(s) 得 I s s s s s s ( ) = + + + − + 1 1 1 5 1 3 2 2 1 = + + + − + 1 1 0 2 1 15 1 05 2 s s s s . . i t e t t = + + − − − (1 0.2 e )A 1 15 1 2 ， t 0 十四. i(0 ) 30 A A 2 2 4 − = − 10 + = − uC (0− ) = 40 − u(0− ) = 80V + _ + + _ _ U (s) Li(0 ) − 30 s 2 u s c (0−) 1 sc sL U s s i s Cu sL sC C ( ) ( ) ( ) = − + + + + − − 30 0 0 1 2 1 = − + + + 40 1 2 10 1 8 10 3 3 ( ) s s u t t t t ( ) = − (e + e ) ( )V − − 40 2 10 8 10 3 3 十五、 1、D 2、A 3、B