第八 2R

§81t=0時，開關(guān)K從1切換2i= uC= i

i= uC= S 初始狀

新穩(wěn) 過渡狀K1R1K1R1iK i

開關(guān)K切換到2RC

US

（R1+R2）i二

p能量 三.穩(wěn)態(tài)分析和動態(tài)（暫態(tài)）

四 動態(tài)電路的分析方 andt

dn1ian1dt

1

iut

r(t)

t t1(t)

tt t1 0t t t n

0t△/

t當n→∞時，即為b

t<0，U=0KRUCutKRUCuC1(t

U tRUC tRUC 1tr(t之間的關(guān)系1 1(t)dr(t)1

ttr(t)

0 tt1()d1()dt

t時移

1(tt)

t1 t10

1(t)E1(t)

tttt0單位沖激函數(shù) 0 (t)

(t(t

(t)dt 1/-/

/

p(t)1[1(t)1(t)] 1limp(t)(2)

(t)tt1(t)

()d

tt3 延時運算：δtt0 (tt) (tt0 (tt

(t- (tt0dt (tt

(t-0E(tt) 0

（E）強度為

(tt000E(tt0)dtδt的

f(t)(t)f(0)(tf(t)(tt0)f(t0)(tt0f(t)(t)dtf(0)(t)dtff(t)(tt0)dtf(t0例 (sintt)(t

)dtsin1 例2F(j)

f(t)ejtdt F(j)

(t)ejtdt§8-21、換路前后瞬間（從0-到0+），若ic(t)有限，則電容電壓 uC(0)uC(0)或q(0)q(0

u(t) i()d

i()d i(

C

C C

uC(0)t

q=Ct0+

q(t)q(0)0iC(u(0)u(0)

當iC()為有限0i( 0

i()dq(0)q(0)0iC(

uC(0+uC(0- q(0+q(0- KRC若突變KRCiC 假如uC(00uC(0)Ui(0)

uC(0) iR而iCpuCiC

pdWd(1Cu2 iCt有限呢？(2) ))US1(.

iL(0iL(0)或(0)uuLi(t)

Lu(

u()d

tu( L

L

L i(0) u( L t=0+時 i(0)i(0)

0u(

當u()

(0)(0) 0u(

0 uL()d0iL(0+)=iL(0-

從電路看:如果電路中沒有 純電感割I(lǐng)SIS1(t iL iL(0-uC(0uC(0iL0iL02 iL(0+)之外的電壓、電流初值，以第二類

duC

例（1）已知：Ｕs=6VR1=10?，R2=5?,L=1H，C=1μ求

uC(0),iL

(0),diL(

(3)

d

d C(0), L(0), L(0), C C u u

uC(0)0 iL(0)0uC(0)uC(0)0,iL(0)iL(0)0第一類非獨立初值：t=0+時的電路 替代定KiCKiCCuL

uC(0) iL(0)duC

i(0) 0.6RC1RCuL(0)US6duC(0)iC(0)6105 取t0+

diL

diL(0)uL(0)6(

第二類非獨立t>0

t>uC(iCiL)R1

Ri Riu

iL)

L

duCR(

diL)

S S

diC(0)diL(0)1duc(0)6

1610559994(A (RR)

R

(RR)diLR

duL

duL(0)RdiC(0)(RR)diL(0)6105(

L Ld ASdAS

1

6105 2 diCdiL1duc d d 1d2u C L

C Cd dt dt dt22diC(0+) iL22

C(0+)6105105(61041 dt RC1

AS26109 AS2例２：已知：激勵為正弦函數(shù)Em 2V,314radei2(0),i4(0),uc1(0),uc3i(0),i(0), (0),di2

e(t)Emsin(te(t)

sin(314t222 R2=1ΩL2=2/ ZRjL

C3 I(C3

) 245o(V

t0i2(0)

7.07(

i1(t)10sin(314t45)i2uC1(t)10sin(314tuC1(0)

/

t0uC3(0)i4(0)uC1(0)uC1(0)7.07(VuC3(0)uC3(0)i2(0)i2(0)7.07(i4(