8-8RLCRiRiSCL

t RiuLuC dRC C d LC CRC C 特征方程為：LCs2RCs1(R)21(R)21 (R)21(R)21 s有三種情LCR LCLCR LCLCR LCLC(一)RLC

設|s|>|s R(R)2(R)21RiSCLuC(t)RiSCL

(t)Aes1tAes2 u(0) AA i(0)i(0) i(0A

U

c(0) 0s1A1s2A2 s

s s

C

(se se A U s 0 s0

s

(ses1tses2t

|s2| s2

s2

u|s2|u

R(R(R)21UCU0

t

s2

(ses1tses1ti-

iC=0,

i

i t=tm時iCiC c (es1 es2 L(s2L Ldi U0 (ses1tses2tL

0<t< t

u (s2s1t uLU0 t,uL

t=2tmuL Ldi (sestsestL L (s2s1

sm(ses1m es2m) 2m

1 1

s1(s2ests2est) 2

s t2

1s 1

s1LCLC|s2|-i0tRiSCLRiSCL0<t<tm uC>0，i<0 t>tm uC>0，RiSCLRiSCL R LC(二)RLC

R(R)(R)212022021令：b 1

(固有振蕩角頻率0

(諧振角頻率s1,2b

Aeb

sin(t uCAeb

sin(td

u(0) Asinc

(0)

i(0C

0A(b)sinAdcosθA ，arctgθωd，ω０，b間的關系

0A0 0

Ueb

sin(t

C ebtsin(t d

duC是其振幅以

0Ue-0C0C

0U0d0 -θ 0U

t=0u uC零點：t-θ，2-θn-uC極值點：t=0， ... θ-θi

Ueb

sin(t C c ebtsin(t)C dCiC零點：t ... ，iC極值點為uL零點 L

Ueb

d d θ-θi

R

L0<t

θ<t<-

-θ<t<RiSCLRiSCLRiRiSCLRiSCLRiSCL <t<

+θ<t<2- 2-θ<t<2則b0

00uC(t)d

0Ueb0

2iSCLCi(t)C ebtsin(tiSCLC du(t)L

Ueb

d dRu(t)Usin(t900)u R i(t)0

sin LCsLCs (三)R u(t)Aes

Ates

AsestAestAtses

u(0) A 由初始條件duC(0)iC(0)0As u(t)U(1st)es

A1

i(t)C

tes 0A 0

u(t)LdiU(1st)est LCRLC1bR 1

2220

θ-θi

CuC

2|>

LCdLCR

b Td

-LCR LCLCCR LCC

u(t)Aes1tAes2 u(t)Aest LCR LCu(0

u(t)Aeb

由初始條件duC(0 uC(0)0i(0)1AC1FL1HL iRRiuLL iRdd LC CRC C （1）R

s u(t)Ae0.382tA i(t)CduC(t)0.382Ae0.382t2.618A CA1A2C

A1A2

2.618A2

u(t)0.447(e0.382te2.618t

LC（2）R LC

s1s2uC(t)(

i(t)Ae

(AAt)e A3

u(t)te (VA4

LC3欠阻尼情 s 1LC3（3）R 33uC(t)Ae2

t233 33iC(t)A e2 t) e2cos(2tAsin

A

33 3

uC(t)

33e2 33 d

dLC CRC C

LC C j uC(t)Asin(0tAsinA0cosuC(t)sin(0t)sin

iC(t)A0cos(0tAR=- 3s 13 3t3uC(t)Ae2 t2333 i(t)A333 Asin

e2

t)

e223A 3

t)3 333A 33

uC(t) e2sin(2+

0.

t=0時打開開(1)iL(0-

uC(0-dd LC CRC C CC

u(t)Ae25tsin(139tCu(0)C Asin

c

104Cu(t)355e25t ) A ，C00+RL+RL+--C

uC(0- dd LC CRC CuC E uCuCP

LCs2RCs1特解 uCu(t)EAestAes (ss u(t)EAeS u(t)EAeb

AteS (s1s22sin(t bj2 u(0 由初值duC(0 K

1/

i1=i-0.5u1=i-0.5(2–i)2=2i-由

2(2i)2i ）ddit t d2i8

12i

K

1/

d2i8di

12i

解的形式為：i(tiP(tih0.i0.i-+2s1=-2，s22i(t)A iP

A

i=0.5u1iP(t)=1A

u1 K

1/

i(t)1Ae2tA

i(0)i(0)Ldi(0)u(0

L L

u(0)0.5u2228(V u(t)2Ae2t6A 1A1A2 A1 A

2A6Ai(t)10.5e2t

+R+R-500. C

求 iR

iR

i CduC d50L d dt LC d

iLP(t)

RLC L RiL +R-500. +R-500. C dt2

RiL

s2200s20000特征根為：s=-100

iLh(t)Ae

sin(100t全

iL(t)iLP(t)iLh(t)1Ae

sin(100t

cos(100t iL(0)u(0)u(0)

1Asin100Asin1