課件上屆第7章二階電路

第7章

特征根判斷二階電路的工作狀態(tài)；1作業(yè)2 +- +-

3 KR+-+L-i已知uC(0-)=KR+-+L-iuR+uL-uC=

uR=- =Ld-

i(0+)=d2

i=-Cdd

uC

)=ULC

(t>uC(0+)=Ud

=-i(0+)=d

4 d2 dLC C+RC C+dt d

(t>+-

-

uC(0+)=U d d

t

C

t

=-i(0+)=CLCdi+RCdi+i=0dt dti(0+)=u(0)=u(0)=Ld

dddd

dtt

R2-4L/R2-4L/

=-R(R(R)2-5

R2-4LR2-4L/

=-R(R)2-(R)2-R>LCR>LC二個(gè)不等負(fù)實(shí)根R=LR=LC二個(gè)相等負(fù)實(shí)根R<LR<LC二個(gè)共軛復(fù)根

6R>R>LC

u( =u( =Ap1pC1+A22uC

p-

(pep1t-pep2t cucp p2-

tp2-

7U uCU0

p2-

(pep1t-pep2t ln

tmt

p2- iC=0,t=￥ =-Cd

2t

ttm時(shí)iC d

-p

0<t<

i增加,u L =Ldi=- (pep1t-pep2tL

t>

i減小,uLd (p2-p1t=

t=￥,uL=

t=2tmuL8uLuLctKR+-+L-iKR+-+L-i0<t< KR+-+L-iKR+-+L-i

9

L(2)RL(2)R< 1w21w2-d20w0

特征根為一對(duì)共軛復(fù)根pd

=Aep1t+AeuC=Ae-dtsin(wt+b) =uC=Ae-dtsin(wt+b) 由初始條件d

uC(0+)=U0fiAsinb= C(0)=0fiA(-d)sinb+Awcosb=d AAdbbw

A=w0w

=0Ue-dtsin(wt+b)1 w1

w0w2w2-d20

Ue-d

sin(wt+b) wu是其振幅以–w0 w w0Ue-

w -w0Ue-w =-Cd

0

iC零點(diǎn)：wt

npuC =Ld

Ue-d

sin(wt- d w

0bp-

p2p-b 0<wt<

b<wt<p-

p-b<wt<

KR+KR+-+L-i-

KR+-KR+-+L-iKR+-+L-i

則d0，ww0

b=2uC=U0sin(wt+90)=i=w

wt

tKRtKR+-+L-i -(3)(3)R=LC

p= =-R=-

dd

uC(0+)=U0fiA1= (0+)=0fiA(-d)+A= =Ue-dt(1+d A1=U0

Cd U0te-d A=Ud

d

uL=Ld

=Ue-dt(1-dt)00LCR LC

=Aep1t+Ae LCR LC

LCR LCuC(0+由初始條件d

CC C(0d d2 LC C+RC C+uC=Eεt)

通解的特征方程為CLCp2+RCp+1=C uC=u

特解

=E+Aep1t+Ae

p uC(0+由初值du C(0 d Et δt

微分方程為d2 LC C+RC C+

(1)t在0－至0＋

d2

Cdt 202

Cdt0

uCdt=

δ(t)d

-LCd

0+udCu0-u0+udCd

t=0-

-

t=0-=LCd

(0+)=1

iiC(0+ i(0)=i(0)=0iL(0+)=iC(0+)=

d2 LC C+RC C+uC= R>LC =Aep1t+AR>LC A

+Ap

A2=-A1

p2-

(ept-e

t2 LC(2

R<LCuC=AR<LC

sin(wt+b)uC

w

uC(0-)=i(0)=100

+-Cu+-C

LC

+uC=uS(t?0+uC(0+)=duC|

=iCt

=i(0+)=10

d2u LC C+RC C+uC=uS (t?0)dt2

uC(t)=uC'+uC4=100+Ae-10tsin(3·104t+bCCuC(0+)=duC

由初始條件求出A167b=-36.9o uC(t)=100+167e sin(3·10t-36.9)V 0p=-d–d2-w0 d>w過阻尼非振蕩放電