电路分析第8章二

§8-1

LC电路的正弦振荡

i

=–C——duCdtuL=uC

=

L——didt设uC(0)=U0

iL(0)=0∵

uL=uC

=

U0≠0

uC

=

U0

iL

=

0didt——≠0∴电流开始上升i↑,电容开始放电uC↓Ⅰ.初始时刻C+uC=U0i=0L–C+uCiL+uL––电流最大

i=ImⅡ.当uC

=

0

，uL

=

0时，didt——=

0电容储存的电场能量全部转化为电感储存的磁场能量因为电感电流不能跃变电感开始输出能量i↓，电容开始反向充电|uC

|↑Ⅲ.当i

=

0

时

,uC

=–

U0磁场能量全部转成电场能量因为uC不能跃变，电容放电|uC

|↓，|

i

|↑C+uC=

0i=

ImL––+uL=

0duCdt——≠0∴i

=–C——duCdtC+uC=–U0i=

0L–+–uLⅣ.当uC

=

0时，i

=

–

I电场能量全部转成磁场能量|uC

|↑，|

i

|↓Ⅴ.当uC

=

U0

时

,i

=

0磁场能量全部转为电场能量，电路回到初始时刻的状态C+uC=

0i=

–

IL–C+uC=

U0i=

0L–§8-2

RLC串联电路的零输入响应求零输入响应

uS=0

LiR+uS-C+uC-uL

+

uR+uC

=

uSdidt

L+Ri

+

uC

=

uS

+

RCd2uCdt2

LCduCdt+

uC

=

uS+

RCd2uCdt2

LCduCdt+

uC

=

0uC(0)

=?两个初始条件i

=CduCdtuL

=

LdidtduCdtt=0

=i(t)Ct=0

=i(0)C

=?R，L，C

取值不同，根号里的值有四种不同情况。

设解为

uC(t)=Kest

代入微分方程LCs2Kest

+

RCsKest+

Kest

=0(LCs2+RCs+1)Kest

=0特征方程的根（固有频率）s1、2=RC±(RC)2

-

4LC2LC2L

R=

-±2L

R()2LC1--+

RCd2uCdt2

LCduCdt+

uC

=

0特征方程LCs2+RCs+1=0R，L，C

取值不同，根号里的值有四种不同情况。

特征方程的根（固有频率）s1,s2为两个不相等的负实数。2L

R1.()2LC1＞s1,s2为两个相等的负实数。2L

R2.()2LC1＝s1,s2为共轭复数。2L

R3.()2LC1＜4.R

=

0s1,s2为共轭虚数。s12=RC±(RC)2

-

4LC2LC2L

R=

-±2L

R()2LC1--

即R＞2LC

即R＝2LC

即R＜2LC

阻尼电阻

Rd＝2LC为过阻尼情况。为临界阻尼情况。为欠阻尼情况。为无阻尼情况。响应是非振荡性的衰减，为过阻尼情况。

s1

=

-2L

R=

-a1+2L

R()2LC1-s2

=

-2L

R=

-a2-2L

R()2LC1-a2

>

a1uC(t)

=

K1es1t+

K2es2t=

K1e-a1t+

K2e-a2t通解的形式K1+

K2

=

uC(0)-

a1K1

-

a2K2

=CiC(0)解出K1

，

K2uC(t)

=K1e–

1t+

K2e–

2t

即R＞2LC为过阻尼情况s1,s2为两个不相等的负实数。2L

R1.()2LC1＞由初始条件确定系数——=–

1K1e–

1t–

2K2e–

2tduCdtduCdtt=0

=i(0)C解：(1)若以uC(t)为求解变量LiR+uS-C+uC-例．已知图示电路中t≥0时uS

=

0R

=

3

L

=12HC

=14F

uC(0)

=

2ViL(0)

=

1A求uC(t)及iL(t)

t≥0+

RCd2uCdt2

LCduCdt+

uC

=

0+d2uCdt2duCdt+

uC

=

01834+

6d2uCdt2duCdt+8

uC

=

0s2

+

6s

+

8

=

0s1,2

=-6±36

-

3222-6±2=s1

=

-2

s2

=

-4过阻尼情况

阻尼电阻

Rd＝2=

2.828

LCR＞

Rd+d2idt2didt+

i

=

01834+

6d2idt2didt+

8i

=

0s2

+

6s

+

8

=

0s1

=

-2

s2

=

-4iL(t)

=

K1e-2t

+

K2e-4t(2)

若以iL(t)为求解变量

uR

+

uL

+

uC

=

0

Ri

+

Ldidt+

uC(0)

+1C∫idt

=

00t两边微分

+

RCd2idt2

LCdidt+

i

=

0例．已知图示电路中t≥0时uS

=

0R

=

3

L

=12HC

=14F

uC(0)

=

2ViL(0)

=

1A求uC(t)及iL(t)

t≥0解：过阻尼情况LiR+uS-C+uC-iL(t)

=

K1e-2t

+

K2e-4t+2V-1A+uL(0)

-3

t

=

0时电路iL(0)

=

K1

+

K2

=

1didt|t=0

=

-2K1

-

4K2

=LuL(0)uL(0)

=

-31-2

=

-5V例．已知图示电路中t≥0时uS

=

0R

=

3

L

=12HC

=14F

uC(0)

=

2ViL(0)

=

1A求uC(t)及iL(t)

t≥0解:-2K1

-

4K2

=

-10得K1

=

-3,K2

=

4iL(t)

=

-3e-2t

-

4e-4tAt≥0uC(t)

=

uC(0)

+1C∫idt0t=

2

+

4(32e-2t

-

e-4t)|0t=

2

+

4(32e-2t

-

e-4t

-12)=

6e-2t

-

4e-4tVt≥0LiR+uS-C+uC-s1,2

=

-2L

R±2L

R()2LC

1-=

-3±9-8=

-3±1s1

=

-2

s2

=

-4uC(t)

=

K1e-2t

+

K2e-4t=

6e-2t

-

4e-4tVt≥0iL(t)

=

CduCdt=

-3e-2t

+

4e-4tAt≥0例．已知图示电路中t≥0时uS

=

0R

=

3

L

=12HC

=14F

uC(0)

=

2ViL(0)

=

1A求uC(t)及iL(t)

t≥0解:（3）不列微分方程

过阻尼情况

阻尼电阻

Rd＝2=

2.828

LCR＞

RdLiR+uS-C+uC-无振荡衰减，临界阻尼

s1

=

s2

=

-2L

R=

-a解的形式uC(t)

=

K1e-at

+

K2te-at=

(K1

+

K2t

)e-at

K1

=

uC(0)

duCdt|t=0

=

[K2e-at

–

a(K1

+

K2t)e-at]|t=0

=

K2

–

aK1

=CiC(0)iC(0)

=

iL(0)K2

–

aK1

=CiL(0)K2

=CiL(0)+

auC(0)CiL(0)+

auC(0)]te-atuC(t)

=

uC(0)e-at

+

[CiL(0)+

auC(0)]t}e-at={uC(0)

+

[s1,s2为两个相等的负实数。2L

R2.()2LC1＝

即R＝2LC为临界阻尼情况。s1

=

-2L

R=

-a

+

jwd+j2L

R()2LC12L

R(-s2

=

-2L

R=

-a

-

jwd-j2L

R()2LC12L

R(-解的形式uC(t)

=

e-at[K1coswdt

+

K2sinwdt]uC(0)

=

K1

CiL(0)=

-aK1

+

wdK2

=duCdt|t=0

=

[-ae-at(K1coswdt

+

K2sinwdt)+e-at(–

wdK1sinwdt

+

wdK2coswdt)]|t=0auC(0)K2

=wdCiL(0)+wds1,s2为共轭复数。3.(2L

R)2LC1＜

即R＜2LC为欠阻尼情况。uC(t)

=

e-at[K1coswdt

+

K2sinwdt]K1=

K12

+

K22e-at[

K12

+

K22coswdt

+K2

K12

+

K22sinwdt

]

K12

+

K22K2K1K2

K12

+

K22sinq

=K1

K12

+

K22cosq

=q

=

arctgK1K2利用公式cos(a

–b)

=

cosacosb

+

sinasinbuC(t)

=

K12

+

K22e-at[cosqcoswdt

+

sinqsinwdt]=

K12

+

K22e-atcos(wdt

-q

)=

Ke-atcos(wdt

+f)K

=

K12

+

K22

f

=

-arctg

K1K2[例]RLC串联电路的零输入响应为uC(t)=5e–2tcos

3tV，已知R=4

，求L和C。–解：由零输入响应的形式可知，此题应为欠阻尼情况。零输入响应的一般形式为uC(t)

=

e-at[K1coswdt

+

K2sinwdt]s1、2

=–—±j

——–(—)2=–

±jd2L

RLC12L

R

=——=22LR

d=

——–(—)2=

3LC12L

R–解得：L=1H，C=—F71固有频率4.

R

=

0

无阻尼特征根s1,s2为共轭虚数s1

=

j=

jw0LC1s2

=

-j=

-jw0LC1解形式uC(t)

=

K1cosw0t

+

K2sinw0tK1

=

uC(0)uC(t)

=

Kcos(w0t

+f)K

=

K12

+

K22

f

=

-arctg

K1K2无衰减等幅振荡Cw0iL(0)K2

=CL(0)duCdt|t=0

=

w0K2

=iuC(t)

=

K1coswO

t

+

K2sinwO

tK1=

K12

+

K22

[

K12

+

K22coswOt

+K2

K12

+

K22sinwOt

]

K12

+

K22K2K1K1

K12

+

K22cosq

=K2

K12

+

K22sinq

=q

=

arctgK1K2利用公式cos(a

–b)

=

cosacosb

+

sinasinbuC(t)

=

K12

+

K22[cosqcoswOt

+

sinqsinwOt]=

K12

+

K22cos(wOt

-q

)=

K

cos(wOt

+f)K

=

K12

+

K22

f

=

-arctg

K1K2§8-3

RLC串联电路的全响应+

RCd2uCdt2

LCduCdt+

uC

=

USuC(0)

=?duCdt|t=0

=?uC(t)

=

uch

+

ucp+

RCd2uchdt2

LCduchdt+

uch

=

0s1

=

-a1

s2

=

-a2如果电路为过阻尼uch(t)

=

K1e-a1t+

K2e-a2tuC(t)

=

K1e-a1t+

K2e-a2t+

US设ucp(t)=Q

与激励形式一样若为直流激励，则Q

=

USK1，K2由初始条件确定根据特征根的四种不同情况，写出齐次方程解的形式USLiR+-C+uC-1Hi1

+US

=

2V-1F+uC-t≥0例．求图示电路中uC(t)t≥0已知uC(0)

=

0iL(0)

=

0+

RCd2uCdt2

LCduCdt+

uC

=

US+d2uCdt2duCdt+

uC

=

2设ucp(t)

=

Q

代入原方程

Q

=

2s2

+

s

+

1

=

0s1,2

=-1±1-42=

-12±

j2

3uch(t)

=

e[K1cos-12t2

3t

+

K2sint

]2

3+

K2sint

]

+

22

3uC(t)

=

e[K1cos-12t2

3tuC(0)

=

K1

+

2

=

0解：为欠阻尼情况CiL(0)duCdt|t=0

=

-12K1

+2

3K2

=23K1

=

-2K2

=

-

3uC(t)

=

e[-2cos-12t2

3t

-t

]

+

22

323

3

sin=

-2.3ecos(-12t2

3t

-

30°)

+

2Vt≥0§8-4

GCL并联电路的分析iC+iG+iL

=

iSCduCdt+

GuC

+

iL

=

iS+

GLd2iLdt2LCdiLdt+

iL

=

iS如果是零输入响应iS

=

0+

GLd2iLdt2LCdiLdt+

iL

=

0iL(0)

=?diLdt|t=0

=?LCs2

+

GLs

+

1

=

0s1,2

=-GL±(GL)2

-

4LC2LC2C

G=

-±2C

G()2LC1-根据固有频率四种情况写出解的形式阻尼电导Gd

=2

—LC–—iS+uC-iCiGiLCGL+

GLd2iLdt2LCdiLdt+

iL

=

ISGCL并联2C

Gs1,2

=

-±2C

G()2LC1-RLC串联s1,2

=

-2L

R±2L

R()2LC1-+

RCd2uCdt2

LCduCdt+

uC

=

US阻尼电导Gd

=2

—LC–—阻尼电阻Rd=2

—CL–—[例]图示电路中，欲使电路产生临界阻尼响应，则C应为何值？iSC1

2H解：Gd

=2

—LC–—阻尼电导欲使电路产生临界阻尼响应，应满足G=Gd

因：G=1S

2

—=1LC–—故：得：C=0.5F解：例：RLC并联电路的零输入响应为uc(t)

=

100e-600tcos400t，若电容初始贮能是130J，求R，L，C以及电感的初始电流。uC(0)

=

100V130wC(0)

=130CuC2(0+)

=12C

=230uC2(0+)=230100=

6.67µF由零输入响应的形式可知，此题应为欠阻尼情况。零输入响应的一般形式为uC(t)

=

e-at[K1coswdt

+

K2sinwdt]=

-a

±

jwd2C

Gs1,2

=

-±2C

G()2LC1-K1=100,K2=0,a=600,wd=400+uC-iCiRiLCRL=

-a

±

jwd2C

Gs1,2

=

-±2C

G()2LC1-a

=2C

G=

600G

=

60026.6710-6

=

80.0410-4R

=G1=

124.9

wd

=

400

=LC1-a2LC1=

4002

+

6002

L

=

0.288HiL(0+)

=

-iR(0+)

-

iC(0+)=

-uC(0+)

R-

CduCdt|t=0=

-100124.9-

6.6710-6

dtd(100e-600tcos400t)|t=0=

-0.8

+

0.4

=

-0.4A+uC-iCiRiLCRL二阶电路分析方法总结

a0dXdtd2Xdt2+

a1+

a2

=

AX(0)

=?dXdt|t=0

=?X(t)

=

Xh(t

)

+

Xp(t

)Xh(t)

=

Kest

代入齐次方程a0s2

+

a1s

+

a2

=

0特征方程s1,2

=-a1±a12

-

4a0a12a0固有频率RLC串联2L2LLCs1,2

=

-

R±

R()21-GCL并联2C

Gs1,2

=

-±2C

G()2LC1-列出非齐次二阶微分方程给定初始条件解的形式s1,s2为两个不相等的负实数s1

=

-a1

s2

=

-a2

无振荡衰减Xh(t)

=

K1e-a1t+

K2e-a2t过阻尼s1,s2为两个相等的负实数s1

=

s2

=

-a

临界阻尼Xh(t)

=

(K1

+

K2t)e-at

无振荡衰减a

–衰减因子wd

–衰减振荡角