電路分析第8章二

§8-1

LC電路的正弦振蕩

i

=–C——duCdtuL=uC

=

L——didt設(shè)uC(0)=U0

iL(0)=0∵

uL=uC

=

U0≠0

uC

=

U0

iL

=

0didt——≠0∴電流開(kāi)始上升i↑,電容開(kāi)始放電uC↓Ⅰ.初始時(shí)刻C+uC=U0i=0L–C+uCiL+uL––電流最大

i=ImⅡ.當(dāng)uC

=

0

，uL

=

0時(shí)，didt——=

0電容儲(chǔ)存的電場(chǎng)能量全部轉(zhuǎn)化為電感儲(chǔ)存的磁場(chǎng)能量因?yàn)殡姼须娏鞑荒苘S變電感開(kāi)始輸出能量i↓，電容開(kāi)始反向充電|uC

|↑Ⅲ.當(dāng)i

=

0

時(shí)

,uC

=–

U0磁場(chǎng)能量全部轉(zhuǎn)成電場(chǎng)能量因?yàn)閡C不能躍變，電容放電|uC

|↓，|

i

|↑C+uC=

0i=

ImL––+uL=

0duCdt——≠0∴i

=–C——duCdtC+uC=–U0i=

0L–+–uLⅣ.當(dāng)uC

=

0時(shí)，i

=

–

I電場(chǎng)能量全部轉(zhuǎn)成磁場(chǎng)能量|uC

|↑，|

i

|↓Ⅴ.當(dāng)uC

=

U0

時(shí)

,i

=

0磁場(chǎng)能量全部轉(zhuǎn)為電場(chǎng)能量，電路回到初始時(shí)刻的狀態(tài)C+uC=

0i=

–

IL–C+uC=

U0i=

0L–§8-2

RLC串聯(lián)電路的零輸入響應(yīng)求零輸入響應(yīng)

uS=0

LiR+uS-C+uC-uL

+

uR+uC

=

uSdidt

L+Ri

+

uC

=

uS

+

RCd2uCdt2

LCduCdt+

uC

=

uS+

RCd2uCdt2

LCduCdt+

uC

=

0uC(0)

=?兩個(gè)初始條件i

=CduCdtuL

=

LdidtduCdtt=0

=i(t)Ct=0

=i(0)C

=?R，L，C

取值不同，根號(hào)里的值有四種不同情況。

設(shè)解為

uC(t)=Kest

代入微分方程LCs2Kest

+

RCsKest+

Kest

=0(LCs2+RCs+1)Kest

=0特征方程的根（固有頻率）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，根號(hào)里的值有四種不同情況。

特征方程的根（固有頻率）s1,s2為兩個(gè)不相等的負(fù)實(shí)數(shù)。2L

R1.()2LC1＞s1,s2為兩個(gè)相等的負(fù)實(shí)數(shù)。2L

R2.()2LC1＝s1,s2為共軛復(fù)數(shù)。2L

R3.()2LC1＜4.R

=

0s1,s2為共軛虛數(shù)。s12=RC±(RC)2

-

4LC2LC2L

R=

-±2L

R()2LC1--

即R＞2LC

即R＝2LC

即R＜2LC

阻尼電阻

Rd＝2LC為過(guò)阻尼情況。為臨界阻尼情況。為欠阻尼情況。為無(wú)阻尼情況。響應(yīng)是非振蕩性的衰減，為過(guò)阻尼情況。

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為過(guò)阻尼情況s1,s2為兩個(gè)不相等的負(fù)實(shí)數(shù)。2L

R1.()2LC1＞由初始條件確定系數(shù)——=–

1K1e–

1t–

2K2e–

2tduCdtduCdtt=0

=i(0)C解：(1)若以u(píng)C(t)為求解變量LiR+uS-C+uC-例．已知圖示電路中t≥0時(shí)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過(guò)阻尼情況

阻尼電阻

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)為求解變量

uR

+

uL

+

uC

=

0

Ri

+

Ldidt+

uC(0)

+1C∫idt

=

00t兩邊微分

+

RCd2idt2

LCdidt+

i

=

0例．已知圖示電路中t≥0時(shí)uS

=

0R

=

3

L

=12HC

=14F

uC(0)

=

2ViL(0)

=

1A求uC(t)及iL(t)

t≥0解：過(guò)阻尼情況LiR+uS-C+uC-iL(t)

=

K1e-2t

+

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

-3

t

=

0時(shí)電路iL(0)

=

K1

+

K2

=

1didt|t=0

=

-2K1

-

4K2

=LuL(0)uL(0)

=

-31-2

=

-5V例．已知圖示電路中t≥0時(shí)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．已知圖示電路中t≥0時(shí)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

過(guò)阻尼情況

阻尼電阻

Rd＝2=

2.828

LCR＞

RdLiR+uS-C+uC-無(wú)振蕩衰減，臨界阻尼

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為兩個(gè)相等的負(fù)實(shí)數(shù)。2L

R2.()2LC1＝

即R＝2LC為臨界阻尼情況。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為共軛復(fù)數(shù)。3.(2L

R)2LC1＜

即R＜2LC為欠阻尼情況。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串聯(lián)電路的零輸入響應(yīng)為uC(t)=5e–2tcos

3tV，已知R=4

，求L和C。–解：由零輸入響應(yīng)的形式可知，此題應(yīng)為欠阻尼情況。零輸入響應(yīng)的一般形式為uC(t)

=

e-at[K1coswdt

+

K2sinwdt]s1、2

=–—±j

——–(—)2=–

±jd2L

RLC12L

R

=——=22LR

d=

——–(—)2=

3LC12L

R–解得：L=1H，C=—F71固有頻率4.

R

=

0

無(wú)阻尼特征根s1,s2為共軛虛數(shù)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無(wú)衰減等幅振蕩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串聯(lián)電路的全響應(yīng)+

RCd2uCdt2

LCduCdt+

uC

=

USuC(0)

=?duCdt|t=0

=?uC(t)

=

uch

+

ucp+

RCd2uchdt2

LCduchdt+

uch

=

0s1

=

-a1

s2

=

-a2如果電路為過(guò)阻尼uch(t)

=

K1e-a1t+

K2e-a2tuC(t)

=

K1e-a1t+

K2e-a2t+

US設(shè)ucp(t)=Q

與激勵(lì)形式一樣若為直流激勵(lì)，則Q

=

USK1，K2由初始條件確定根據(jù)特征根的四種不同情況，寫(xiě)出齊次方程解的形式USLiR+-C+uC-1Hi1

+US

=

2V-1F+uC-t≥0例．求圖示電路中uC(t)t≥0已知uC(0)

=

0iL(0)

=

0+

RCd2uCdt2

LCduCdt+

uC

=

US+d2uCdt2duCdt+

uC

=

2設(shè)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：為欠阻尼情況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并聯(lián)電路的分析iC+iG+iL

=

iSCduCdt+

GuC

+

iL

=

iS+

GLd2iLdt2LCdiLdt+

iL

=

iS如果是零輸入響應(yīng)iS

=

0+

GLd2iLdt2LCdiLdt+

iL

=

0iL(0)

=?diLdt|t=0

=?LCs2

+

GLs

+

1

=

0s1,2

=-GL±(GL)2

-

4LC2LC2C

G=

-±2C

G()2LC1-根據(jù)固有頻率四種情況寫(xiě)出解的形式阻尼電導(dǎo)Gd

=2

—LC–—iS+uC-iCiGiLCGL+

GLd2iLdt2LCdiLdt+

iL

=

ISGCL并聯(lián)2C

Gs1,2

=

-±2C

G()2LC1-RLC串聯(lián)s1,2

=

-2L

R±2L

R()2LC1-+

RCd2uCdt2

LCduCdt+

uC

=

US阻尼電導(dǎo)Gd

=2

—LC–—阻尼電阻Rd=2

—CL–—[例]圖示電路中，欲使電路產(chǎn)生臨界阻尼響應(yīng)，則C應(yīng)為何值？iSC1

2H解：Gd

=2

—LC–—阻尼電導(dǎo)欲使電路產(chǎn)生臨界阻尼響應(yīng)，應(yīng)滿足G=Gd

因：G=1S

2

—=1LC–—故：得：C=0.5F解：例：RLC并聯(lián)電路的零輸入響應(yīng)為uc(t)

=

100e-600tcos400t，若電容初始貯能是130J，求R，L，C以及電感的初始電流。uC(0)

=

100V130wC(0)

=130CuC2(0+)

=12C

=230uC2(0+)=230100=

6.67μF由零輸入響應(yīng)的形式可知，此題應(yīng)為欠阻尼情況。零輸入響應(yīng)的一般形式為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二階電路分析方法總結(jié)

a0dXdtd2Xdt2+

a1+

a2

=

AX(0)

=?dXdt|t=0

=?X(t)

=

Xh(t

)

+

Xp(t

)Xh(t)

=

Kest

代入齊次方程a0s2

+

a1s

+

a2

=

0特征方程s1,2

=-a1±a12

-

4a0a12a0固有頻率RLC串聯(lián)2L2LLCs1,2

=

-

R±

R()21-GCL并聯(lián)2C

Gs1,2

=

-±2C

G()2LC1-列出非齊次二階微分方程給定初始條件解的形式s1,s2為兩個(gè)不相等的負(fù)實(shí)數(shù)s1

=

-a1

s2

=

-a2

無(wú)振蕩衰減Xh(t)

=

K1e-a1t+

K2e-a2t過(guò)阻尼s1,s2為兩個(gè)相等的負(fù)實(shí)數(shù)s1

=

s2

=

-a

臨界阻尼Xh(t)

=

(K1

+

K2t)e-at

無(wú)振蕩衰減a

–衰減因子wd

–衰減振蕩角