電路第十三章：拉普拉斯變換

第十三章拉普拉斯變換內(nèi)容提要

本章介紹拉普拉斯變換法在線性電路分析中的應(yīng)用。主要內(nèi)容有：拉普拉斯變換的定義，拉普拉斯變換與電路分析有關(guān)的一些基本性質(zhì)，求拉普拉斯反變換的部分分式法（分解定理），還將介紹KCL和KVL的運(yùn)算形式，運(yùn)算阻抗，運(yùn)算導(dǎo)納及運(yùn)算電路，并通過實(shí)例說明它們在電路分析中的應(yīng)用。目錄§13—1拉普拉斯變換的定義§13—2拉普拉斯變換的基本性質(zhì)§13—3拉普拉斯反變換的部分分式展開§13—4運(yùn)算電路§13—5應(yīng)用拉普拉斯變換法分析線性電路§13—1拉普拉斯變換的定義拉普拉斯變換是一種數(shù)學(xué)變換。定義：F(s)=f(t)e–stdt

0–

拉普拉斯正變換f(t)=F(S)estds2j1

+j

–j拉普拉斯反變換S=+jf(t)原函數(shù)F(s)象函數(shù)拉氏正變換拉氏反變換以一對應(yīng)簡寫符號F(s)=L[f(t)]f(t)=L–1[F(s)]例：解：F(s)=f(t)e–stdt

0–

1.F(s)=L[(t)]=(t)e–stdt

0–

=e–stdt

0–

=1s=–

e–st1s

0–2.F(s)=L[(t)]=(t)e–stdt

0–

=(t)dt

0–0+=1計(jì)算下列原函數(shù)的象函數(shù)；1.f(t)=(t)2.f(t)=(t)3.f(t)=e–t

(t)

4.

f(t)=t(t)3.F(s)=L[e–t

(t)]=e–t

e–stdt

0–

0–=–+s1e–(+s)t=+s14.F(s)=L[t(t)]=te–stdt

0–

=–[te–st

]1s

0–

–e–stdt

0–

=1s2同理：

F(s)=L[tn

(t)]=n

Sn+1§13—2拉普拉斯變換的基本性質(zhì)若：L[f1(t)]=F1(s)L[f2(t)]=F2(s)則：L[A1f1(t)+A2f2(t)]=A1F1(s)+A2F2(s)證：L[A1f1(t)+A2f2(t)]=[A1f1(t)+A2f2(t)]e–stdt

0–

=A1f1(t)e–stdt+A2f2(t)e–stdt

0–

=A1f1(t)e–stdt+A2f2(t)e–stdt

0–

0–

=A1F1(s)+A2F2(s)一、線性性質(zhì)例：計(jì)算下列原函數(shù)的象函數(shù)；1、常數(shù)U2、A(1–e–t)3、sint解：1、L[U]2、L[A(1–e–t)]

3、L[sint]=s2+

2

=L[A]–L[Ae–t]

As(s+)=AsAs+

=–=L[U(t)]Us=12j12j=L[ejt–e–jt]12js–j

1–12js+j

1=同理：L[cost]=s2+

2s二、(時(shí)域)微分性質(zhì)設(shè)：L[f(t)]=F(s)則：L[f(t)]=sF(s)–f(0–)證：L[f(t)]=

0–

df(t)dte–stdt=

0–

e–stdf(t)=e–stf(t)–f(t)(–s)e–stdt

0–

0–

=–f(0–)+sf(t)e–stdt

0–

=sF(s)–f(0–)

導(dǎo)數(shù)性質(zhì)的意義在于把原函數(shù)求導(dǎo)數(shù)的運(yùn)算轉(zhuǎn)換為象函數(shù)乘以s再減去初始值的代數(shù)運(yùn)算。推廣：L[f(t)]=s2F(s)–sf(0–)–f

(0–)L[fn(t)]=snF(s)–sn–1f(0–)–sn–2f

(0–)……f(n–1)(0–)例：(t)RCuC求：uc(t)的沖擊相應(yīng)解：cducdt+uc=(t)1R等式兩邊進(jìn)行拉普拉斯變換L[c]+L[uc]=L[(t)]ducdt1RscUC(s)–Cuc(0–)+UC(s)=11R(sc+

)UC(s)=11RUC(s)=sc+1Rsc+1=1Cs+1RC1進(jìn)行拉氏反變換uc(t)=L–1[

]1Cs+1RC11C=

e–

t三、(時(shí)域)積分性質(zhì)設(shè)：L[f(t)]=F(s)則：L[f(

)d(

)]=

0–tF(s)s

積分性質(zhì)的意義在于把時(shí)域中原函數(shù)的積分運(yùn)算轉(zhuǎn)換為復(fù)頻域中象函數(shù)除以s的代數(shù)運(yùn)算。證：ddt

0–tf(

)d(

)=f(t)兩邊進(jìn)行拉氏變換ddt

0–tf(

)d(

)L[

]=L[f(t)]根據(jù)導(dǎo)數(shù)性質(zhì)因此：L[f(

)d(

)]=

0–tF(s)ssL[f(

)d(

)]–f(

)d(

)=F( s)

0–t

0–tt=0=0四、(時(shí)域平移)延遲性質(zhì)tf(t)時(shí)域平移tf(t–t0)t0設(shè)：L[f(t)]=F(s)則：L[f(t–t0)(t–t0)]=e

F(s)–st0例：求單個正弦波的象函數(shù)。tf(t)Ttf(t

)Ttf(t

)Tf(t)=sint(t)–sin(t–T)(t–T)F(s)=L[f(t)]–e–sT=s2+

2

s2+

2

s2+

2

=(1–e–sT)五、(頻域)導(dǎo)數(shù)性質(zhì)設(shè)：L[f(t)]=F(s)則：L[tf(t)]=–dF(s)ds推廣：L[tn

f(t)]=(–1)ndnF(s)dsn六、(頻域)平移性質(zhì)設(shè)：L[f(t)]=F(s)則：L[e–

tf(t)]=F(s+

)例：求：L[e

tsint]解：

L[tn]=n

sn+1L[tnet]=n

(s–

)n+1例：求：L[tnet]=s2+

2

解：L[sint]L[e–

tf(t)]=

(s–

)2+

2常用函數(shù)的拉氏變換表原函數(shù)f(t)象函數(shù)F(s)原函數(shù)f(t)象函數(shù)F(s)A(t)Ae–

tcos(t)s+(s+

)2+

2A(t)A/ste–t(s+

)21Ae–ts+

Ats211–e–ts(s+)

sinh(t)s2–

2

sin(t)s2+

2

cosh(t)s2–

2scos(t)s2+

2s(1–t)e–t(s+

)2ssin(t+)s2+

2ssin+cos

t221s31cos(t+)s2+

2scos+sin

tnn！

1sn+11e–

tsin(t)(s+

)2+

2

tne–tn！

1(s+

)n+11§13—3拉普拉斯反變換的部分分式展開f(t)=F(S)estds2j1

+j

–j求：L–1[

](s+

)21解：L–1[

]=ts21L–1[

]=te–t(s+

)21頻域平移性質(zhì)例1：例2：求：L–1[(1–2e–t+e–2t

)/s2]解：L–1[(1–2e–t+e–2t

)/s2]=L–1[–e–t+e–2t

]s21s22s21=t–2(t–

)

(t–

)+(t–2

)

(t–2

)時(shí)域平移性質(zhì)部分分式展開法F(s)一般可以寫成關(guān)于s的兩個多項(xiàng)式之比。F(s)=N(s)D(s)N(s)、D(s)是關(guān)于s的多項(xiàng)式

=a0sm+a1sm–1+‥‥‥+am–1s+am

b0sn+b1sn–1+‥‥‥+bn–1s+bn

設(shè)：F(s)為有理式(

n>m)=(s–p1)(s–p2)‥‥‥(s–pn)N(s)對分母進(jìn)行因式分解式中p1、p2、‥‥‥pn為D(s)=0的根，稱為F(s)的極點(diǎn)。一、F(s)的極點(diǎn)為各不相等的實(shí)數(shù)根F(s)=(s–p1)(s–p2)‥‥‥(s–pn)N(s)p1p2

‥‥‥

pnp1、p2‥‥‥pn為實(shí)數(shù)=++‥‥‥+s–p1k1s–p2k2s–pnkn則：L–1[F(s)]=k1ept+k2ept+‥‥‥+knept1n2如何求k?用(s–p1)乘以上面等式兩邊(s–p2)‥‥‥(s–pn)N(s)=k1+

(s–p1)+‥‥‥+

(s–p1)s–p2k2s–pnkn令s–p1(s–p2)‥‥‥(s–pn)N(s)k1=s=p1即k1=(s–p1)F(s)s=p1k2=(s–p2)F(s)s=p2kn=(s–pn)F(s)s=pn例1:求:L–1[

]s(s+2)(s+3)S2+2s–2解:F(s)=s(s+2)(s+3)S2+2s–2=++sk1s+2k2s+3k3k1=[

]=–(s+2)(s+3)S2+2s–2S=013k2=[

]=1s(s+3)S2+2s–2S=–213k3=F(s)=–

++131ss+2113s+31L–1[F(s)]=+e–2t+e–3t1313例2:求:L–1[

]s3+6s2+15s+11s2+5s+6解:F(s)=s+1+s2+5s+64s+5=s+1++s+2k1s+3k2=s+1++s+2–3s+37L–1[F(s)]=(t)+(t)–3e–2t+7e–3t二、F(s)有共軛復(fù)極點(diǎn)F(s)=(s––j)(s–+j)N(s)=s––j+s–+jk1k2k1=s–+jN(s)s=+j=N(+j)j2=|k1|ej

1k2=s––jN(s)s=–j=N(–j)–j2=|k1|e–j

1k1、k2共軛L–1[F(s)]=L–1[+

]s––j|k1|ej

1s–+j|k1|e–j

1=|k1|ej

1e(+j)t+|k1|e–j

1e(–j)t=|k1|et[ej(+t)+e–j(+t)]11=2|k1|e

tcos(t+1)波形f(t)=2|k1|e

tcos(t+1)tf(t)=0>0tf(t)<0tf(t)例：求：L–1[](s2+2s+5)(s+2)s2+3解：s2+2s+5=(s+1–j2)(s+1+j2)則：=–1=2F(s)=++s+1–j2k1s+1+j2k2s+2k3k1=(s+1+j2)(s+2)s2+3s=–1+j2=–1+j25=0.45ej116.6o則：|k1|=0.45

1=116.6os2+2s+5s2+3k3=s=–2=1.4f(t)=0.9e–tcos(2t+116.6o)+1.4e–2t三、F(s)有重極點(diǎn)F(s)=F2(s)(s–p1)mN(s)=s–p1k1m+(s–p1)2k1m–1+……+(s–p1)mk11+……++……k11=[(s–p1)mF(s)]s=p1k12=[(s–p1)mF(s)]dsds=p1k13=

[(s–p1)mF(s)]ds2d2s=p112k1m=

[(s–p1)mF(s)]dsm–1dm–1s=p11(m–1)！

……例：求：L–1[]s(s+1)3s–2解：F(s)=sk1+s+1k23+(s+1)2k22+(s+1)3k21k1=(s+1)3s–2s=0=–2s=–1k21=ss–2=3ss–2k22=[]dsds=–1=s22s=–1=2s=–1k21=12dsd[]s22=2F(s)=s–2+s+12+(s+1)22+(s+1)33L–1[F(s)]=–2+2e–t+2te–t+t2e–t32L[tn]=n

sn+1部分分式展開法的一般步驟：1、若n<m，則先將F(s)化為真分式和多項(xiàng)式之和。2、對真分式的分母進(jìn)行因式分解，求其極點(diǎn)。3、將真分式展開成部分分式，確定其系數(shù)。4、對每個部分分式和多項(xiàng)式逐項(xiàng)求拉氏反變換。§13—4運(yùn)算電路一、元件的運(yùn)算等效電路1、電阻u

iRu=iR時(shí)域電路模型兩邊進(jìn)行拉氏變換U(s)=I(s)RRU(s)I(s)運(yùn)算電路模型2、電容Cuii=Cdudt兩邊進(jìn)行拉氏變換I(s)=sCU(s)–Cuc(0–)Cuc(0–)I(s)U(s)sC1或：初始狀態(tài)引起的附加電流源U(s)=

I(s)+sC1suc(0–)–+sC1suc(0–)U(s)I(s)初始狀態(tài)引起的附加電壓源運(yùn)算電路模型3、電感Luiu=Ldidt兩邊進(jìn)行拉氏變換U(s)=sLI(s)–Li(0–)U(s)I(s)sL–+Li(0–)或：I(s)=

U(s)+sL1si(0–)si(0–)I(s)U(s)sL四、受控源u=

iU(s)=I(s)5、含有耦合電感的電路i1L1L2Mi2u2u1L1di1dtu1=+Mdi2dtL2di2dtu2=+Mdi1dtU1(s)=sL1I1(s)–L1i1(0–)+sMI2(s)–Mi2(0–)自感電壓

自感附加電壓源互感電壓

互感附加電壓源U2(s)=sL2I2(s)–L2i2(0–)+sMI1(s)–Mi1(0–)電路模型：U1(s)=sL1I1(s)–L1i1(0–)+sMI2(s)–Mi2(0–)U2(s)=sL2I2(s)–L2i2(0–)+sMI1(s)–Mi1(0–)sL1I1(s)U1(s)+–L1i1(0–)+–sMI2(s)+–+–+–I2(s)U2(s)sL2L2i2(0–)sMI1(s)Mi1(0–)sMI1(s)I2(s)U1(s)U2(s)+–+–L2i2(0–)Mi1(0–)+–L1i1(0–)+–Mi2(0–)Mi2(0–)+–二、運(yùn)算電路+–2ARCL1L2S1ViL1iL2uC

已知：電路原已達(dá)穩(wěn)態(tài)，電容無儲存能量，t=0時(shí)將開關(guān)S合上。試畫出運(yùn)算形式電路圖。解：iL1(0+)=iL1(0–)=2AiL2(0+)=iL2(0–)=2AuC(0+)=uC(0–)=0AA2sV1sR+–sL1sL2sC1IL1(s)IL2(s)UC(s)例：2L1–+2L2+–三、運(yùn)算形式的電路定律線性無源U(s)I(s)=Z(s)U(s)I(s)運(yùn)算形式歐姆定律Z(s)運(yùn)算阻抗

基爾霍夫定律的運(yùn)算形式KCL:I(s)=0KVL:U(s)=0提示：電阻電路的各種分析方法及定理都可以在運(yùn)算電路中應(yīng)用?！?3—5應(yīng)用拉普拉斯變換法分析線性電路運(yùn)算法1、確定iL(0–)、uC(0–)。步驟：2、對uS(t)、iS(t)進(jìn)行拉普拉斯正變換。3、畫出運(yùn)算形式電路圖。4、應(yīng)用電路分析的方法求出響應(yīng)的象函數(shù)。5、對象函數(shù)進(jìn)行拉普拉斯反變換，求出響應(yīng)的時(shí)域解。例1：+–RLCS12(t=0)2Vei(t)2

1H12F

已知：t=0–時(shí)，iL(0–)=1A，uC(0–)=1V，t=0時(shí)將開關(guān)S從12。求：t≥0時(shí)i(t)。解：i(t)為二階電路全響應(yīng)，應(yīng)用運(yùn)算法。V1s2

2s+–+––+V2sI(s)RsLsC1uC(0–)sLiL(0–)s1VI(s)=2+s+s2s2+1–s1=s2+2s+2s+1i(t)=e–tcost(A)進(jìn)行拉氏反變換響應(yīng)為欠阻尼震蕩衰減。例2：+–10VR1R2L2L1S(t=0)i(t)2

3

0.3H0.1H

已知：電路原已達(dá)穩(wěn)態(tài)，求S打開后電路中的電流和電感上的電壓。解：iL1(0–)=5AiL2(0–)=0AS閉合后，L1與L2串聯(lián)，此時(shí)iL1(0+)=iL2(0+)，電感中的電流將發(fā)生跳變，電路仍為二階電路，應(yīng)用運(yùn)算法。

注意：+––+2

0.3s3

0.1s1.5V10VsI(s)I(s)=2+0.3s+3+0.1s10s+1.5=2s+s+12.51.75+––+2

0.3s3

0.1s1.5V10VsI(s)拉氏反變換i(t)=2+1.75e–12.5tAti3.752i(t)iL1(t)電感中的電流發(fā)生跳變，說明電路中必有沖擊電壓。UL1(s)UL2(s)UL1(s)=0.3sI(s)–1.5=s+12.5–6.6–0.38uL1(t)=–6.6e–12.5t–0.38(t)VUL2(s)=0.1sI(s)=s+12.5–2.2+0.38uL2(t)=–2.2e–12.5t+0.38(t)ViL2(t)續(xù)例3：i(t)u(t)CLRiL已知：R=,L=0.2H,C=0.5F,uC(0–)=2V,iL(0–)=3A,i(t)=10sin5t(t)。求：u(t)3.51I(s)U(s)sLR+–+–sC1s20.6解：I(s)=L[10sin5t]=s2+2550列結(jié)點(diǎn)方程

(++sC)U(S)=I(s)+–sL1R1sC1s2sL0.6代入數(shù)據(jù),整理:U(s)=(s2+25