华中科技大学电路理论课件(汪建版)ch11讲稿

y(0+)，y(1)(0+)，···，y(n–1)(0+)第十一章动态网络的复频域分析1、动态网络的描述引言对正弦稳态,x(t),y(t),jddtX.Y.问题：一般动态网络的分析(时域分析)[an(j)n+an–1(j)n–1++a1(j)+a0]Y……=[bm(j)m+bm–1(j)m–1++b1(j)+b0]X……••dnydtndn–1ydtn–1dn–2ydtn–2dydtyanan–1an–2a1a0+++•••++dxdtdmxdtmdm–1xdtm–1dm–2xdtm–2bmbm–1bm–2b1b0x+++•••++=*2、为什么要将拉普拉斯变换引入动态网络分析？11-1拉普拉斯变换11-1-1拉普拉斯变换的定义0-£[f(t)]=f(t)e–Stdt=F(S)关于积分下限0–例0-£[K]=Ke–Stdt=Ke–St–S10-=KSS=+j£[1(t)]=1(t)e–Stdt0-£[(t)]=(t)e–Stdt0-=e–Stdt0+=1S=(t)dt0-0+=1£[e–t]=e–te–Stdt0-e–(+S)tdt0-=e–(+S)t–(S+)1=0-S+1=£[]=SF(S)–f(0-)df(t)dt£[1f1(t)+2f2(t)]=1F1(S)+2F2(S)11-1拉普拉斯变换11-1-2拉普拉斯变换的基本性质设£[f1(t)]=F1(S)£[f2(t)]=F2(S)1、线性性质2、微分性质£[kcost]=£[0.5k(ejt+e–jt)]=0.5k()S–jS+j11+=kS2+2S设£[f(t)]=F(S)uCCR+-iLus(t)+-£[f(t)dt]=F(S)0-t1S11-1拉普拉斯变换11-1-2拉普拉斯变换的基本性质3、积分性质设£[f(t)]=F(S)£[i(t)]=I(S)£[uS(t)]=US(S)Ri+L+uC(0–)+idtdidtC10–t=uS(t)R£[i(t)]+L£[]++£[]=£[uS(t)]didtC1idt0–tuC(0–)S(R+SL+)I(S)–Li(0–)+=US(S)SCuC(0–)S1I(S)=SCUS(S)+SLCi(0–)–CuC(0–)S2LC+SRC+1(R+SL+)I(S)–Li(0–)+=US(S)SCuC(0–)S111-1拉普拉斯变换11-1-2拉普拉斯变换的基本性质11-1-3部分分式法求拉普拉斯反变换出发点£[ke–t]S+k=£–1[]=ke–tS+k集中参数电路中响应变换式的特点F1(S)F2(S)F(S)=bmSm+bm–1Sm–1++b1S+b0•••anSn+an–1Sn–1++a1S+a0•••=变换式在一般情况下为S的实系数有理函数F1(S)F2(S)F(S)=bmSm+bm–1Sm–1++b1S+b0•••anSn+an–1Sn–1++a1S+a0•••=11-1拉普拉斯变换11-1-3部分分式法求拉普拉斯反变换F(S)=H0(S–zi)mi=1(S–pj)j=1nH0实数常数ziF(S)的零点pjF(S)的极点(1)nm(2)nmF(S)=Q(S)+F2(S)R(S)F(S)可展开为部分分式之和例F(S)=S3+1S2+2S+2=S–2+S2+2S+22S+5其中，£–1(S–2)=(t)2(t)F(S)的极点单极点重极点实数复数复数实数1、F(S)只含实数单极点F(S)=S–p1A1S–p2A2S–pkAkS–pnAn+•••+•••+++f(t)=£–1[F(S)]=Akepktk=1n问题归结为求F(S)的极点和确定相应的常数Ak11-1拉普拉斯变换11-1-3部分分式法求拉普拉斯反变换Ak=(S–pk)F(S)S=pkF(S)=S–p1A1S–p2A2S–pkAkS–pnAn+•••+•••+++(S+1)(S+2)(S+3)S2+3S+5F(S)=例求的反变换S3+6S2+11S+6S2+3S+5F(S)=S+1S+2S+3A1A2A3++=A1=(S+1)F(S)=(S+2)(S+3)S2+3S+5S=–1=1.5A2=(S+2)F(S)=(S+1)(S+3)S2+3S+5S=–2=–3A3=(S+3)F(S)=(S+1)(S+2)S2+3S+5S=–3=2.5(S+1)(S+2)(S+3)S2+3S+5F(S)=S+1S+2S+31.5–32.5++=11-1拉普拉斯变换11-1-3部分分式法求拉普拉斯反变换f(t)=£–1[F(S)]=1.5e–t–3e–2t+2.5e–3tt02、F(S)除含实数单极点外，还含有复数单极点1、F(S)只含实数单极点(1)复数极点是共轭形式成对出现的F(S)=S–(+j)A1+S–(–j)+A2•••(2)与复数极点对应的两个常数也互为共轭复数A2=A1~A1=A1ej令A2=A1e–j则11-1拉普拉斯变换11-1-3部分分式法求拉普拉斯反变换2、F(S)除含实数单极点外，还含有复数单极点F(S)=S–(+j)A1+S–(–j)+A2•••A1=A1ej令A2=A1e–j则f(t)=A1eje(+j)t+A1e–je(–j)t+•••=A1et[ej(t+)+e–j(t+)]+•••=2A1etcos(t+)+•••注意A1是虚部为正的极点对应的那个常数方程*£S域代数方程（初始条件含在其中）（复频域）Y(S)£–1y(t)初始条件（时域）例求的反变换[(S+2)2+4](S+1)S2+3S+7F(S)=F(S)=S–(–2+j2)S–(–2–j2)S+1A1A1A3~++A1=S=–2+j2[S–(–2–j2)](S+1)S2+3S+7=0.25ej90°(S+2)2+4S2+3S+7A3=S=–1=1f(t)=£–1[F(S)]=0.5e–2tcos(2t+90°)+e–tt011-2运算法（讨论电路基本定律，元件特性方程的复频域形式）获得复频域代数方程的途径时域电路微分方程(初始条件)£频域(S)代数方程频域电路（运算模型）11-2-1KCL与KVL的运算形式1、KCL（运算电流）Ik(S)=02、KVL–I1(S)+I2(S)–I3(S)=0ik(t)=00-£[ik(t)]=ik(t)e–Stdt=Ik(S)£线性性质I1(S)I3(S)I2(S)i1i3i2Uk(S)=011-2运算法11-2-2电路元件的运算模型1、线性时不变电阻元件2、线性时不变电感元件Li(t)+-u(t)SLI(S)+-U(S)+-Li(0-)u(t)=Ldi(t)dtU(S)=SLI(S)–Li(0-)I(S)=U(S)+1SLi(0-)SI(S)+-U(S)1SLi(0-)S£微分性质Ri(t)+-u(t)RI(S)+-U(S)3、线性时不变电容元件I(S)+U(S)SC–cu(0-)U(S)=I(S)+1SCu(0-)SI(S)=SCU(S)–Cu(0-)4、线性时不变耦合电感元件11-2-2电路元件的运算模型I(S)+U(S)u(0-)/S1SC+––u1=L1di1dtdi2dt+–Mdi1dt+–u2=+L2Mdi2dtU1(S)=SL1I1(S)SMI2(S)–L1i1(0–)Mi2(0–)+––+U2(S)=SL2I2(S)SMI1(S)–L2i2(0–)Mi1(0–)+––+讨论:1）初具电源(附加电源)由uC(0-)、iL(0-)提供，参考方向，UL(S),UC(S)等的计算2）考虑零状态情况运算阻抗与运算导纳11-2-2电路元件的运算模型U(S)=I(S)1SCI(S)=SCU(S)U(S)=SLI(S)I(S)=U(S)1SLU(S)=RI(S)I(S)=GU(S)U=RI••I=GU••U=jLI••I=U1jL••U=I1jC••I=jCU••SLI(S)+-U(S)+-Li(0-)I(S)+U(S)u(0-)/S1SC+––11-2-3运算电路，电阻性网络各种解法的适用性电路基本定律、元件特性的描述uS(t)、iS(t)uk(t)、ik(t)R、L、C等元件时域电路运算电路(频域电路)£US(S)、IS(S)£Uk(S)、Ik(S)运算阻抗(或导纳)和初具电源Ik(S)=0Uk(S)=0ik(t)=0uk(t)=0U(S)=RI(S)U(S)=I(S)+1SCu(0-)SU(S)=SLI(S)–Li(0-)u(t)=Ri(t)例1求图示电路的冲激响应(t)11Fu+–1F–+时域分析的困难节点方程(2S+1)U(S)=SU(S)=S2S+1=1214(S+1/2)–11-2-3运算电路，电阻性网络各种解法的适用性1U(S)+–1S11S–+u(t)=£–1[U(S)]=(t)–e1214t2–1(t)例2100F100+–50viL+–uck1000.4HiL(0-)=0.25AuC(0-)=25v100+–+0.4SS50–S25IL(S)+–0.1104/S11-2-3运算电路，电阻性网络各种解法的适用性5025SIL(S)=+0.1–S100+0.4s+104/SIL(S)=0.25S+62.5S2+250S+25000IL(S)=+AS+125–j96.8AS+125+j96.8*0.25S+62.5(S+125)2+9375=A=|S=–125+j96.80.25S+62.5S+125+j96.8=0.204–52.2ºiL(t)=0.408e–125tcos(96.8t–52.2º)(t0)100F100+–50viL+–uck1000.4HiL(0-)=0.25AuC(0-)=25vIL(S)=+0.204–52.2ºS+125–j96.80.20452.2ºS+125+j96.8203040V-+25HiL0.01FuC+-例3图示电路在开关闭合前处于稳态，t=0时将开关闭合，求开关闭合后uC(t)和iL(t)的变化规律。iL(0-)==0.8A4050uC(0-)=0.820=16V11-2-3运算电路，电阻性网络各种解法的适用性(S3+5S2+4S)UC=16S2+80S+160UC(S)=16S2+80S+160S(S+1)(S+4)2020-+25SILUC+--+40S16S100S+-IL(S)=20S2+124S+20025S(S+1)(S+4)203040V-+25HiL0.01FuC+-iL(0-)==0.8A4050uC(0-)=0.820=16V120125S(0.01S++)UC40S+2025S=0.16+IL(S)=20+40/S–UC25SUC(S)=16S2+80S+160S(S+1)(S+4)IL(S)=20S2+124S+20025S(S+1)(S+4)UC=SA1S+1A2A3S+4++A1=UC(S)SS=0=16S2+80S+160(S+1)(S+4)S=0=40iL(t)=2–1.28e–t+0.08e–4tt0uC(t)=40–32e–t+8e–4tt0A2=(S+1)UC(S)S=–1=–32A3=(S+4)UC(S)S=–4=816S2+80S+160S(S+4)S=–1=16S2+80S+160S(S+1)S=–4=u1(0-)=15=9V35u2(0-)=6VUOC(S)=–+=–9S6S3S例4求i+–15V152F2F3F3Fi––++u2u1+–9S6S12S13S12S13S6S9S+–+–+–+–15S10I(S)解法一、应用戴维南定理+–9S6S12S13S12S13S6S9S+–+–+–+–15S–+U0Ci(t)=–0.3e–0.04t–10I(S)+3S25S–13S12SZ0(S)=2=·13S12S+25S12S13S12S13SUOC(S)=–3S+–9S6S12S13