拉普拉斯变换

§14-3拉普拉斯反变换由象函数求原函数的方法：(1)利用公式dseSFjtfstjj)(21)((2)分解定理：对F(S)进行部分分式展开12()()()()()()nNSFSFSFSFSDS象函数的一般形式：101101()()()()mmmnnnaSaSaNSFSnmDSbSbSb12.()0nnmDSnSS设，的根为个单根利用部分分式F(S)分解为：nnSSkSSkSSkSF2211)(tsntstsnekekektf2121)()()()()(110110mnbSbSbaSaSaSDSNSFnnnmmm111()()SSkFSSS222()()SSkFSSS()()nnnSSkFSSS0()()()NSFSADS1.nm设，将有理分式化为真分式iiSSiSSiiSDSSSNSFSSk)())(()()()()())((limSDSNSSSNissi)()(iiSDSN不定式00例6。的原函数求)(10712)(23tfsssssF解：令D(s)=0，则s1=0，s2=－2，s3=－51.01014312)()(0211sssssssDsNK20.5K=tteetf526.05.01.0)(30.6K=-jS2,1)())(()()()()(SQjSjSSNSDSNSF)()(21SQSPjSkjSk1()[()()]()sjsjNsksjFsDs2()[()()]()sjsjNsksjFsDs()()12()jtjtftkeke3.()0nmDS设，的根为共轭复根k1、k2也是一对共轭复根111211jjekkekk，则设)cos(2)(11)(1)(1)(2)(111tekeekeekekektfttjjtjjtjtj237()()25SFSftSS例求的原函数。21,0)(12jssD则解：令4525.0223)()(2121'1jSjssssDsNk)452cos(2)452cos(2)(1tetektftt4525.0223)()(2121'2jSjssssDsNk)()(1110nmmmSSaSaSaSFnnnnSSkSSkSSkSSkSF)()()()(11111121121111)]()[(11SSnnSFSSk1)]()[(111SSnnSFSSdsdk1)]()[(!2112221SSnnSFSSdsdk1)]()[()!1(111111SSnnnSFSSdsdnk4.()0nmDSn设，有重根121222F(s)(1)(1)KKKSSS=++++4)1(4)0(021SssssK3)1(4)1(12222SssssK1221)]()1[(SSFSdsdK4]4[1SSSdsdttteetf344)(248F(s)=()(1)SftSS例求的原函数。小结：1.)n=m时将F(S)化成真分式0()()()NSFSADS=+1.由F(S)求f(t)的步骤2.)求真分式分母的根，确定分解单元3.)求各部分分式的系数4.)对每个部分分式和多项式逐项求拉氏反变换。2.拉氏变换法分析电路)()(titu正变换反变换U(S)I(S)相量形式KCL、KVL元件复阻抗、复导纳相量形式电路模型i§14-4运算电路类似地)()(titu元件运算阻抗、运算导纳运算形式KCL、KVL运算形式电路模型uIUI(S)U(S)2.电路元件的运算形式R：u=Ri)()(SGUSI)()(SRISU1.运算形式的电路定律00uKVLiKCL0)(SU0(S)I+u-iR+U(S)-I(S)RL:dtdiLu)0()()(LiSSLISUSiSLSUSI)0()()(1/SLi(0-)/S+U(S)-I(S)I(S)Li(0-)+U(S)-SLi+u-L+u-iC:SuSISCSUccc)0()(1)()0(10ctccudtiCu)0()()(cCCCuSSCUSIIC(S)1/SCuc(0-)/S+UC(S)-＋－+UC(S)-Cuc(0-)SCIC(S)dtdiMdtdiLudtdiMdtdiLu12222111)0()()0()()()0()()0()()(11222222211111MiSSMIiLSISLSUMiSSMIiLSISLSUML1L2i1i2+u1-+u2-L1i1(0-）Mi2(0-）Mi1(0-）L2i2(0-）+U2(S)-+U1(S)-I1(S)I2(S)SL1SL2+－SM＋－－+－+121uuiRu)()()()(121SUSURSISU(s)U+1(s)-RI(S)+U2-U1(S)+u1-+u2-u1Ri＋－0)0(0)0(Lciu)0(10ctcudtiCdtdiLiRusuSISCLiSSLIRSISUc)0()(1)0()()()(suLisUSCSLRSIc)0()0()()1)((运算阻抗SCSLRSZ1)(+u-iRLC+U(S)-I(S)RSL1/SC－＋＋－uc(0-)/sLi(0-)3.运算电路运算电路如L、C有初值时，初值应考虑为附加电源RRLLCi1i2Ee(t)时域电路0)0(0)0(Lciu物理量用象函数表示元件用运算形式表示RRLSL1/SCI1(S)E/SI2(S)＋－注意（1）借助拉氏变换，将时域电压－电流（微积分关系）变换为复频域的运算形式（代数关系）；（2）若将各元件复频域VCR中的s换为jω，且令各初始状态为0，则得到其频域VCR(相量形式)；（3）对任一电路，将其所有元件变换为运算形式，而保持连接关系不变，则得到该电路运算电路。§14.5拉普拉斯变换法分析线性电路步骤：1.由换路前电路计算uc(0-),iL(0-)2.画运算电路图3.应用电路分析方法求象函数4.反变换求原函数例1：200V30Ω0.1H10Ω-uc+1000μFiLt=0时闭合k,求iL，uL。100)0(cu已知：VAiL5)0()1(解：(2)画运算电路SSL1.0SSSC1000101000116200/S300.1s0.5101000/S100/SIL(S)I2(S)Vuc100)0(例1：200V30Ω0.1H10Ω-uc+1000μFiL)3(回路法221)200()40000700(5)(SSSSSI5.0200)(10)1.040)((21SSISSISSISSI100)()100010()(10-21200/S300.1s0.5101000/s100/SIL(S)I2(S)I1(S)I2(S)2222111)200(200)(SKSKSKSI(4)反变换求原函数200030)(321SSSSD，个根有221)200()40000700(5)(SSSSSI01)(SSSFK5200400)40000700(50222SSSSS1500)200)((200222SSSFK2222111)200(200)(SKSKSKSI0)]()200[(200221SSFSdsdK21)200(1500)200(05)(SSSSI2001()(51500)titteA200200()()(15030000)()ttLLditutLetetVdte0tSLSISUL)()(1求UL(S)5.0)()(1SLSISUL2)200(30000200150SS200200()(15030000)()ttLutetetVeUL(S)200/S300.1s0.5101000/S100/SIL(S)I2(S)?例2在电路中，已知R1=1Ω,L1=1H,R2=1Ω,L2=4H,开关K原是闭合的,电路已经稳定,t=0时把开关打开。求i(t),uL1(t)和uL2(t)。解R1L1R2L2100ViLuL1uL2KsIL(s)100s1Ω1Ω4s400UL1(s)UL2(s)（1）求初值:0)0(1LiA100)0(2Li（2）画运算电路图UL1(s)UL2(s)sIL(s)100s1Ω1Ω4s400（3）选择适当的方法求解.sssIL52400100)()4.0(2080sss120.4KKss014.02080sssK50304.022080sssK,0.4()(5030)A(0)tLitet)4.0(2080)(1sssUL4.01280s400)(4)(2ssIsULL4.04880s0.41()80()12()VtLuttete0.42()80()48()VtLutteteR1L1R2L2100ViLuL1uL2KUL1(s)UL2(s)sIL(s)100s1Ω1Ω4s400分析:（1）从结构上看此电路为跃变电路,从所得结果中看有δ(t)；（2）在计算之初,并未考虑电路是否跃变，用复频域法计算跃变电路,只计算0-时刻值即可,非常方便。0.4()(5030)A(0)tLitet0.41()80()12()VtLuttete0.42()80()48()VtLutteteRC+ucis(t)例3求冲激响应0)0(CuR1/SC+Uc(S)IS1SCSISCRRSUC1)(/1)()/1(RCSRCR()()1CCRCSISSCUSRCS1111RCSRCSRCS/1()tRCcuetCe/1()()tRCcitetRCetuc(V)C10ticRC1)(t例4图示电路，已知R1=1,R2=3，L1=2H,L2=3H，M=1H，us=6V，试求：t=0时开关打开后的电流i1(t)和i2(t)。R1L1R2L2usK**Mi1i2R1L1sR2L2sUs(s)L1i1(0-)Mi2(0-)L2i2(0-)Mi1(0-)**Ms例5图示电路已处于稳态，t=0时将开关S闭合，已知us1=2e-2tV,us2=5V,R1=R2=5,L=1H,求t≥0时的uL(t).22]2[][21seLuLtssLuLs5]5[][2ARuisL1)0(22SR1R2＋iL+＋US1LuLUS2－－－R1R2＋+＋UL(s)－－－sL－Li(0-)+22ss5①1212(0)1112151()()2LLiUSRRSLSRSRSL)52)(2(2)(SSSSULVeetuttL)54()(5.220t