自动控制原理及其应用课后习题解答.ppt

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1、第二章习题课　(2-1a)uoi2=R2C＋－＋－uiuoR1R22-1(a)试建立图所示电路的动态微分方程。解：输入量为ui，输出量为uo。ui=u1+uou1=i1R1ic=Cducdt=dtd(ui-uo)i1=i2-icu1=[]R1+uouo-Cd(ui-uo)dtR2R2ui=uoR1-CR1R2+CR1R2+uoR2duidtdtduouoR1+CR1R2+uoR2=R2ui+CR1R2duoduidtdt＋－＋－Cuc＋R1R2uii1i2－＋－uoicCEvaluationonly.CreatedwithAspose.Slides

2、for.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.第二章习题课　(2-1b)2-1(b)试建立图所示电路的动态微分方程。uo＋＋－－uiR1LR2Ci1=iL+icuL=LdiLdtuoiL=i2=R2uL=LR2duodtic==+CducdtCLR2d2uodt2duodt+uoR2CLR2d2uodt2duodti1=+Cuoi2=R2输入量为ui，输出量为uo。ui=u1+uou1=i1R1ic=Cducdt=dtd(ui-uo)Evaluationonly.Crea

3、tedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.习题课一(2-2)求下列函数的拉氏变换。(1)f(t)=sin4t+cos4t解:∵L[sinwt]=ww2+s2sw2+s2∴L[sin4t+cos4t]=4s2+16ss2+16=s+4s2+16+L[coswt]=Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2

4、011AsposePtyLtd.(2)f(t)=t3+e4t3!解:L[t3+e4t]=+=+3!s3+11s-4s41s-4(3)f(t)=tneat解:L[tneat]=n!(s-a)n+1(4)f(t)=(t-1)2e2t解:L[(t-1)2e2t]=e-(s-2)2(s-2)3Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.2-3-1函数的拉氏变换。F(s)=s+1(s+1)(s+3)解：

5、A1=(s+2)s+1(s+1)(s+3)s=-2=-1(s+1)(s+3)A2=(s+3)s+1s=-3=2F(s)=-2s+31s+2f(t)=2e-3t-e-2tEvaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.F(s)=s(s+1)2(s+2)2-3-2函数的拉氏变换。解:f(t)=est+lim[est]s(s+1)2s=-2ddsss+2s-1=-2e-2t+lim(est+est)s-

6、1sts+22(s+2)2=-2e-2t-te-t+2e-t=(2-t)e-t-2e-2tEvaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.F(s)=2s2-5s+1s(s2+1)2-3-3函数的拉氏变换。解:F(s)(s2+1)s=+j=A1s+A2s=+jA1=1,A2=-5A3=F(s)s=1s=0∴f(t)=1+cost-5sintF(s)=++1ss2+1s-5s2+1Evaluation

7、only.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.2-3-4函数的拉氏变换。(4)F(s)=s+2s(s+1)2(s+3)解:f(t)=est+ests+2(s+1)2(s+3)s=0s+2s(s+1)2s=-3+lims-1d[est]s+2s(s+3)ds=+e-3t+lim[+]23112s-1(-s2-4s-6)est(s2+3)2(s+2)tests2+3s=+e-3t-e-t-e-t2311234t2Eva

8、luationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.