《自动控制原理》黄坚课后习题答案

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1、《自动控制原理》黄坚课后习题答案2-1试建立图所示电路的动态微分方程C-+-uiuoR1R2i1ii2C-+-uiuoR1R2i1ii2C-+-uiuoR1R2i1ii2Lu1C-+-uiuoR1R2i1ii2LC-+-uiuoR1R2i1ii2Lu1解u1=ui-uoi2=Cdu1dti2=Cdu1dti1=i-i2uoi=R2uoi=R2u1i1=R1u1i1=R1=ui-uoR1=ui-uoR1dtd(ui-uo)=Cdtd(ui-uo)=C(a)Cd(ui-uo)dtuo-R2=ui-uoR

2、1Cd(ui-uo)dtuo-R2=ui-uoR1i=i1+i2i2=Cdu1dti2=Cdu1dtuoi1=R2uoi1=R2u1-uo=LR2duodtu1-uo=LR2duodtR1i=(ui-u1)R1i=(ui-u1)(b)解)-R2(ui-uo)=R1u0-CR1R2(duidtdtduo)-R2(ui-uo)=R1u0-CR1R2(duidtdtduoCR1R2duodtduidt+R1uo+R2u0=CR1R2+R2uiCR1R2duodtduodtduidt+R1uo+R2u0=CR1R2+

3、R2ui=R1ui-u1uo+CR2du1dt=R1ui-u1uo+CR2du1dtu1=uo+LR2duodtu1=uo+LR2duodtduodtR1R2Lduodt+CLR2d2uodt2=--uiR1uoR1uoR2+CduodtR1R2Lduodt+CLR2d2uodt2=--uiR1uoR1uoR2+C)uoR1R2Lduodt)CLR2d2uodt2=++(uiR11R11R2+(C+)uoR1R2Lduodt)CLR2d2uodt2=++(uiR11R11R2+(C+2-2求下列函数的拉氏变换。

4、(1)f(t)=sin4t+cos4tL[sinωt]=ωω2+s2L[sinωt]=ωω2+s2=s+4s2+16=s+4s2+16L[sin4t+cos4t]=4s2+16ss2+16+L[sin4t+cos4t]=4s2+16ss2+16+sω2+s2L[cosωt]=sω2+s2L[cosωt]=解www.docin.com(2)f(t)=t3+e4t解L[t3+e4t]=3!s41s-4+L[t3+e4t]=3!s41s-4+6s+24+s4s4(s+4)=6s+24+s4s4(s+4)=(3)f

5、(t)=tneatL[tneat]=n!(s-a)n+1L[tneat]=n!(s-a)n+1解(4)f(t)=(t-1)2e2tL[(t-1)2e2t]=e-(s-2)2(s-2)3L[(t-1)2e2t]=e-(s-2)2(s-2)3解2-3求下列函数的拉氏反变换。A1=(s+2)s+1(s+2)(s+3)s=-2A1=(s+2)s+1(s+2)(s+3)s=-2=-1=2f(t)=2e-3t-e-2t(1)F(s)=s+1(s+2)(s+3)(1)F(s)=s+1(s+2)(s+3)解A2=(s+3

6、)s+1(s+2)(s+3)s=-3A2=(s+3)s+1(s+2)(s+3)s=-3F(s)=2s+31s+2-F(s)=2s+31s+2-=A1s+2s+3+A2=A1s+2s+3+A2(2)F(s)=s(s+1)2(s+2)(2)F(s)=s(s+1)2(s+2)f(t)=-2e-2t-te-t+2e-t解=A2s+1s+2+A3+A1(s+1)2=A2s+1s+2+A3+A1(s+1)2A1=(s+1)2s(s+1)2(s+2)s=-1A1=(s+1)2s(s+1)2(s+2)s=-1A3=(s+2)

7、s(s+1)2(s+2)s=-2A3=(s+2)s(s+1)2(s+2)s=-2ddsss+2][A2=s=-1ddsss+2][A2=s=-1=-1=2=-2(3)F(s)=2s2-5s+1s(s2+1)(3)F(s)=2s2-5s+1s(s2+1)F(s)(s2+1)s=+j=A1s+A2s=+jF(s)(s2+1)s=+j=A1s+A2s=+jA2=-5A3=F(s)ss=0A3=F(s)ss=0f(t)=1+cost-5sint解=s+A3s2+1A1s+A2=s+A3s2+1A1s+A2=1s2s2

8、-5s+1=A1s+A2s=js=js2s2-5s+1=A1s+A2s=js=jj-2-5j+1