《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-uoR1Cd(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+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。(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：(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(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)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)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=1s2s2-5s+1=A1s+A2s=js=js2s2-5s+1=A1s+A2s=js=jj-2-5j+1=jA1+A2j-2-5j+1=jA1+A2-5j-1=-A1+jA2A1=1F(s)=1ss2+1s-5s2+1++F(s)=1ss2+1s-5s2+1++(4)F(s)=s+2s(s+1)2(s+3)(4)F(s)=s+2s(s+1)2(s+3)解：=+s+1A1s+3A2(s+1)2+sA3+A4=+s+1A1s+3A2(s+1)2+sA3+A4-12A1=-12A1=23A3=23A3=112A4=112A4=A2=d[s=-1ds](s+2)s(s+3)A2=d[s=-1ds](s+2)s(s+3)(s+2)s(s+3)-34=-34=-34A2=-34A2=+-43+f(t)=e-t32e-3t2-te-t121+-43+f(t)=e-t32e-3t2-te-t121=s=-1[s(s+3)]2[s(s+3)-(s+2)(2s+3)]=s=-1[s(s+3)]2[s(s+3)-(s+2)(2s+3)](2-4)求解下列微分方程。y(0)=y(0)=2·y(0)=y(0)=2·+6y(t)=6+5d2y(t)dt2dy(t)dt(1)+6y(t)=6+5d2y(t)dt2dy(t)dt(1)解：s2Y(s)-sy(0)-y'(0)+5sY(s)-5y(0)+6Y(s)=6ss2Y(s)-sy(0)-y'(0)+5sY(s)-5y(0)+6Y(s)=6sA1=1y(t)=1+5e-2t-4e-3tA2=5A3=-4Y(s)=6+2s2+12ss(s2+5s+6)Y(s)=6+2s2+12ss(s2+5s+6)=A1s+2s+3+A3s+A2=A1s+2s+3+A3s+A22-5试画题图所示电路的动态结构图，并求传递函数。(1)ii2+-uruc+-R2R1c+-uruc+-R2R1ci1解：I2(s)I1(s)+Uc(s)Ur(s)_Cs1R11R1+R2Uc(s)I(s)Ur(s)Uc(s)=1R1(1+(+sC)R21R1+sC)R2Ur(s)Uc(s)=1R1(1+(+sC)R21R1+sC)R2=R2+R1R2sCR1+R2+R1R2sC=R2+R1R2sCR1+R2+R1R2sC(2)C＋-+-urucR1R2Lu1C＋-+-urucR1R2Lu1I(s)Ur(s)_1R11R1U1(s)解：I1(s)-I2(s)L31Cs1CsU1(s)Uc(s)-1Ls1LsR2I1(s)Uc(s)L1L2L1=-R2/LsL2=-/LCs2L3=-1/sCR1Δ1=1L1L3=R2/LCR1s2P1=R2/LCR1s2=R1CLs2+(R1R2C+L)s+R1+R2Ur(s)Uc(s)R2=R1CLs2+(R1R2C+L)s+R1+R2Ur(s)Uc(s)R22-8设有一个初始条件为零的系统，系统的输入、输出曲线如图，求G(s)。c(t)t0c(t)t0TKδ(t)c(t)t0TKδ(t)c(t)t0c(t)t0TKδ(t)解:c(t)=KTt-(t-T)KTc(t)=KTt-(t-T)KTC(s)=K(1-e)Ts2-TSC(s)=K(1-e)Ts2-TSC(s)=G(s)2-9若系统在单位阶跃输入作用时，已知初始条件为零的条件下系统的输出响应，求系统的传递函数和脉冲响应。r(t)=I(t)c(t)=1-e+e-2t-tc(t)=1-e+e-2t-tR(s)=1sR(s)=1s解:G(s)=C(s)/R(s)1s+21s-C(s)=1s+1+1s+21s-C(s)=1s+1+=s(s+1)(s+2)(s2+4s+2)=s(s+1)(s+2)(s2+4s+2)=(s+1)(s+2)(s2+4s+2)=(s+1)(s+2)(s2+4s+2)C(s)=(s+1)(s+2)(s2+4s+2)C(s)=(s+1)(s+2)(s2+4s+2)c(t)=δ(t)+2e-2t-e-tc(t)=δ(t)+2e-2t-e-t2-10已知系统的拉氏变换式，试画出系统的动态结构图并求传递函数。解:X1(s)=R(s)G1(s)-G1(s)[G7(s)-G8(s)]C(s)X2(s)=G2(s)[X1(s)-G6(s)X3(s)]X3(s)=G3(s)[X2(s)-C(s)G5(s)]C(s)=G4(s)X3(s)={R(s)-C(s)[G7(s)-G8(s)]}G1(s)G1G1G1G2G2G2G3G3G3G5G5G5---C(s)-R(s)G4G4G4G6G6G6G8G7G8G8G7G7C(s)[G7(s)-G8(s)]G6(s)X3(s)X1(s)X2(s)C(s)G5(s)X3(s)G1G2G5-C(s)-R(s)G7-G81+G3G2G6G3G4G1G2G5G5-C(s)-R(s)G7-G81+G3G2G6G3G41+G3G2G6G3G41+G3G2G6+G3G4G5+G1G2G3G4(G7-G8)G1G2G3G4R(s)C(s)=1+G3G2G6+G3G4G5+G1G2G3G4(G7-G8)G1G2G3G4R(s)C(s)=2-11求系统的传递函数(a)G1(s)G2(s)G3(s)H1(s)__+R(s)C(s)H2(s)G1(s)G2(s)G3(s)H1(s)__+R(s)C(s)H2(s)解:L1=-G2H1L2=-G1G2H2P1=G1G2P2=G3G2Δ1=1Δ2=1R(s)C(s)=Σnk=1PkΔkΔR(s)C(s)=Σnk=1PkΔkΔΔ=1+G2H1+G1G2H21+G2H1+G1G2H2G2G1+G2G3=1+G2H1+G1G2H2G2G1+G2G3=(b)G1(s)G2(s)G3(s)G4(s)_++R(s)C(s)H(s)G1(s)G2(s)G3(s)G4(s)_++R(s)C(s)H(s)解:R(s)C(s)=1+G1G2H+G1G4HG1G2+G2G3+G1G2G3G4HR(s)C(s)=1+G1G2H+G1G4HG1G2+G2G3+G1G2G3G4HL1=-G1G2HL2=-G1G4HP1=G1G2Δ1=1P2=G3G2Δ=1+G1G4H+G1G2HΔ2=1+G1G4HH1_+++G1+C(s)R(s)G3G2H1H1_+++G1G1+C(s)R(s)G3G3G2G2(c)H1_+G1+C(s)R(s)G3G2H1H1H1_+G1G1+C(s)R(s)G3G3G2G2H1H1R(s)C(s)1+G1G2+G1H1–G3H1G1G2(1–G3H1)=R(s)C(s)1+G1G2+G1H1–G3H1G1G2(1–G3H1)=H_G1+C(s)R(s)G2HH_G1G1+C(s)R(s)G2G2(d)解:L1=-G2HP1=G1Δ1=1P2=G2Δ2=11+G2H1(G1+G2)R(s)C(s)=1+G2H1(G1+G2)R(s)C(s)=-_G1+C(s)R(s)G2G3G4-_G1G1+C(s)R(s)G2G2G3G3G4G4(e)解:L2=G1G4L3=-G2G3L4=G2G4L1=-G1G3P1=G1Δ1=1P2=G2Δ2=11+G1G3+G2G3–G1G4-G2G4=(G1+G2)C(s)R(s)1+G1G3+G2G3–G1G4-G2G4=(G1+G2)1+G1G3+G2G3–G1G4-G2G4=(G1+G2)C(s)R(s)_G1+C(s)R(s)G2_G1G1+C(s)R(s)G2G2(f)L1L2解:L1=-G1G2L2=G2P1=G1Δ1=1-G2Δ=1+G1G2-G2C(s)R(s)1+G1G2–G2G1(1–G2)=C(s)R(s)1+G1G2–G2G1(1–G2)=2-12(a)R(s)