自动控制原理及其应用(第二版)答案黄坚

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－＋－uoicC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)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]=(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)32-3-1函数的拉氏变换。F(s)=s+1(s+1)(s+3)解：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-2tF(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-1sts+22(s+2)2=-2e-2t-te-t+2e-t=(2-t)e-t-2e-2tF(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+12-3-4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-t2311234t2(2-4-1)求下列微分方程。d2y(t)dt2+5+6y(t)=6，初始条件：dy(t)dty(0)=y(0)=2。·解:s2Y(s)-sY(0)-Y(0)+5sY(s)-5Y(0)+6Y(s)=1s′A1=sY(s)s=0∴y(t)=1+5e-2t-4e-3tA2=(s+2)Y(s)s=-2A3=(s+3)Y(s)s=-3A1=1,A2=5,A3=-4∴Y(s)=6+2s2+12ss(s2+5s+6)(2-4-2)求下列微分方程。d3y(t)dt3+4+29=29,d2y(t)dt2dy(t)dt初始条件:y(0)=0,y(0)=17,y(0)=-122···解:2-5-a试画题2-1图所示电路的动态结构图,并求传递函数。＋－＋－Cuc＋R1R2uii1i2－＋－uoicC解:ui=R1i1+uo，i2=ic+i1UI(s)=R1I1(s)+UO(s)ducic=CdtI2(s)=IC(s)+I1(s)IC(s)=CsUC(s)即:=I1(s)UI(s)-UO(s)R1[UI(s)-UO(s)]Cs=IC(s)UO(s)UI(s)=1R1＋(sC)R21+1R1＋(sC)R2=R2+R1R2sCR1+R2+R1R2sC1R1sCR2UI(s)-UO(s)IC(s)I1(s)I2(s)＋＋1R1sCR2＋()UI(s)-UO(s)2-5-b试画出题2-1图所示的电路的动态结构图,并求传递函数。uo＋＋－－uiR1LR2C解：ui=R1I1+ucuc=uo+uLuL=LdiLdtiL=uoR2i1=iL+icic=CducdtUi(s)=R1I1(s)+UC(s)UC(s)=UO(s)+UL(s)UL(s)=sLIL(s)I1(s)=IL(s)+IC(s)∴1R1CssLR2I1＋UOUiIC--UC=UO+ULILULI2(s)=UO(s)R2IC(s)=CsUC(s)I1(s)=UO(s)R2I1(s)=UI(s)+UC(S)R1即：IL(s)=I1(s)-IC(s)IC(s)=UC(s)Cs解:电路等效为:2-6-a用运算放大器组成的有源电网络如图所示,试采用复数阻抗法写出它们的传递函数。UO=－R3＋SCR2R2＋1UIR1UOR3SCR2R21SC·＋1＋=－R1+R3+R2R3CS=－R1(R2SC+1)R2R3=－(+)R1(R2SC+1)R1R1R2=－(+R3)(R2SC+1)1=R21R3R2SC＋＋R1－C(S)=UO(S)UI(S)∴－＋＋∞CR1R2R3uiuo－＋＋∞CR1R2R3uiuo－＋＋∞CR1R2R3uiuoR4R52-6-b用运算放大器组成的有源电网络如力所示,试采用复数阻抗法写出它们的传递函数。=－R5R4+R5UO(R3SC+1)R2R3SC+R2+R3UOUI=(R2R3SC+R2+R3)(R4+R5)R1(R3SC+1)R5－=－(R4+R5)(R2+R3)(SC+1)R2R3R2+R3R1R5(R3SC+1)UIR1=R5R4+R5UOR2＋R3SCSCR3SC＋1－R5R4+R5UOR2＋R3R3SC＋1－c(t)t0TKδ(t)2-8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)t0TKδ(t)c(t)=KTt-(t-T)KTC(s)=K(1-e)Ts2-TSC(s)=G(S)第二章习题课(2-8)解: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-t解:R(s)=1sG(S)=C(s)/R(s)1s+21s-C(s)=1s+1+=s(s+1)(s+2)(s2+4s+2)=(s+1)(s+2)(s2+4s+2)C(s)=(s+1)(s+2)(s2+4s+2)脉冲响应:2s+2=1+1s+1-c(t)=δ(t)+2e+e-2t-t第二章习题课(2-９)2-10已知系统的微分方程组的拉氏变换式，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)G1G2G3G5---C(s)-R(s)G4G6G8G7X1(s)={R(s)-C(s)[G7(s)-G8(s)]}G1(s)C(s)[G7(s)-G8(s)]G6(s)X3(s)X1(s)X2(s)C(s)G5(s)X3(s)G1G2G3G5---C(s)-R(s)G4G2G6G8G7G1G2G5-C(s)-R(s)G7-G81+G3G2G6G3G4-C(s)R(s)G7-G81+G3G2G6+G3G4G5G1G2G3G41+G3G2G6+G3G4G5+G1G2G3G4(G7-G8)G1G2G3G4R(s)C(s)=第二章习题课(2-10)解:2-11(a)G1(s)G2(s)G3(s)H1(s)__+R(s)C(s)H2(s)G1(s)G2(s)H1(s)__+R(s)C(s)H2(s)G3(s)求系统的传递函数1+G2H1G2G1+G31+G1H21+G2H1G21+G2H1G2=1+G2H1+G1G2H2G2R(s)C(s)=1+G2H1+G1G2H2G2G1+G2G3G1(s)G2(s)G3(s)H1(s)__+R(s)C(s)G1(s)H2(s)第二章习题课(2-11a)2-11(a)G1(s)G2(s)G3(s)H1(s)__+R(s)C(s)H2(s)求系统的传递函数解:L1L1=-G2H1L2L2=-G1G2H1P1=G1G2P2=G3G2Δ1=1Δ2=1R(s)C(s)=Σnk=1PkΔkΔΔ=1+G2H1+G1G2H21+G2H1+G1G2H2G2G1+G2G3=第二章习题课(2-11a)解:2-11(b)G1(s)G2(s)G3(s)G4(s)_++R(s)C(s)H(s)求系统的传递函数G1(s)G2(s)G3(s)G4H_+R(s)C(s)H(s)1+G4G1HG1G2(s)G3(s)_+R(s)C(s)H(s)1+G4HG1G1G2G3_+R(s)C(s)1+G4HG1G1HG11+G4HG1G1+G3(1+HG1G4)1+G4HG1G2(1+HG1G4)1+G4G1H+G1G2HR(s)C(s)=1+G1G2H+G1G4HG1G2+G2G3+G1G2G3G4H第二章习题课(2-11b)解:2-11(b)G1(s)G2(s)G3(s)G4(s)_++R(s)C(s)H(s)求系统的传递函数R(s)C(s)=1+G1G2H+G1G4HG1G2+G2G3+G1G2G3G4HL1L1=-G1G2HL1=-G1G4HL2P1=G1G2Δ1=1P2=G3G2Δ=1+G4G2H+G1G2HΔ2=1+G1G4H第二章习题课(2-11b)H1_+++G1+C(s)R(s)G3G22-11c求系统的闭环传递函数。解:H1_+G1+C(s)R(s)G3G2H1R(s)C(s)1+G1G2+G1H1–G3H1G1G2(1–G3H1)=_G1C(s)R(s)G2H1+G21-G3H11第二章习题课(2-11c)H_G1+C(s)R(s)G22-11d求系统的闭环传递函数。解:(1)_G1+C(s)R(s)G2HG21+G2H1(G1+G2)R(s)C(s)=(2)L1L1=-G2HP1=G1Δ1=1P2=G2Δ2=1第二章习题课(2-11d)-_G1+C(s)R(s)G2G3G42-11e求系统的闭环传递函数。解:(1)_C(s)R(s)G1+G2G3-G4C(s)=R(s)1+(G1+G2)(G3-G4)(G1+G2)1+G1G3+G2G3–G1G4-G2G4=(G1+G2)第二章习题课(2-11e)L1L2L3L4L2=G1G4L3=-G2G3L4=G2G4(2)L1=-G1G3P1=G1Δ1=1P2=G2Δ2=11+G1G3+G2G3–G1G4-G2G4=(G1+G2)C(s)R(s)_G1+C(s)R(s)G22-11f求系统的闭环传递函数。_C(s)R(s)G11-G2G2C(s)=R(s)1+1-G2G1G1G21+G1G2–G2G1(1–G2)=第二章习题课(2-11f)解:(1)(2)L1L1=-G1G2L2L2=G2P1=G1Δ1=1-G2Δ=1+G1G2-G2C(s)R(s)1+G1G2–G2G1(1–G2)=2-12(a)R(s)G1(s)G2(s)H2(s)_+C(s)H3(s)H1(s)_+D(s)解:求:D(s)C(s)R(s)C(s)D(s)=01-G2H2G2G(s)=1-G2H2G1G2C(s)=R(s)1+1-G2H2G1G2H31-G2H2G1G21-G2H2+G1G2H3G2G1=R(s)=0结构图变换成：G2(s)H2(s)_+C(s)G1H3G1H1_D(s)1-G2H2G21-G1H1C(s)=D(s)1+1-G2H2G21-G2H