自动控制原理黄坚第二版第二章习题答案

第二章习题课第二章习题课(2-1a)uoi2=R2CuiuoR1R22-1(a)试建立图所示电路的动态微分方试建立图所示电路的动态微分方程。程。解：解：输入量为输入量为ui，输出量为，输出量为uo。ui=u1+uou1=i1R1ic=Cducdt=dtd(ui-uo)i1=i2-icu1=R1+uouo-Cd(ui-uo)dtR2R2ui=uoR1-C R1R2+C R1R2+uoR2duidtdtduouoR1+C R1R2+uoR2=R2ui+C R1R2duoduidtdtCucR1R2uii1i2uoicC第二章习题课第二章习题课(2-1b)2-1(b)试建立图所示电路的动态微分方试建立图所示电路的动态微分方程。程。uouiR1LR2Ci1=iL+icuL=LdiLdtuoiL=i2=R2uL=LR2duodtic=+CducdtCLR2d2uodt2duodt+uoR2CLR2d2uodt2duodti1=+Cuoi2=R2输入量为输入量为ui，输出量为，输出量为uo。ui=u1+uou1=i1R1ic=Cducdt=dtd(ui-uo)习题课一习题课一 (2-2)求下列函数的拉氏变换。求下列函数的拉氏变换。(1)f(t)=sin4t+cos4t解解:Lsinwt=ww2+s2sw2+s2Lsin4t+cos4t=4s2+16ss2+16 =s+4s2+16+Lcoswt=(2)f(t)=t3+e4t3!解解:Lt3+e4t=+=+3!s3+11s-4s41s-4(3)f(t)=tneat解解:Ltneat=n!(s-a)n+1(4)f(t)=(t-1)2e2t解解:L(t-1)2e2t=e-(s-2)2(s-2)32-3-1 函数的拉氏变换。函数的拉氏变换。F(s)=s+1(s+1)(s+3)解解：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。函数的拉氏变换。解解:f(t)=est +lim ests(s+1)2s=-2ddsss+2s -1=-2e-2t+lim(est+est)s -1 sts+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。函数的拉氏变换。解解: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。函数的拉氏变换。(4)F(s)=s+2s(s+1)2(s+3)解解:f(t)=est +ests+2(s+1)2(s+3)s=0s+2s(s+1)2s=-3+lim s -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。求下列微分方程。d2y(t)dt2+5 +6y(t)=6，初始条件：初始条件：dy(t)dty(0)=y(0)=2。解解:s2Y(s)-sY(0)-Y(0)+5sY(s)-5Y(0)+6Y(s)=1sA1=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。求下列微分方程。d3y(t)dt3 +4 +29 =29,d2y(t)dt2dy(t)dt初始条件初始条件:y(0)=0 ,y(0)=17 ,y(0)=-122 解解:2-5-a 试画题试画题2-1图所示电路的动态结构图图所示电路的动态结构图,并并求传递函数。求传递函数。CucR1R2uii1i2uoicC解解:ui=R1i1+uo ，i2=ic+i1UI(s)=R1I1(s)+UO(s)ducic=CdtI2(s)=IC(s)+I1(s)IC(s)=CsUC(s)即即:=I1(s)UI(s)-UO(s)R1UI(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)1R1sC R2()UI(s)-UO(s)2-5-b 试画出题试画出题2-1图所示的电路图所示的电路的动态结构图的动态结构图,并并求传递函数。求传递函数。uouiR1LR2C解：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)1R1CssLR2I1UOUiIC-UC=UO+ULILULI2(s)=UO(s)R2IC(s)=CsUC(s)I1(s)=UO(s)R2I1(s)=UI(s)+UC(S)R1即：即：IL(s)=I1(s)-IC(s)IC(s)=UC(s)Cs解解:电路等效为电路等效为:2-6-a 用运算放大器组成的有源电网络如图用运算放大器组成的有源电网络如图所示所示,试采用复数阻抗法写出它们的传递函数。试采用复数阻抗法写出它们的传递函数。UO=R3SCR2R21UIR1UOR3SCR2R21SC1=R1+R3+R2R3CS=R1(R2SC+1)R2R3=(+)R1(R2SC+1)R1R1R2=(+R3)(R2SC+1)1=R21R3R2SCR1C(S)=UO(S)UI(S)CR1R2R3uiuoCR1R2R3uiuoCR1R2R3uiuoR4R52-6-b 用运算放大器组成的有源电网络如用运算放大器组成的有源电网络如力所示力所示,试采用复数阻抗法写出它们的传试采用复数阻抗法写出它们的传递函数。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+R5UOR2R3SCSCR3SC1R5R4+R5UOR2R3R3SC 1c(t)t0TK(t)(t)2-8 设有一个初始条件为零的系统，系设有一个初始条件为零的系统，系统的输入、输出曲线如图，求统的输入、输出曲线如图，求G(s)。c(t)t0TK(t)(t)c(t)=KTt-(t-T)KT C(s)=K(1-e )Ts2-TSC(s)=G(S)第二章习题课第二章习题课(2-8)解解:2-9 若系统在单位阶跃输入作用时，已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。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)脉冲响应脉冲响应:2s+2=1+1s+1-c(t)=(t t)+2e +e-2t-t第二章习题课第二章习题课(2-)2-10 已知系统的微分方程组的拉氏变换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式，试画出系统的动态结构图并求传递函数。函数。解解: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)=第二章习题课第二章习题课(2-10)解解: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)求系统的求系统的传递函数传递函数1+G2H1G2 G1+G31+G1H21+G2H1G2 1+G2H1G2=1+G2H1+G1G2H2G2 R(s)C(s)=1+G2H1+G1G2H2G2G1+G2G3G1(s)G2(s)G3(s)H1(s)_+R(s)C(s)G 1(s)H2(s)第二章习题课第二章习题课(2-11a)2-11(a)G1(s)G2(s)G3(s)H1(s)_+R(s)C(s)H2(s)求系统的求系统的传递函数传递函数解解:L1L1=-G2H1L2L2=-G1G2H1P1=G1G2P2=G3G21=12=1R(s)C(s)=nk=1Pkk=1+G2H1+G1G2H21+G2H1+G1G2H2G2G1+G2G3=第二章习题课第二章习题课(2-11a)解解:2-11(b)G1(s)G2(s)G3(s)G4(s)_+R(s)C(s)H(s)求系统的求系统的传递函数传递函数G1(s)G2(s)G3(s)G4H_+R(s)C(s)H(s)1+G4G1HG1 G2(s)G3(s)_+R(s)C(s)H(s)1+G4HG1G1 G2G3_+R(s)C(s)1+G4HG1G1 HG1 1+G4HG1G1+G3(1+HG1G4)1+G4HG1G2(1+HG1G4)1+G4G1H+G1G2HR(s)C(s)=1+G1G2H+G1G4HG1G2+G2G3+G1G2G3G4 H第二章习题课第二章习题课(2-11b)解解:2-11(b)G1(s)G2(s)G3(s)G4(s)_+R(s)C(s)H(s)求系统的求系统的传递函数传递函数R(s)C(s)=1+G1G2H+G1G4HG1G2+G2G3+G1G2G3G4 HL1L1=-G1G2HL1=-G1G4HL2P1=G1G21=1P2=G3G2=1+G4G2H+G1G2H2=1+G1G4H第二章习题课第二章习题课(2-11b)H1_+G1+C(s)R(s)G3G22-11c 求系统的闭环传递函数求系统的闭环传递函数。解解:H1_+G1+C(s)R(s)G3G2H1R(s)C(s)1+G1G2+G1H1G3H1G1G2(1 G3H1)=_G1C(s)R(s)G2H1+G21-G3H11第二章习题课第二章习题课(2-11c)H_G1+C(s)R(s)G22-11d 求系统的闭环传递函数求系统的闭环传递函数。解解:(1)_G1+C(s)R(s)G2HG21+G2H1(G1+G2)R(s)C(s)=(2)L1L1=-G2HP1=G11=1P2=G22=1第二章习题课第二章习题课(2-11d)-_G1+C(s)R(s)G2G3G42-11e 求系统的闭环传递函数求系统的闭环传递函数。解解:(1)_C(s)R(s)G1+G2G3-G4C(s)=R(s)1+(G1+G2)(G3-G4)(G1+G2)1+G1G3+G2G3G1G4-G2G4=(G1+G2)第二章习题课第二章习题课(2-11e)L1L2L3L4L2=G1G4L3=-G2G3L4=G2G4(2)L1=-G1G3P1=G11=1P2=G22=11+G1G3+G2G3G1G4-G2G4=(G1+G2)C(s)R(s)_G1+C(s)R(s)G22-11f 求系统的闭环传递函数求系统的闭环传递函数。_C(s)R(s)G11-G2G2C(s)=R(s)1+1-G2G1G1G21+G1G2G2G1(1 G2)=第二章习题课第二章习题课(2-11f)解解:(1)(2)L1L1=-G1G2L2L2=G2P1=G11=1-G2=1+G1G2-G2C(s)R(s)1+G1G2G2G1(1 G2)=2-12(a)R(s)G1(s)G2(s)H2(s)_+C(s)H3(s)H1(s)_+D(s)解解:求求:D(s)C(s)R(s)C(s)D(s)=01-G2H2G2 G(s)=1-G2H2G1G2 C(s)=R(s)1+1-G2H2G1G2H31-G2H2G1G2 1-G2H2+G1G2H3G2G1=R(s)=0结构图变结构图变 换成：换成：G2(s)H2(s)_+C(s)G1H3G1H1_D(s)1-G2H2G2 1-G1H1C(s)=D(s)1+1-G2H2G21-G2H2G2 G1H3(1-G1H1)1-G2H2+G1G2H3G2(1-G1H1)=第二章习题课第二章习题课(2-12a)2-12(b)求求:D(s)C(s)R(s)C(s)R(s)Gn+D(s)解解:D(s)=0G(s)=1+G1G2HG1G2 G1G2H_C(s)C(s)=R(s)1+1+G1G2HG1G2 1+G1G2HG1G2 1+G1G2H+G1G2G1G2=R(s)=0Gn+D(s)结构图变结构图变 换成：换成：G1G2H-C(s)Gn+D(s)G1G2H-C(s)Gn/G1+D(s)1+G1G2HG1G2-C(s)+D(s)1+G1G2HG2Gn 1+G1G2HG2G1系统的传递函数系统的传递函数:)C(s)=D(s)1+1(1+1+G1G2HG1G2 1+G1G2HGnG2 1+G1G2+G1G2H=1+GnG2+G1G2H第二章习题课第二章习题课(2-12b)2-13(a)求求:R(s)E(s)R(s)C(s)C(s)E(s)G1G2G3_+R(s)解解:L1L1=-G2L2L2=-G1G2G3P1=G2G3P2=G1G2G3R(s)C(s)=1+G2+G1G2G3G2G3+G1G2G31=12=1E(s)结构图变结构图变 换成：换成：G1G2+-E(s)G3-R(s)G1+-E(s)R(s)1+G2G3G2-E(s)-R(s)1+G2G2G3G1G2G31+G2系统的传递函数系统的传递函数:)E(s)=R(s)1+1(1-G1G2G31+G2 1+G2G2G31+G2+G1G2G3=1+G2-G2G3第二章习题课第二章习题课(2-13a)R(s)G4(s)+C(s)E(s)G1(s)G2(s)-+G3(s)第二章习题课第二章习题课(2-14)D(s)X(s)2-14 求求:R(s)C(s)解解:D(s)=0结构图结构图变换为变换为 R(s)G4(s)+C(s)G1(s)G2(s)-+G3(s)G3(s)(G1+G2)(G3+G4)1+(G1+G2)G3G1+G2C(s)R(s)=1+G3(G1+G2)(G1+G2)(G3+G4)D(s)第二章习题课第二章习题课(2-14)R(s)+-E(s)G3G2G1E(s)R(s)=1+G3(G1+G2)1R(s)G4(s)+C(s)E(s)G1(s)G2(s)-+G3(s)X(s)求求:R(s)E(s)2-14解解:D(s)=0结构图变换为结构图变换为 G3(G1+G2)D(s)C(s)D(s)R(s)C(s)G4(s)+E(s)G1(s)G2(s)-+G3(s)X(s)求求:2-14解解:R(s)=0D(s)C(s)=1E(s)X(s)=G2(s)E(s)X(s)第二章习题课第二章习题课(2-14)C1(s)R1(s)第二章习题课第二章习题课(2-15)求求:2-15+G1G2G3C1(s)R1(s)+-H2H1G4G5-G6C2(s)R2(s)解解:结构图结构图变换为变换为+G1G2G3C1(s)R1(s)-H2H1G4G5-1+G4G4G5H1H21+G4G4G5H1H21+G1G1G21+G4+G1G4G5H1H2-G1G2-G1G2G4G1G2G3(1+G4)=C1(s)R1(s)1+G4+G1G4G5H1H2G1G2(1+G4)=C2(s)R2(s)G3C1(s)R1(s)+G1G2-H2H1G4G5G6-C2(s)R2(s)求求:2-14解解:结构图结构图变换为变换为+G4G5G6C2(s)R2(s)-H1H2G2G1-第二章习题课第二章习题课(2-15)1+G4G4G5G1H1H21-G1G21+G4+G1G4G5H1H2-G1G2-G1G2G4G4G5G6(1-G1G2)=1+1+G4G6G4G5G1H1H21-G2G21+G4G4G5C2(s)R2(s)C2(s)G4G5G6-+G1G2-H2H1R1(s)G3C1(s)R2(s)求求:2-14解解:结构图变换为结构图变换为 G4G5G6-+-G1G2H2H1R1(s)C2(s)第二章习题课第二章习题课(2-15)C2(s)R1(s)1+G4H2G4G5G11-G1G21+G4+G1G4G5H1H2-G1G2-G1G2G4G1G4G5G6H2=1+G6H1C2(s)R1(s)G11-G1G21+G4H2G4G5G11-G1G21+G4H2G4G5C2(s)G6R1(s)G4G5-+G1G2-H2H1G3C1(s)R2(s)求求:2-14解解:结构图变换为结构图变换为 第二章习题课第二章习题课(2-15)G1G2G3+G5G4-H1R2(s)C1(s)H2-C1(s)R2(s)G1G2G3+G5G4-H1R2(s)C1(s)H2-G21+G4G4G51-G1G2-G1H11+G4+G1G4G5H1H2-G1G2-G1G2G4-G1G2G3G4G5H1=1+1+G4G4G5G1H11-G2G21+G4G4G5C1(s)R2(s)H2-G1H11-G2G2G2G3G3C1(s)R1(s)+G1G2+-H2H1G4G5-G6C2(s)R2(s)C1(s)R1(s)求求2-15第二章习题课第二章习题课(2-15)解解:L1=G1G2L3=-G4L2=-G1G4G5H1H2P1=G1G2G3=1-G1G2+G1G4G35H1H2+G4-G1G2G41=1+G41+G4+G1G4G5H1H2-G1G2-G1G2G4G1G2G3(1+G4)=C1(s)R1(s)G3C1(s)R1(s)+G1G2+-H2H1G4G5-G6C2(s)R2(s)C2(s)R2(s)求求2-15解解:L1=G1G2L3=-G4L2=-G1G4G5H1H2P1=G4G5G6=1-G1G2+G1G4G5H1H2+G4-G1G2G41=1-G1G21+G4+G1G4G5H1H2-G1G2-G1G2G4G4G5G6(1-G1G2)=C2(s)R2(s)第二章习题课第二章习题课(2-15)G3C1(s)R1(s)+G1G2+-H2H1G4G5-G6C2(s)R2(s)C1(s)R2(s)求求2-15解解:L1=G1G2L3=-G4L2=-G1G4G5H1H2=1-G1G2+G1G4G5H1H2+G4-G1G2G41=1P1=-G1G2G3G4G5H11+G4+G1G4G5H1H2-G1G2-G1G2G4-G1G2G3G4G5H1=C1(s)R2(s)第二章习题课第二章习题课(2-15)G3C1(s)R1(s)+G1G2+-H2H1G4G5-G6C2(s)R2(s)C2(s)R1(s)求求2-15解解:L1=G1G2L3=-G4L2=-G1G4G5H1H2=1-G1G2+G1G4G5H1H2+G4-G1G2G41=1P1=G1G4G5G6H21+G4+G1G4G5H1H2-G1G2-G1G2G4G1G4G5G6H2=C2(s)R1(s)第二章习题课第二章习题课(2-15)返回返回