自动控制原理第2章习题课答案

1、自动控制原理第2章习题课答案第二章习题课 (2-1b)2-1(b) 试建立图所示电路的动态微分方程。uouiR1LR2Ci1=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:Lsinwt= ww2+s2sw2+s2Lsin4t+cos4t= 4s2+16ss2+1

2、6 =s+4s2+16+ Lcoswt=(2) f(t)=t3+e4t3!解:Lt3+e4t= + = +3!s3+11s-4s41s-4(3) f(t)=tneat解:Ltneat=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。

3、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。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

4、+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。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)=1sA1=sY(s)s=0y(t)=1+5e-2t-4e-3tA2=(s+2)Y(s) s=-2A3=(s+3)Y(s)

5、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。CucR1R2uii1i2uoicC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)R1UI(s)-UO(s)Cs

6、=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。 uouiR1LR2C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=U

7、O+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= R3SCR2R21UIR1UOR3SCR2R21SC1= R1+R3+R2R3CS=R1(R2SC+1)R2R3=( + )R1(R2SC+1)R1R1R2= ( +R3 )(R2SC+1)1=R21R3R2SCR1C(S)=UO(S)UI(S)CR1R2R3uiuoCR1R2R3uiuoCR

8、1R2R3uiuoR4R52-6-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+ R5UOR2R3SCSCR3SC1R5R4+ R5UOR2R3R3SC1c(t)t0TK(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)t0TK(t) c(t)= KTt-(t-T)KT C(s)=

9、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)=(t)+2e +e-2t-t第二章习题课 (2-)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

10、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+G3G

11、2G6 +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+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

12、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=G3G21 =12 =1R(s)C(s)=nk=1Pkk=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+G4G1HG1 G2(s)G3(s)_+R(s)C(s)H(s)1+G4HG1G1

13、 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 (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+G1G2G3G4 HL1L1=-G1G2HL1=-G1G4HL2P1=G1G21 =1P2=G3G2=1+G4G2H+G1G2H2=1+G1G4H第二章习题

14、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+G1H1G3H1G1G2 (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=G11 =1P2=G22 =1第二章习题课 (2-11d)-_G1+C(s)R(s)G2G3G42-11e 求系统的闭环

15、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+G2G3G1G4-G2G4=(G1+G2)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 。_C(s)R(s)G11-G2G2C(s)=R(s)1+1-G2G1G1G21+G1G2G2G1 (1 G2)=第二章习题课 (2-1

16、1f)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: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 换成：G2(s)H2(s)_+C(s) G1H3G1H1_D(s)1-G2H2G2 1-G1H1C(s)=D(s)1+1

17、-G2H2G21-G2H2G2 G1H3(1-G1H1 )1-G2H2+G1G2H3G2(1-G1H1 )=第二章习题课 (2-12a)2-12(b)求:D(s)C(s)R(s)C(s)R(s)Gn+D(s)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 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

18、数:)C(s)=D(s)1+1(1+1+G1G2HG1G2 1+G1G2HGnG2 1+G1G2+G1G2H=1+GnG2+G1G2H第二章习题课 (2-12b)2-13(a)求:R(s)E(s)R(s)C(s)C(s)E(s)G1G2G3_+R(s)解:L1L1=-G2L2L2=-G1G2G3P1=G2G3P2=G1G2G3R(s)C(s)=1+G2+G1G2G3G2G3+G1G2G31 =12 =1E(s)结构图变 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:)E(s)=R(s)1

19、+1(1-G1G2G31+G2 1+G2G2G31+G2+G1G2G3=1+G2-G2G3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 (2-14)D(s)X(s)2-14求:R(s)C(s)解:D(s)=0结构图变换为 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 (2-14)R(s)+-E(s)G3G2G1E(s)R(s)=1+G3(G1+G2)1R(

20、s)G4(s)+C(s)E(s)G1(s)G2(s)-+G3(s)X(s)求:R(s)E(s)2-14解:D(s)=0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:2-14解:R(s)=0D(s)C(s)=1E(s)X(s)=G2(s)E(s)X(s)第二章习题课 (2-14)C1(s)R1(s)第二章习题课 (2-15)求:2-15+G1G2G3C1(s)R1(s)+-H2H1G4G5-G6C2(s)R2(s)解:结构图变换为 +G1G2G3C1(s)R1(s)-H2H1G4G5-1+G4G4G5H1

21、H21+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:2-14解:结构图变换为 +G4G5G6C2(s)R2(s)-H1H2G2G1-第二章习题课 (2-15)1+G4G4G5G1H1H21-G1G21+G4+G1G4G5H1H2-G1G2-G1G2G4G4G5G6(1-G1G2)=1+1+G4G6G4G5G1H1H21-G2G21+G4G4G5C2

22、(s)R2(s)C2(s)G4G5G6-+G1G2-H2H1R1(s)G3C1(s)R2(s)求:2-14解:结构图变换为 G4G5G6-+-G1G2H2H1R1(s)C2(s)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:2-14解:结构图变换为 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+-H2H