自动控制原理及其应用答案第二版_黄坚_课后答案

1、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

2、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

3、+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

4、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+2

5、s(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

6、=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)=R

7、1I1(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=LdiLd

8、tiL=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

9、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)(

10、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

11、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

12、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)

13、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+G1G2H2G2G

14、1+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

15、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+G1G

16、2H+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

17、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=-G1G

18、3P1=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

19、: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)

20、=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

21、: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)

22、+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(

23、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

24、+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

25、(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

26、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=-G1G4G5H1

27、H2P1=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 (2-1

28、5)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-G1G

29、2+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)3-1 设温度计需要在一分钟内指示出响设温度计需要在一分钟内指示出响应值的应值的98%，并且假设温度计为一阶系并且假设温度计为一阶系统，求时间常数统，求时间常数T。如果将温度计放在。如果将温度计放在澡盆内，澡盆的温度以澡盆内，澡盆的温度以10oC/min的速度的速度线性变化，求温度计的误差。线性变化，求温度计的误差。第三章习题课第三章习题课 (3-1)解解: c(t)=c()98%t=

30、4T=1 min r(t)=10t e(t)=r(t)-c(t) c(t)=10(t-T+e)-t/T=10(T-e)-t/Tess=lim t e(t)=10T=2.5T=0.253-2 电路如图，设系统初始状态为领电路如图，设系统初始状态为领第三章习题课第三章习题课 (3-2)解解:(1)求系统的单位阶跃响应求系统的单位阶跃响应,及及uc(t1)=8 时的时的t1值值R1Cs+1R1/R0G(s)= R0=20 k R1=200 k C=2.5F uc(t)=K(1 e tT-)KTs + 1=T=R1C=0.5 K=R1/R0=10 =10(1 e -2t) 8=10(1 e -2t)0

31、.8=1 e -2te -2t =0.2 t=0.8 -+R1R0Curucg(t)=e-t/TTK t1=0.8 =4 (2) 求系统的单位脉冲响应，单位斜坡求系统的单位脉冲响应，单位斜坡 响应响应,及单位抛物响应在及单位抛物响应在t1时刻的值时刻的值第三章习题课第三章习题课 (3-2)解解:uc(t)=K(t-T+Te-t/T) =4 R(s)=1s2R(s)=1R(s)=1s3T2=K(ss+1/T+Ts2-1s3-T2)=1.2 1s3KTs + 1Uc(s)= -0.5t+0.25-0.25e-2t)12t2uc(t)=10( 3-3 已知单位负反馈系统的开环传递函已知单位负反馈系统

32、的开环传递函数，求系统的单位阶跃响应。数，求系统的单位阶跃响应。4s(s+5)G(s)=第三章习题课第三章习题课 (3-3)解解:= s2+5s+4 C(s)R(s)4s(s+1)(s+4)C(s)=4R(s)=s11/3s+41s+=4/3s+1-13c(t)=1+ -4t-t43-e e 3-4 已知单位负反馈系统的开环传递函已知单位负反馈系统的开环传递函数，求系统的上升时间数，求系统的上升时间tr、峰值时间、峰值时间tp、超调量超调量% % 和调整时间和调整时间ts。1s(s+1)G(s)=第三章习题课第三章习题课 (3-4)解解:=s2+s+1 C(s)R(s)12 = 1n 2n =

33、1=0.5=1n =0.866d = n2 1- =60o-1=tg 21- tr=d - = 3.14-3.14/30.866=2.42 tp=d3.140.866 = =3.63%=100%e-1-2=16%-1.8ets= 3n =6 ts= 4n =83-6 已知系统的单位阶跃响应已知系统的单位阶跃响应:c(t)=1+0.2e -60t-10t-1.2e(1) 求求系统的闭环传递函数。系统的闭环传递函数。(2) 求求系统的阻尼比和无阻尼振荡频率。系统的阻尼比和无阻尼振荡频率。第三章习题课第三章习题课 (3-6)解解:0.2s+601s+C(s)=1.2s+10-s(s+60)(s+10

34、)=600= s2+70s+600 C(s)R(s)600R(s)=s12 =600n 2n =70=1.43=24.5n 3-7 设二阶系统的单位阶跃响应曲线如图，设二阶系统的单位阶跃响应曲线如图，系统的为单位反馈，求系统的传递函数。系统的为单位反馈，求系统的传递函数。tc(t)010.11.3解解:第三章习题课第三章习题课 (3-7)tp=0.121- n =0.3e-1-2e1-2=3.3n2 1- 3.140.1=31.421- /=ln3.3 =1.1921- )2/ (=1.429.862=1.42-1.422=0.35=33.4n nn s(s+2 )G(s)=21115.6s(

35、s+22.7)=3-8 已知单位负反馈系统的开环传递函已知单位负反馈系统的开环传递函数，求系统数，求系统K、T值以满足动态指标：值以满足动态指标：%30%30%，ts0.3(5%)0.3(5%)。Ks(Ts+1)G(s)=第三章习题课第三章习题课 (3-8)解解:=Ts2+s+K C(s)R(s)Ks2+=T1s+TKTKT12n =ts= 3n 0.3n 10T0.050.3e-1-20.3528.6n 2 =n TK=817.96K40.93-11 已知闭环系统的特征方程式，试用已知闭环系统的特征方程式，试用劳斯判据判断系统的稳定性。劳斯判据判断系统的稳定性。第三章习题课第三章习题课 (3

36、-11)(1) s3+20s2+9s+100=0解：解： 劳斯表如下：劳斯表如下：s1 s0 s3 s2 1 9 20 100 4 100系统稳定。系统稳定。(3) s4+8s3+18s2+16s+5=0 1 18 5 s4 s3 8 16 劳斯表如下：劳斯表如下：s2 16 5 s1 216 16 s0 5系统稳定。系统稳定。3-12 已知单位负反馈系统的开环传递函已知单位负反馈系统的开环传递函数，试确定系统稳定时数，试确定系统稳定时K值范围。值范围。第三章习题课第三章习题课 (3-12)G(s)=s(s+1)(0.5s2+s+1)K(0.5s+1)解解:0.5s4+1.5s3+2s2+s+

37、0.5Ks+K=0 0.5 2 K s4 s3 1.5 1+0.5K s2 b31 b31= 1.5*2-0,5(1+0.5K) 1.5 3-0,5-0.25K0 0.25K003-14 已知系统结构如图，试确定系统稳已知系统结构如图，试确定系统稳定时定时值范围。值范围。第三章习题课第三章习题课 (3-14)R(sR(s) )- s+1s10s(s+1)C(s)C(s)解解:G(s)=s2(s+1)10(s+1)(s)=s3 +s2+10 s+1010(s+1)s3 s2 1 10 1 10 s1 b31 s0 10 b31= 10 -10 1 01r(t)=I(t)+2t+t23-16 已知

38、单位反馈系统的开环传递函数，已知单位反馈系统的开环传递函数，试求试求K p、Kv和和Ka 并求稳态误差并求稳态误差ess第三章习题课第三章习题课 (3-16)解：解： (0.1s+1)(0.2s+)(1) G(s)=20Kp=20 =0ess1=R01+Ks2R(s)=1s2+s32+=121K =0 ess2=Ka=0 ess3= ess=1K =10 Ka=0 ess3= ess=s(s+2)(s+10)(2) G(s)=200s(0.5s+1)(0.1s+1)=10Kp= ess1=0ess2=K 2=210s2(s2+4s+10)(3) G(s)=10(2s+1)=2Kp= ess1=

39、0s2(0.1s2+0.4s+1)=(2s+1)K = ess2=0Ka=1 ess3=2 ess=23-17 已知系统结构如图。已知系统结构如图。第三章习题课第三章习题课 (3-17)解解:-R(sR(s) )-K1ss2C(s)C(s)(1) 单位阶跃输入单位阶跃输入:G(s)=s2+K1sK1(s)=s2+K1s+K1K1确定确定K1 和和值值 。 %=20% %=20% ts =1.8(5%)2n =K12 =K1n =0.2e-1-2ts= 3n =1.8=0.45n31.8*0.45=3.72 n K1=13.7=0.24(2)求系统的稳态误差：求系统的稳态误差：r(t)=I(t)

40、, t ,12t2 解解:G(s)=s2+K1sK1=s(s+1)K111=1Kp= ess1=0R(s)=1sR(s)=s21K =K ess2=0.24R(s)=s31Ka=0 ess3=第三章习题课第三章习题课 (3-17)3-18 已知系统结构如图。为使已知系统结构如图。为使=0.7时时 单位斜坡输入的稳态误差单位斜坡输入的稳态误差ess=0.25第三章习题课第三章习题课 (3-18)解解:确定确定 K 和和值值 。-s-Ks(s+2)R(sR(s) )C(s)C(s)G(s)=s2+2s+K sK=s(s+1)2+K12+KK(s)=s2+(2+K )s+KK2n =2+K2 =Kn

41、 =2*0.7 K ess= 2+K K =0.25 = 0.25K-2 K K K=31.6=0.186+R(s)G(s)F(s)C(s)+-D1(s)D2(s)E(s)3-19 系统结构如图。系统结构如图。第三章习题课第三章习题课 (3-19)解解: r(t)=d1(t)=d2(t)=I(t)(1) 求求r(t)作下的稳态误差作下的稳态误差essr=lim s 1+G(s)F(s)s0s1= 1+G(0)F(0)1(2) 求求d1(t)和和d2(t)同时作用下的稳态误差同时作用下的稳态误差essd= lim s -F(s)1+G(s)F(s)s0-11+G(s)F(s)+s1Ed(s)=

42、-G2(s)H(s)1+G1(s)G2(s)H(s)D(s)= 1+G(0)F(0)-1+F(s)(3) 求求d1(t)作用下的稳态误差作用下的稳态误差G(s)=Kp +KsJs1F(s)=essd= lim s -F(s)1+G(s)F(s)s0s1-s0s1= lim s 1+(Kp+KsJs1)Js1=0返回返回4-1 已知系统的零、极点分布如图，大已知系统的零、极点分布如图，大致绘制出系统的根轨迹。致绘制出系统的根轨迹。第四章习题课第四章习题课 (4-1)解解:j0(1)(2)j0(3)j0(4)j0600900600第四章习题课第四章习题课 (4-1)(5)j0(6)j0j0(7)(

43、8)j0600450135036010804-2 已知开环传递函数，试用解析法绘制已知开环传递函数，试用解析法绘制出系统的根轨迹，并判断点出系统的根轨迹，并判断点(-2+j0),(0+j1),(-3+j2)是否在根轨迹上。是否在根轨迹上。第四章习题课第四章习题课 (4-2)解解:Kr(s+1)G(s)=Kr(s)=s+1+KrKr=0s=-1-Krj0系统的根轨迹系统的根轨迹 s=-1-1Kr=s=-s=-2+j0-2s=0+j10+j1-3+j2s=-3+j2j04-3 已知系统的开环传递函数，试绘制出已知系统的开环传递函数，试绘制出根轨迹图。根轨迹图。第四章习题课第四章习题课 (4-3)解

44、解:s(s+1)(s+5)(1) G(s)=Kr(s+1.5)(s+5.5) 1）开环零、极点）开环零、极点 p1 p1=0p2p2=-1p3 p3=-52）实轴上根轨迹段）实轴上根轨迹段 p1p2z1 z1=-1.5z2 z2=-5.5z1p3z2- 3）根轨迹的渐近线）根轨迹的渐近线 n-m= 1= +180o4）分离点和会合点）分离点和会合点A(s)B(s)=A(s)B(s)A(s)=s3+6s2+5sB(s)=s2+7s+8.25A(s)=3s2+12s+5B(s)=2s+7解得解得 s1=-0.63s2=-2.5s3=-3.6s4=-7.28第四章习题课第四章习题课 (4-3)s(s

45、+1)(s+4)(2) G(s)=Kr(s+1.5) 1）开环零、极点）开环零、极点 p1 p1=0p2p2=-1p3 p3=-4j02）实轴上根轨迹段）实轴上根轨迹段 p1p2z1 z1=-1.5p3z1 3）根轨迹的渐近线）根轨迹的渐近线 n-m= 2= +90o2= -1-4+1.5 =-1.75-1.754）分离点和会合点）分离点和会合点A(s)=s3+5s2+4sB(s)=s+1.5A(s)=3s2+10s+4B(s)=1解得解得 s=-0.625）系统根轨迹）系统根轨迹第四章习题课第四章习题课 (4-3)s(s+1)2(3) G(s)=Kr 1）开环零、极点）开环零、极点 p1=0

46、p2=-1p3 p3=-12）实轴上根轨迹段）实轴上根轨迹段 p1p2p3- 3）根轨迹的渐近线）根轨迹的渐近线 n-m=3j0p1 p23= -1-1 =-0.67-0.67 4）根轨迹与虚轴的交点）根轨迹与虚轴的交点= +180o+60o , 闭环特征方程为闭环特征方程为 s3+2s2+s+Kr=0Kr=0 Kr=2 2,3=11=01-15）分离点和会合点）分离点和会合点A(s)=s3+2s2+sB(s)=1A(s)=3s2+4s+1B(s)=0解得解得 s=-0.336）系统根轨迹）系统根轨迹第四章习题课第四章习题课 (4-3) 1）开环零、极点）开环零、极点 p1=0p2=-3 p3

47、=-7j02）实轴上根轨迹段）实轴上根轨迹段 p1p2p4 p4=-15z1 z1=-8p3z1p4- 3）根轨迹的渐近线）根轨迹的渐近线 n-m=3s(s+3)(s+7)(s+15)(4) G(s)=Kr(s+8)p1 p2p3 3= -3-7-15+8 =-5.67= +180o+60o , -5.67 4）根轨迹与虚轴的交点）根轨迹与虚轴的交点闭环特征方程为闭环特征方程为 s4+25s3+171s2+323s+8Kr=0Kr=0 1=0Kr=638 2,3=6.26.2-6.25）分离点和会合点）分离点和会合点A(s)=s4+25s3+171s2+315sB(s)=s+8A(s)=4s3

48、+75s2+342s+315B(s)=2s+7解得解得 s=-1.46）系统根轨迹）系统根轨迹第四章习题课第四章习题课 (4-4)4-5 已知系统的开环传递函数。已知系统的开环传递函数。(1)试绘制试绘制出根轨迹图。出根轨迹图。(2)增益增益Kr为何值时，复数特为何值时，复数特征根的实部为征根的实部为-2。第四章习题课第四章习题课 (4-5)解解:，s(s+1)Kr(s+2)G(s)=j0p1=0p1 p2=-1p2 z1=-2z1 p1p2z1-分离点和会合点分离点和会合点s2+4s+2=0s1=-3.41s2=-0.59s2+s+Krs+2Kr=0闭环特征方程式闭环特征方程式系统根轨迹系统

49、根轨迹s=-2+j(-2+j)2+(-2+j )(1+Kr)+2Kr=0-4 +(1+Kr ) =04-2-2(1+Kr )+2Kr=0Kr=3=1.414-6 已知系统的开环传递函数，试确定闭已知系统的开环传递函数，试确定闭环极点环极点=0.5时的时的Kr值。值。第四章习题课第四章习题课 (4-6)解解:p1=0 p2=-1 p3=-3p1p2p3 -8+60o= +180o, 3= -1-3=-1.3根轨迹的分离点：根轨迹的分离点：A(s)B(s)=A(s)B(s)3s2+8s+3=0s1=-0.45s2=-2.2s2没有位于根轨迹上，舍去。没有位于根轨迹上，舍去。js(s+1)(s+3)

50、KrG(s)H(s)=与虚轴交点与虚轴交点 s3+4s2+3s+Kr=02=0Kr-43+3=0-Kr=0 Kr=12 2,3=1.71=01.7-1.7p3-30p1 p2-1系统根轨迹系统根轨迹s1=0.5得得 s1=-0.37+j0.8Kr=|s3|s3+1|s3+3|s3=-4+0.372=-3.26=3.262.260.26=1.9s34-7 已知系统的开环传递函数，已知系统的开环传递函数，(1) 试绘制试绘制出根轨迹图。出根轨迹图。第四章习题课第四章习题课 (4-7)解解:p1=0 p2=-2 p3=-4p1p2p3 -8+60o= +180o, 3= -2-4=-2根轨迹的分离点

51、：根轨迹的分离点：A(s)B(s)=A(s)B(s)3s2+12s+8=0s1=-0.85s2=-3.15s2没有位于根轨迹上，舍去。没有位于根轨迹上，舍去。j(3)与虚轴交点与虚轴交点 s3+6s2+8s+Kr=02=0Kr-63+8=0-Kr=0 Kr=48 2,3=2.81=02.8-2.8p3-40p1 p2-2系统根轨迹系统根轨迹s1s3s(s+2)(s+4)KrG(s)H(s)=(2) 阻尼振荡响应的阻尼振荡响应的Kr值范围值范围s=-0.85Kr=0.851.153.15=3.1s=j2.8Kr=48 (4)=0.5s1=-0.7+j1.2s3=-6+0.72=-4.6Kr=4.

52、62.60.6=7.2第四章习题课第四章习题课 (4-8)第四章习题课第四章习题课 (4-9)第四章习题课第四章习题课 (4-10)第四章习题课第四章习题课 (4-11)第五章习题课第五章习题课 (5-1)5-1 已知单位负反馈系统开环传递函数，已知单位负反馈系统开环传递函数，当输入信号当输入信号r(t)=sin(t+30o)，试求系统的，试求系统的稳态输出。稳态输出。(s+1) G(s)=10解：解：(s+11) (s)=10 )=A( 112+( )210=0.905=112+110=12210=-5.2o ( )11 =-tg-1 111 =-tg-1 cs(t)= 0.9sin(t+2

53、4.8o)第五章习题课第五章习题课 (5-2)5-2 已知单位负反馈系统开环传递函数，已知单位负反馈系统开环传递函数， 试绘制系统开环幅相频率特性曲线试绘制系统开环幅相频率特性曲线 。s(s+5)(s+15)(1) G(s)=750 解：解：n-m=3I型系统型系统=0=)= A( -90o ( )=-270o ( )=0)= A( ReIm0=0=(2s+1)(8s+1)(3) G(s)=10 解：解：n-m=20型系统型系统=0=)=10 A( -180o ( )=0)= A( ReIm0=0=0o ( )=s(s-1)(5) G(s)=10 解：解：n-m=2I型系统型系统=0=)= A

54、( -270o ( )=-180o ( )=0)= A( ReIm0=0=第五章习题课第五章习题课 (5-2)s2(s+0.1)(s+15)(7) G(s)=10(s+0.2) 解：解：n-m=3II型系统型系统=0=)= A( -180o ( )=-270o ( )=0)= A( ReIm0=0=第五章习题课第五章习题课 (5-2)第五章习题课第五章习题课 (5-2)5-2 已知单位负反馈系统开环传递函数，已知单位负反馈系统开环传递函数， 试绘制系统开环对数频率特性曲线。试绘制系统开环对数频率特性曲线。s(s+5)(s+15)(1) G(s)=750解：解：-20020400-180 -90

55、-270 dB L( ) ( s( G(s)=1051s+1)s+1)(1511=52=15低频段曲线：低频段曲线：-20dB/dec20lgK=20dB15-40dB/dec15-60dB/dec=0=-90o ( )=-270o ( )=相频特性曲线：相频特性曲线：第五章习题课第五章习题课 (5-2)(2s+1)(8s+1)(3) G(s)=10解：解：-200200-180 -90dB L( ) ( 低频段曲线：低频段曲线：20lgK=20dB20lgK1=0.1250.1252=0.50.5-20dB/dec-40dB/dec相频特性曲线：相频特性曲线：=0=0o ( )=-180o

56、( )=第五章习题课第五章习题课 (5-2)s(s-1)(5) G(s)=10解：解：-20020400-180 -90-270 dB L( ) ( 低频段曲线：低频段曲线：20lgK=20dB1-20dB/dec1=1-40dB/dec=0=-270o ( )=-180o ( )=相频特性曲线：相频特性曲线：第五章习题课第五章习题课 (5-2)s2(s+0.1)(s+15)(7) G(s)=10(s+0.2)解：解：-20020400-180 -90-270 dB L( ) ( s2(10s+1)(0.67s+1)=1.33(5s+1)低频段曲线：低频段曲线：20lgK=2.5dB1-40d

57、B/dec0.11=0.1-60dB/dec2=0.20.23=15-40dB/dec15-60dB/dec=0=-180o ( )=-270o ( )=相频特性曲线：相频特性曲线：第五章习题课第五章习题课 (5-4)5-4 已知系统的开环幅频率特性曲线，已知系统的开环幅频率特性曲线， 写出传递函数并画出对数相频特性曲线。写出传递函数并画出对数相频特性曲线。20lgK0-20dB/decdB L( )c20(a)1020lgK=20K=1010G(s)=(0.1s+1)020dB/decdB L( )-20(b)2020lgK=-20K=0.10.1sG(s)=(0.05s+1)101第五章习

58、题课第五章习题课 (5-4)(c)0.01-20dB/dec-60dB/dec-40dB/decdB L( )1000s100G(s)=(100s+1)K=100(0.01s+1)(d)20lgK-40dB/dec0-20dB/decdB L( )48110100-60dB/dec5020lgK=48K=251251G(s)=(s+1)(0.1s+1)(0.01s+1)第五章习题课第五章习题课 (5-4)100-20dB/dec-60dB/dec4.58dBdB L( )=45.3r(e)0由图可得：由图可得：20lgMr=4.58dBMr=1.7得：得：=11- 2 2 1=0.942=0.32得得=0.3r =1-2 2 n根据根据得得=50n由频率曲线得由频率曲线得s100G(s)=(0.02s)2+0.01s+1)=1000K=2T =0.011)2 T2=(=0.022n第五章习题课第五章习题课 (5-7)5-7 已知奈氏曲线，已知奈氏曲线，p为不稳定极点个数，为