2022年《自动控制原理》黄坚课后习题答案 .pdf

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1、2-1试建立图所示电路的动态微分方程+C-uiuoR1R2i1ii2+C-uiuoR1R2i1ii2Lu1解：u1=ui-uoi2=Cdu1dti1=i-i2uoi=R2u1i1=R1=ui-uoR1dtd(ui-uo)=C(a)uCd(ui-uo)dtuo-R2=i-uoR1i=i1+i2i2=Cdu1dtuoi1=R2u1-uo=LR2duodtR1i=(ui-u1)(b)解：)-R2(ui-uo )=R1u0-CR1R2(duidtdtduoCR1R2duodtduidt+R1uo+R2u0=CR1R2+R2uiu=R1i-u1uo+CR2du1dtu1=uo+LR2duodtuduod

2、tR1R2Lduodt+ CLR2d2uodt2=-iR1uoR1uoR2+C)uoR1R2Lduodt) CLR2d2uodt2=+(uiR11R11R2+(C+2-2 求下列函数的拉氏变换。(1) f(t)=sin4t+cos4tLsint= 2+s2=s+4s2+16Lsin4t+cos4t = 4s2+16ss2+16+s2+s2Lcost=解：精品资料 - - - 欢迎下载 - - - - - - - - - - - 欢迎下载 名师归纳 - - - - - - - - - -第 1 页，共 19 页 - - - - - - - - - - (2) f(t)=t3+e4t解：Lt3+e

3、4t= 3!s41s-4+ 6s+24+s4s4(s+4)= (3) f(t)=tneatLtneat=n!(s-a)n+1解：(4) f(t)=(t-1)2e2tL(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=-2=-1=2f(t)=2e-3t-e-2t(1) F(s)=s+1(s+2)(s+3)解：A2=(s+3)s+1(s+2)(s+3)s=-3F(s)= 2s+31s+2-= A1s+2 s+3+ A2(2) F(s)=s(s+1)2(s+2)f(t)=-2e-2t-te-t+2e-t解：= A2s+1

4、s+2+ A3+ A1(s+1)2A1=(s+1)2s(s+1)2(s+2)s=-1A3=(s+2)s(s+1)2(s+2)s=-2ddsss+2A2= s=-1=-1=2=-2(3) F(s)=2s2-5s+1s(s2+1)F(s)(s2+1)s=+j=A1s+A2s=+jA2=-5A3=F(s)ss=0f(t)=1+cost-5sint解：= s+ A3s2+1A1s+A2=12ss 2-5s+1=A1s+A2 s=js=jj -2-5j+1=jA1+A2 -5j-1=-A1+jA2 A1=1F(s)= 1ss2+1s-5s2+1+(4) F(s)=s+2s(s+1)2(s+3)解：=+s

5、+1A1s+3A2(s+1)2+sA3+A4-12A1= 23A3= 112A4= A2= ds=-1ds(s+2)s(s+3) -34= -34A2= +-43+f(t)=e-t32e-3t2-te-t121= s=-1 s(s+3)2s(s+3)-(s+2)(2s+3) 精品资料 - - - 欢迎下载 - - - - - - - - - - - 欢迎下载 名师归纳 - - - - - - - - - -第 2 页，共 19 页 - - - - - - - - - - (2-4) 求解下列微分方程。y(0)=y(0)=2 +6y(t)=6+5d2y(t)dt2dy(t)dt(1)解：s2Y(

6、s)-sy(0)-y(0)+5sY(s)-5y(0)+6Y(s)= 6sA1=1y(t)=1+5e-2t-4e-3tA2=5 A3=-4Y(s)=6+2s2+12ss(s2+5s+6)= A1s+2 s+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+-R2R1ci1解：I2(s)I1(s)+Uc(s)Ur(s)_Cs1R1+R2Uc(s)I(s)( Ur(s)Uc(s)=1R11+(+sC)R21R1+sC)R2=R2+R1R2sCR1+R2+R1R2sC(2)+C-urucR1R2Lu1I(s)Ur(s)_1R1U1(s)解：I1(s)-I2

7、(s)L31CsU1(s)Uc(s)-1LsR2I1(s)Uc(s)L1L2L1=-R2 /Ls L2=-/LCs2L3=-1/sCR11=1L1L3=R2/LCR1s2P1=R2/LCR1s2=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)t0TK(t)c(t)t0TK(t)解:Kt-Tc(t)=T(t-T)KC(s)=TsK(1-e )2-TSC(s)=G(s)精品资料 - - - 欢迎下载 - - - - - - - - - - - 欢迎下载 名师归纳 - - - - - - - -

8、 - -第 3 页，共 19 页 - - - - - - - - - - 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-tR(s)=1s解:G(s)=C(s)/R(s)s+21-1sC(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)c(t)= (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)

9、-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)G1G2G3G5-C(s)-R(s)G4G6G8G7C(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-G8G1+G3G2G63G4C1+G3G2G6 +G3G4G5+G1G2G3G4(G7 -G8)G1G2G3G4R(s)(s)=2-11求系统的传递函数(a)G1(s)G2(s)G3(s)H1(s)_+R

10、(s)C(s)H2(s)解: L1=-G2H1L2=-G1G2H2P1=G1G2P2=G3G21 =12 =1R(s)C(s)=nk=1Pkk=1+G2H1+G1G2H21+G2H1+G1G2H2G2G1+G2G3=(b)G1(s)G2(s)G3(s)G4(s)_+R(s)C(s)H(s)解:R(s)C(s)=1+G1G2H+G1G4HG1G2+G2G3+G1G2G3G4 HL1=-G1G2H L2=-G1G4HP1=G1G21 =1P2=G3G2=1+G1G4H+G1G2H2=1+G1G4HH1_+G1+C(s)R(s)G3G2(c)H1_+G1+C(s)R(s)G3G2H1C(s)R(s)

11、1+G1G2+G1H1 G3H1G1G2(1 G3H1)=精品资料 - - - 欢迎下载 - - - - - - - - - - - 欢迎下载 名师归纳 - - - - - - - - - -第 4 页，共 19 页 - - - - - - - - - - H_G1+C(s)R(s)G2(d) 解: L1=-G2HP1=G11 =1P2=G22 =11+G2H1(G1+G2 )R(s)C(s)=-_G1+C(s)R(s)G2G3G4(e)解: L2=G1G4L3=-G2G3L4=G2G4L1=-G1G3P1=G11=1 P2=G22=11+G1G3+G2G3 G1G4-G2G4=(G1+G2)

12、C(s)R(s)_G1+C(s)R(s)G2(f) L1L2解: L1=-G1G2L2=G2P1=G11=1-G2=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)L1L2L1=G2H2L2=-G1G2H31=1P1=G1G21-G2H2+G1G2H3G2G1=R(s)C(s)P1=G21=1 P2=-G1G2H12=11-G2H2+G1G2H3G2(1-G1H1 )=D(s)C(s)(b)C(s)R(s)G1G2H_解: L1=-G1G2L2=-G1G2H1=1P1=G1

13、G21+G1G2H+G1G2G1G2=R(s)C(s)P1=GnG21=1P2=12=1+G1G2HD(s)C(s)1+G1G2+G1G2H=1+GnG2+G1G2H2-13 (a)C(s)E(s)G1G2G3_+R(s)L1L2解: L1=-G2L2=-G1G2G3P1=G2G3P2=G1G2G3R(s)C(s)=1+G2+G1G2G3G2G3+G1G2G31=12=1P1=-G2G32=1+G21=1P2=1R(s)C(s)=1+G2+G1G2G3-G2G3+1+G2(b)C(s)E(s)G1G2-+R(s)G3G4G5解: L1=-G3G4L2=-G2G3G51=1P1=G1G5P1=G

14、1G51=1P2=12=1+G3G4P2=G2G3G52=11+G2G3G5+G3G4=R(s)C(s)G1G2G5+G1G51+G2G3G5+G3G4=R(s)E(s)G1G5+(1+G1G5 )精品资料 - - - 欢迎下载 - - - - - - - - - - - 欢迎下载 名师归纳 - - - - - - - - - -第 5 页，共 19 页 - - - - - - - - - - E(s)G(s)X(s)R(s)G4(s)+C(s)1(s)G2(s)-+G3D(s)2-14C(s)R(s)=1+G3(G1+G2)(G1+G2)(G3+G4)解:L1=-G1G3L2=-G2G31=

15、1P1=G1G3P2=G2G32=1P3=G1G43=1P4=G2G44=1E(s)R(s)=1+G3(G1+G2)1(CD s)(s)=1=G2(s)E(s)X(s)G3C1(s)R1(s)+G1G2+-H2H1G4G5-(s)G6C2R2(s)2-15解: L1=G1G2L3=-G4L2=-G1G4G5H1H2P1=G1G2G3=1-G1G2+G1G4G35H1H2+G4 -G1G2G41=1+G41+G4+G1G4G5H1H2-G1G2-G1G2G4G1G2G3(1+G4 )=C1(s)R1(s)1+G4+G1G4G5H1H2-G1G2-G1G2G4G4G5G6(1-G1G2)=C2(s

16、)R2(s)1+G4+G1G4G5H1H2-G1G2-G1G2G4-G1G2G3G4G5H1=C1(s)R2(s)1+G4+G1G4G5H1H2-G1G2-G1G2G4G1G4G5G6H2=C2(s)R1(s)解:c(t)=c()98%t=4T=1 minr(t)=10te(t)=r(t)-c(t)c(t)=10(t-T+ e)-t/T=10(T-e)-t/Tess=limte(t)=10T =2.5T=0.25-+R1R0Curuc解:R1Cs+1R1/R0G(s)= uc(t)=K(1e tT-)KTs+ 1=T=R1C=0.5 K=R1/R0=10 =10(1e -2t)8=10(1e

17、-2t)0.8=1e -2te -2t=0.2 t=0.83-1 设温度计需要在一分钟内指示出响应值的98%，并且假设温度计为一阶系统，求时间常数 T。如果将温度计放在澡盆内， 澡盆的温度以 10oC/min 的速度线性变化，3-2 电路如图，设系统初始状态为零。（1）求系统的单位阶跃响应 ,及 uc(t1)=8 时的 t1值R0=20 kR1=200 kC=2.5F (2) 求系统的单位脉冲响应，单位斜坡响应,及单位抛物响应在 t1时刻的值精品资料 - - - 欢迎下载 - - - - - - - - - - - 欢迎下载 名师归纳 - - - - - - - - - -第 6 页，共 19

18、 页 - - - - - - - - - - g(t)=e-t/TTKt1=0.8=4解:uc(t)=K(t-T+Te-t/T)=4R(s)=1s2R(s)=1R(s)=1s3ss+1/T+T2=K(Ts2-1s3-T2)=1.2Ts1s3K+ 1Uc(s)= -0.5t+0.25-0.25e-2t)12t2uc(t)=10( 4s(s+5)G(s)=解:R(s)=s2+5s+4C(s)4s(s+1)(s+4)C(s)=4R(s)=s1s+41+1/3s=4/3s+1-13c(t)=1+ -4t-t43-e e G(s)=1s(s+1)解:C(s)=s2+s+1R(s)12= 1n 2 n =

19、1=0.5 =1n =0.866d= n2 1-=60o=tg-121- tr=d -= 3.14-3.14/30.866=2.42tp=d3.140.866= =3.63%=100%e-1-2=16%-1.8ets= 3n =6ts= 4n =83-6 已知系统的单位阶跃响应:c(t)=1+0.2e -60t-10t-1.2e(1) 求系统的闭环传递函数。(2) 求系统的阻尼比和无阻尼振荡频率。解:s+60+0.21sC(s)=1.2s+10-s(s+60)(s+10)=600=s2+70s+600C(s)R(s)600R(s)=s12=600n 2n =70=1.43 =24.5n 1.3

20、tc(t)010.1解: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 s(s+2 nn )G(s)=21115.6s(s+22.7)=3-3 已知单位负反馈系统的开环传递函数， 求系统3-4 已知单位负反馈系统的开环传递函数，求系统的上升时间 t 、峰值时间 t 、超调量 % 和调整时间 t。3-7 设二阶系统的单位阶跃响应曲线如图，系统的为单位反3-11 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。精品资料 - - -

21、 欢迎下载 - - - - - - - - - - - 欢迎下载 名师归纳 - - - - - - - - - -第 7 页，共 19 页 - - - - - - - - - - (1) s3+20s2+9s+100=0解：劳斯表如下：s1s0 s3 s2 1 9 20 100 4100系统稳定。(3) s4+8s3+18s2+16s+5=0118 5 s4 s3 8 16 劳斯表如下：s2 16 5 s121616s0 5系统稳定。G(s)=s(s+1)(0.5s2+s+1)K(0.5s+1)解: s4+3s3+4s2+2s+Ks+2K=014 2K s4 s3 3 2+K s2 b31b3

22、1= 10-2K3 (K-1.7)(K+11.7)0K00R(s)-s+1s10s(s+1)C(s)解: G(s)=s2(s+1)10(s+1)(s)=s3 +s2+10 s+1010(s+1)s3 s2 1 10110 s1 b31s0 10b31= 10 -10 101r(t)=I(t)+2t+t2s2R(s)=1s2+s32+3-12 已知单位负反馈系统的开环传递函3-13 已知系统结构如图， 试确定系统稳定时 值范围。3-14 已知系统结构如图，试确定系3-16 已知单位反馈系统的开环传递函数，精品资料 - - - 欢迎下载 - - - - - - - - - - - 欢迎下载 名师归

23、纳 - - - - - - - - - -第 8 页，共 19 页 - - - - - - - - - - 解：(0.1s+1)(0.2s+)(1) G(s)=20Kp=20=0ess1=R01+K=211K =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=K2=10210(2s+1)s2(s2+4s+10)(3) G(s)=2Kp=ess1=0s2(0.1s2+0.4s+1)=(2s+1)K =ess2=0Ka=1ess3=2ess=23-

24、17 已知系统结构如图。-R(s)-K1ss2C(s)(1) 单位阶跃输入 :确定 K1 和值 。%=20%ts=1.8(5%)解:sKG(s)=s2+K11(s)=s2+K1s+K1K12 n =K12=K1n =0.2e- 1-2ts= 3n =1.8=0.45n31.8*0.45=3.72n K1=13.7=0.24(2) 求系统的稳态误差：1tr(t)=I(t), t ,22解: G(s)=s2+K1sK1=s(s+1)K111=1Kp=ess1=0R(s)=1sR(s)=s21=K K ess2= =0.24R(s)=s31Ka=0 ess3=-s-Ks(s+2)R(s)C(s)解:

25、KG(s)=s2+2s+K s=s(s+1)2+K12+KK(s)=s2+(2+K )s+KK2n =2+K 2=Kn =2*0.7 Kess= 2+K K=0.25= 0.25K-2 KK=31.6=0.1863-18 已知系统结构如图。为使 =0.7 时,单位斜坡输入的稳态误差ess=0.25 确定 K和 值 。精品资料 - - - 欢迎下载 - - - - - - - - - - - 欢迎下载 名师归纳 - - - - - - - - - -第 9 页，共 19 页 - - - - - - - - - - +R(s)G(s)F(s)C(s)+-D1(s)D2(s)E(s)3-19 系统结

26、构如图。r(t)=d1(t)=d2(t)=I(t)(1) 求r(t)作下的稳态误差解: essr=lim s 1+G(s)F(s)s0s1=1+G(0)F(0)1(2) 求d1(t)和d2(t)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)= -G2(s)H(s)1+G1(s)G2(s)H(s) D(s)=1+G(0)F(0)-1+F(s)(3) 求d1(t)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)J

27、s1=0解:j0(1)(2)j0600(3)j0(4)j0900600(5)j0(6)j0600j0(7)(8)j045013503601080Kr(s+1)G(s)=解:Kr(s)=s+1+KrKr=0s=-1-Krj0s=-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，试绘制出根轨迹图。解:s(s+1)(s+5)(1) G(s)=Kr(s+1.5)(s+5.5)p1p2p3z1z24-1 已知系统的零、极点分布如图，大致绘制出系统的根轨迹。4-2 已知开环传递函数，试用解析法绘制出系统的根轨精品资料 - - - 欢迎下载

28、 - - - - - - - - - - - 欢迎下载 名师归纳 - - - - - - - - - -第 10 页，共 19 页 - - - - - - - - - - 1）开环零、极点p1=0p2=-1 p3=-52）实轴上根轨迹段p1p2z1=-1.5z2=-5.5z1p3z2-3）根轨迹的渐近线n-m= 1= +180o4）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+7s1=-0.63 s2=-2.5s3=-3.6s4=-7.28Ks(s+1)(s+4)(2) G(s)=r(s+1

29、.5)1）开环零、极点p1=0p2=-1p3=-42）实轴上根轨迹段p1p2z1=-1.5p3z13）根轨迹的渐近线n-m= 2= +90o2=-1-4+1.5=-1.75p1p2p3j0z1-1.754）分离点和会合点A(s)=s3+5s2+4sB(s)=s+1.5A(s)=3s2+10s+4B(s)=1s=-0.62Ks(s+1)2(3) G(s)=r1）开环零、极点p1=0p2=-1p3=-12）实轴上根轨迹段p1p2p3-3）根轨迹的渐近线n-m=33=-1-1=-0.674）根轨迹与虚轴的交点= +180o+60o, s3+2s2+s+Kr=0Kr=0 Kr=2 2,3=11=0p3

30、j0p1p2-0.671-15）分离点和会合点A(s)=s3+2s2+sB(s)=1A(s)=3s2+4s+1B(s)=0s=-0.33精品资料 - - - 欢迎下载 - - - - - - - - - - - 欢迎下载 名师归纳 - - - - - - - - - -第 11 页，共 19 页 - - - - - - - - - - 1）开环零、极点p1=0p2=-3p3=-72）实轴上根轨迹段p1p2p4=-15 z1=-8p3z1p4-3）根轨迹的渐近线n-m=3s(s+3)(s+7)(s+15)(4) G(s)=Kr(s+8)3=-3-7-15+8=-5.67= +180o+60o,

31、4）根轨迹与虚轴的交点s4+25s3+171s2+323s+8Kr=0Kr=0 1=0Kr=638 2,3=6.2j0p4z1p1p2p3-5.676.2-6.25）分离点和会合点A(s)=s4+25s3+171s2+315sB(s)=s+8A(s)=4s3+75s2+342s+315B(s)=2s+7s=-1.4解:K，s(s+1)r(s+2)G(s)=p1=0p2=-1z1=-2p1p2z1-分离点和会合点s2+4s+2=0s1=-3.41 s2=-0.59s2+s+Krs+2Kr=0闭环特征方程式s=-2+j=0(-2+j )2+(-2+j)(1+Kr)+2Krj0p1p2z1-4 +(

32、1+Kr ) =04-2-2(1+Kr )+2Kr=0Kr=3=1.41解:p1=0 p2=-1 p3=-3p1p28p3 -+60o= +180o, 3= -1-3=-1.3根轨迹的分离点：A(s)B(s)=A(s)B(s)3s2+8s+3=0s1=-0.45s2=-2.2舍去s(s+1)(s+3)KrG(s)H(s)=与虚轴交点s3+4s2+3s+Kr=02=0Kr-43+3 =0-Kr=0 Kr=12 2,3=1.71=0(1)j1.7-1.7p3-30p1p2-1s1=0.5 得 s1=-0.37+j0.8Kr=|s3|s3+1|s3+3|s3=-4+0.372=-3.26=3.262

33、.260.26=1.9s34-5 已知系统的开环传递函数。(1)试绘制出根轨迹图。 (2)增益 Kr为何值时，复数特征根的实部为 -2。4-6 已知系统的开环传递函数，试确定闭环极点 =0.5 时的 Kr值。精品资料 - - - 欢迎下载 - - - - - - - - - - - 欢迎下载 名师归纳 - - - - - - - - - -第 12 页，共 19 页 - - - - - - - - - - s(s+3)(sG(s)H(s)=2+2s+2)Kr(2)解：p3.4=-1jp2=-3p1=0p1p24=-3-1-1=-1.25 = +135o+45o, 根轨迹的出射角3=+1-2-4

34、-=+-135o -90o -26.6o=-71.6o与虚轴的交点s(s+3)(s2+2s+2)+Kr=0s4+5s3+8s2+6s+Kr=0(j)4+5(j)3+8(j)2+j6+Kr=04-8 2+Kr=0-5 3+6 =0Kr=0 1=0Kr=8.16 2,3=1.1j0-1.2p1p2p3p413526.690-71.61.1-1.1分离点和会合点4s3+15s2+16s+6=0解得s=-2.3-2.3=0.5s1得s1=-0.36+j0.75Kr=|s1|s1+3|s1+1+j|s1+1-j|=2.92 4-7 已知系统的开环传递函数，(1) 试绘制出根轨迹图。解:p1=0 p2=-

35、2 p3=-4p1p28p3 -+60o= +180o, 3= -2-4=-2根轨迹的分离点：A(s)B(s)=A(s)B(s)s1=-0.85s2=-3.15舍去(3)与虚轴交点s3+6s2+8s+Kr=0s(s+2)(s+4)KrG(s)H(s)=(2) 阻尼振荡响应的 Kr值范围s=-0.85Kr=0.851.153.15=3.1s=j2.8Kr=48 j2=0Kr-6-3+8 =0Kr=0 Kr=48 2,3=2.81=02.8-2.8p3-40p1p2-2s1s3(4)=0.5s1=-0.7+j1.2s3=-6+0.72=-4.6Kr=4.62.60.6=7.2(s+1)G(s)=1

36、0(1) 解：(s+11)(s)=10A(10)=112+( )2=0.905=112+110=12210=-5.2o( )11 =-tg-1111 =-tg-1cs(t)= 0.9sin(t+24.8o)5-1已知单位负反馈系统开环传递函数，当输入信号r(t)=sin(t+30o)，试求系统的稳态输出。5-2 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。精品资料 - - - 欢迎下载 - - - - - - - - - - - 欢迎下载 名师归纳 - - - - - - - - - -第 13 页，共 19 页 - - - - - - - - - - s(s+5)(s+

37、15)(1) G(s)=750解：n-m=3I型系统=0=A( )=-90)=-90o( )=-270)=-270o( )=0)= A( ReIm0=0=(2s+1)(8s+1)(3) G(s)=10解：n-m=20型系统=0=A( )=10 -180)=-180o( )=0)= A( ReIm0=0=0)=0o( )=解：s(s-1)(5) G(s)=10解：n-m=2I型系统=0=)=A( -270)=-270o( )=-180)=-180o( )=0)= A(ReIm0=0=s2(s+0.1)(s+15)(7) G(s)=10(s+0.2)解：n-m=3II 型系统=0=A()=-180

38、)=-180o( )=-270)=-270o( )=0A()= ReIm0=0=s(s+5)(s+15)(1) G(s)=750解：-2002040-180 0-90-270 dBL( )()s(G(s)=1051s+1)s+1)(1511=52=15-20dB/dec20lgK=20dB15-40dB/dec15-60dB/dec=0=-90)=-90o( )=-270)=-270o( )=(2s+1)(8s+1)(3) G(s)=10解：20-2000-180 -90L(dB)()20lgK=20dB20lgK1=0.1250.1252=0.50.5-20dB/dec-40dB/dec=0

39、=0)=0o( )=-180)=-180o( )=s(s-1)(5) G(s)=10解：20-20040-180 0-90-270 L( dB)()20lgK=20dB1-20dB/dec1=1-40dB/dec=0=-270)=-270o( )=-180)=-180o( )=5-2 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。精品资料 - - - 欢迎下载 - - - - - - - - - - - 欢迎下载 名师归纳 - - - - - - - - - -第 14 页，共 19 页 - - - - - - - - - - s2(s+0.1)(s+15)(7) G(s)

40、=10(s+0.2)解：20-20040-180 0-90-270 L(dB )()s2(10s+1)(0.67s+1)=1.33(5s+1)20lgK=2.5dB1-40dB/dec0.11=0.1-60dB/dec2=0.20.23=15-40dB/dec15-60dB/dec=0=-180)=-180o()=-270( )=-270o)=20lgK0-20dB/decL(dB)c20(a)1020lgK=20K=1010G(s)=(0.1s+1)020dB/decdBL()-20(b)2020lgK=-20K=0.10.1sG(s)=(0.05s+1)101(d)20lgK-40dB/d

41、ec0-20dB/decL( dB)48110100-60dB/dec5020lgK=48K=251251G(s)=(s+1)(0.1s+1)(0.01s+1)(c)0.01-20dB/dec-60dB/dec-40dB/decL( dB)1000s100G(s)=(100s+1)K=100(0.01s+1)(c)0.01-20dB/dec-60dB/dec-40dB/decL(dB)1000s100G(s)=(100s+1)K=100(0.01s+1)100-20dB/dec-60dB/dec4.58dBdBL()=45.3r(e)0由图可得：20lgMr=4.58dBMr=1.7得：1=1

42、- 2 2 1=0.942=0.32=0.3r =1-22 n =50ns100G(s)=(0.02s)2+0.01s+1)=1000K=2T =0.011)2T2=(=0.022np=0-1(a)ReIm0=0=0系统不稳定p=0-1(b)ReIm0=0=2=0+系统稳定5-4 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。5-7 已知奈氏曲线， p为不稳定极点个数， 为积分环节个数，试判别系统稳定性。精品资料 - - - 欢迎下载 - - - - - - - - - - - 欢迎下载 名师归纳 - - - - - - - - - -第 15 页，共 19 页 - - -

43、 - - - - - - - p=0-1(c)ReIm0=0=2=0+系统不稳定p=0-1(d)ReIm0=0=3=0+系统稳定p=0-1(e)ReIm0=0=1=0+系统稳定p=1-1(f)ReIm0=0=0系统稳定p=0-1(g)ReIm0=0=1=0+系统稳定p=1-1(h)ReIm0=0=0系统不稳定解：s10G(s)=(10s+1)K=10(0.05s+1)c101012=1c=180o-90o-tg-110-tg-10.05=180o+)( c=90o-84.3o-2.9o= 2.8o0.120-20dB/dec-60dB/dec-40dB/decL(dB)c0100-180 -9

44、0()51R(s)-20.5s+1s(0.02s+1)C(s)5-17 已知系统开环幅频率特性曲线(1)写出传递函数。 (2)利用相位裕量判断稳定性 (3)将对数幅频特性向右平移十倍频程，讨论对系统性能的影响。5-18 已知系统结构， 试绘制系统的开环对数幅频特性曲线，并计算相角稳定裕量。精品资料 - - - 欢迎下载 - - - - - - - - - - - 欢迎下载 名师归纳 - - - - - - - - - -第 16 页，共 19 页 - - - - - - - - - - s(0.5s+1)(0.02s+1)G(s)=10解：c0.51012=4.5c-tg-10.024.5=1

45、80o-90o-tg-1 0.54.5=90o-66o-2.6o= 21.4o20-20040L(dB)-20dB/dec12-40dB/dec50-60dB/deccG0(s)=s(s+5)500Ko解： K =Kv=100G0(s)=s(0.2s+1)10020lgK=40dB=12.6取=5.6o=45o 12.6o+5.6o =381+sinm1sinma=m=-+=4.2c0.210012=22.4c90-900-180-2002040L05-20dB/dec-40dB/dec408224LcLL(dB )()c0cc=6.2dBaT=0.04Gc(s)=1+0.04s1+0.01s

46、G(s)=G0(s)Gc(s)=10lgaL0( m= m=40ca=82= 2maT1=23.8= 1821T=0.01G0(s)=s(0.5s+1)(0.2s+1)K解：K=Kv=1010G0(s)=s(0.5s+1)(0.2s+1)20lg10=20dB=-18=-180o+= -180o+50o+10o=-120oc0.51012=4.5c040-2020-270-180-900L0250.5260.0050.1LcLL(dB )( ccc06-1 已知单位负反馈系统开环传递函数，采用超前校正满足 K =100, 45o。6-5 已知单位负反馈系统开环传递函数，试设计串联校正装置以满足

47、K =10, 50o。精品资料 - - - 欢迎下载 - - - - - - - - - - - 欢迎下载 名师归纳 - - - - - - - - - -第 17 页，共 19 页 - - - - - - - - - - =0.05=0.11 = 15Tc21= 2=0.005取Gc(s)=1+Ts1+ Ts=200s+110s+1G(s)=G0(s)Gc(s)=26dBL0( c=-20lg 取=0.5c解：(a) G0(s)=s(0.1s+1)20Gc(s)=10s+1s+120(s+1)G(s)=s(0.1s+1)(10s+1)-2002040L010200.11LcLdBL()c(b

48、)G0(s)=s(0.1s+1)20Gc(s)=0.01s+10.1s+1G(s)=s(0.01s+1)20-20020L010-20dB/dec100LcLL(dB)206-12 已知系统G0(s) 和校正装置Gc(s)的对数频率特性曲线，要求绘制校正后系统的对数频率特性曲线，并写出开环传递函数。精品资料 - - - 欢迎下载 - - - - - - - - - - - 欢迎下载 名师归纳 - - - - - - - - - -第 18 页，共 19 页 - - - - - - - - - - 文档编码：KDHSIBDSUFVBSUDHSIDHSIBF-SDSD587FCDCVDCJUH 欢迎下载 精美文档欢迎下载 精品资料 - - - 欢迎下载 - - - - - - - - - - - 欢迎下载 名师归纳 - - - - - - - - - -第 19 页，共 19 页 - - - - - - - - - -