《自动控制原理》黄坚课后习题答案.docx

1、2-1 试建立图所示电路的动态微分方程ui1R1 R-i2i+u-Ci2ou1L+CRuiR1i1uiio-(a) 解：i =i-i22-u =u -u121iouu -uui =R1= Rioi=Ro1112i =C1=Cdud(u -u )iou -uu-Rio = RoCd(u -u )dito2(b) 解：dtdt12(u -u )i=i +ii=Ri1121duduudui = Roi =C dt 1R (u -u2io)=Ru -CR1 01R2( dt i- dt o )212L duCR Rduo+Ru +Ru =CRR dui+R uu -u =Rdt o12 dt1 o2

2、01 2 dt2 i1o2u -uuduui - uo - L duo = uo+C duo+ CLd2uo Ri1 = Ro +Cdt 1RRR R dtR111 22dtRdt22L12 duoui = CLd2uo+(C+ L )duo +( 1 + 1 )u u =u +R dtRRdt2R RdtRRo1o2121 2122-2 求下列函数的拉氏变换。(1) f(t)=sin4t+cos4t解：Lsint=Lcost= s2+s22+s2Lsin4t+cos4t=4+s= s+4 s2+16s2+16 s2+16(2) f(t)=t3+e4t解：Lt3+e4t=3!+1 = 6s+2

3、4+s4s4(3) f(t)=tneat解：Ltneat=s-4n!s4(s+4)(4) f(t)=(t-1)2e2t(s-a)n+12(s-2)解：L(t-1)2e2t=e-(s-2)3s+12-3 求下列函数的拉氏反变换。(1) F(s)=AA=1 +2 (s+2)(s+3)s+2s+3解：A =(s+2)s+11(s+2)(s+3) s=-2=-1A =(s+3)s+1=22F(s)= 2(s+2)(s+3) s=-3-1f(t)=2e-3t-e-2ts+3ss+2AAA(2) F(s)=(s+1)2(s+2)=(s+11)2+ s+12 + s+23解：A =(s+1)2s=-11(s

4、+1)2(s+2) s=-1A = d s =22 dss+2s=-1A =(s+2)s=-23(s+1)2(s+2) s=-2f(t)=-2e-2t-te-t+2e-t(4) F(s)=s+2 s(s+1)2(s+3)2s2-5s+1 A s+AA AAAA12 +34 (3) F(s)= s(s2+1) = s1 +1 2 + s32解：=(s+1)2 + s+1s + s+3-121-3解： F(s)(s2+1)s=+j =A s+AA = 2A = 3A4=12A = 42s2-5s+11=A s+A2 s=+j1(s+32)2d s(s+3)ss=j12 s=jA =ds-2-5j+

5、1=jA +A-5j-1=-A +jA2s=-1j1212= s(s+3)-(s+2)(2s+3)= -3123A =1A =-5A =F(s)s s=0 =1s(s+3)2s=-14+1-5 f(t)= -t e-t-3 e-t+ 2+ 1 e-3tF(s)= s +ss2+1s2+1f(t)=1+cost-5sint24312(2-4) 求解下列微分方程。+5+6y(t)=6(1) d2y(t)dy(t)y(0)=y(0)=2dt2dt6解：s2Y(s)-sy(0)-y(0)+5sY(s)-5y(0)+6Y(s)= sY(s)=1 +6+2s2+12sAs(s2+5s+6)sA s+22

6、+A s+33A =1A =5A =-4123y(t)=1+5e-2t-4e-3t2-5试画题图所示电路的动态结构图，并求传递函数。i+1iu-2R1i Rur2(1) c解：+U (s)r_I (s)Cs 1+1+I(s)R2U (s)ccU (s)RI (s)-c1 2( 1 +sC)RU (s)R2R +R R sCUr(s)=11=R +R2 +R1 R2 sCc1+(R +sC)R121 2(2) ur-12解：U (s)r_1I(s)I (s) 1 U (s) cU (s) L1I (s)2U (s)1-R-CsL1Ls1Rc2U (s) L1I (s)1213Ru1L+c1 CR

7、u2-L =-R /Ls L =-/LCs2 L =-1/sCRL L =R /LCR s2122311 321P =R /LCR s2=1U (s)R1211U r(s)= R CLs2+(R R2C+L)s+R +Rc11 2122-8 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)。 (t)c(t)K0T解: (t)c(t)Kt0Ttc(t)=Kt- K(t-T) C(s)=K (1-e-T)S C(s)=G(s)TTTs22-9 若系统在单位阶跃输入作用时，已知初始s条件为零的条件下系统的输出响应，求系统的r(t)=I(t) c(t)=1-e-2t+e-t传递函数和

8、脉冲响应。R(s)= 1解: C(s)=111(s2+4s+2)-+=ss+2 s+1s(s+1)(s+2)(s2+4s+2)C(s)=(s+1)(s+2)G(s)=C(s)/R(s)=(s2+4s+2)(s+1)(s+2)c(t)= (t)+2e-2t-e-tR(s)-G (s)X (s)G663C(s)-GGGG1 X (s)2-3 X (s)41X (s)23C(s)G (s)-G (s)C(s)G (s)G5785-G7G2-10 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 : X (s)=R(s)G (s)-G (s)G (s)-G (s)C(s)11178=R(s

9、)-C(s)G (s)-G (s)G (s)781X (s)=G (s)X (s)-G (s)X (s)22163X (s)=G (s)X (s)-C(s)G (s)C(s)=G (s)X (s)3325438R(s)G GG G 1+G3 G4 GC(s)-1 2-3 2 6G5C(s)G G G GG -GR(s) =1+G G G +G G1G2+G3 G4G G (G -G )782-11 求系统的传递函数3 2 63 4 51 2 3 478解: L =-G H =1+G H +G G H12 12 11 2 2(a)R(s)G (s)3_G (s)+G (s)C(s)1_2H (s

10、)1H (s)L =-G G H2P =G G =11 2 211 21P nC(s)P =G G =123 22G G +G GR(s) = k=1 kk=1+G2H1+G2G3H2 121 2 2(b)R(s)G (s)3C(s)解 : L =-G G H L =-G G H P =G G =1_ G (s)+G (s)11 221 411 2112 =1+G G H+G G H P =G G =1+G G HG (s)+1 41 223 221 44C(s) G G +G G +G G G G HH(s)R(s) =1 12+G 2G 3H+G1G2 H3 41 21 4R(s)C(s)

11、_G1G2G+3+H1R(s)_GG C(s)12G3H+1+H1(c)C(s)G G (1 G H )R(s) = 1+G 1G 2+G H 3G1 H1 21 13 1R(s)GC(s)1+_G2H(d)P =G =1解: L =-G H11112C(s)P =G =1222(e)R(s)=(G +G )11+G H122R(s)GC(s)-1+G2_G3G解: L =-G GL =G GL =-G G11 321 432 3L =G G P =G =1 P =G =142 4111222C(s)(G +G )R(s) =1+G G +G G1 G2G -G G41 32 31 42 4(

12、f)解: L =-G GL =G P =G =1-GR(s)C(s)_LLG11 G22+11 2221 G1 (1 12 =1+G G -GC(s) =1 G )21 22R(s) 1+G G G1 222-12 (a)H (s)D(s)1R(s)_G (s)1+G (s) C(s)+2LLH (s)122H (s)3L =G HL =-G G H P =G G =112 221 2 311 211C(s) =G2G R(s) 1-G H +G G H2 21 2 3P =G =1 P =-G G H =112121 2 12C(s) =G (1-G H )21 1D(s) 1-G H +G

13、 G H2 21 2 3解 : L =-G GL =-G G H P =G G =111 221 211 21R(s)_ G1GC(s)2H(b)C(s)G GR(s)= 1+G G1 H2+G GP =G G1 21 2 =11n 21P =1 =1+G G H221 2=n 21 2C(s) 1+G G +G G H D(s) 1+G G +G G H1 21 2=1 C(s)解 : L =-G L =-G G GP =G G =11221 2 312 312-13 (a)P =G G G G G +G G G=2 31 2 3 R(s)E(s)_G+1_GGC(s)L23L1221 2

14、32R(s)1+G +G G GP =-G G =1P =1 =12+G 1 2 312 31222C(s)-G G +1+G R(s) = 1+G2 +3G G G221 2 3解 : L =-G G L =-G G GP =G G =113 422 3 5 G 1G G1 +5G G1E(s) 1GR(s)-G2-G+GC(s)35G4(b)P =G G G2 =1 C(s)=1 2 51 52 3 52R(s) 1+G G G +G GP =G G =1P =12 =3 1+5 G G3 411 51223 4E(s) G G +(1+G G )= 1 51 5 R(s) 1+G G G

15、 +G G2 3 53 42-14G (s)解 : P =G G =1 P =G G =1411 3122 32R(s) E(s)G (s)G (s) + C(s)P =G G =1 P =G G =1-1+331 4342 44G (s)X(s)L =-G GL =-G GC(s) (G +G )(G +G )=1234 2D(s)11 3 22 3 R(s)1+G (G +G )E(s)= 1X(s) =G (s)C(s)=1312R(s) 1+G (G +G )E(s)2D(s)312R (s)1+-GGGC (s)1123H2H1R (s)+2-GGGC (s)24562-15解 :

16、L =G GL =-G G G H HL =-G11 221 4 5 1 234 =1-G G +G G G H H +G-G G G1 21 4 35 1 241 2 4P =G G G =1+G11 2 314C (s)G G G (1+G )1R1(s) =1+G +G G G2 H3 H -G4 G -G G G141 4 5 1 21 21 2 4C (s)G G G (1-G G )C (s)-G G G G G H1R2(s) =1+G +G G4 G5 H6 H -G1 G2-G G GR (s) =1+G +G G G1 H23H 4-G5 G1-G G G241 4 5 1

17、 21 21 2 4241 4 5 1 21 21 2 4C (s)G G G G HR2(s) =1+G +G G1G4H5H6 -G2 G -G G G141 4 5 1 21 21 2 43-1 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，解: c(t)=c( )98%t=4T=1 minT=0.25r(t)=10t c(t)=10(t-T+e-t/T) =10(T- e-t/T)e(t)=r(t)-c(t)e=lime(t)=10T=2.5ss tR1u Rr0-+C+u

18、c3-2 电路如图，设系统初始状态为零。C=2.5 FR =20 kR =200 k01（1）求系统R的/R单位阶跃响应,及 uc(t1)=8 时的 t1 值解: G(s)=R C1 s+01 =KT=R C=0.5K=R /R =10Ts + 11101- t-2t8=10(1 e-2tu (t)=K(1 e T)=10(1 e)c0.8=1 e-2te-2t=0.2t=0.8(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T解:t =0.8R(s)=1g(t)=1K e-t/T=4s2R(s)= 1u (t)=K(t-T+Te-t/T)=4cR(s)= 1U (

19、s)=K1 =K(1 - T + T2 - T2)s3cTs + 1 s3s3s2ss+1/Tu (t)=10( 1 t2-0.5t+0.25-0.25e-2t)=1.2c2G(s)=43-3 已知单位负反馈系统的开环传递函数，求系统s(s+5)解: C(s)=4R(s)= 1C(s)=4= 1 + 1/3 - 4/3 R(s)s2+5s+4ss(s+1)(s+4)ss+4s+1c(t)=1+ 1 e-4t- 4 e-t33G(s)=13-4 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 。s(s+1)解C(s)12 =1 =1d = n

20、1- 2 =0.866: R(s)= s2+s+1n 2 = 1nn1- 2 =0.5 =tg-1 =60ot = - = 3.14-3.14/3=2.42t = = 3.14 =3.63r d0.866p d0.866o %=e- 1- 2100% =16%t =3 =6 t =4 =8e-1.8s ns n3-6 已知系统的单位阶跃响应: c(t)=1+0.2e-60t-1.2e-10t(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: C(s)=+-=10.21.2600ss+60 s+10 s(s+60)(s+10)R(s)= 1C(s)=600 2n=70 =

21、24.5nsR(s) s2+70s+600 2 =600n3-7设二阶系统的单位阶跃响应曲线如图，系统的为单位反 =1.43c(t)1.3100.1t解:t = =0.1 / 1- 2 =ln3.3 =1.19 n 1- 2 = 3.14 =31.4pn 1- 20.1- 1- =0.3( )2/1- 2 =1.42 =33.4e2 1- =3.39.86 2=1.42-1.42 2n 21115.6ne 2 =0.35G(s)=s(s+2n) = s(s+22.7)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。解：(1) s3+20s2+9s+100=0劳斯表如下：s319

22、s220100s14s0 100(3) s4+8s3+18s2+16s+5=0劳斯表如下：s41185s3816s2165s121616系统稳定。G(s)=s05系统稳定。K(0.5s+1)3-12 已知单位负反馈系统的s开(s环+1传)(0递.5函s2+s+1)=b解: s4+3s3+4s2+2s+Ks+2K=0s4142Ks332+Ks2 b2K31s1 b4110-2K313=bK2+10K-204110-K(K-1.7)(K+11.7)0 K0s1b31s010311+10 0R(s)- s+1ss(s+1)10C(s)3-14 已知系统结构如图，试确定系解: G(s)= 10( s+

23、1) (s)=10( s+1)s2(s+1)s3110s3 +s2+10 s+10110 -10s2110b =031s1b31s010 1r(t)=I(t)+2t+t23-16 已知单位反馈系统的开环传递函数，R(s)=122+ss2s3pK =20R1e=0 = (1) G(s)=20=0ss1 1+K21解：(0.1s+1)(0.2s+)K =0e=ss2K =0ae =sse=ss3K =e=0(2) G(s)=200=10=1pss122 s(s+2)(s+10)s(0.5s+1)(0.1s+1)K =10ess2= K =10K =0ae =sse=ss3(3) G(s)=10(2

24、s+1)=(2s+1)K =pe=0ss1s2(s2+4s+10)s2(0.1s2+0.4s+1) =2K =e=0ss2K =1ae =2sse=2ss33-17 已知系统结构如图。(1) 单位阶跃输入:R(s) - K C(s) s21%=20%ts=1.8(5%) 确定K1和值。s1(s)=KK2n=Ke- 1- 2=0.2s +Ks解: G(s)=121s2+K s+K11 2 =K1t = =1.8 =0.45 =3n3=3.7K =1s n =0.24n 1.8*0.452=13.71n(2) 求系统的稳态误差： r(t)=I(t), t ,1 t221K =K1R(s)= spe

25、=0ss11s2解: G(s)=s2+Ks =s(1 s+1)=1R(s)= 1K =Ke= =0.24ss2s31K1R(s)= 1K =0ea=ss3R(s)-s(s+2) sKC(s)3-18 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 KK解: G(s)=K=K2+K (s)=和值 。s2+2s+K ss(1s+1)s2+(2+K )s+K2+K2 =2+K =2*0.7 Ke = 2+K =0.25K =31.6nssKK 2 =Kn = 0.25K-2 =0.1863-19 系统结构如图。D (s)1D (s)2r(t)=d (t)=d (t

26、)=I(t)R(s)E(s) G(s)+ F(s) +C(s)12-(1) 求r(t)作下的稳态误差1解: e=lim ss=1ssrs01+G(s)F(s) 1+G(0)F(0)(2) 求d (t)和d (t)同时作用下的稳态误差1-2G (s)H(s)Ed(s)= 1+G (s2)G (s)H(s)D(s)1e= lim s 2-F(s)+-1 1 = -1+F(s)ssds01+G(s)F(s)1+G(s)F(s) s1+G(0)F(0)(3) 求d (t)作用下的稳态误差1KF(s)= 1G(s)=Kp + sJs- 1)e= lim s-F(s)1 = lim sJs1 =0 ssd

27、s0 1+G(s)F(s) ss0 1+(K +pK 1 ss Js4-1已知系统的零、极点分布如图，大致绘制出系统的根轨迹。解: (1)j0 (3)j(4)j60009000 (2)j0 600(5)j6000 (6)j0 (7)1350j4500 (8)1080j36004-2已知开环传递函数，试用解析法绘制出系统的根轨G(s)=K (s+1)rK解: (s)= s+1+r KrK =0s=-1s=-1-Kj-3+j20+j1-2 -10 rs=-2+j0K =r s=-rs=0+j1 s=-3+j2jp3p2pzz01 214-3 已知系统的开环传递函数， 试绘制出根轨迹图。解:K (s

28、+1.5)(s+5.5)(1) G(s)=rs(s+1)(s+5)1）开环零、极点p =0p =-1 p =-54）分离点和会合点A(s)B(s)=A(s)B(s)123z =-1.5z =-5.5212）实轴上根轨迹段p pz pz -A(s)=s3+6s2+5s B(s)=s2+7s+8.25 A(s)=3s2+12s+512132B(s)=2s+73） 根轨迹的渐近线n-m= 1 = +180os =-0.63 s =-2.5s1=-3.6s2=-7.2834jp3-1.75pp21z10K (s+1.5) (2) G(s)= s(sr+1)(s+4)1） 开环零、极点p =0p =-1

29、p =-41z =2 -1.5312） 实轴上根轨迹段p pp z12313） 根轨迹的渐近线n-m= 2 = +90o2 =-1-4+1.5 =-1.75(3) G(s)=Ks(s+r1)21）开环零、极点j1p2pp -0.6701 3-1p =0p =-1 p =-14） 分离点和会合点A(s)=s3+5s2+4sB(s)=s+1.5 A(s)=3s2+10s+4B(s)=1s=-0.621232） 实轴上根轨迹段p pp -1233） 根轨迹的渐近线n-m=3 = +60o , +180o3 =-1-1 =-0.674） 根轨迹与虚轴的交点s3+2s2+s+K =05） 分离点和会合点

30、A(s)=s3+2s2+s B(s)=1 A(s)=3s2+4s+1K =0rK =2=0r1=1B(s)=0s=-0.33r2,3j6.2p4p32ppz -5.67011-6.2K (s+8)(4) G(s)= s(s+3)(sr+7)(s+15)1） 开环零、极点p =0p =-3 p =-7 p =-15 z =-8123412） 实轴上根轨迹段p pp zp -123143） 根轨迹的渐近线n-m=3 = -3-7-15+8 =-5.67o180 = +60 , +3o4） 根轨迹与虚轴的交点s4+25s3+171s2+323s+8K =05） 分离点和会合点A(s)=s4+25s3+171s2+315sA(s)=4s3+75s2+342s+315 B(s)=s+8B(s)=2s+7K =0 r=0K =638 1rr =6.22,3s=-1.4jp21pz0 14-5 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。( )=K (s+2)G ss(rs+，1)1解:p =0p =-1 z =-21 p p 2 z -121分离点和会合点s2+4s+2=0 s =-3.41 s =-0.59-4 +(1+K) =012r闭环特征方程式4-2-2(1+Kr)+2K =0rs2+s+K s+2K =0s=-2