《自动控制原理》黄坚课后习题答案

1、2-1试建立图所示电路的动态微分方程UiU1 Li1R27 i2Uo(a)解： i1=i-i2U1=Ui-U。i1=zR1= R1i=Uodu1 Li2=C dt=Cd(Ui-Uo)dtUi-Uo Uo 厂 d(Ui-U。)RT=R2-C dt(b)解:(i=i1+i2.(Ui-ud i=.U。ROR1i2=CL dUoL U1-Uo=RLdfR2(Ui-Uo)=RiU0-CRiR2(dUi.dUo)CRiRd0dU+R1Uo+R2Uo=CR1R2dti+R2UidUo Ci d2UUi.U L dUo= Uo+C dUo. CL R1 R1 R1R2 dt = R2+C dt + R2 dt

2、2UiR1=C2 器+（。+）帶+（R/ 走）Uo2- 2求下列函数的拉氏变换。(1) f(t)=sin4t+cos4t解:Lsinwt=j-2+s2 Lco&o4 s s+4Lsin4t+cos4|l=s2+16 + s2+16=s2+16(2) f(t)=t 3+e4t解: Lt3+e4t=送+缶気瓣(3) f(t)=t neat解：Lfeal=崗E f(t)=(t-1)2e2t解：L(t-1)2e2t=e-(s-2)(s2)32-3求下列函数的拉氏反变换。(1) F(s)= (s+2)(s+3)= sA2+sA3 解: A1=(s+2)(s+2)(sl+3) s=-2=A2=(s+3)(

3、s+2)(；1+3) s=-3=2F(s)= s+3 -勇 f(t)=2e-3t-e-2t(2) F(s)= (s+1S(s+2)=(SAp+iA1+SA2 解: Ai=(s+1)2(s+i)2Ss+2)：s=-i=-1a2= dss+2s=-i=2A3=(s+2)(s+1)2ss+2) s=-2 =-2 f(t)=-2-te-t+2e-t F(s2=5s+1=As+A2+牛解：F(s)(s2+1Hs=+j=A1S+A2s=+j2s/s=j=A1s+A24=jjh+A2A1=1A2=-5 A3=F(s)ss=o=1F(s)= s+s + s+f f(t)=1+cost-5sint尸(9=53苹

4、+3) 解:=為？+缶+售+翕 A1= 21 A3=舟 A4=12 A2= 4 人2=督s=-1=s(s+3)-(s+2)(2s+3).3=s(s+3)2s=-1= 4f(t)=ye-t-ye_t+ +1：2e-3t(2-4)求解下列微分方程。(1) 豁 +5 聲+6y(t)=6 y(0)=y(0)=2 解：s2Y(s)-sy(0)-y(0)+5sY(s)-5y(0)+6Y(s)=_6s6+2s2+12s Ai A2 . A3 丫(s)=s(V+5s+6)=+ s+2+s+3A1=1A2=5A3=-4y(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。=+U

5、cUr(s) ( R1 +sC)R2R2+R1R2SCUc(s)=i+(+sC)R=R1+R2+R1R2sCr*C :R2*L+UUcL4=-R2 /Ls L2=-/LCs L3二 1/sCRq L止3=R2/LCR1s2P1=R2/LCR1s2 1=1Ur(s=R2Uc(s)= R1CLs2+(R1R 2C+L)s+RR22-8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)3 (t) c(t)0 T t解：30 Tt环)=尹(订)c(s)=TK2(1-e-TS) C(s)=G(s)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r(t)=l

6、(t) c(t)=1-e2t +传递函数和脉冲响应。R(s)=S解:c(s)=4 爲+由=s(s+4)(+2)c(s)=(s2+4sS?2)G(s)=C(s)/R(s)=(S+4(S+2)c(t)=3 (t)+2e_2td2-10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：Xi(s)=R(s)Gi(s)-Gi(s)G7(s)-G8(s)C(s) =R(s)-C(s)Gz(s)-G8(s)Gi(s) X2(S)=G2(S)Xi(S)-G6(S)X3(S)X3(S)=G3(S)X2(S)-C(S)G5(S) C(s)=G4(s)X3(s)C(S)_G1G2G3G4R(s) 1+

7、G3G2G6 +G3G4G5+G1 G2G3G4(G7 -G8)2-11求系统的传递函数解：L1=-G2H1 =1+G2H1+G1G2H2R(s)|H 1(s|-L2=-G 1 G?H 2 P=G 1G2 1 =1nP2=G3G2 2 =1C(S) _ 掐 k = G2G1+G2G3 R(s) 1+G 2H1+G1G2H 2(b)R(s)G2(s)C(s)H(s)解：l1_-g1g2h L2_-GG4H p1_g1g2 1 _1 _1+G 1 G4H+G 1G2H P?_G3G2 2_1 +G1G4HC(s) G1G2+G2G3+G1G2G3G4 HR(s)_1+G1G2H+G1G4H(c)C

8、(s)G1G2(1G3H1)R(s)_ 1+G1G2+G1H1 -G3H1解：Li=-G2H Pi=Gi 1 =1P 2=G2 2 =1 謂=(Gi+g2)i+G2h解：Li=-GiG3 L2=Gi G4L3=-G 2G3L4=G2G4 Pi=G 1 1 = 1 P2=G2 2=1C(s) =_ _ (G1+G0R(S)=1+G1G3+G2G3 J1G4-G2G4解：L1=-GtG2 L2=G 2 P1 =G1 1 = 1-G2 =1+GtG2-G2C(s)_ G1 (1 -G2)R(s)=1+G辰2L1 =G2H 2 L2=-G 1G2H 3 P1=GtG2 1 = 1C(s)G2G1R(S

9、)=1-g2h2+g1g2h3P1 =G2 1 = 1 P2 = -G 1G2H 1 2=1C(s)G2(1-G1H1 )d(s)=gh+GGH；解：L1 =-G 1G2 L2=-G 1G2 H P1=G1 G2 1 =1(b)C(s) =G1G2R(s)=1+G1G2H+G 1G2P1 =GnG2 1 = 1 P2=1C(s) 1+GnG2+G1G2HD(s)=1+G1G2+G1G2H解：L1=-G2 L2=-GjG2G3 P1=G2G3 1=12-13 (a)P2=GjG2G3 2=1P1=-G2G3 1=1C(s) = G2G3+G1G2G3R(s) = 1+G2+G1G2G3 P2=1

10、 2=1+G2C(s)= -G2G3+1+G2R(s) = 1+G2+G1G2G3解： = -4 L2 = -Gb3G5 P1=GG 1 = 1P2=G2G3G5 2=1 3= G1G2G5 + GG5R(s) 1+G2G3G5+G3G4 P1 =G1 G5 1=1P2 = 1E(s) = GjG5+(1+G1G5 )R(s) =1+626365+6364Li=-GiG3R|=1+G3(&+G2)解：Pi=GiG3 1=1 P2=G2G3 2=1P3=G1G4 3=1 P4=G2G4 4=1L 2=-G 2G 3C(s)= (G1+G2)(G3+G4)R(s)= 1+G3(G1+G2)X(S)

11、=G2(S)鴿=1H1解：l1=g1g2 l2=-g1g4g5h1h2 l3=-g4囱IL：叵輕A =1-G 1G2+G1G4G35H1H 2+G4 -G1G2G4P1=GtG2G3 A 1=1 +G4G(S)=G1G2G3(1+G4 )R1(s) =1+G4+G 1G4G5H1H 2-G1G2-G1G2G4C2(s)GaGsGsH-GhG?) S(s)-G1G2G3G4G5H1R2(s) =1+G 4+G 1G4G5H 1H 2-G1 G2-G1G2G4 R2(s)1+G4+G 1G4G5H 1H 2-G1G2-G 1G2G4C2(s) G1G4G5G6H 2R1(s) -1+G4+G1G4

12、G5H 1H2-G 1G2-G1G2G43-1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10C/min的速度线性变化, 解：C(t)=O()98%t=4T=1 min T=0.25r(t)=10tc(t)=10(t-T+ e-t/T) =10(Te-t/T)e(t)=r(t)-c(t)ess=limie(t)=10T =2.53-2电路如图，设系统初始状态为零C=2.5 卩 FRo=20 kQR1=200 kQ(1 )求系统的单位阶跃响应，及uc(ti)=8时的ti值.解：G(s)= RRCs+1 =TsK1 T

13、=R1C=05 K=rJR=10Uc(t)=K(4- *)=10(1 -e2t)8=10(1 -e2t)0.8=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ti时刻的值.解：t1=0.8 r(s)=1g(t)=e-t/T=4R(s)=S2Uc(t)=K(t-T+T et/T)=4r(s)=S3uc(s)=T=k( i-t+：S-+17T)Uc(t)=10( 212-0 5t+0.25-0.25e-2t)=1.23-3已知单位负反馈系统的开环传递函数，求系G(s)=S(s+5)解: C(s)= 4 r(S)= 1 c(S)=4+ 1/3 4/

14、3解:R(s) s2+5s+4 民(s)= s C(S)=s(s+1)(s+4)= s + s+4 - s+ 1 c(t)=1 + 彳 e-# e-t13- 4已知单位负反馈系统的开环传递函数，求系统的上)=可阿升时间t、峰值时间t、超调量0%和调整时间t OC(S)1 近伽=1R(s)=s2+s+1 Lf= 1rn=1k =0.5W d =3 n.1z 2 =0.866 卩=tcj1p=60on -B = 3.14-3.14/3 tr= wd =0.866=2.42tP=nnd=o.8.16=3.63o %= *金100%6% ts=z3n=6 ts=zfn=8-1.8pp3-6已知系统的单

15、位阶跃响应：c(t)=1+0.-2e1.2e-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)=4 +0.2 1.2 600s+60s+10=s(s+60)(s+10)R(s)=1 龍)600S2+70s+6003-7设二阶系统的单位阶跃响应曲线如图，系统的为单位反/ tP=Wn，n-Z2=0.1eZnF =0.3Z n / 1# 2z M-Z2=ln3.=1.19 (z )2/1-Z2=1.42 9.8区 2=1.42-1.42匸 2 z =0.35Zwn=70Iw =600W n=33.4G(s)=ssi為=黑鬆)3-11已知闭环系统的特征方程式，试

16、用劳斯判据判断系统的稳定性。解： s3+20s2+9s+100=0 劳斯表如下：S319S220100s14S0 100系统稳定。(3) s4+8s3+18s2+16s+5=0 劳斯表如下：S41185S3816s2 165S1 216S 16s0 5 系统稳定。3-12已知单位负反馈系统的开环传1递函+1)解：S4+3s3+4s2+2s+Ks+2K=0S4142Ks3 32+Ks2 b31 2Ks1 b41b31=警Kb = K2+10K-20b41=10-K(K- 1.7)(K+11.7)0K03-14已知系统结构如图，试确定系解:6伍)=器闭(s)=s3 +S0T(TS+10S3s2S1

17、S011T110b3110b31 =严 0T 13-16已知单位反馈系统的开环传递函数,r(t)=l(t)+2t+t 2R(s)=+_J+_Jr Kp=20解: (1) G(s)=(0.1s+2)(0.2s+) u =0 & =0I Ka=0=Ro = 1ess1=1+K =21ess2=xess-Fess=x(2) G(s)=s(s+2)(is+l0)= s(0.5s+11)00.1s+1)U =1 YKp=oK =10I Ka=0essi=0- 2 ess2 K =10ess-= 8(3) G(s)=10(2s+1)(2s+1)s2(s2+4s+10) = s2(0.1s2+0.4s+1)

18、ess=8Kp= 8甩=8Ka=1ess=2essi=0ess2=0ess：=23-17已知系统结构如图。(1)单位阶跃输入：譽尸严可 国C(s)(T %=20% ts=1.8(5%)确定 K1 和 t 值。 L士4 |解：G(s)=Ki2Wn = KiTLn2=Ki(s)=s2+Kts+KiZ =0.45wn=i 8*0 45 =3.7 0=亦=13.7e传=0.2=0.24=盘=1.8(2)解:求系统的稳态误差:r(t)=l(t), t ,壬t2g(s)=?+K11s(kTs+1)4 一戶戶 =2 2pppo 2 -L s 乐es24o3-18已知系统结构如图。为使 Z0.7时,单位斜坡输

19、入的稳态误差和解直g(s)=S+2S+tS=确定K竽幺(s)=s(2+Kr S+1)TKS2+(2+Kc)s+KKls(s+2)|t Zn=2+Kr=2*0wn2=Kess= 2+KT=0250.25K-2 T =K=31.6t =0.1863-佃 系统结构如图。r(t)=d1(t)=d2(t)=l(t)(1)求r(t)作下的稳态误差.sDi(s)D2(s)解：essr=SbJl S 1+G(s)F(s) = 1+G(0)F(0)(2) 求d1(t )和d2(t)同时作用下的稳态误差.E (s)=-G 2(s)H(s)D(s)Ed(S)= 1+G 1(s)G2 (s)H(s) D(s)es奸S

20、1+g(S)F)(s) +(3) 求d1(t)作用下的稳态误差.G(s)=Kp+-1 丄-1+F(s) 1+G(s)F(s) S = 1+G(0)F(0)s F(s)= Js丄S1+GS謫=四 S1Jj S =04-1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)490：60jw(7)vj135任丫j453o 迭 厂J0(T600(8)6 ,(T4-2已知开环传递函数，试用解析法绘制出系统的根轨G(s)=命1)解：(s)=s+1+K, Kr=0s=-1Kr=s=-xS=-1-Kr-3+j24-3已知系统的开环传递函数，试绘制出根轨迹图。解：(1) G(s)= Kr(S+1 5)

21、(s+5.5)解-(1) G(s)=s(s+1)(s+5)jw*+ji-2 -1 /(聖7j ws=-2+j0 s=0+j1 s=-3+j2o1 /1) 开环零、极点Pl=0 P2=-1 P3=-5 Z4=-15 Z2=-5 52) 实轴上根轨迹段PlP2 ZiP3 Z2-3) 根轨迹的渐近线n-m= 10 = +1804 )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+5i B(s)=2s+7S=0.63 S2=2.5S3=-3.6S4=-7.28(2) g(s)= KS+,1.5(2) g(s)= s(

22、S+1)(s+4)1) 开环零、极点P1=0 P2=-1 P3=-44=1.52) 实轴上根轨迹段P1P2 P3Z13)根轨迹的渐近线n-m= 20 = 90(T141.52=-1.754 )分离点和会合点A(s)=s3+5s2+4sB(s)=s+1 .5Ia(s)=3s2+10s+4 LB(s)=1 s=-0.62jl3P2 /Pl ,P3-0.6T0(T95)分离点和会合点A(s)=s3+2s2S,B(s)=1A(s)=3s2+4s+1 lB(s)=0 s=-0.33(3) G(S)=S(S1)21) 开环零、极点P1=0 P2=-1 P3=-12) 实轴上根轨迹段P1P2P3-X3) 根

23、轨迹的渐近线 n-m=3o = 60o, 180o(T =-131=-0.674) 根轨迹与虚轴的交点s3+2s2+S+K0Kr=03 1=0-Kr=23 2,3= 1Kr(S+8)(4) G(s)= s(s+3)(s+7)(s+15)1) 开环零、极点P1 =0 p2=-3 p3=-7 P4=-15 Z1=-82) 实轴上根轨迹段P1P2 P3Z1 P4-X3) 根轨迹的渐近线n-m=3 t = -3-7-35+8 =-5.679 = 60, 土1804) 根轨迹与虚轴的交点s4+25s3+ 仃1s2+323s+8Kr=0Kr=0 3 1=0 Kr=638 3 2,3= 6.25 )分离点和

24、会合点I A(s)=S4+25s3+171s2+315s电 A(s)=4s3+75s2+342s+315B(s)=s+8B(s)=2s+7s=-1.44-5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。G(S)=Kr(S+2)G(S)= s(s+1) 解：P1=0 P2=-1 Z1=-2P1P2 Z1-X分离点和会合点S2+4s+2=0 S1=-3.41 S2=-0.59闭环特征方程式S2+s+Krs+2Kr=0 s=-2+3 (-2+jco )2+(-2+j)(1+Kr)+2Kr=0f -4w +(1+Kr 3 =04 32-2(1+k+2K

25、r=03 = + 1.41L Kr=34-6已知系统的开环传递函数，试确定闭环极点Z0.5时的Kr值。(1) G(s)H(s)=s(s+版s+3)解：P!=0 p2=-1 p3=-3 p1p2 p3 -8 (T = 133 = -1.3 9 =+600 +180o根轨迹的分离点：A(s)B(s)=A(s)B(s)q=-0.45s2=-2.2与虚轴交点?+4s2+3s+Kr=03+3w =0 Kr=0丨 Kr=1223s +8s+3=0舍去j 3F1.7S3 P3P2-3-1Pl 0莎-1.7s Kr-4 2=03 1=03 2,3= 4.7Z =0.5 得 Si =-037+j0.8 S3=4

26、+0.37 x 2=326 Kr=|S3|S3+1|S3+3|=3.26 X2.26 x 0.26=1.9jo分离点和会合点4s3+15s2+16s+6=0解得 s=-2.3 z =0.5得 s!=-0.36+j0.75Kr=|S1|S13|S11j|S11-j|=2.92(2) G(s)H(s)=s(s+3)(S+2s+2)解：Pi=0 P2=-3 P3.4=-1 j PiP2 a =341= -1.25 o = 45, 135 根轨迹的出射角03= 士冗 O i-6 2O 4=+n -135o90o26.6o= 71.6o 与虚轴的交点 s(s3)(S22s2)Kr=0 s45s38s26

27、sKr=0 (jw )4+5(j co )3+8(j co )2+j6s +Kr=0 4-8w 2+Kr=0 J Kr=0o 1 =01-5 3+6w =0Kr=8.16 o 2,3= 土 仁1Jo 3o =0 J Kr=0 o 1=0 Kr-63 2=0 I Kr=48 o 2,3= 2.84- 7已知系统的幵环传递函数，(1)试绘制出根轨迹图。G Hess+s 解：P1=0 P2= -2 P3=-4 p1p2 p3 -8a = -2-4 =-2 O = +60, +180o根轨迹的分离点：A(s)B(s)=A (s) B(s)S1=-0.85 s2=-3.15 舍去(2) 阻尼振荡响应的K

28、r值范围 z =0.5s1=-0.7j1.2s5=-6+0.7 x 2=-4.6Kr=4.6 X2.6 X0.6=7.2s=-0.85 Kr=0.85x 1.15X 3.15=3.1 s= j2.8 Kr=48(3) 与虚轴交点s3+6s28sK05-1已知单位负反馈系统开环传递函数，当输入信号r(t)=sin(t30)，试求系解统的稳态输+出。e(s)=(s?1) Ao )= 0 )=-tg-1H=-tg-*=-5.2 Cs(t)=0.9sin(t24.8)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解： G(s)= s(s+5)(s+15) 1 型系统3 =0 A

29、(3 )= 8(3 )=33 = 8 a(3 )=0 3 )=-2703 =0 A(3 )=103 =8 A(3 )=0=0解:(3) G(s)=(2s+11)08s+1)0型系统n-m=2 3 )=0 3 )=-180解：I型系统3 =0 A(w )= 83 =8 A(3 )=0II型系统解: (5) G(s)= sis0!)n-m=2 3 )=-270 3 )=-1810(s+0.2)解: G(s)=s(s+0.1)(s+15)3 =0A3 )= 8 3 )=-1803 =03 = 8A3 )= 0 3 )=-270n-m=3ImRe1 40dB/decs(s+5)(s+15)4010 2

30、00-201530-90-180-270-60dB/dec 3 )=-90 3 )=-270oL(3 )0-200.1250.53Ld . Liiha J I.LLS1 1 bau JJ -4lec3 =8 3 )=-H8roo5-2已知单位负反馈系统开环传递函数, 解：(1) G(S)=s7s+750+T5) 40 UdBdecG(s)=sgs+1)(15S+1)20lgK=201=5 32=153 =03 =试绘制系统开环对数频率特性曲线。 G(s)=(2s+i)s+1)L(o)解: 20lgK=20 203产0 1252=o.53 =0 3 )=0o-180(7)G(s)= = 2 13

31、3(5s+1)(7) G(s) s2(s+0.1)(s+15rs2(10s+l)(0.67s+1) 解：20lgK=2.5dB5=0.132=0.2 33=15w =0 e 3)=-i8oo3 = 00 e 3 )=-270o40200-200*-90l(w)dB-180 -270 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。20lgK=20K=10G(s)=(0?1s+1)20-20dB/dec33c20lgK10L(3 dB(b) 20lgK=-20L典K=0.10G(s)=0.1s -20G(s)=(0.05s+1)B720320dB/dec1(d) 20lgK=4848K=2510fL(vdB4_系统稳定”一i4512(f)Imw =0系统稳定p=1u =0(h)p=1 tIm u =0系统不稳定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