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Uii1R1 DR2卩 i UoUi LUiii2Uo(a)解：ii=i-i2Ui=Ui-Uoii=詐肾 i=R：du1 i2=Cdt1 =Cd(Ui-Uo)dtUi-UoUo 厂 d(Ui-Uo)D = D - CRi R2dt(b)解：f i=ii+i2 i=(UoRiii=R72=時R2(Ui-Uo)=RiU0-CRiR2(dUi-dUo)Ui-Uo亠业R2 dtCRiR2djto+RiUo+R2U0=CRiR2djUi+R2UiUi-Ui 鱼亠cdUi= R2+CdtRiUi=U+R2dUodUo.CL 匹Ul.Uo L dUo= Uo.C dUo.

2、 CL Ri Ri RiR2 dt = R2+c dt + R2 dt2Ri=Ct 器+(c+RlR2)dUo+( Ri+RpUo2- 2求下列函数的拉氏变换。(1) f(t)=sin4t+cos4t解：Lsin3t=ji+s2 Lcoaj t= _4s s+4Lsin4t+cos4=S2+i6 + s2+i6=ss+i6(2) f(t)=t 3+e4t解：Lt3+e4t=爭+缶鸯辦(3) f(t)=t neat 解：Ltneat=(s-a)n+T f(t)=(t-1)2e2t解：L(t-1)2e2t=e-(s-2)(s-232- 3求下列函数的拉氏反变换。 F(s)=(s+2+1+3) =梟

3、+ s+3 解: A1=(S+2)(s+2j(S1+3)/s=-2=-1A2=(S+3)(s+2S(+1+3s=-3=2F(s)= s+3- s+2 f(t)=2e-3t-e-2t F(s)=(s+1j2(s+2)=s+12+备+s+2解: A1=(s+1)2s+1辰)/s=-1=ja2=dds+2s=-1=2A3=(s+2)(s+1)2ss+2)/s=-2 =-2f(t)=-2(2t-te-t+2e-t F(s2臨1=As+A2 + 售解：F(s)(s2+1)/s=+j=A1S+A2s=+j 2s-fs+1s=j=A 1s+A2 4=j jh+A? -5j-1=-A1+jA2A1=1A2=-

4、5 A3=F(s)ss=0=1F(s)= s+s2+1 + s+1 f(t)=1+cost-5sint F(s)=s(s+s+(s+3)解: =(s+lj2+s+l+ + s+3 Ai=号 A3=孑 A4=12 A2=于A =曲碍A2=ds ls=-1=s(s+3)-(s+2)(2s+3)?.3=s(s+3)2&-1= 4f(t)=fe-t-f et+f+12e-3t(2-4)求解下列微分方程。豁+5 聲+6y(t)=6 y(0)=y(0)=2 解：s2Y(s)-sy(0)-y(0)+5sY(s)-5y(0)+6Y(s)=、“-、_6+2s2+12s A1 . A2 A3Y(s)=s(s2+5

5、s+6) = -s + 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。(1)+0Ur i2 R1 典UcqQ -Ri+ Ur(sl解：Uc(s)|Ii(S)RJlCs)R2+R1R2SCUr(s) ( +sC)RUc(s)=i+氏+sC)VRi+R2+RiR2sCL1=-R2 /Ls L2=-/LCs L3=-1/sCR1 L止3=R2/LCR1s2P1=R2/LCR1s2 A i=1Ur(j= r1cls2+(r1r2C+L)s+Pr22-8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)。2-

6、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(1) c(t)=1-e2t +e-tR(s)=-S解：丄-1 亠 1 一 (sJqs+Z)解:C(s)=s s+2+s+1 = s(s+1)(S+2)G(s)=C(s)/R(s)= (S+4S：%C(s)=c(t)=5 (t)+2e2td2-10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）-G8（s）C（s） =R（s）-C（s）G7（s）-G8（s）G1（s） X2（S）=G2（S）X1（S）-G6（S）X

7、3（S）X3（S）=G3（S）X2（S）-C（S）G5（S） C（s）=G4（s）X3（s）R(s)C(s)gC(s)G7 (s)-G8(s)(s)3(SC(s)G 5(S)n 彳 G8 5G1G2G3G4G3G4 1+G3G2G6謂=1+G3G2G6+G3G4G5+G1G2G3G4(G7-G8)2-11求系统的传递函数g3(s R(s)G1(sC(s)解：L1=-G2H 1L2=-G 1 GqH 2 =1+G2H1+G1G2H2P1=G1G2 1 =1P 2=6362 2 =1G2G1+G2G3nCM=掐 Pk k =R(s) 1+G 2H 1+G 1G2H 2(b)R(s)uLjG4(s=

8、+63h(s) k11解:L1=-G1G2H L2=-(4H P1=G1G2 1 =1 =1+G1G4H+G 1G2H P2=G3G2 2=1+G1G4HC(s) G1G2+G2G3+G1G2G3G4 HR(s) = 1+G1G2H+G1G4H(c) R(su(Gn4GnCjs)G3 hGiG2(1 GHi)Rsi=1+GiG2+GiHiGHi(d) Rs)C(s)_ G1 (1 -G2) R(s)-1+G辰飞2解：L1=-G 2H P 1=G1 A 1 =1P2=3? A 2 =1Rs)=(G1+G2 Jt+Gh解：L1=-G1G3 L2=G1 G4L3=-G 2G3L4=G2G4 P 1=

9、G1 a 1 = 1 P2=G2 a 2=1C(s) =_ _ (G1+GJR(s) =1+G1G3+G2G3 G1G4-G2G4L1 =-G1G2 L2=G2 P1 =G1 A 1 = 1-G2 A =1+G1G2-G22-12唇呜D(s)Gi(s)L2H 2(S)&|H3(S)=l解：L1 =-G 1G2R(s)=1+G1G2H+G 1G2P1 =GnG2 A 1 = 1 p2=1A 2=1+11S2HC(s) 1+GnG2+G1G2HD(s)=1+G1G2+G1G2H2-13 (a)石严GD解：L1=-G2 L2=-G1G2G3 P1=G2G3 A 1=1P =G G G A =1 Cs

10、) = G2G3+G1G2G3P2=G1G2G3 a 2=1 R(s)=1+G2+G1G2G3P 1=-G2G3 a 1=1 P2=1 A 2=1+G2C(s) -G2G3+1+G2R(s)= 1+G2+G1G2G3(b)C(sG5G3R(s)解：11 = -4 L2 = -2jG3G5 P1=G1G5 a 1 = 1P2=235 A 2=1 3燈中恿 R(s) 1+G2G3G5+G3G4P1=G1 G5 A 1=1P2=1A 2=1+334E(s) = 15+(1+10K0s3 s s1 s01 10 (1+1T) 10鸟110T0R(* T3-14已知系统结构如图，试确定系1pTs+1)

11、r s+10解:G(s)=1s2(s+i)(s)=s3+2TT11T1 10b3110s3 s2 s1 s0b31 = 10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 + $=21e = R0 ess11+Kess2=8ess-=8Kp=20 解： G(s)=(0.1s+2)(0.2s+) u =0 K =0I Ka=0 G(s)=s(s+2)(00-1O)= s(O.5s+1)0O.1s+1)(2s+1)(3) G(s)= s2(12+4ss101) s2(0.1s2+0.4s+1)ess=8Kp=8)K. =10Ka=0 ess=Kp

12、= 8K. = 8Ka=1ess=2ess1=02 2 ess2=T2 = Wess-= 8ooess1=0ess2=0ess：=23-17已知系统结构如图。(1)单位阶跃输入：& %=20% ts=1.8(5%)确定 K1 和 T 值。解：G(s)=小 K1rZcon=KT reznk=0.2s+Kn (s)=s2+kts+k1 1扁戒1仏=盘=1.8Z =0.45cjn=1 8*045 =3.7 0=3(=13.7 t =0.24求系统的稳态误差：r(t)=I(t), t ,才t2Ki解:G(s)=s2+KT=s)R(s)FR(s)=H R(s)=Kp= 8 甩=KKa=0ess1=0es

13、s2 =0.24ess-=3-18已知系统结构如图。为使Z0.7 时,单位斜坡输入的稳态误差和解值g(s)=s2+2s+tess=(K25 确定 K2+Krs=1(s)=s(2+K： s+1)J Z3n=2+Kr =2*07Kess=0.25K=31.6r2 = KT = 0.25K-2 T =0.186Ks2+(2+Kr)s+KDi(s)D2(s)90； WK3(4)0(7(8)J08；忒60 一(73-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.解:essr=Sim S 1+G(sSF(s) = 1+G(1)F(0)求d1(t )和d2(t)

14、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)eSS= |im S -F(s) + -1_丄=-1+F(s)】 essd艸 S1+G(S)F(S) + 1+G(s)F(s)】S= 1+G(0)F(0)(3)求d1(t)作用下的稳态误差.G(s)=Kp+#F(S)= JS丄essd= SiG S1+(G(SiF(1= SJO S1+(K jSK) 1 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。G(s)= (S匸 1)K解：(s)=s+1+KrKr=0s=-1Kr=S=-1-Krs=-s=-2

15、+j0 S=0+j1 s=-3+j2j30+j1-2 -10 T-3+j24- 2已知开环传递函数，试用解析法绘制出系统的根轨4-3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)(s+5.5r解： G(S)= s(S+1)(S+5)1) 开环零、极点Pi=0 p 2=-1 P3=-5 z1=-1.5 Z2=-5.52) 实轴上根轨迹段P 1P2 ZiP3 百3) 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+5 B(s)

16、=2s+7Si =-0.63 S2=-2.5S3=-3.6S4=-7.284 )分离点和会合点” A(s)=s3+5s2+4sB(s)=s+1.5Ia(s)=3s2+10s+4 Ib(s)=1s=-0.62 G(s)= sJS+S+f)1) 开环零、极点P i=0 P2=-1 P3=-4z1=-1.52) 实轴上根轨迹段P 1P2 P3Zi3) 根轨迹的渐近线n-m= 20 = +90(T =-1-4+1.5 =-1.75K G(s)=s(s+121) 开环零、极点p i=0 P2=-1 P3=-12) 实轴上根轨迹段PiP2P3-X3) 根轨迹的渐近线 n-m=30 = +60, 180T

17、= 131-0.674) 根轨迹与虚轴的交点s32s2+S+K0Kr=03 i=0Kr=23 2,3= 1P2 / f rP-o-ef 0p75)分离点和会合点A(s)=s3+2s2+S 弓 B(s)=1A(s)=3s2+4s+1 、B(s)=0 s=-0.33Kr(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 = -3-7-15+8 =-5.670 = 60, +1804) 根轨迹与虚轴的交点s425s3+ 仃1s2

18、323s8Kr=0Kr=0 W 1=0 Kr=638 W 2,3= 士 6.25 )分离点和会合点r A(s)=S425s3171s2315s电 A(s)=4s375s2342s315B(s)=s8B(s)=2s7s=-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= s=-2+W (-2+jw )2+(-2+J)(1+Kr

19、)+2Kr=0jwpi:-43 +(1+Kr W =0l4-w2-2(1+K)+2Kr=0w = + 1.41I 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+紿) 解：P1=0 P2=-1 P3=-3 p1p2 p3 8T = 1g3=-1.3 0 =60 180根轨迹的分离点：A(s)B(s)=A(s)B(s) s1=-0.45s2=-2.2与虚轴交点s?+4s2+3s+Kr=0w SId =0 f Kr=0Kr-4w 2=0 lKr=123s2+8s+3=0舍去w 1=0w 2,3= 1.7S3 p3I *-34-1.7Z =0

20、.5 得 S1=0.37+j0.8 S3=-4+0.37 X 2=-3.26 Kr=|S3|Sj+1|S3+3|=3.26 X2.26 X 0.26=1.9K(2) G(s)H(s)=s(s3)(S22s2) 解：P 1=0 P2=-3 P3.4=-1 j P1P2 J =341z1=-1.250 = 45, 135根轨迹的出射角03= 士冗-0 10 20 4=n -13590o26.6ot-71.6 与虚轴的交点s(s3)(S22s2)Kr=0s4 5s38s26sKr=0(jw )45(j 3 )38(j 3 )2j63 Kr=03 4-8w 2Kr=0 jKr=03 1=053 36w

21、 =0 LKr=8.16 3 2,3= 土 1.1P4135*0 (J-1.1分离点和会合点4s3+15s2+16s+6=0解得 s=-2.3 z =0.5得 Si=-0.36+j0.75Kr=|Si|Si+3|Si + 1+j|Si + 1-j|=2.924-7已知系统的幵环传递函数,(1)试绘制出根轨迹图。G H(s)=s(s+2)(s+4) 解：P1=0 P2=-2 P3=-4 P 1P2 P3(J = -234 =-20 = 60, 180。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Kr值范围s=-0.85 Kr

22、=0.85x 1.15x 3.15=3.1 S= j2.8 Kr=48(3) 与虚轴交点s36s28sK0 Z =0.5S3=-6+0.7 X 2=-4.6Kr=4.6 X2.6 X0.6=7.23 1=03 2,3= 2.8s1=-0.7+j1.25-1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)，试求系解统的稳态输出。e(S)=(s1r) Ab)咼代吊墙議=0.905 e 3 戶-tg-1祎=40*=-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：(1) G(s)=3 =0解:(3) G(

23、s)=(2s+108s+1)0型系统n-m=23 =0 A3 )=10 e3 )=03 =O A(w )=0 e 3 )=-180OIm=0Re解：(5) Gs10!)解：I型系统3 =0 A3 )= O3 =O A(3 )=0n-m=2e 3 )=-270O e 3 )=-103 =0Im解： G(s)=s2(s+0s1+0s+)15) II 型系统3 =0A(3 )= 乂 e 3 )=180o3 =03 OOA3 )= 0e 3 )=-270On-m=3Im3 =O 一Re5-2已知单位负反馈系统开环传递函数,解:(1) G(s)=75 L3 牌s(s+5)(s+15)4010L3 )dB

24、0dB/dec 20 卜1 LG(s)=sgs+1)15s+1)20lgK=201=5 32=153 =03 =Oe 3 )=900 e 3 )=-2700-20,e 3)0*-90*-180 * -270*15-40dB/dec-60dB/dec试绘制系统开环对数频率特性曲线。(3) G(s)= 10解: 20lgKndi2031=0.1252=0.53 =0 e 3 )=0O3 =o e 3 )=-1i8ffio璋,性曲线。 l(3 ydB(2s+1)(8s+1) I ,20 20lgK :趣鲨-20AOdB/dece 3 )0-90-180解:(5) G(s)=s(S01)31=1 20

25、lgK=2B)3 =040200O -203 OOe 3 )=-270e 3 )=-180 0-90-180“-270L(vdBfOdB/dec40dB/dec1、3 G(s)=s2(s1+0s+(5+l5)s2(10s+1 20lgK=2.5dB 01=0.102=0.2 03=15co =0 0 )=-180oo = 00 0 )=-270o解:1.33(5s+1)L(o1)(0.67s+1)40、200-20L(o ydB-40dB/dec -60dB/dec 0 )0-90-180-270co-40dB/dec-60dB/dec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

26、性曲线。5-4(b)20lgK=-20K=0.10G(s)=0.1s -20G(s)=(0.05s+1)UvdB4_-20dB/dec20lgK XdB/dec 1 110 sSJ00-60dB/dec0dB/dec20 0(d) 20lgK=4848K=2510251G(s)=(s+1X0-1s+(0)01s+1)(c) K=100100G(s)= s(100s+100.01s+1)(c)K=1000LydB-20dB/dec0.01-40dB/dec-60dB/decG(s)=s(1OOs+1)01s+1)L(o 闻B-20dB/dec-60dB/dec0.01 -40dB/dec(e)由

27、图可得：20lgMr=4.58dB Mr=1.7 =乙 /1弋 2Z1= 0.94 z2= 0.32Z =0.3or =on122 on=50K=O0=100T2=on)2=0.0222Z =0.01得:L(o vdB 国忑叫丄4.58dB100 :-60dB/decG(s)= s(0.02s)1+00.01s+1)5-7已知奈氏曲线，P为不稳定极点个数,U 为积分环节个数,试判别系统稳定性）p=0|Im(b)系统不稳定p=0系统稳定“5讪u =2o =0TRe(c)P=03 =0+ -1系统不稳定Im u =23 =00nRed.z(d)3 =0Im p=0/ /u =3-4-M3 =0(f

28、)3 =0系统稳定DP=1u =00ReIm(h)P=1Imu =0=0 Re系统不稳定-2705- 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。解：K=10G(s)=s(10s+1X0.05s+1)3c=113計Y =佃0。+ 3c)=180O-90O-tg-110-tg-10.05 =90。-84.3。-2.9。= 2.8。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。s0.5S+1s(O.o2s+i)解：G(s)=s(05s+110002s+1)05bF 3c=4-5Y =180-90-tg-1 05x 4,5-tg-1002x 4.5=90o-66o-26o= 21.4L(3 JdB-20dB/dec-60dB/de40200-20采 G0(s)=00K)50:32 3c-