自动控制原理习题答案

1、解:.UiUi-Uoi1=Ri=Ri+UoR2(a)输入Ui输出ii=i-i2Ui=Ui-UoRi一也d(Ui-Uo)Li2=Cdt=CdtUi-UoUod(Ui-Uo)Ri=R2-CdtR2(Ui-Uo)=RiUo-CRiR2(望-蹈)dUodUiCRiR2dt+RiUo+R2Uo=CRiR2dt+R2Ui解：(b)，i=i1+i2i=(Ui-Ui)Ri%Uo.“亚iii=R2i2=Cdt1dUoIUi-U。苞dt辿1=R2+cdtLdUoJiJo+R2dtUiUpldUo四dUojQL色Ri-RjRiR2dt=R2+cdt+R2dt22R；=CL需+(C+RL2请+(需噫)Uo2-2求下列

2、函数的拉氏变换。(1)f=sin4tcos4t(4)f(t)=(t-1)2e2t解：(1)f(t)=sin4t+cos4tLsinwt=w2Lcoswt尸5r4ss+4Lsin4t+cosfl=s2+16+16f(t)=t3+e4t解：Lge+¥+s14=蟹普等(3) f(t)=tneat解：Lfeat=(s_a+i(4) f(t)=(t-1)解：L(t-1)2e2t=e(s-2)/-4pF(S)=s+1(s+2)(s+3)2-3求下列函数的拉氏反变换。(2)F(s)=1(s+1)(s+2)(3)F(s)=2s2-5s+1s(s2+1)(4) F(s)=-s+2s2+4s+3)(s+

3、2)F(s恒济1s+3)（2）F=肃*>AiA2A3(s+ir(sn+ij2+s+i+s+2解：ai=(s+2s+(3)/s=.2=-1人2=伯+3)恒+据+3)附2F(s)=Ts12f(t)=2e3t-e2t斛：Ai=(s+1)(s+iF：F+2，s=-i=-1A2=：sfefelLi=243=伯+2)伯+1)21+2产2=或F(s)=-2e2t-tet+2et(3)F(s)=W=F+s解：Fd+)+j=AS+A£=+j2s2-5s±J=j=AiS+A2s=jF(s)=s(s+；胃S+3)解：A-1Ai=2_AiA2A3A4-(s+1)2+s+1+s+s+321Aa

4、=3A4I2Az=4-5j-1=-Ai+jA2A产1A2=-5A3=F(s)ss=o=1F(s)=t+新+焉A尸喈（擒伊2）华3）-t-t3-t21-3tf(t)=fe4e+f+i2ef(t)=1+cost-5sint.22-4解下列微分方程一yP5dy夕6y(t)=6dt2dt初始条件：y(0)=y(0)=2解：s2Y(s)-sy(0)-y'(0)+5sY(s)-5y(0)+6Y(s)=6s2Y(s)-2s-2+5sY(s)-10+6Y(s)=sY(s)=6+2s2+12s_AiA2A3s(s2+5s+6)="s+s+2+s+3Ai=1A2=5A3=-4y(t)=1+5e-

5、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)解：1Ur.(R1+sC)R2R2+R1R2sCUc(s)"1+(R1+sC)R2=R1+R2+R1R2sC解：L1=-Rz/LsL2=-/LCs2L3=-1/sCRU1L十R1i.Ur1C.R2i2-P1=R2/LCR1s2+UcL1L3=R2/LC&s2A1=1rfU!s=5_R2Uc(s厂R1cLs2+(R1R2C+L)s+R1+R22-6用运算放大器组成的有源电网络如图所示，试采用复数阻抗法写出它们的传递函数。解:电路等效为:UI.篇一Uo1R2SC+r2+cL2SCUo_R_R2sC+l+R3一,

6、R2一.(R(R2SC+1+2-8设有一个初始条件为零的系统,如图所示。试求系统的传递函数。解:1(R2R八Ri(R2SC+1)R3)R2,R二RzSJi+R3=一RiR1+R3+R2R3CSR1(R2SC+1)当其输入端作用一个脉冲函数8(t)时，它的输出响应c(t)c(t)=辛t-¥(t-T)C(s)=备(I-*C(s)=G(s)2-9若某系统在阶跃输入作用r(t)=1(t)时，系统在零初始条件下的输出响应为:-2t_t一、c(t)=1-ee,试求系统的传递函数。解：(S2+4S+2)G(s)-C(s)/R(s)-(S+1)(s+2)(S2+4S+2)2_1：C(s)-(s+1)

7、(s+2)=1+s+2-s+1C(s)Rc(t)=s(t)+23t-e-t2-10已知系统的微分方程组的拉氏变换式，试画出系统的动态结构图并求传递函数Xi(s)=R(s)Gi(s)-Gi(s)G7(s)-G8(s)C(s)X2(s)=G2(s)X1(s)-G6(s)X3(s)X3(s)=X2(s)-C(s)Gs(s)G3(s)C(s)=G/s)X3(s)解：Xi(s)=R(s)Gi(s)-Gi(s)G7(s)-G8(s)C(s)=R(s)-C(s)G7(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）R(S)G

8、1G2G3G4C(s)1+G3G2GG5G7-G82-11已知控制系统结构图如图所示，试分别用结构图等效变换和梅逊公式求系统传递函数C（s）0R（s）超5)解：(a)(a)求系爨)传递函数G1+G3G3(s)GG2(s)Gi(s)H2(s+1瑞则»G3(S)R(s)G1(s)解：R(s)1+G1H0s)岛施F42H1+G21G2H¥C(s)_G2G1+G2G6)R(s)1+G2H1+Gs)iG2H2(c)(b)求系统的传递函教G3(S)R(s)Gi(s)解：L产-G1G2HL2=-G1G4HP产G1G2A1=1LiG4(s)H(S)C(s)G2(s)+2I+G1G4H+G1

9、G2HP2=G3G2A2=1+GiG4H+heH1C解:R(s)：:反!=R(s)H1H1+G2卜C(s)=GiG2+G2G3+G1G2G3G4HR(s)1+G1G2H+G1G4HC(s)GiG2(1-G3H1)R(s)=1+G1G2+G1H103H1G2(d)解：(1)RIG1C(s)g2C+i(s)_G2_R(s)=(G1+G2)1+G2H(2)L1=-G2HP1=G1P2=G2解:(1)R(s)C(s)G1+G2G3-G4C(S)_(G1+G2)R(s)L3aL4A1=1A2=1R(S)=1+(Gi+G2)(G3-G4)(2)Li=-03L2=GiG4G1G2G3G4C(s)+LL2l_

10、3="G2G3Pz=GzA2=1R(s)=(G1+G2)1+G2HL4=G2G4Pi=GiA产1C(s)_(G1+G2)R(s)=1+GiG3+G2G39iG4-G2G4C(s)_Gi_Gi(1G)R(s)=.G1G2-1+G1G201+1-G7(2) L产-G1G2L2=G2P1=G1产1-G2A=1+G1G2-G2C(s)=G1(1-G2)1 22R(s)1+G1G2c22-12求图所示系统的传递函数，R(s)D(s)解：（a）C(s)丽L产G2H2L2=-G22H3P产G1G2A1=1C(s)_G2G1C(s)函R(s)FG2H2+G1G2H3L1=G2H2L2=-G1G2H3

11、P1=G2A1=1P2=-G1G2H1C(s)=G2(1-GH1)A2=1D(s)1-G2H2+G1G2H3(b)C(s)R(s)解：L产-G1G2L2=-G1G2HP产G1G2产1C(s)G1G2R(s)=1+G1G2H+G1G2P产GnG2A1=1P2=1A2=1+G1G2HC(s)1+GnG2+G1G2HD(s)1+G1G2+G1G2H2-13求图所示系统的传递函数C(s)E(s)R(s)'R(s)P产G2G3A1=1P2=GiG2G3A2=1CS)_G2G3+G1G2G3R(s)=1+G2+G1G2G3P产-G2G3A1=1P2=1A2=1+G2C(s)_-G2G3+1+G2R

12、(s)=1+G2+G1G2G3R(s)R(s)U=U解：L1=-G3G4l_2=-G2G3G5P1=G1G5A1=1P2=G2G3G5A2=1C(s)_G1G2G5+G1G5R(s)=1+G2G3G5+G3G4P产G1G5A1=1P2=1A2=1+G3G4E(s)G1G5+(1+G1G5)R(s)=1+G2G3G5+G3G42-14求图所示系统的传递函数C(s)E(s)Cjs)Xs),R(s)R(s)D(s)E(s)2-14解：L产-G1G3L2=-G2G3P产G1G3A1=1P2=G2G3A2=1P3=G1G4a3=1P4=G2G4A4=1P1=1A1=1C(s)(G+GEG+G)R(s)=

13、1+G3(G1+G2)E(s)_1R(s)=1+G3(G1+G2)X(s)=G(E(s)=G2(s)Cs)D(s)=12-15求图所示系统的传递函数Ci(s)Ci(s)C2(s)C2(s),Ri(s)R2(s)Ri(s)R2(s)解:求C1(s)Ri(s)GiRi(s)H2G2G3Ci(s)IHiL产G1G2L2=-GiG4G5H1H2RC06L3=-G4a=i-G1G2+G1G4G35H1H2+G4-G1G2G4P产GiG2G3Ai=1+G4Ci(s)_GiG2G3(g)Rsj=1+G4+G1G4G5H1H2-G1G2-G1G2G4求酒承R2(s)L1=G1G2L2=-GG4G5H1H2L3=-G4A=1-G1G2+G1G4G5H1H2+G4-G1G2G4Pi=G4G5G6Ai=1-GiG2C#)_G4G5G6(1-GiG2)R=1+G4+G1G4G5H1H2-G1G2G1G2G4+CiCs)求RL产G1G2L2=-GiG4G5H1H2l3=-g4a=1-G1g2+g