自動(dòng)控制原理第二章習(xí)題答案

1、2-1試建立如圖 所示電路的動(dòng)態(tài)微分方程。解:(a) 輸入U(xiǎn)i輸出ii=i-i2Ui = UUoh=詐Ui-Uoi= UoUi-Uo= Uod(Ui-Uo)Ri = R2 -C dtdUi i2=C d"=CctUi-Uo)dtR2(Ui-Uo )=RiUo-CRiR2(器-器)dUo亠 dUCR1R2 dt +Ri Uo+R2Uo=CRiR2 dt +R2Ui解: (b)r i=i1+i2 i=(Ui-Ui)：=止 i =C 如1Ui-Ui 蟲(chóng) dUii1= R2 i2=C dtRi = R2+C dtL dUo1_l dUoUi-Uo卞2 dtUi=Uo+R2 dtUi Up l

2、 dUo-比dUo jCLRi- Ri-RiR2 dt = R2+c dt +R2 dt22齊護(hù)+(6眾踹 +( Ri+ R2)Uo2-2求下列函數(shù)的拉氏變換。(1) f (t) = sin 4t cos4t(2) f (t)二t3 - e4t(3) f(t)十 te(4) f(t) =(t1)2e2t解:(1) f(t)=sin4t+cos4t解: Lisin® t=3 2；$Lcosw t=224ss+4Lsin4t+cos4=s2+16 + s"+16 =s2+16(2) f(t)=t3+e4t解：L件e4t= g+si = 6s+s鬻(3) f(t)=t neat解

3、：Ltneat=(s-anU(4) f(t)=(t-1) 2e2t解：L(t-1)2e2t=e-(s-2)(s-2)3 F(S)二S 1(S 2)(s 3)F(s)二s(s 1)2(s 2)F(s)二2s2 -5s 1s(s21)F(s)二s + 2(s24s 3)(s 2)(1) F(S)=2)(s+3)解：人1=伍+2(+2+1+3=-2=-1A2=(s+3)(s+J)(S+3)s=-3=2F(s)= s+3 - s+2f(t)=2e-3t-e-2t F(s)=(s+1s(s+2)=(s+A)2+ 釦 sA2 解:A1=(s+1(s+1)2ss+2) s=-1 =-1A2=】丿備；=-1=

4、2A3=(s+2)(s+1)2ss+2)s=-2=-2F(s)=-2e'2t-te"t+2e"t(3)F(s)= 2Sss5s+= Ass+A+A3 解：F(s)(s2+1s)+j =AiS+AsS=+j 2s2-SS+1S=j=AiS+A2 s=j F(s)=s(s+1)+2s+3)解:A1 =-12=A1+ A2 + A3 + A4= (s+1)2 + s+1 + s + s+3-5j-1=-A 1+jA2A1=1A2=-5嚴(yán)(s+2) A2=A3=F(s)s s=0 =1F(s)= s+s+r+s+1d【s(s+3)】_ s(s+3)-(s+2)(2s+3)

5、dss=-1=s(s+3)2s=-1t -tf(t)= 2 e -e-t+N+ 丄 e + 3 +123t e2-3求下列函數(shù)的拉氏反變換。f (t) =1 cost -5 sin t22-4解下列微分方程d y(t) 5dy(t)dt2dt6y(t) = 6初始條件：y(0) = y(0) =2解：SY(s)-sy(0)-y'(0)+5sY(s)-5y(0)+6Y(s)= Ss2Y(s)-2s-2+5sY(s)-10+6Y(s)= gY(s)=6+2s2+12s_ ai . A2 + A3s(sz+5s+6) = s + s+2 + s+3Ai=1A2=5A3=-4y(t)=1+5e

6、-2t-4e-3t2-5試畫(huà)題圖所示電路的動(dòng)態(tài)結(jié)構(gòu)圖,C(1)并求傳遞函數(shù)。解：Ur(s+RiJc(s)+urUc22Ur(s)= ( Ri+sC)R2R2+R1R2sCUc(s)= 1+( +sC)R2 = Ri+R2+RiR2sc解：Li=R2 /LsL2=-/ LCs2L3=-1/sCRLi L3=R2/LCRiS2UrRicUi Liii2Pi=R2/LCRiS2 i=ic(s) LUi(s) L3"LSi(s)Uc(s)+UcURg+C+Ds+R2-6用運(yùn)算放大器組成的有源電網(wǎng)絡(luò)如圖所示，試采用復(fù)數(shù)阻抗法寫(xiě)出它們的 傳遞函數(shù)。R2R3卜解:電路等效為:UiUoR；=- R

7、1_Uo2SC +R3 =亙+rR2+ SCR2SC1+R3R2R1 (R2 S C+1)+1r2Ri(R2SC+1)+R )Rzr7Sc1+R3r1+r3+r2r3csR1(R2SC+1)2-8設(shè)有一個(gè)初始條件為零的系統(tǒng), 如圖所示。試求系統(tǒng)的傳遞函數(shù)。當(dāng)其輸入端作用一個(gè)脈沖函數(shù)S 時(shí)，它的輸出響應(yīng)c(t)C(t)=辛t-半(t-T)C(s)= £2(1 -e-TS)C(s)=G(s)2-9若某系統(tǒng)在階躍輸入作用r(t)=1(t)時(shí)，系統(tǒng)在零初始條件下的輸出響應(yīng)為:c(t) =1 -e't e，試求系統(tǒng)的傳遞函數(shù)。解：G(s)=C(s)/R(s)=(S+4(S；%脈沖響應(yīng)：

8、c(s)=(y+4(s+2)=1+爲(wèi)-詁1c(t) = (t)+2et-e-t2-10已知系統(tǒng)的微分方程組的拉氏變換式，試畫(huà)出系統(tǒng)的動(dòng)態(tài)結(jié)構(gòu)圖并求傳遞函數(shù)C(s)R(s)。X's)二 R(s)G(s) -G(s)G7(s)-G8(s)C(s)X2（s） =G2（s）Xi（s）-G6（s）X3（s）X3（s） =X2（s） -C（s）G5（s）G3（s）C（s）二 G4（s）X3（s）G5G7-G8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（

9、s）=G3（s）X2（s）-C（s）G5（s）2-11已知控制系統(tǒng)結(jié)構(gòu)圖如圖所示，試分別用結(jié)構(gòu)圖等效變換和梅遜公式求系統(tǒng)傳遞函數(shù)c（s）R（S）°(0解：巧+輪,=1+6中覽1眾R(+G1HC(s)= GG+G2G31R(s) = 1+G2HHG1G2H2(b)求系統(tǒng)的傳遞函數(shù)解：l1=-g1g2hL?=-G 1G4HP1=G1G2 1 =1Q(S)R(s)Gi(s)L1G4(S)H(s)G2(s) C(s)+(c) 解：GT =1+G 1G4H+G 1G2HP2=G 3G2 2=1 +G1G4HR(s)+H1 >H1 c(S)G1R(s)!_+ GGgGi1GIH1H1+G2

10、”C(s) = G1G2+G2G3+G1G2G3G4 H R(s) =1+G1G2H +G1G4HC(s)G1G2(1 -G3H1)R(s) 1+G1G2+G1H133H1G2(d)解:(1)R(s) IG1C(s)(e)解:(1)R(s)R(s)g2 C(s)C(s)C(s) = (G1+G2)1+G2H(2)L1=-G 2HP1=G1P2=G2R(s)=(G1+G2 )G2H 1 =1 2 =1_ G1+G2 C(s)G3-G4C(s)=(G1+G2)G1L G2 .3 G3 L4G4C(s)+R(s)=1+(G1+G2)(G3-G4) L1=-(13 L2=G 1G4 L4=G2G4 P

11、1=G1 1 = 1C(s) =(G1+G2)R(s) 1+G1G3+G2G3 31G4-G2G4L3=-G 2G3P?=G2 2=1C(s)= Gi = Gi (1 -G2)R(s)G1G2 1+G1G2 J21+i-G2(2) Li=GiG? L?=G2 Pi=Gi i=1-G2A =i+GiG2-G2C(s)= Gi(i G)i 2 2 R(s) i+GiG2 -G22-12求圖所示系統(tǒng)的傳遞函數(shù)C© , C(S)R(s) D(s)解：(a)CLi=G2H2L2=-GiG2H3Pi=GiG2 A iC(s)_G2GiR(s) = i-G2H 2+GiG2H3C(s)=G2(i-

12、GiHi )D(s) i-G 2H 2+G iG?H 3Li=G2H2 L?=G iG?H3 Pi=G? A i =i P2=-GiG2HiA 2=i(b)求：RC(s)Cs)R(s)D(s)解：Li =-G iGGi2HCG3=D(s)L2=-GiG2H Pi=GiG? A i=iGiG2Cs) =R(s) = i+G iG2H+GiG2Pi=G nG2 A i=iP2=iA 2=i+G iG2HC(s)= i+GnG2+GiG2HD(s) 1+GiG2+GiG2H2-13求圖所示系統(tǒng)的傳遞函數(shù)C(s) E(s)R(s) R(s)Pl=G2G3 1=1P2=GlG2G3 2 = 1C(S)=

13、 G2G3+G1G2G3R(s) 1+G2+G1G2G3Pl=-G2G3 1=1P?=1 2 = 1+G?C(s) = -G2G3+1+G2(b)C(s)R(s)Es!R(s)UG2G3G4R(s) 1+G2+G1G2G3P2=G2G3G5 2=12-14求圖所示系統(tǒng)的傳遞函數(shù)X(s)E(s)P 1=G 1G3A 1=1P2=G?G3 A 2=1P3=GqG4 A 3=1P4=G2G4 A 4=1P1=1A 1=1G4(s)_=i4(s>jn_G2(s) 一 IX(S)11 DC(s) E(s) C(s)R(s) ' R(s)，D(s)2-14解：盧乜占彳L2=- G2G3C(s

14、) (61+62)(63+64)R(s)= 1+G3(Gs+G2)E(s)=1R(s) =1+Gs(G1+G2)Esj=G2（s）謁可2-15求圖所示系統(tǒng)的傳遞函數(shù)Ci (s)Ci (s) C2 (s)Ri (s) ' R2 (s) ' Ri (s)C2(s)R2G)解：L1=-G3G4 L2=-G 2G3G5 P1=G1G5 1=1C(s)= G1G2G5+G1G5 R(s)=1+G 2G3G5+G3G4P1=G1G5 1=1P2=1 A 2=1+G3G4Es)= GG+d+G 1G5 )R(s) =1+G 2G 3G 5+G3G4解：求 C1(s)Ri(s)R1(冷G1 G

15、2'G3H2=|-H14Ci(s)I b.G4 亍 G5Li=GiG2L2=-GiG4G5HiH2RCL3=-G4 rl-GiGz+GiGqGssH iH 2+G4 -G1G2G4 Pi=GiG2G3 1=1+G4C1(s)= GiG2G3(1+G4 )Ri(s) I+G4+G1G4G5H iH 2-G1G2-G1G2G4求 r2(s) l1=g1g2l2=-GJg4g5h1h2L3=-G4 =1-G 1G2+G1G4G5H1H 2+G4 -G1G2G4 P 1=G4G5G6 1 = 1 -G1G2C2(s)G4G5G6(1-GqG2)R2(s) =1+G4+GqG4G5H 1H 2-G1G2-G 1G2G4十 Cl?)求 R2(s) L1=G1G2L2=-G1G4G5H1H2L3=-G4 =1-C2+G1G