自動(dòng)控制原理及其應(yīng)用答案第二版_課后答案

1、第二章習(xí)題課第二章習(xí)題課 (2-1a)uoi2=R2CuiuoR1R22-1(a) 試建立圖所示電路的動(dòng)態(tài)微分方試建立圖所示電路的動(dòng)態(tài)微分方程。程。解：解：輸入量為輸入量為ui，輸出量為，輸出量為uo。ui=u1+uou1=i1R1ic=Cducdt=dtd(ui-uo)i1=i2-icu1= R1+uouo-Cd(ui-uo)dtR2R2ui=uoR1-C R1R2+C R1R2+uoR2duidtdtduouoR1+C R1R2+uoR2=R2ui+C R1R2duoduidtdtCucR1R2uii1i2uoicC第二章習(xí)題課第二章習(xí)題課 (2-1b)2-1(b) 試建立圖所示電路的動(dòng)態(tài)

2、微分方試建立圖所示電路的動(dòng)態(tài)微分方程。程。uouiR1LR2Ci1=iL+icuL=LdiLdtuoiL=i2=R2uL=LR2duodtic= = +CducdtCLR2d2uodt2duodt + uoR2CLR2d2uodt2duodti1= +Cuoi2=R2輸入量為輸入量為ui，輸出量為，輸出量為uo。ui=u1+uou1=i1R1ic=Cducdt=dtd(ui-uo)習(xí)題課一習(xí)題課一 (2-2)求下列函數(shù)的拉氏變換。求下列函數(shù)的拉氏變換。(1) f(t)=sin4t+cos4t解解:Lsinwt= ww2+s2sw2+s2Lsin4t+cos4t= 4s2+16ss2+16 =s

3、+4s2+16+ Lcoswt=(2) f(t)=t3+e4t3!解解:Lt3+e4t= + = +3!s3+11s-4s41s-4(3) f(t)=tneat解解:Ltneat=n!(s-a)n+1(4) f(t)=(t-1)2e2t解解:L(t-1)2e2t=e-(s-2)2(s-2)32-3-1 函數(shù)的拉氏變換。函數(shù)的拉氏變換。F(s)=s+1(s+1)(s+3)解解：A1=(s+2)s+1(s+1)(s+3)s=-2= -1(s+1)(s+3)A2=(s+3)s+1s=-3= 2F(s)= - 2s+31s+2f(t)=2e-3t-e-2tF(s)=s(s+1)2(s+2)2-3-2

4、函數(shù)的拉氏變換。函數(shù)的拉氏變換。解解:f(t)= est +lim ests(s+1)2s=-2ddsss+2s -1 =-2e-2t+lim( est+ est)s -1 sts+22(s+2)2=-2e-2t-te-t+2e-t=(2-t)e-t-2e-2tF(s)=2s2-5s+1s(s2+1)2-3-3 函數(shù)的拉氏變換。函數(shù)的拉氏變換。解解:F(s)(s2+1)s=+j=A1s+A2s=+jA1=1, A2=-5A3=F(s)s =1s=0 f(t)=1+cost-5sintF(s)= + +1ss2+1s-5s2+12-3-4 函數(shù)的拉氏變換。函數(shù)的拉氏變換。(4) F(s)=s+2

5、s(s+1)2(s+3)解解:f(t)= est + ests+2(s+1)2(s+3)s=0s+2s(s+1)2s=-3+ lim s -1d est s+2s(s+3)ds= + e-3t+lim + 23112s -1(-s2-4s-6)est(s2+3)2(s+2)tests2+3s= + e-3t- e-t- e-t2311234t2 (2-4-1)求下列微分方程。求下列微分方程。d2y(t)dt2+5 +6y(t)=6 ，初始條件：初始條件：dy(t)dty(0)=y(0)=2 。解解:s2Y(s)-sY(0)-Y(0)+5sY(s)-5Y(0)+6Y(s)=1sA1=sY(s)s

6、=0 y(t)=1+5e-2t-4e-3tA2=(s+2)Y(s) s=-2A3=(s+3)Y(s)s=-3A1=1 , A2=5 , A3=-4 Y(s)=6+2s2+12ss(s2+5s+6) (2-4-2)求下列微分方程。求下列微分方程。d3y(t)dt3 +4 +29 =29,d2y(t)dt2dy(t)dt初始條件初始條件:y(0)=0 , y(0)=17 , y(0)=-122 解解:2-5-a 試畫題試畫題2-1圖所示電路的動(dòng)態(tài)結(jié)構(gòu)圖圖所示電路的動(dòng)態(tài)結(jié)構(gòu)圖,并并求傳遞函數(shù)。求傳遞函數(shù)。CucR1R2uii1i2uoicC解解:ui=R1i1+uo ，i2=ic+i1UI(s)=R

7、1I1(s)+UO(s)ducic=CdtI2(s)=IC(s)+I1(s)IC(s)=CsUC(s)即即: =I1(s)UI(s)-UO(s)R1UI(s)-UO(s)Cs=IC(s)UO(s)UI(s)=1R1( sC)R21+1R1( sC)R2=R2+R1R2sCR1+R2+R1R2sC1R1sCR2UI(s)-UO(s)IC(s)I1(s)I2(s)1R1sC R2( )UI(s)-UO(s)2-5-b 試畫出題試畫出題2-1圖所示的電路圖所示的電路的動(dòng)態(tài)結(jié)構(gòu)圖的動(dòng)態(tài)結(jié)構(gòu)圖,并并求傳遞函數(shù)。求傳遞函數(shù)。 uouiR1LR2C解：解：ui=R1I1+ucuc=uo+uLuL=LdiLd

8、tiL=uoR2i1=iL+icic=CducdtUi(s)=R1I1(s)+UC(s)UC(s)=UO(s)+UL(s)UL(s)=sLIL(s)I1(s)=IL(s)+IC(s)1R1CssLR2I1UOUiIC-UC=UO+ULILULI2(s)=UO(s)R2IC(s)=CsUC(s)I1(s)=UO(s)R2I1(s)=UI(s)+UC(S)R1即：即：IL(s)=I1(s)-IC(s)IC(s)=UC(s)Cs解解:電路等效為電路等效為:2-6-a 用運(yùn)算放大器組成的有源電網(wǎng)絡(luò)如圖用運(yùn)算放大器組成的有源電網(wǎng)絡(luò)如圖所示所示,試采用復(fù)數(shù)阻抗法寫出它們的傳遞函數(shù)。試采用復(fù)數(shù)阻抗法寫出它們

9、的傳遞函數(shù)。UO= R3SCR2R21UIR1UOR3SCR2R21SC1= R1+R3+R2R3CS=R1(R2SC+1)R2R3=( + )R1(R2SC+1)R1R1R2= ( +R3 )(R2SC+1)1=R21R3R2SCR1C(S)=UO(S)UI(S)CR1R2R3uiuoCR1R2R3uiuoCR1R2R3uiuoR4R52-6-b 用運(yùn)算放大器組成的有源電網(wǎng)絡(luò)如用運(yùn)算放大器組成的有源電網(wǎng)絡(luò)如力所示力所示,試采用復(fù)數(shù)阻抗法寫出它們的傳試采用復(fù)數(shù)阻抗法寫出它們的傳遞函數(shù)。遞函數(shù)。=R5R4+ R5UO(R3SC+1)R2R3SC+R2+R3UOUI=(R2R3SC+R2+R3)(

10、R4+R5)R1(R3SC+1)R5=(R4+R5)(R2+R3)( SC+1) R2R3R2+R3R1R5(R3SC+1)UIR1=R5R4+ R5UOR2R3SCSCR3SC1R5R4+ R5UOR2R3R3SC 1c(t)t0TK(t)(t)2-8 設(shè)有一個(gè)初始條件為零的系統(tǒng)，系設(shè)有一個(gè)初始條件為零的系統(tǒng)，系統(tǒng)的輸入、輸出曲線如圖，求統(tǒng)的輸入、輸出曲線如圖，求G(s)。c(t)t0TK(t)(t) c(t)= KTt-(t-T)KT C(s)= K(1-e )Ts2-TSC(s)=G(S)第二章習(xí)題課第二章習(xí)題課 (2-8)解解:2-9 若系統(tǒng)在單位階躍輸入作用時(shí)，已若系統(tǒng)在單位階躍輸入

11、作用時(shí)，已知初始條件為零的條件下系統(tǒng)的輸出響知初始條件為零的條件下系統(tǒng)的輸出響應(yīng)，求系統(tǒng)的傳遞函數(shù)和脈沖響應(yīng)。應(yīng)，求系統(tǒng)的傳遞函數(shù)和脈沖響應(yīng)。r(t)=I(t)c(t)=1-e +e-2t-t解解:R(s)=1sG(S)=C(s)/R(s)1s+21s-C(s)=1s+1+=s(s+1)(s+2)(s2+4s+2) =(s+1)(s+2)(s2+4s+2) C(s)=(s+1)(s+2)(s2+4s+2) 脈沖響應(yīng)脈沖響應(yīng):2s+2=1+1s+1-c(t)=( (t t) )+2e +e-2t-t第二章習(xí)題課第二章習(xí)題課 (2-)2-10 已知系統(tǒng)的微分方程組的拉氏變換已知系統(tǒng)的微分方程組的拉

12、氏變換式，試畫出系統(tǒng)的動(dòng)態(tài)結(jié)構(gòu)圖并求傳遞式，試畫出系統(tǒng)的動(dòng)態(tài)結(jié)構(gòu)圖并求傳遞函數(shù)。函數(shù)。解解:X1(s)=R(s)G1(s)-G1(s)G7(s)-G8(s)C(s)X2(s)=G2(s)X1(s)-G6(s)X3(s)X3(s)=G3(s)X2(s)-C(s)G5(s)C(s)=G4(s)X3(s)G1G2G3G5-C(s)-R(s)G4G6G8G7X1(s)=R(s)-C(s)G7(s)-G8(s)G1(s)C(s)G7(s)-G8(s)G6(s)X3(s)X1(s)X2(s)C(s)G5(s)X3(s)G1G2G3G5-C(s)-R(s)G4G2G6G8G7G1G2G5-C(s)-R(s)

13、G7-G81+G3G2G6G3G4-C(s)R(s)G7-G81+G3G2G6 +G3G4G5G1G2G3G41+G3G2G6 +G3G4G5+G1G2G3G4(G7 -G8)G1G2G3G4R(s)C(s)=第二章習(xí)題課第二章習(xí)題課 (2-10)解解:2-11(a)G1(s)G2(s)G3(s)H1(s)_+R(s)C(s)H2(s)G1(s)G2(s)H1(s)_+R(s)C(s)H2(s)G3(s)求系統(tǒng)的求系統(tǒng)的傳遞函數(shù)傳遞函數(shù)1+G2H1G2 G1+G31+G1H21+G2H1G2 1+G2H1G2 =1+G2H1+G1G2H2G2 R(s)C(s)=1+G2H1+G1G2H2G2G

14、1+G2G3G1(s)G2(s)G3(s)H1(s)_+R(s)C(s)G 1(s)H2(s)第二章習(xí)題課第二章習(xí)題課 (2-11a)2-11(a)G1(s)G2(s)G3(s)H1(s)_+R(s)C(s)H2(s)求系統(tǒng)的求系統(tǒng)的傳遞函數(shù)傳遞函數(shù)解解:L1L1=-G2H1L2L2=-G1G2H1P1=G1G2P2=G3G21 =12 =1R(s)C(s)=nk=1Pkk=1+G2H1+G1G2H21+G2H1+G1G2H2G2G1+G2G3 =第二章習(xí)題課第二章習(xí)題課 (2-11a)解解:2-11(b)G1(s)G2(s)G3(s)G4(s)_+R(s)C(s)H(s)求系統(tǒng)的求系統(tǒng)的傳遞

15、函數(shù)傳遞函數(shù)G1(s)G2(s)G3(s)G4H_+R(s)C(s)H(s)1+G4G1HG1 G2(s)G3(s)_+R(s)C(s)H(s)1+G4HG1G1 G2G3_+R(s)C(s)1+G4HG1G1 HG1 1+G4HG1G1+G3 (1+HG1G4)1+G4HG1G2 (1+HG1G4)1+G4G1H+G1G2HR(s)C(s)=1+G1G2H+G1G4HG1G2+G2G3+G1G2G3G4 H第二章習(xí)題課第二章習(xí)題課 (2-11b)解解:2-11(b)G1(s)G2(s)G3(s)G4(s)_+R(s)C(s)H(s)求系統(tǒng)的求系統(tǒng)的傳遞函數(shù)傳遞函數(shù)R(s)C(s)=1+G1G

16、2H+G1G4HG1G2+G2G3+G1G2G3G4 HL1L1=-G1G2HL1=-G1G4HL2P1=G1G21 =1P2=G3G2=1+G4G2H+G1G2H2=1+G1G4H第二章習(xí)題課第二章習(xí)題課 (2-11b)H1_+G1+C(s)R(s)G3G22-11c 求系統(tǒng)的閉環(huán)傳遞函數(shù)求系統(tǒng)的閉環(huán)傳遞函數(shù) 。解解: H1_+G1+C(s)R(s)G3G2H1R(s)C(s)1+G1G2+G1H1G3H1G1G2 (1 G3H1)=_G1C(s)R(s)G2H1+G21-G3H11第二章習(xí)題課第二章習(xí)題課 (2-11c)H_G1+C(s)R(s)G22-11d 求系統(tǒng)的閉環(huán)傳遞函數(shù)求系統(tǒng)的

17、閉環(huán)傳遞函數(shù) 。解解: (1)_G1+C(s)R(s)G2HG21+G2H1(G1+G2 )R(s)C(s)=(2)L1L1=-G2HP1=G11 =1P2=G22 =1第二章習(xí)題課第二章習(xí)題課 (2-11d)-_G1+C(s)R(s)G2G3G42-11e 求系統(tǒng)的閉環(huán)傳遞函數(shù)求系統(tǒng)的閉環(huán)傳遞函數(shù) 。解解: (1) _C(s)R(s)G1+G2G3-G4C(s)=R(s) 1+(G1+G2)(G3-G4)(G1+G2)1+G1G3+G2G3G1G4-G2G4=(G1+G2)第二章習(xí)題課第二章習(xí)題課 (2-11e)L1L2L3L4L2=G1G4L3=-G2G3L4=G2G4(2)L1=-G1G

18、3P1=G11 =1P2=G22 =11+G1G3+G2G3G1G4-G2G4=(G1+G2)C(s)R(s)_G1+C(s)R(s)G22-11f 求系統(tǒng)的閉環(huán)傳遞函數(shù)求系統(tǒng)的閉環(huán)傳遞函數(shù) 。_C(s)R(s)G11-G2G2C(s)=R(s)1+1-G2G1G1G21+G1G2G2G1 (1 G2)=第二章習(xí)題課第二章習(xí)題課 (2-11f)解解: (1) (2)L1L1=-G1G2L2L2=G2P1=G11 =1-G2=1+G1G2-G2C(s)R(s) 1+G1G2G2G1 (1 G2)=2-12(a)R(s)G1(s)G2(s)H2(s)_+C(s)H3(s)H1(s)_+D(s)解解

19、:求求:D(s)C(s)R(s)C(s)D(s)=01-G2H2G2 G(s)=1-G2H2G1G2 C(s)=R(s)1+1-G2H2G1G2H31-G2H2G1G2 1-G2H2+G1G2H3G2G1=R(s)=0結(jié)構(gòu)圖變結(jié)構(gòu)圖變 換成：換成：G2(s)H2(s)_+C(s) G1H3G1H1_D(s)1-G2H2G2 1-G1H1C(s)=D(s)1+1-G2H2G21-G2H2G2 G1H3(1-G1H1 )1-G2H2+G1G2H3G2(1-G1H1 )=第二章習(xí)題課第二章習(xí)題課 (2-12a)2-12(b)求求:D(s)C(s)R(s)C(s)R(s)Gn+D(s)解解: D(s)

20、=0G(s)=1+G1G2HG1G2 G1G2H_C(s)C(s)=R(s)1+1+G1G2HG1G2 1+G1G2HG1G2 1+G1G2H+G1G2G1G2=R(s)=0Gn+D(s)結(jié)構(gòu)圖變結(jié)構(gòu)圖變 換成：換成：G1G2H-C(s)Gn+D(s)G1G2H-C(s)Gn/G1+D(s)1+G1G2HG1G2 -C(s)+D(s)1+G1G2HG2Gn 1+G1G2HG2G1系統(tǒng)的傳遞函數(shù)系統(tǒng)的傳遞函數(shù):)C(s)=D(s)1+1(1+1+G1G2HG1G2 1+G1G2HGnG2 1+G1G2+G1G2H=1+GnG2+G1G2H第二章習(xí)題課第二章習(xí)題課 (2-12b)2-13(a)求求

21、:R(s)E(s)R(s)C(s)C(s)E(s)G1G2G3_+R(s)解解:L1L1=-G2L2L2=-G1G2G3P1=G2G3P2=G1G2G3R(s)C(s)=1+G2+G1G2G3G2G3+G1G2G31 =12 =1E(s)結(jié)構(gòu)圖變結(jié)構(gòu)圖變 換成：換成：G1G2+-E(s)G3-R(s)G1+-E(s)R(s)1+G2G3G2 -E(s)-R(s)1+G2G2G3G1G2G31+G2系統(tǒng)的傳遞函數(shù)系統(tǒng)的傳遞函數(shù):)E(s)=R(s)1+1(1-G1G2G31+G2 1+G2G2G31+G2+G1G2G3=1+G2-G2G3第二章習(xí)題課第二章習(xí)題課 (2-13a)R(s)G4(s)

22、+C(s)E(s)G1(s)G2(s)-+G3(s)第二章習(xí)題課第二章習(xí)題課 (2-14)D(s)X(s)2-14 求求:R(s)C(s)解解: D(s)=0結(jié)構(gòu)圖結(jié)構(gòu)圖變換為變換為 R(s)G4(s)+C(s)G1(s)G2(s)-+G3(s)G3(s)(G1+G2)(G3+G4)1+(G1+G2)G3G1+G2C(s)R(s)=1+G3(G1+G2)(G1+G2)(G3+G4)D(s)第二章習(xí)題課第二章習(xí)題課 (2-14)R(s)+-E(s)G3G2G1E(s)R(s)=1+G3(G1+G2)1R(s)G4(s)+C(s)E(s)G1(s)G2(s)-+G3(s)X(s)求求:R(s)E(

23、s)2-14解解: D(s)=0結(jié)構(gòu)圖變換為結(jié)構(gòu)圖變換為 G3(G1+G2)D(s)C(s)D(s)R(s)C(s)G4(s)+E(s)G1(s)G2(s)-+G3(s)X(s)求求:2-14解解:R(s)=0D(s)C(s)=1E(s)X(s)=G2(s)E(s)X(s)第二章習(xí)題課第二章習(xí)題課 (2-14)C1(s)R1(s)第二章習(xí)題課第二章習(xí)題課 (2-15)求求:2-15+G1G2G3C1(s)R1(s)+-H2H1G4G5-G6C2(s)R2(s)解解:結(jié)構(gòu)圖結(jié)構(gòu)圖變換為變換為 +G1G2G3C1(s)R1(s)-H2H1G4G5-1+G4G4G5H1H21+G4G4G5H1H21

24、+G1G1G21+G4+G1G4G5H1H2-G1G2-G1G2G4G1G2G3(1+G4 )=C1(s)R1(s)1+G4+G1G4G5H1H2G1G2(1+G4 )=C2(s)R2(s)G3C1(s)R1(s)+G1G2-H2H1G4G5G6-C2(s)R2(s)求求:2-14解解: 結(jié)構(gòu)圖結(jié)構(gòu)圖變換為變換為 +G4G5G6C2(s)R2(s)-H1H2G2G1-第二章習(xí)題課第二章習(xí)題課 (2-15)1+G4G4G5G1H1H21-G1G21+G4+G1G4G5H1H2-G1G2-G1G2G4G4G5G6(1-G1G2)=1+1+G4G6G4G5G1H1H21-G2G21+G4G4G5C2

25、(s)R2(s)C2(s)G4G5G6-+G1G2-H2H1R1(s)G3C1(s)R2(s)求求:2-14解解:結(jié)構(gòu)圖變換為結(jié)構(gòu)圖變換為 G4G5G6-+-G1G2H2H1R1(s)C2(s)第二章習(xí)題課第二章習(xí)題課 (2-15)C2(s)R1(s)1+G4H2G4G5G11-G1G21+G4+G1G4G5H1H2-G1G2-G1G2G4G1G4G5G6H2=1+G6H1C2(s)R1(s)G11-G1G21+G4H2G4G5G11-G1G21+G4H2G4G5C2(s)G6R1(s)G4G5-+G1G2-H2H1G3C1(s)R2(s)求求:2-14解解:結(jié)構(gòu)圖變換為結(jié)構(gòu)圖變換為 第二章習(xí)

26、題課第二章習(xí)題課 (2-15)G1G2G3+G5G4-H1R2(s)C1(s)H2-C1(s)R2(s)G1G2G3+G5G4-H1R2(s)C1(s)H2-G21+G4G4G51-G1G2-G1H11+G4+G1G4G5H1H2-G1G2-G1G2G4-G1G2G3G4G5H1=1+1+G4G4G5G1H11-G2G21+G4G4G5C1(s)R2(s)H2-G1H11-G2G2G2G3G3C1(s)R1(s) +G1G2+-H2H1G4G5-G6C2(s)R2(s)C1(s)R1(s)求求2-15第二章習(xí)題課第二章習(xí)題課 (2-15)解解:L1=G1G2L3=-G4L2=-G1G4G5H1

27、H2P1=G1G2G3=1-G1G2+G1G4G35H1H2+G4 -G1G2G41=1+G41+G4+G1G4G5H1H2-G1G2-G1G2G4G1G2G3(1+G4 )=C1(s)R1(s)G3C1(s)R1(s) +G1G2+-H2H1G4G5-G6C2(s)R2(s)C2(s)R2(s)求求2-15解解:L1=G1G2L3=-G4L2=-G1G4G5H1H2P1=G4G5G6=1-G1G2+G1G4G5H1H2+G4 -G1G2G41=1-G1G21+G4+G1G4G5H1H2-G1G2-G1G2G4G4G5G6(1-G1G2)=C2(s)R2(s)第二章習(xí)題課第二章習(xí)題課 (2-1

28、5)G3C1(s)R1(s) +G1G2+-H2H1G4G5-G6C2(s)R2(s)C1(s)R2(s)求求2-15解解:L1=G1G2L3=-G4L2=-G1G4G5H1H2=1-G1G2+G1G4G5H1H2+G4 -G1G2G41=1P1=-G1G2G3G4G5H11+G4+G1G4G5H1H2-G1G2-G1G2G4-G1G2G3G4G5H1=C1(s)R2(s)第二章習(xí)題課第二章習(xí)題課 (2-15)G3C1(s)R1(s) +G1G2+-H2H1G4G5-G6C2(s)R2(s)C2(s)R1(s)求求2-15解解:L1=G1G2L3=-G4L2=-G1G4G5H1H2=1-G1G

29、2+G1G4G5H1H2+G4 -G1G2G41=1P1=G1G4G5G6H21+G4+G1G4G5H1H2-G1G2-G1G2G4G1G4G5G6H2=C2(s)R1(s)第二章習(xí)題課第二章習(xí)題課 (2-15)3-1 設(shè)溫度計(jì)需要在一分鐘內(nèi)指示出響設(shè)溫度計(jì)需要在一分鐘內(nèi)指示出響應(yīng)值的應(yīng)值的98%，并且假設(shè)溫度計(jì)為一階系并且假設(shè)溫度計(jì)為一階系統(tǒng)，求時(shí)間常數(shù)統(tǒng)，求時(shí)間常數(shù)T。如果將溫度計(jì)放在。如果將溫度計(jì)放在澡盆內(nèi)，澡盆的溫度以澡盆內(nèi)，澡盆的溫度以10oC/min的速度的速度線性變化，求溫度計(jì)的誤差。線性變化，求溫度計(jì)的誤差。第三章習(xí)題課第三章習(xí)題課 (3-1)解解: c(t)=c()98%t=

30、4T=1 min r(t)=10t e(t)=r(t)-c(t) c(t)=10(t-T+e)-t/T=10(T-e)-t/Tess=lim t e(t)=10T=2.5T=0.253-2 電路如圖，設(shè)系統(tǒng)初始狀態(tài)為領(lǐng)電路如圖，設(shè)系統(tǒng)初始狀態(tài)為領(lǐng)第三章習(xí)題課第三章習(xí)題課 (3-2)解解:(1)求系統(tǒng)的單位階躍響應(yīng)求系統(tǒng)的單位階躍響應(yīng),及及uc(t1)=8 時(shí)的時(shí)的t1值值R1Cs+1R1/R0G(s)= R0=20 k R1=200 k C=2.5F uc(t)=K(1 e tT-)KTs + 1=T=R1C=0.5 K=R1/R0=10 =10(1 e -2t) 8=10(1 e -2t)0

31、.8=1 e -2te -2t =0.2 t=0.8 -+R1R0Curucg(t)=e-t/TTK t1=0.8 =4 (2) 求系統(tǒng)的單位脈沖響應(yīng)，單位斜坡求系統(tǒng)的單位脈沖響應(yīng)，單位斜坡 響應(yīng)響應(yīng),及單位拋物響應(yīng)在及單位拋物響應(yīng)在t1時(shí)刻的值時(shí)刻的值第三章習(xí)題課第三章習(xí)題課 (3-2)解解:uc(t)=K(t-T+Te-t/T) =4 R(s)=1s2R(s)=1R(s)=1s3T2=K(ss+1/T+Ts2-1s3-T2)=1.2 1s3KTs + 1Uc(s)= -0.5t+0.25-0.25e-2t)12t2uc(t)=10( 3-3 已知單位負(fù)反饋系統(tǒng)的開環(huán)傳遞函已知單位負(fù)反饋系統(tǒng)

32、的開環(huán)傳遞函數(shù)，求系統(tǒng)的單位階躍響應(yīng)。數(shù)，求系統(tǒng)的單位階躍響應(yīng)。4s(s+5)G(s)=第三章習(xí)題課第三章習(xí)題課 (3-3)解解:= s2+5s+4 C(s)R(s)4s(s+1)(s+4)C(s)=4R(s)=s11/3s+41s+=4/3s+1-13c(t)=1+ -4t-t43-e e 3-4 已知單位負(fù)反饋系統(tǒng)的開環(huán)傳遞函已知單位負(fù)反饋系統(tǒng)的開環(huán)傳遞函數(shù)，求系統(tǒng)的上升時(shí)間數(shù)，求系統(tǒng)的上升時(shí)間tr、峰值時(shí)間、峰值時(shí)間tp、超調(diào)量超調(diào)量% % 和調(diào)整時(shí)間和調(diào)整時(shí)間ts。1s(s+1)G(s)=第三章習(xí)題課第三章習(xí)題課 (3-4)解解:=s2+s+1 C(s)R(s)12 = 1n 2n =

33、1=0.5=1n =0.866d = n2 1- =60o-1=tg 21- tr=d - = 3.14-3.14/30.866=2.42 tp=d3.140.866 = =3.63%=100%e-1-2=16%-1.8ets= 3n =6 ts= 4n =83-6 已知系統(tǒng)的單位階躍響應(yīng)已知系統(tǒng)的單位階躍響應(yīng):c(t)=1+0.2e -60t-10t-1.2e(1) 求求系統(tǒng)的閉環(huán)傳遞函數(shù)。系統(tǒng)的閉環(huán)傳遞函數(shù)。(2) 求求系統(tǒng)的阻尼比和無阻尼振蕩頻率。系統(tǒng)的阻尼比和無阻尼振蕩頻率。第三章習(xí)題課第三章習(xí)題課 (3-6)解解:0.2s+601s+C(s)=1.2s+10-s(s+60)(s+10

34、)=600= s2+70s+600 C(s)R(s)600R(s)=s12 =600n 2n =70=1.43=24.5n 3-7 設(shè)二階系統(tǒng)的單位階躍響應(yīng)曲線如圖，設(shè)二階系統(tǒng)的單位階躍響應(yīng)曲線如圖，系統(tǒng)的為單位反饋，求系統(tǒng)的傳遞函數(shù)。系統(tǒng)的為單位反饋，求系統(tǒng)的傳遞函數(shù)。tc(t)010.11.3解解:第三章習(xí)題課第三章習(xí)題課 (3-7)tp=0.121- n =0.3e-1-2e1-2=3.3n2 1- 3.140.1=31.421- /=ln3.3 =1.1921- )2/ (=1.429.862=1.42-1.422=0.35=33.4n nn s(s+2 )G(s)=21115.6s(

35、s+22.7)=3-8 已知單位負(fù)反饋系統(tǒng)的開環(huán)傳遞函已知單位負(fù)反饋系統(tǒng)的開環(huán)傳遞函數(shù)，求系統(tǒng)數(shù)，求系統(tǒng)K、T值以滿足動(dòng)態(tài)指標(biāo)：值以滿足動(dòng)態(tài)指標(biāo)：%30%30%，ts0.3(5%)0.3(5%)。Ks(Ts+1)G(s)=第三章習(xí)題課第三章習(xí)題課 (3-8)解解:=Ts2+s+K C(s)R(s)Ks2+=T1s+TKTKT12n =ts= 3n 0.3n 10T0.050.3e-1-20.3528.6n 2 =n TK=817.96K40.93-11 已知閉環(huán)系統(tǒng)的特征方程式，試用已知閉環(huán)系統(tǒng)的特征方程式，試用勞斯判據(jù)判斷系統(tǒng)的穩(wěn)定性。勞斯判據(jù)判斷系統(tǒng)的穩(wěn)定性。第三章習(xí)題課第三章習(xí)題課 (3

36、-11)(1) s3+20s2+9s+100=0解：解： 勞斯表如下：勞斯表如下：s1 s0 s3 s2 1 9 20 100 4 100系統(tǒng)穩(wěn)定。系統(tǒng)穩(wěn)定。(3) s4+8s3+18s2+16s+5=0 1 18 5 s4 s3 8 16 勞斯表如下：勞斯表如下：s2 16 5 s1 216 16 s0 5系統(tǒng)穩(wěn)定。系統(tǒng)穩(wěn)定。3-12 已知單位負(fù)反饋系統(tǒng)的開環(huán)傳遞函已知單位負(fù)反饋系統(tǒng)的開環(huán)傳遞函數(shù)，試確定系統(tǒng)穩(wěn)定時(shí)數(shù)，試確定系統(tǒng)穩(wěn)定時(shí)K值范圍。值范圍。第三章習(xí)題課第三章習(xí)題課 (3-12)G(s)=s(s+1)(0.5s2+s+1)K(0.5s+1)解解:0.5s4+1.5s3+2s2+s+

37、0.5Ks+K=0 0.5 2 K s4 s3 1.5 1+0.5K s2 b31 b31= 1.5*2-0,5(1+0.5K) 1.5 3-0,5-0.25K0 0.25K003-14 已知系統(tǒng)結(jié)構(gòu)如圖，試確定系統(tǒng)穩(wěn)已知系統(tǒng)結(jié)構(gòu)如圖，試確定系統(tǒng)穩(wěn)定時(shí)定時(shí)值范圍。值范圍。第三章習(xí)題課第三章習(xí)題課 (3-14)R(sR(s) )- s+1s10s(s+1)C(s)C(s)解解:G(s)=s2(s+1)10(s+1)(s)=s3 +s2+10 s+1010(s+1)s3 s2 1 10 1 10 s1 b31 s0 10 b31= 10 -10 1 01r(t)=I(t)+2t+t23-16 已知

38、單位反饋系統(tǒng)的開環(huán)傳遞函數(shù)，已知單位反饋系統(tǒng)的開環(huán)傳遞函數(shù)，試求試求K p、Kv和和Ka 并求穩(wěn)態(tài)誤差并求穩(wěn)態(tài)誤差ess第三章習(xí)題課第三章習(xí)題課 (3-16)解：解： (0.1s+1)(0.2s+)(1) G(s)=20Kp=20 =0ess1=R01+Ks2R(s)=1s2+s32+=121K =0 ess2=Ka=0 ess3= ess=1K =10 Ka=0 ess3= ess=s(s+2)(s+10)(2) G(s)=200s(0.5s+1)(0.1s+1)=10Kp= ess1=0ess2=K 2=210s2(s2+4s+10)(3) G(s)=10(2s+1)=2Kp= ess1=

39、0s2(0.1s2+0.4s+1)=(2s+1)K = ess2=0Ka=1 ess3=2 ess=23-17 已知系統(tǒng)結(jié)構(gòu)如圖。已知系統(tǒng)結(jié)構(gòu)如圖。第三章習(xí)題課第三章習(xí)題課 (3-17)解解:-R(sR(s) )-K1ss2C(s)C(s)(1) 單位階躍輸入單位階躍輸入:G(s)=s2+K1sK1(s)=s2+K1s+K1K1確定確定K1 和和值值 。 %=20% %=20% ts =1.8(5%)2n =K12 =K1n =0.2e-1-2ts= 3n =1.8=0.45n31.8*0.45=3.72 n K1=13.7=0.24(2)求系統(tǒng)的穩(wěn)態(tài)誤差：求系統(tǒng)的穩(wěn)態(tài)誤差：r(t)=I(t)

40、, t ,12t2 解解:G(s)=s2+K1sK1=s(s+1)K111=1Kp= ess1=0R(s)=1sR(s)=s21K =K ess2=0.24R(s)=s31Ka=0 ess3=第三章習(xí)題課第三章習(xí)題課 (3-17)3-18 已知系統(tǒng)結(jié)構(gòu)如圖。為使已知系統(tǒng)結(jié)構(gòu)如圖。為使=0.7時(shí)時(shí) 單位斜坡輸入的穩(wěn)態(tài)誤差單位斜坡輸入的穩(wěn)態(tài)誤差ess=0.25第三章習(xí)題課第三章習(xí)題課 (3-18)解解:確定確定 K 和和值值 。-s-Ks(s+2)R(sR(s) )C(s)C(s)G(s)=s2+2s+K sK=s(s+1)2+K12+KK(s)=s2+(2+K )s+KK2n =2+K2 =Kn

41、 =2*0.7 K ess= 2+K K =0.25 = 0.25K-2 K K K=31.6=0.186+R(s)G(s)F(s)C(s)+-D1(s)D2(s)E(s)3-19 系統(tǒng)結(jié)構(gòu)如圖。系統(tǒng)結(jié)構(gòu)如圖。第三章習(xí)題課第三章習(xí)題課 (3-19)解解: r(t)=d1(t)=d2(t)=I(t)(1) 求求r(t)作下的穩(wěn)態(tài)誤差作下的穩(wěn)態(tài)誤差essr=lim s 1+G(s)F(s)s0s1= 1+G(0)F(0)1(2) 求求d1(t)和和d2(t)同時(shí)作用下的穩(wěn)態(tài)誤差同時(shí)作用下的穩(wěn)態(tài)誤差essd= lim s -F(s)1+G(s)F(s)s0-11+G(s)F(s)+s1Ed(s)=

42、-G2(s)H(s)1+G1(s)G2(s)H(s)D(s)= 1+G(0)F(0)-1+F(s)(3) 求求d1(t)作用下的穩(wěn)態(tài)誤差作用下的穩(wěn)態(tài)誤差G(s)=Kp +KsJs1F(s)=essd= lim s -F(s)1+G(s)F(s)s0s1-s0s1= lim s 1+(Kp+KsJs1)Js1=0返回返回4-1 已知系統(tǒng)的零、極點(diǎn)分布如圖，大已知系統(tǒng)的零、極點(diǎn)分布如圖，大致繪制出系統(tǒng)的根軌跡。致繪制出系統(tǒng)的根軌跡。第四章習(xí)題課第四章習(xí)題課 (4-1)解解:j0(1)(2)j0(3)j0(4)j0600900600第四章習(xí)題課第四章習(xí)題課 (4-1)(5)j0(6)j0j0(7)(

43、8)j0600450135036010804-2 已知開環(huán)傳遞函數(shù)，試用解析法繪制已知開環(huán)傳遞函數(shù)，試用解析法繪制出系統(tǒng)的根軌跡，并判斷點(diǎn)出系統(tǒng)的根軌跡，并判斷點(diǎn)(-2+j0),(0+j1),(-3+j2)是否在根軌跡上。是否在根軌跡上。第四章習(xí)題課第四章習(xí)題課 (4-2)解解:Kr(s+1)G(s)=Kr(s)=s+1+KrKr=0s=-1-Krj0系統(tǒng)的根軌跡系統(tǒng)的根軌跡 s=-1-1Kr=s=-s=-2+j0-2s=0+j10+j1-3+j2s=-3+j2j04-3 已知系統(tǒng)的開環(huán)傳遞函數(shù)，試?yán)L制出已知系統(tǒng)的開環(huán)傳遞函數(shù)，試?yán)L制出根軌跡圖。根軌跡圖。第四章習(xí)題課第四章習(xí)題課 (4-3)解

44、解:s(s+1)(s+5)(1) G(s)=Kr(s+1.5)(s+5.5) 1）開環(huán)零、極點(diǎn)）開環(huán)零、極點(diǎn) p1 p1=0p2p2=-1p3 p3=-52）實(shí)軸上根軌跡段）實(shí)軸上根軌跡段 p1p2z1 z1=-1.5z2 z2=-5.5z1p3z2- 3）根軌跡的漸近線）根軌跡的漸近線 n-m= 1= +180o4）分離點(diǎn)和會(huì)合點(diǎn)）分離點(diǎn)和會(huì)合點(diǎn)A(s)B(s)=A(s)B(s)A(s)=s3+6s2+5sB(s)=s2+7s+8.25A(s)=3s2+12s+5B(s)=2s+7解得解得 s1=-0.63s2=-2.5s3=-3.6s4=-7.28第四章習(xí)題課第四章習(xí)題課 (4-3)s(s

45、+1)(s+4)(2) G(s)=Kr(s+1.5) 1）開環(huán)零、極點(diǎn)）開環(huán)零、極點(diǎn) p1 p1=0p2p2=-1p3 p3=-4j02）實(shí)軸上根軌跡段）實(shí)軸上根軌跡段 p1p2z1 z1=-1.5p3z1 3）根軌跡的漸近線）根軌跡的漸近線 n-m= 2= +90o2= -1-4+1.5 =-1.75-1.754）分離點(diǎn)和會(huì)合點(diǎn)）分離點(diǎn)和會(huì)合點(diǎn)A(s)=s3+5s2+4sB(s)=s+1.5A(s)=3s2+10s+4B(s)=1解得解得 s=-0.625）系統(tǒng)根軌跡）系統(tǒng)根軌跡第四章習(xí)題課第四章習(xí)題課 (4-3)s(s+1)2(3) G(s)=Kr 1）開環(huán)零、極點(diǎn)）開環(huán)零、極點(diǎn) p1=0

46、p2=-1p3 p3=-12）實(shí)軸上根軌跡段）實(shí)軸上根軌跡段 p1p2p3- 3）根軌跡的漸近線）根軌跡的漸近線 n-m=3j0p1 p23= -1-1 =-0.67-0.67 4）根軌跡與虛軸的交點(diǎn)）根軌跡與虛軸的交點(diǎn)= +180o+60o , 閉環(huán)特征方程為閉環(huán)特征方程為 s3+2s2+s+Kr=0Kr=0 Kr=2 2,3=11=01-15）分離點(diǎn)和會(huì)合點(diǎn)）分離點(diǎn)和會(huì)合點(diǎn)A(s)=s3+2s2+sB(s)=1A(s)=3s2+4s+1B(s)=0解得解得 s=-0.336）系統(tǒng)根軌跡）系統(tǒng)根軌跡第四章習(xí)題課第四章習(xí)題課 (4-3) 1）開環(huán)零、極點(diǎn)）開環(huán)零、極點(diǎn) p1=0p2=-3 p3

47、=-7j02）實(shí)軸上根軌跡段）實(shí)軸上根軌跡段 p1p2p4 p4=-15z1 z1=-8p3z1p4- 3）根軌跡的漸近線）根軌跡的漸近線 n-m=3s(s+3)(s+7)(s+15)(4) G(s)=Kr(s+8)p1 p2p3 3= -3-7-15+8 =-5.67= +180o+60o , -5.67 4）根軌跡與虛軸的交點(diǎn)）根軌跡與虛軸的交點(diǎn)閉環(huán)特征方程為閉環(huán)特征方程為 s4+25s3+171s2+323s+8Kr=0Kr=0 1=0Kr=638 2,3=6.26.2-6.25）分離點(diǎn)和會(huì)合點(diǎn)）分離點(diǎn)和會(huì)合點(diǎn)A(s)=s4+25s3+171s2+315sB(s)=s+8A(s)=4s3

48、+75s2+342s+315B(s)=2s+7解得解得 s=-1.46）系統(tǒng)根軌跡）系統(tǒng)根軌跡第四章習(xí)題課第四章習(xí)題課 (4-4)4-5 已知系統(tǒng)的開環(huán)傳遞函數(shù)。已知系統(tǒng)的開環(huán)傳遞函數(shù)。(1)試?yán)L制試?yán)L制出根軌跡圖。出根軌跡圖。(2)增益增益Kr為何值時(shí)，復(fù)數(shù)特為何值時(shí)，復(fù)數(shù)特征根的實(shí)部為征根的實(shí)部為-2。第四章習(xí)題課第四章習(xí)題課 (4-5)解解:，s(s+1)Kr(s+2)G(s)=j0p1=0p1 p2=-1p2 z1=-2z1 p1p2z1-分離點(diǎn)和會(huì)合點(diǎn)分離點(diǎn)和會(huì)合點(diǎn)s2+4s+2=0s1=-3.41s2=-0.59s2+s+Krs+2Kr=0閉環(huán)特征方程式閉環(huán)特征方程式系統(tǒng)根軌跡系統(tǒng)

49、根軌跡s=-2+j(-2+j)2+(-2+j )(1+Kr)+2Kr=0-4 +(1+Kr ) =04-2-2(1+Kr )+2Kr=0Kr=3=1.414-6 已知系統(tǒng)的開環(huán)傳遞函數(shù)，試確定閉已知系統(tǒng)的開環(huán)傳遞函數(shù)，試確定閉環(huán)極點(diǎn)環(huán)極點(diǎn)=0.5時(shí)的時(shí)的Kr值。值。第四章習(xí)題課第四章習(xí)題課 (4-6)解解:p1=0 p2=-1 p3=-3p1p2p3 -8+60o= +180o, 3= -1-3=-1.3根軌跡的分離點(diǎn)：根軌跡的分離點(diǎn)：A(s)B(s)=A(s)B(s)3s2+8s+3=0s1=-0.45s2=-2.2s2沒有位于根軌跡上，舍去。沒有位于根軌跡上，舍去。js(s+1)(s+3)

50、KrG(s)H(s)=與虛軸交點(diǎn)與虛軸交點(diǎn) s3+4s2+3s+Kr=02=0Kr-43+3=0-Kr=0 Kr=12 2,3=1.71=01.7-1.7p3-30p1 p2-1系統(tǒng)根軌跡系統(tǒng)根軌跡s1=0.5得得 s1=-0.37+j0.8Kr=|s3|s3+1|s3+3|s3=-4+0.372=-3.26=3.262.260.26=1.9s34-7 已知系統(tǒng)的開環(huán)傳遞函數(shù)，已知系統(tǒng)的開環(huán)傳遞函數(shù)，(1) 試?yán)L制試?yán)L制出根軌跡圖。出根軌跡圖。第四章習(xí)題課第四章習(xí)題課 (4-7)解解:p1=0 p2=-2 p3=-4p1p2p3 -8+60o= +180o, 3= -2-4=-2根軌跡的分離點(diǎn)

51、：根軌跡的分離點(diǎn)：A(s)B(s)=A(s)B(s)3s2+12s+8=0s1=-0.85s2=-3.15s2沒有位于根軌跡上，舍去。沒有位于根軌跡上，舍去。j(3)與虛軸交點(diǎn)與虛軸交點(diǎn) s3+6s2+8s+Kr=02=0Kr-63+8=0-Kr=0 Kr=48 2,3=2.81=02.8-2.8p3-40p1 p2-2系統(tǒng)根軌跡系統(tǒng)根軌跡s1s3s(s+2)(s+4)KrG(s)H(s)=(2) 阻尼振蕩響應(yīng)的阻尼振蕩響應(yīng)的Kr值范圍值范圍s=-0.85Kr=0.851.153.15=3.1s=j2.8Kr=48 (4)=0.5s1=-0.7+j1.2s3=-6+0.72=-4.6Kr=4.

52、62.60.6=7.2第四章習(xí)題課第四章習(xí)題課 (4-8)第四章習(xí)題課第四章習(xí)題課 (4-9)第四章習(xí)題課第四章習(xí)題課 (4-10)第四章習(xí)題課第四章習(xí)題課 (4-11)第五章習(xí)題課第五章習(xí)題課 (5-1)5-1 已知單位負(fù)反饋系統(tǒng)開環(huán)傳遞函數(shù)，已知單位負(fù)反饋系統(tǒng)開環(huán)傳遞函數(shù)，當(dāng)輸入信號(hào)當(dāng)輸入信號(hào)r(t)=sin(t+30o)，試求系統(tǒng)的，試求系統(tǒng)的穩(wěn)態(tài)輸出。穩(wěn)態(tài)輸出。(s+1) G(s)=10解：解：(s+11) (s)=10 )=A( 112+( )210=0.905=112+110=12210=-5.2o ( )11 =-tg-1 111 =-tg-1 cs(t)= 0.9sin(t+2

53、4.8o)第五章習(xí)題課第五章習(xí)題課 (5-2)5-2 已知單位負(fù)反饋系統(tǒng)開環(huán)傳遞函數(shù)，已知單位負(fù)反饋系統(tǒng)開環(huán)傳遞函數(shù)， 試?yán)L制系統(tǒng)開環(huán)幅相頻率特性曲線試?yán)L制系統(tǒng)開環(huán)幅相頻率特性曲線 。s(s+5)(s+15)(1) G(s)=750 解：解：n-m=3I型系統(tǒng)型系統(tǒng)=0=)= A( -90o ( )=-270o ( )=0)= A( ReIm0=0=(2s+1)(8s+1)(3) G(s)=10 解：解：n-m=20型系統(tǒng)型系統(tǒng)=0=)=10 A( -180o ( )=0)= A( ReIm0=0=0o ( )=s(s-1)(5) G(s)=10 解：解：n-m=2I型系統(tǒng)型系統(tǒng)=0=)= A

54、( -270o ( )=-180o ( )=0)= A( ReIm0=0=第五章習(xí)題課第五章習(xí)題課 (5-2)s2(s+0.1)(s+15)(7) G(s)=10(s+0.2) 解：解：n-m=3II型系統(tǒng)型系統(tǒng)=0=)= A( -180o ( )=-270o ( )=0)= A( ReIm0=0=第五章習(xí)題課第五章習(xí)題課 (5-2)第五章習(xí)題課第五章習(xí)題課 (5-2)5-2 已知單位負(fù)反饋系統(tǒng)開環(huán)傳遞函數(shù)，已知單位負(fù)反饋系統(tǒng)開環(huán)傳遞函數(shù)， 試?yán)L制系統(tǒng)開環(huán)對(duì)數(shù)頻率特性曲線。試?yán)L制系統(tǒng)開環(huán)對(duì)數(shù)頻率特性曲線。s(s+5)(s+15)(1) G(s)=750解：解：-20020400-180 -90

55、-270 dB L( ) ( s( G(s)=1051s+1)s+1)(1511=52=15低頻段曲線：低頻段曲線：-20dB/dec20lgK=20dB15-40dB/dec15-60dB/dec=0=-90o ( )=-270o ( )=相頻特性曲線：相頻特性曲線：第五章習(xí)題課第五章習(xí)題課 (5-2)(2s+1)(8s+1)(3) G(s)=10解：解：-200200-180 -90dB L( ) ( 低頻段曲線：低頻段曲線：20lgK=20dB20lgK1=0.1250.1252=0.50.5-20dB/dec-40dB/dec相頻特性曲線：相頻特性曲線：=0=0o ( )=-180o

56、( )=第五章習(xí)題課第五章習(xí)題課 (5-2)s(s-1)(5) G(s)=10解：解：-20020400-180 -90-270 dB L( ) ( 低頻段曲線：低頻段曲線：20lgK=20dB1-20dB/dec1=1-40dB/dec=0=-270o ( )=-180o ( )=相頻特性曲線：相頻特性曲線：第五章習(xí)題課第五章習(xí)題課 (5-2)s2(s+0.1)(s+15)(7) G(s)=10(s+0.2)解：解：-20020400-180 -90-270 dB L( ) ( s2(10s+1)(0.67s+1)=1.33(5s+1)低頻段曲線：低頻段曲線：20lgK=2.5dB1-40d

57、B/dec0.11=0.1-60dB/dec2=0.20.23=15-40dB/dec15-60dB/dec=0=-180o ( )=-270o ( )=相頻特性曲線：相頻特性曲線：第五章習(xí)題課第五章習(xí)題課 (5-4)5-4 已知系統(tǒng)的開環(huán)幅頻率特性曲線，已知系統(tǒng)的開環(huán)幅頻率特性曲線， 寫出傳遞函數(shù)并畫出對(duì)數(shù)相頻特性曲線。寫出傳遞函數(shù)并畫出對(duì)數(shù)相頻特性曲線。20lgK0-20dB/decdB L( )c20(a)1020lgK=20K=1010G(s)=(0.1s+1)020dB/decdB L( )-20(b)2020lgK=-20K=0.10.1sG(s)=(0.05s+1)101第五章習(xí)

58、題課第五章習(xí)題課 (5-4)(c)0.01-20dB/dec-60dB/dec-40dB/decdB L( )1000s100G(s)=(100s+1)K=100(0.01s+1)(d)20lgK-40dB/dec0-20dB/decdB L( )48110100-60dB/dec5020lgK=48K=251251G(s)=(s+1)(0.1s+1)(0.01s+1)第五章習(xí)題課第五章習(xí)題課 (5-4)100-20dB/dec-60dB/dec4.58dBdB L( )=45.3r(e)0由圖可得：由圖可得：20lgMr=4.58dBMr=1.7得：得：=11- 2 2 1=0.942=0.32得得=0.3r =1-2 2 n根據(jù)根據(jù)得得=50n由頻率曲線得由頻率曲線得s100G(s)=(0.02s)2+0.01s+1)=1000K=2T =0.011)2 T2=(=0.022n第五章習(xí)題課第五章習(xí)題課 (5-7)5-7 已知奈氏曲線，已知奈氏曲線，p為不穩(wěn)定極點(diǎn)個(gè)數(shù)，為