自動控制原理及其應用答案第二版課后答案

1、第二章習題課第二章習題課 (2-1a)uoi2=r2cuiuor1r22-1(a) 試建立圖所示電路的動態(tài)微分方試建立圖所示電路的動態(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第二章習題課第二章習題課 (2-1b)2-1(b) 試建立圖所示電路的動態(tài)

2、微分方試建立圖所示電路的動態(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)習題課一習題課一 (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圖所示電路的動態(tài)結(jié)構(gòu)圖圖所示電路的動態(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態(tài)結(jié)構(gòu)圖的動態(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 用運算放大器組成的有源電網(wǎng)絡如圖用運算放大器組成的有源電網(wǎng)絡如圖所示所示,試采用復數(shù)阻抗法寫出它們的傳遞函數(shù)。試采用復數(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 用運算放大器組成的有源電網(wǎng)絡如用運算放大器組成的有源電網(wǎng)絡如力所示力所示,試采用復數(shù)阻抗法寫出它們的傳試采用復數(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 設有一個初始條件為零的系統(tǒng)，系設有一個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)第二章習題課第二章習題課 (2-8)解解:2-9 若系統(tǒng)在單位階躍輸入作用時，已若系統(tǒng)在單位階躍輸入

11、作用時，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，求系統(tǒng)的傳遞函數(shù)和脈沖響應。應，求系統(tǒng)的傳遞函數(shù)和脈沖響應。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) 脈沖響應脈沖響應:2s+2=1+1s+1-c(t)=( (t t) )+2e +e-2t-t第二章習題課第二章習題課 (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)的動態(tài)結(jié)構(gòu)圖并求傳遞式，試畫出系統(tǒng)的動態(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)=第二章習題課第二章習題課 (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)第二章習題課第二章習題課 (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 =第二章習題課第二章習題課 (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第二章習題課第二章習題課 (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第二章習題課第二章習題課 (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第二章習題課第二章習題課 (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第二章習題課第二章習題課 (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)第二章習題課第二章習題課 (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)=第二章習題課第二章習題課 (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 )=第二章習題課第二章習題課 (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第二章習題課第二章習題課 (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第二章習題課第二章習題課 (2-13a)r(s)g4(s)

22、+c(s)e(s)g1(s)g2(s)-+g3(s)第二章習題課第二章習題課 (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)第二章習題課第二章習題課 (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)第二章習題課第二章習題課 (2-14)c1(s)r1(s)第二章習題課第二章習題課 (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-第二章習題課第二章習題課 (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)第二章習題課第二章習題課 (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

26、題課第二章習題課 (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第二章習題課第二章習題課 (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)第二章習題課第二章習題課 (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)第二章習題課第二章習題課 (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)第二章習題課第二章習題課 (2-15)3-1 設溫度計需要在一分鐘內(nèi)指示出響設溫度計需要在一分鐘內(nèi)指示出響應值的應值的98%，并且假設溫度計為一階系并且假設溫度計為一階系統(tǒng)，求時間常數(shù)統(tǒng)，求時間常數(shù)t。如果將溫度計放在。如果將溫度計放在澡盆內(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，求溫度計的誤差。線性變化，求溫度計的誤差。第三章習題課第三章習題課 (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，設系統(tǒng)初始狀態(tài)為領電路如圖，設系統(tǒng)初始狀態(tài)為領第三章習題課第三章習題課 (3-2)解解:(1)求系統(tǒng)的單位階躍響應求系統(tǒng)的單位階躍響應,及及uc(t1)=8 時的時的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，單位斜坡求系統(tǒng)的單位脈沖響應，單位斜坡 響應響應,及單位拋物響應在及單位拋物響應在t1時刻的值時刻的值第三章習題課第三章習題課 (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 已知單位負反饋系統(tǒng)的開環(huán)傳遞函已知單位負反饋系統(tǒng)

32、的開環(huán)傳遞函數(shù)，求系統(tǒng)的單位階躍響應。數(shù)，求系統(tǒng)的單位階躍響應。4s(s+5)g(s)=第三章習題課第三章習題課 (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 已知單位負反饋系統(tǒng)的開環(huán)傳遞函已知單位負反饋系統(tǒng)的開環(huán)傳遞函數(shù)，求系統(tǒng)的上升時間數(shù)，求系統(tǒng)的上升時間tr、峰值時間、峰值時間tp、超調(diào)量超調(diào)量% % 和調(diào)整時間和調(diào)整時間ts。1s(s+1)g(s)=第三章習題課第三章習題課 (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)的單位階躍響應已知系統(tǒng)的單位階躍響應: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。第三章習題課第三章習題課 (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 設二階系統(tǒng)的單位階躍響應曲線如圖，設二階系統(tǒng)的單位階躍響應曲線如圖，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:第三章習題課第三章習題課 (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 已知單位負反饋系統(tǒng)的開環(huán)傳遞函已知單位負反饋系統(tǒng)的開環(huán)傳遞函數(shù)，求系統(tǒng)數(shù)，求系統(tǒng)k、t值以滿足動態(tài)指標：值以滿足動態(tài)指標：%30%30%，ts0.3(5%)0.3(5%)。ks(ts+1)g(s)=第三章習題課第三章習題課 (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。第三章習題課第三章習題課 (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 已知單位負反饋系統(tǒng)的開環(huán)傳遞函已知單位負反饋系統(tǒng)的開環(huán)傳遞函數(shù)，試確定系統(tǒng)穩(wěn)定時數(shù)，試確定系統(tǒng)穩(wěn)定時k值范圍。值范圍。第三章習題課第三章習題課 (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)定時定時值范圍。值范圍。第三章習題課第三章習題課 (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第三章習題課第三章習題課 (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。第三章習題課第三章習題課 (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=第三章習題課第三章習題課 (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時時 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第三章習題課第三章習題課 (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。第三章習題課第三章習題課 (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)同時作用下的穩(wěn)態(tài)誤差同時作用下的穩(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、極點分布如圖，大已知系統(tǒng)的零、極點分布如圖，大致繪制出系統(tǒng)的根軌跡。致繪制出系統(tǒng)的根軌跡。第四章習題課第四章習題課 (4-1)解解:j0(1)(2)j0(3)j0(4)j0600900600第四章習題課第四章習題課 (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，并判斷點出系統(tǒng)的根軌跡，并判斷點(-2+j0),(0+j1),(-3+j2)是否在根軌跡上。是否在根軌跡上。第四章習題課第四章習題課 (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)，試繪制出已知系統(tǒng)的開環(huán)傳遞函數(shù)，試繪制出根軌跡圖。根軌跡圖。第四章習題課第四章習題課 (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）開環(huán)零、極點 p1 p1=0p2p2=-1p3 p3=-52）實軸上根軌跡段）實軸上根軌跡段 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）分離點和會合點）分離點和會合點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第四章習題課第四章習題課 (4-3)s(s

45、+1)(s+4)(2) g(s)=kr(s+1.5) 1）開環(huán)零、極點）開環(huán)零、極點 p1 p1=0p2p2=-1p3 p3=-4j02）實軸上根軌跡段）實軸上根軌跡段 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）分離點和會合點）分離點和會合點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)根軌跡第四章習題課第四章習題課 (4-3)s(s+1)2(3) g(s)=kr 1）開環(huán)零、極點）開環(huán)零、極點 p1=0

46、p2=-1p3 p3=-12）實軸上根軌跡段）實軸上根軌跡段 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）根軌跡與虛軸的交點= +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）分離點和會合點）分離點和會合點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)根軌跡第四章習題課第四章習題課 (4-3) 1）開環(huán)零、極點）開環(huán)零、極點 p1=0p2=-3 p3

47、=-7j02）實軸上根軌跡段）實軸上根軌跡段 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）根軌跡與虛軸的交點閉環(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）分離點和會合點）分離點和會合點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)根軌跡第四章習題課第四章習題課 (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)試繪制試繪制出根軌跡圖。出根軌跡圖。(2)增益增益kr為何值時，復數(shù)特為何值時，復數(shù)特征根的實部為征根的實部為-2。第四章習題課第四章習題課 (4-5)解解:，s(s+1)kr(s+2)g(s)=j0p1=0p1 p2=-1p2 z1=-2z1 p1p2z1-分離點和會合點分離點和會合點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)極點環(huán)極點=0.5時的時的kr值。值。第四章習題課第四章習題課 (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：根軌跡的分離點：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)=與虛軸交點與虛軸交點 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) 試繪制試繪制出根軌跡圖。出根軌跡圖。第四章習題課第四章習題課 (4-7)解解:p1=0 p2=-2 p3=-4p1p2p3 -8+60o= +180o, 3= -2-4=-2根軌跡的分離點

51、：根軌跡的分離點：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)與虛軸交點與虛軸交點 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) 阻尼振蕩響應的阻尼振蕩響應的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第四章習題課第四章習題課 (4-8)第四章習題課第四章習題課 (4-9)第四章習題課第四章習題課 (4-10)第四章習題課第四章習題課 (4-11)第五章習題課第五章習題課 (5-1)5-1 已知單位負反饋系統(tǒng)開環(huán)傳遞函數(shù)，已知單位負反饋系統(tǒng)開環(huán)傳遞函數(shù)，當輸入信號當輸入信號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)第五章習題課第五章習題課 (5-2)5-2 已知單位負反饋系統(tǒng)開環(huán)傳遞函數(shù)，已知單位負反饋系統(tǒng)開環(huán)傳遞函數(shù)， 試繪制系統(tǒng)開環(huán)幅相頻率特性曲線試繪制系統(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=第五章習題課第五章習題課 (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=第五章習題課第五章習題課 (5-2)第五章習題課第五章習題課 (5-2)5-2 已知單位負反饋系統(tǒng)開環(huán)傳遞函數(shù)，已知單位負反饋系統(tǒng)開環(huán)傳遞函數(shù)， 試繪制系統(tǒng)開環(huán)對數(shù)頻率特性曲線。試繪制系統(tǒng)開環(huán)對數(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：第五章習題課第五章習題課 (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、( )=第五章習題課第五章習題課 (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：第五章習題課第五章習題課 (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：第五章習題課第五章習題課 (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)并畫出對數(shù)相頻特性曲線。寫出傳遞函數(shù)并畫出對數(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

58、題課第五章習題課 (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)第五章習題課第五章習題課 (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第五章習題課第五章習題課 (5-7)5-7 已知奈氏曲線，已知奈氏曲線，p為不穩(wěn)定極點個數(shù)，為