自動控制原理及其應(yīng)用_課后習(xí)題答案_第二章_圖文

1、第二章習(xí)題課 (2-1au o i 2=R 2C+-+-u i u oR1R 22-1(a 試建立圖所示電路的動態(tài)微分方程。解:輸入量為u i ,輸出量為u o 。u i =u 1+u o u 1=i 1R 1i c =C duc dt =dtd (u i -u o i 1=i 2-i cu 1= R 1+u o u o -C d (u i -u o dt R 2R 2u i =u o R 1-C R 1R 2+C R 1R 2+u o R 2du i dt dt du o u o R 1+C R 1R 2+u o R 2=R 2u i +C R 1R 2du o du i dt dt+-C

2、uc+R 1R 2u i i 1i 2-+-u o i cC第二章習(xí)題課 (2-1b2-1(b 試建立圖所示電路的動態(tài)微分方程。u o +-u i R 1L R 2C i 1=i L +i cu L =L di Ldtu oi L =i 2=R 2u L =L R 2du o dt i c = = +C du c dt CL R 2d 2u o dt 2du odt + u o R 2CL R 2d 2u o dt 2du o dt i 1= +C u o i 2=R 2輸入量為u i ,輸出量為u o 。u i =u 1+u o u 1=i 1R 1i c =C du c dt =dtd (

3、u i -u o 習(xí)題課一(2-2(2-2求下列函數(shù)的拉氏變換。(1 (1 f(t f(t f(t=sin4t+cos4t=sin4t+cos4t 解:L sinwt = w w 2+s 2s w 2+s 2L sin4t+cos4t =4s 2+16s s 2+16=s+4s 2+16+ LL coswt =(2 (2 f(t f(t f(t=t=t 3+e 4t 3!解:L t 3+e 4t = + = +3!s 3+11s-4s 41s-4(3 (3 f(t f(t f(t=t t n e at 解:L t n e at =n!(s-an+1(4 (4 f(t f(t f(t=(t-1=

4、(t-12e 2t 解:L (t-12e 2t =e -(s-22(s-232-3-1 函數(shù)的拉氏變換。F(s F(s=s+1(s+1(s+3解:A 1=(s+2s+1(s+1(s+3s=-2= -1(s+1(s+3A 2=(s+3s+1s=-3= 2F(s F(s= - = - 2s+31s+2f(t f(t=2e=2e -3t -e -2tF(s F(s=s (s+12(s+22-3-2 函數(shù)的拉氏變換。解:f(t f(t= = = e e st st +lim lim e st s (s+12s=-2d ds s s+2s -1 =-2e -2t +lim( +lim( e e st +

5、 + e e st s -1 st s+22(s+22=-2e -2t -te -t +2e -t=(2-te -t -2e -2tF(s F(s=2s 2-5s+1s(s 2+12-3-3 函數(shù)的拉氏變換。解:F(s(s 2+1s=+j =A 1s+A 2s=+j A 1=1, A 2=-5A 3=F(ss =1s=0 f(t f(t f(t=1+cost-5sint=1+cost-5sint F(s F(s= + += + +1s s 2+1s -5s 2+12-3-4 函數(shù)的拉氏變換。(4 (4 F(s F(s F(s=s+2s(s+12(s+3解:f(t f(t= = = e e st

6、 st + + e e st s+2(s+12(s+3s=0s+2s(s+12s=-3+ + lim lim lim s -1d e st st s+2s(s+3ds= + e -3t +lim + 23112s -1(-s 2-4s-6e st (s 2+32(s+2te st s 2+3s= + e -3t - - e e -t - - e e -t 2311234t 2(2-4-1求下列微分方程。d 2y(tdt2+5 +6y(t=6 ,初始條件:dy(t dy(tdt y(0=y(0=2 。·解:s 2Y(s-sY(0-Y(0+5sY(s-5Y(0+6Y(s=1s A 1=s

7、Y(s sY(ss=0y(t y(t=1+5e=1+5e -2t -4e -3t A 2=(s+2Y(s s=-2A 3=(s+3Y(ss=-3A 1=1 , A 2=5 , A 3=-4 Y(s Y(s=6+2s 2+12s s(s 2+5s+6(2-4-2(2-4-2求下列微分方程。d 3y(tdt 3+4 +29 =29,+4 +29 =29,d 2y(tdt 2dy(t dy(tdt 初始條件:y(0=0 , y(0=17 , y(0=-122···解:2-5-a 試畫題2-1圖所示電路的動態(tài)結(jié)構(gòu)圖,并求傳遞函數(shù)。+-Cu c +R 1R 2u i i 1

8、i 2-+-u o i cC 解:u i =R 1i 1+u o ,i 2=i c +i 1U I (s (s=R =R 1I 1(s+U O (sdu ci c =CdtI 2(s=I C (s+I 1(sI C (s (s=CsU CsU C (s (s即: =I 1(sU I (s-U O (s (sR 1U I (s-U O (sCs Cs=I C (s (sU O(s(s U I(s(s=1R1+( sCR21+1R1+( sCR2=R2+R1R2sCR1+R2+R1R2sC 1R1sCR2U I(s(s-U O(s(s I C(s(sI1(sI2(s+1R1sC R2+( U I(

9、s(s-U O(s(s2-5-b 試畫出題2-1圖所示的電路的動態(tài)結(jié)構(gòu)圖,并求傳遞函數(shù)。u o +-u i R 1LR 2C解:u i =R 1I 1+u cu c =u o +u L u L =Ldi Ldt i L =u o R 2i 1=i L +i ci c =Cdu c dtU i (s (s=R =R 1I 1(s+U C (sU C (s (s=U U O (s+U L (s (sU L (s (s=sLI sLI L (s (sI 1(s=(s=I I L (s+I C (s (s1R 1CssL R 2I 1+U OU iIC-U C =U O +U LI LU L I 2(

10、s=U O (s (sR2I C (s (s=CsU CsU C (s (sI 1(s=U O (s (sR 2I 1(s=U I (s+U C (S (SR 1即:I L (s (s=I=I 1(s-I C (sI C (s (s=U C (s (sCs解: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)如圖所示,試采用復(fù)數(shù)阻抗法寫出它們的傳遞函數(shù)。U O=- R 3+SC R 2R 2+1U I R 1U OR 3SCR 2R 21SC ·+1+=-R 1+R 3+R 2R 3CS=-R 1(R 2SC+1R 2R 3=-( + R 1(R 2SC+1R 1R 1R 2=-

11、( +R ( +R 3 (R 2SC+11=R 21R 3R 2SC +R 1-C(S=U O (SU I(S-+CR 1R2R3u iu o-+CR 1R2R3u iu o-+C R 1R2R3u iu o R 4R 52-6-b 用運(yùn)算放大器組成的有源電網(wǎng)絡(luò)如力所示,試采用復(fù)數(shù)阻抗法寫出它們的傳遞函數(shù)。=-R 5R 4+ R 5U O (R 3SC+1R 2R 3SC+R 2+R 3U O U I =(R 2R 3SC+R 2+R 3(R 4+R 5R 1(R 3SC+1R 5-=-(R 4+R 5(R 2+R 3( SC+1R 2R 3R 2+R 3R 1R 5(R 3SC+1U I R

12、 1=R 5R 4+ R 5U O R 2+R 3SC R 3SC+1-R 5R 4+ R 5U O R 2+R 3R 3SC +1-c(t c(ttTK(t2-8 設(shè)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G(sG(s。c(t c(ttTK(tc(t c(t= K T t-(t-TK T C(s C(s= K (1-e Ts 2-TS C(s C(s=G(S=G(S解:2-9 若系統(tǒng)在單位階躍輸入作用時,已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)。r(t r(t=I(t I(t I(tc(t c(t=1-e +e =1-e +e -2t-t 解:R (s

13、 =1s G(S=G(S=C(s/R(s C(s/R(s C(s/R(s1s+21s -C(s C(s=1s +1+=s(s+1(s+2(s 2+4s+2 =(s+1(s+2(s 2+4s+2 C(s C(s=(s+1(s+2(s 2+4s+2 脈沖響應(yīng):2s +2=1+1s +1-c(t c(t=(t +2e +e -2t-t2-10 已知系統(tǒng)的微分方程組的拉氏變換式,試畫出系統(tǒng)的動態(tài)結(jié)構(gòu)圖并求傳遞函數(shù)。解:X 1(s=R (sG 1(s-G 1(sG 7(s-G 8(sC(sX 2(s=G 2(sX 1(s-G 6(sX 3(sX 3(s=G 3(sX 2(s-C (sG 5(sC (s

14、(s=G 4(sX 3(sG 1G 2G 3G 5-C(s-R(sG 4G 6G 8G 7X 1(s=R (s-C(sG 7(s-G 8(s(sG G 1(sC(sG 7(s-G 8(sG 6(sX 3(sX 1(sX 2(sC (sG 5(sX 3(sG 1G 2G 3G 5-C(s-R(sG 4G 2G 6G 8G 7G 1G 2G5-C(s-R(sG 7-G 81+G 3G 2G6G 3G 4-C(sR(sG 7-G81+G 3G 2G 6 +G 3G 4G 5G 1G 2G 3G 41+1+G G 3G 2G 6 +G 3G 4G 5+G 1G 2G 3G 4(G 7 -G 8G 1G

15、 2G 3G 4R (s C (s =解:2-11(aG 1(sG 2(sG 3(sH 1(s_+R(sC(sH 2(sG 1(sG 2(sH 1(s_+R(sC(sH 2(sG 3(s求系統(tǒng)的傳遞函數(shù)1+G 2H 1G 2 G 1+G 31+G 1H 21+G 2H 1G 2 1+G 2H 1G 2=1+G 2H 1+G 1G 2H 2G 2 R (s C (s =1+G 2H 1+G 1G 2H 2G 2G 1+G 2G 3G 1(sG 2(sG 3(sH 1(s_+R(sC(sG 1(sH 2(s2-11(aG 1(sG 2(sG 3(sH 1(s_+R(sC(sH 2(s求系統(tǒng)的傳遞函

16、數(shù)解:L 1L 1=-G 2H 1L 2L 2=-G 1G 2H 1P 1=G 1G 2P 2=G 3G 21 =12 =1R (s C (s =nk=1P k k=1+G 2H 1+G 1G 2H 21+G 2H 1+G 1G 2H 2G 2G 1+G 2G 3 =解:2-11(bG 1(sG 2(sG 3(sG 4(s_+R(sC(sH(s H(s H(s求系統(tǒng)的傳遞函數(shù)G 1(sG 2(sG 3(sG 4H _+R(sC(sH(s H(s1+G 4G 1HG1 G 2(sG 3(s_+R(sC(sH(s H(s1+G 4HG1G 1 G 2G 3_+R(sC(s1+G 4HG 1G 1H

17、G 1 1+G 4HG 1G 1+G 3 (1+HG 1G 41+G 4HG 1G 2 (1+HG 1G 41+G 4G 1H +G 1G 2H R (s C (s =1+G 1G 2H +G 1G 4H G 1G 2+G 2G 3+G 1G 2G 3G 4 H解:2-11(bG 1(sG 2(sG 3(sG 4(s_+R(sC(sH(s H(s求系統(tǒng)的傳遞函數(shù)R (s C (s =1+G 1G 2H +G 1G 4H G 1G 2+G 2G 3+G 1G 2G 3G 4 HL 1L 1=-G 1G 2HL 1=-G 1G 4H L 2P 1=G 1G 21 =1P 2=G 3G 2=1+G

18、4G 2H +G 1G 2H 2=1+G 1G 4HH 1_+G 1+C(sR(sG 3G 22-11c 求系統(tǒng)的閉環(huán)傳遞函數(shù) 。解: H 1_+G 1+C(sR(sG 3G 2H 1R (s C (s 1+G 1G 2+G 1H 1G 3H 1G 1G 2 (1(1 G 3H 1=_G 1C(sR(sG 2H 1+G 21-G 3H 11H_G 1+C(sR(sG 22-11d 求系統(tǒng)的閉環(huán)傳遞函數(shù) 。解: (1_G 1+C(sR(sG 2HG 21+G 2H 1(G 1+G 2 R (s C (s =(2L 1L 1=-G 2H P 1=G 11 =1P 2=G 22 =1-_G 1+C(

19、sR(sG 2G 3G 42-11e 求系統(tǒng)的閉環(huán)傳遞函數(shù) 。解: (1(1(1 _C(sR(sG 1+G 2G 3-G 4C (s =R (s 1+(G 1+G 2(G 3-G 4(G +G 21+G 1G 3+G 2G 3G 1G 4-G 2G 4=(G 1+G 2L 1L 2L 3L 4L 2=G 1G 4L 3=-G 2G 3L 4=G 2G 4(2L 1=-G 1G 3P 1=G 11 =1P 2=G 22 =11+G 1G 3+G 2G 3G 1G 4-G 2G 4=(G 1+G 2C (s (sR (s (s_G 1+C(sR(sG 22-11f 求系統(tǒng)的閉環(huán)傳遞函數(shù) 。_C(s

20、R(sG 11-G 2G 2C (s =R (s 1+1-G2G 1G 1G 21+G 1G 2G 2G 1 (1(1 G 2=解: : (1(1(1(2L 1L 1=-G 1G 2L 2L 2=G 2P 1=G 11 =1-G 2=1+G 1G 2-G 2C (s R (s 1+G 1G 2G 2G 1 (1(1 G 2=2-12(aR(sG 1(sG 2(sH 2(s_+C(sH 3(sH 1(s_+D(s D(s解:求:D (s C (s R (s C (s D (s =01-G 2H 2G 2 G (s =1-G 2H 2G 1G 2 C (s =R (s 1+1-G 2H 2G 1G

21、 2H 31-G 2H 2G 1G 2 1-G 2H 2+G 1G 2H 3G 2G 1=R (s =0結(jié)構(gòu)圖變 換成:G 2(sH 2(s_+C(s G 1H 3G 1H 1_D(s D(s1-G 2H 2G 21-G 1H 1C (s =D (s 1+1-G 2H 2G 21-G 2H 2G 2 G 1H 3(1-G 1H 1 1-G 2H 2+G 1G 2H 3G 2(1-G 1H 1 =2-12(b求:D (s C (s R (s C (s R(sn +D (s 解:D (s =0G (s =1+G 1G 2H G 1G 2 G 1G 2H _C(sC (s =R (s 1+1+G 1

22、G 2H G 1G 2 1+G 1G 2H G 1G 2 1+G 1G 2H+H+G G 1G 2G 1G 2=R (s =0G 結(jié)構(gòu)圖變 換成:G 1G 2H -C(s G n+D (s G 1G 2H -C(s C(sG n /G 1+D (s 1+G 1G 2H G 1G 2 -C(s C(s+D (s 1+G 1G 2H G 2G n 1+G 1G 2H G 2G 1系統(tǒng)的傳遞函數(shù):C (s =D (s 1+1(1+1+G 1G 2HG 1G 2 1+G 1G 2H G n G 2 1+G 1G 2+G 1G 2H =1+G n G 2+G 1G 2H2-13(a求:R (s E (s

23、 R (s C (s C(sE (s G 1G 2G 3_+R(s解:L 1L 1=-G 2L 2L 2=-G 1G 2G 3P 1=G 2G 3P 2=G 1G 2G 3R (s C (s =1+G 2+G 1G 2G3G 2G 3+G 1G 2G 31 =12 =1結(jié)構(gòu)圖變 換成:G 1G 2+-E(s E(sG 3-R (s G 1+-E(s E(sR (s 1+G 2G 3G 2 -E(s E(s-R (s 1+G 2G 2G 3G 1G 2G 31+G 2系統(tǒng)的傳遞函數(shù):E (s =R (s 1+1(1-G 1G 2G 31+G 1+G21+G 2G 2G 31+G 2+G 1G 2

24、G3=1+G 2-G 2G 3R(sG 4(s+C(sE(s E(sG 1(sG 2(s-+G 3(sD(s D(sX(s X(s2-14求:R (s C (s 解:D (s =0結(jié)構(gòu)圖變換為 R(sG 4(s+C(sG 1(sG 2(s-+G 3(sG 3(s(G 1+G 2(G 3+G 41+(G 1+G 2G 3G 1+G 2C (s R (s =1+G 3(G 1+G 2(G 1+G 2(G 3+G 4D(s D(sR(s R(s+-E(s E(sG 3G 2G 1E (s R (s =1+G 3(G 1+G 21R(sG 4(s+C(sE(s E(sG 1(sG 2(s-+G 3(s

25、X(s X(s求:R (s E (s 2-14解:D (s =0結(jié)構(gòu)圖變換為G 3(G 1+G 2D (s C (s D(s D(sR(sC(sG 4(s+E(s E(sG 1(sG 2(s-+G 3(sX(s X(s求:2-14解:R (s =0D (s C (s =1E (s X (s =G 2(sE (s X (s C 1(sR 1(s求:2-15+G 1G 2G 3C 1(sR 1(s+-H 2H 1G 4G 5-G 6C 2(sR 2(s解:結(jié)構(gòu)圖變換為 +G 1G 2G 3C 1(sR 1(s-H 2H 1G 4G 5-1+G 4G 4G 5H 1H 21+G 4G 4G 5H 1

26、H 21+G 1G 1G 21+G 4+G 1G 4G 5H 1H 2-G 1G 2-G 1G 2G 4G 1G 2G 3(1+G 4 =C 1(sR 1(s1+G 4+G 1G 4G 5H 1H 2G 1G 2(1+G 4 =C 2(sR 2(sG 3C 1(sR 1(s+G 1G 2-H 2H 1G 4G 5G 6-C 2(sR 2(s求:2-14解:結(jié)構(gòu)圖變換為 +G 4G 5G 6C 2(sR 2(s-H 1H 2G 2G 1-1+G 4G 4G 5G 1H 1H 21-G 1G 21+G 4+G 1G 4G 5H 1H 2-G 1G 2-G 1G 2G 4G 4G 5G 6(1-G

27、1G 2=1+1+G 4G 6G 4G 5G 1H 1H21-G 2G 21+G 4G 4G 5C 2(sR 2(sC 2(sG 4G 5G 6-+G 1G 2-H 2H 1R 1(sG 3C 1(sR 2(s求:2-14解:結(jié)構(gòu)圖變換為G 4G 5G 6-+-G 1G 2H 2H 1R 1(s C 2(sC 2(sR 1(s1+G 4H 2G 4G 5G 11-G 1G 21+G 4+G 1G 4G 5H 1H 2-G 1G 2-G 1G 2G4G 1G 4G 5G 6H 2=1+G 6H 1C 2(sR 1(sG 11-G 1G 21+G 4H 2G 4G 5G 11-G 1G 21+G

28、4H 2G 4G 5C 2(sG 6R 1(sG 4G 5-+G 1G 2-H 2H 1G 3C 1(sR 2(s求:2-14解:結(jié)構(gòu)圖變換為 G 1G 2G 3+G 5G 4-H 1R 2(sC 1(sH 2-C 1(sR 2(sG 1G 2G 3+G 5G 4-H 1R 2(sC 1(sH 2-G21+G 4G 4G 51-G 1G 2-G 1H 11+G 4+G 1G 4G 5H 1H 2-G 1G 2-G 1G 2G 4-G 1G 2G 3G 4G 5H 1=1+1+G 4G 4G 5G 1H 11-G 2G 21+G 4G 4G 5C 1(sR 2(sH 2-G 1H11-G 2G 2G 2G 3