《自動(dòng)控制原理》黃堅(jiān)課后習(xí)題答案

1、2-1試建立圖所示電路的動(dòng)態(tài)微分方程(a)解:ii=i-i 2%二UM。_ ui ui_u。- uoii=R1= R1 i=R2.dui d(ui-uo)ui-uo uo d d(ui-uo)i2=CdF=C dfRi =R2-C dt(b)解：(Ui-Ui) "i = ii+i2 i= Rii 二 2°: _p du iii R2 i2=C dtL dUoI Ui-Uo=FL2dt-R2(ui-uo)=Riuo-CRiR2(給-第)CRiR2Cdto+Riuo+R2u0=CRiR2dFi+R2uiUi-Ui_ Uo曲iRi = R2 +Cdt ue+R;翳蟲(chóng)分。_送Ri

2、Ri RiR2 dt =R2+c dt + R 2 dt2±_CL,Ci i、“Ri= R2 dt2 +(C+RiR2) dt +(Ri+R2)Uo2-2求下列函數(shù)的拉氏變換。(i) f(t)=sin4t+cos4t解：Lsin(ot= op+s2 Lcosw t=Lsin4t+cos4t = s2+i6 +3 2+s2s+4s2+i6=s2+i6(2) f(t)=t 3+e4t解：Lt3+e4t =魯+ s1L- 6s,+24+s4 s-4S4(s+4) f(t)=t neat解：Ltneat=&-a*T(4) f(t)=(t-1) 2e2t解：L(t-1)2e2t=e-(

3、s-2)_2_(s-2)32-3求下列函數(shù)的拉氏反變換。(1) F(s)= (s+s+1s+3) = s+1 + s+3解：Ai=(s+2)(s+2s&3)行.A2=(s+3)(s+2)(s+3) ,s=-3=2 21 F(s)= s+3 -s+2 f(t)=2e -3t-e-2t(2) F(s)=sA i + 2A2+(s+1)2(s+2)=(s+1)2 +s+1 +A3s+2解：A1=(s+1)2(s+1)2ss+2)s=-1 =-1a2=忠島&=2A3=(s+2)(s+1)2ss+2)：s=-2 =-2f(t)=-2(-2t-te-t+2e-tF(s尸 1+s2+1 +f

4、(t)=1+cost-5sintF(s聒計(jì)Ass爭(zhēng)+年 解：F(s)(s2+1)/s=+j=A1s+A2s=+j2sM3s=j=A1s+A2 4=j2-jj=jA 1+A 2 -5j-1=-A 1+jA 2A1 = 1A2=-5 A3=F(s)s/s=0=1(4) F(s尸s(s+薦2s+3)京 A1 A2A3A4斛：二后仔+前+可+及A1= 2 A3= _3 A4=f2 A2= i4dr_(s+21d _d s(s+3)iA2=ds s=-1_ s(s+3)-(s+2)(2s+3)/ _ _3=s(s+3)2,，s=-1= 4s1s+。)f(t)=力工標(biāo)協(xié)經(jīng)+12e-3tcUEucQ -Ur

5、(s)( R；+sC)1R2+RiR2sCL3=-1/sCR1 L1L3=R2/LCR 1s2Uc(s)=i+(太+sC)R2一Ri+R2+RiR2sCL i=-R2 /Ls L2=-/LCs P1=R2/LCR 1s2 A1=1R2Ur(s)=Uc(s)= R1cLs2+(R 1R2C+L)s+R+R(2-4)求解下列微分方程。 贊 +5 誓+6y(t)=6y(0)=y(0)=2解：s2Y(s)-sy(0)-y'(0)+5sY(s)-5y(0)+6Y(s尸-6Y(s)=6+2s2+12s A +與二 Y(s)= s(s2+5s+6) = 丁 + s+2 +s+3 5 y(t)=lA2

6、5="A3=-42-5試畫(huà)題圖所示電路的動(dòng)態(tài)結(jié)構(gòu)圖，并求傳遞函數(shù)。+/ Riur 2小2-0G(s)。解：Mt) I c(t)K0T2-8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c(t尸苧JK(t-T) C(s)= Ts2(1-e-TS) C(s)=G(s)2-9若系統(tǒng)在單位階躍輸入作用時(shí)，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)的口c(t)=1-e2t+et傳遞函數(shù)和脈沖響應(yīng)。R(s)=J解：CgJ+，_(s2+4s+2)用午: C(s)= s s+2+s+1 =s(s+僅s+2)G(s)=C(s)/R(s)= (S+4S+2)C(s尸摩懶2)c(t)= 6

7、(t)+2e-2t-e-t2-10已知系統(tǒng)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)。解：X1 (s)=R(s)G 1 (s)-G 1 (s)G 7 (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尸G3(S)X2(S)-C(S)G5(S)C(S尸G4(S)X3(S)R(s)s)X3(s.G<-X1Gy<(sGecC(s)G 7(s)-G8(s)C(s)G5(s)G5G7G8G4R(S)G1G21+G3G2G6G5| =G3G4G7-GG1G2G3G41+G3G2G6 +G3G4G

8、5+G1 G2G3G4(G 7 -G8)2-11求系統(tǒng)的傳遞函數(shù)(a)；恒螞R(s)(b)R(s)G2歸Gi(s)=H 1(s)：；_: H 2(s)二= G3(S)GQ)解：L產(chǎn)-G2HlL2=-G1G2H2n_Cs)= k=1P"R(s) F"A=1+G2Hi+G£2H2 P1=G1G2 A 1=1 P2 = G3G2 A 2=1 k_ G2G1+G2G3 -=1+G2H1+G1G2H2解：L1=-G1G2H L2=-04H P1=G1G2 1=1 =1+G1G4H+G 1G2H P2=G3G2 A2=1+G1G4HC(s) _ G1G2+G2G3+G1G2G

9、3G4 HR(s) =1+G1G2H+G1G4HC(s)G2(s)H(s)C(s)_GiG2(1GHi)R(s)=1+GiG2+GiHi03Hi(d) R(s)G2解：L1=-G2HPi=Gi A 1=1P2=G2 、2 =1C(s)=(G1+G2)1+G2 H(e) R(sC+1+G1G3+G2G3 e1G4-G2G4(f)+DR(s)+G2(s)+1(b)H+C(s)(b)C(s)E+R(s)；G4C(s) R(s)3=1Gi(s)2-12 (a)G1H1(s)G2L2L1L1L1L 23G5G1P1L2G1L 1D(s) 1-G2H 2+G1G2H 31 +G2+G1G2 G3A 2=1

10、+G224 L2P=GG5 1=1 G1G2G5+GG5 =1+G1G2-G2C(s)一R(s) =P1=G2 1=1 C(s)1-G2H 2+G1G2H 3 P2=-G1G2Hl A2=1 G2(1-G1H1 )L2=-GiG2H P產(chǎn)G1G2 Ai=11G1G2P2=GiG2G3 2=12 L2=-G 13 A 1 = 1 P2=1 , >(s) -G2G3+1+GP1=G1 Ai = 1 P2=G2、2 (G1+G2)2=1.2 R(s)C(s)_ G1(1 G) R(s)=1+GiG2 c2R(s)-1+GiG2H+G 1G2P二GnG2 A 1=1 P2=12=1+1G2HC(

11、s) 1+GnG2+G£2HD(s)=1+G1G2+G1G2HG1 sE2-13 (a)Os= H3(s) P1=G1G2 A 1=1Pi=GiG5 Ai=1P2=1A2=1+3G4E(s) GiG5+(1+GiG5 ) R(s)=1+G2G3G5+G3G4L1=-G1G3 L2=-G2G3(G1+G2)(G 3+G4)1+G3(G1+GT解：Pi=GiG3 A 1=1 P2=G2G3 A2=1 P3=G1G4 A 3=1 P4=G2G4 A 4=1E(s)_X(s) C(s).R(s)=T+G3(G1+G2)E(S)=G2(s) D(S)=1解：L1=G1G2 L2=-G1G4G5

12、H1H2 L3=-G4 =1-G 1G2+G1G4G35Hl H 2+G4-G1G2G4P1=G1G2G3 A1=1+G4C1(s)=G£2G3(1+G4 )R1(s) 1+G 4+G1G4G5H1H 2G1G2-G1G2G4C2(S)= G4G5G6(1-G1G2) C1(s) _ -G1G2G3G4G5H1R2(s) 1+G4+G1G4G5H 1H2-G1G2-G1G2G4 R2(s) 1+G4+G1G4G5H 1H 2-G1G2-G1G2G4C2G£4G5G6H2R1(s) 1+G4+G1G4G5H 1H2-G1G2-G1G2G3-1設(shè)溫度計(jì)需要在一分鐘內(nèi)指示出響應(yīng)值

13、的 98%,并且假設(shè)溫度計(jì)為一階系統(tǒng),求時(shí)間常數(shù)To如果將溫度計(jì)放在澡盆內(nèi)，澡盆的溫度以10oC/min的速度線性變化,解：C(t)=C()98%t=4T=1 min T=0.25r(t)=10tc(t)=10(t-T+ et/T) =10(誼,)e(t)=r(t)-c(t)ess=lime(t)=10T =2.53-2電路如圖，設(shè)系統(tǒng)初始狀態(tài)為零。C=2.5 仙 FR0=20 k QRi=200 k Q(1)求系統(tǒng)R勺思位階躍響應(yīng)，及uc(t 1)=8時(shí)的t1值. 解：G(s)=R1cs+01 =tsKi T=RiC=0.5k=Ri/R=10Uc(t尸K(1e+) = 10(1 -e-2t)

14、8=10(1 -e")0.8=1 -e-2te-2t=0.2t=0.8(2)求系統(tǒng)的單位脈沖響應(yīng)，單位斜坡響應(yīng)，及單位拋物響應(yīng)在t1時(shí)刻的值.解：t1=0.8 R(s)=1g(t尸千 e-t/T=41R(s尸合uc(t尸K( t-T+T et/T)=4R"Uc(s)=TsK7=K( 4 v S+T12T)Uc(t)=10( 2- t2-0.5t+0.25-0.25e -2t)=1.23-3已知單位負(fù)反饋系統(tǒng)的開(kāi)環(huán)傳遞函數(shù)，求系室(s尸s(s+5)的年： Cs)二4 R 1 C(s)=443 R(s)= s2+5s+4R(s)= s C(s)=s(s+1)(s+4)= s +

15、 s+4 - s+11 -4t 4 -tc(t)=1+ 3e - 3t eG(s)=，1 /、3-4已知單位負(fù)反饋系統(tǒng)的開(kāi)環(huán)傳遞函數(shù)，求系統(tǒng)的上s(s+1)升時(shí)間tr、峰值時(shí)間tp、超調(diào)量%和調(diào)整時(shí)間ts。解:C(s) 1J2；Wn = 1 f Wn=1W= nJ1=0.866R(s) s+s+1n2= 1 卜=0.5B=tg-14-2 =60°3.14-3.14/3 r= wd =0.866=2.42«%=廣九七 100=16%-1.8tp=工 _ 3.14 d =0.866=3.633-6已知系統(tǒng)的單位階躍響應(yīng)：c(t)=1+0.2e1.2e10t(1)求系統(tǒng)的閉環(huán)傳遞

16、函數(shù)。(2)求系統(tǒng)的阻尼比和無(wú)阻尼振蕩頻率。前犁：1*_02_-2_600_：C(s)=s +s+60 s+10=s(s+60)(s+10)R(s)=1CCs)= s2+70s+6003-7設(shè)二階系統(tǒng)的單位階躍響應(yīng)曲線如圖，系統(tǒng)的為單位反解：jtp=ojn-p=0.1(九 /行=ln3.31.19:產(chǎn)避 =03(立)2/1<2 = 1.42e> k=3 39.86 2=1.42-1.421 2(=0.35JN 3n=70 13n2 =600Wn=33.4G(s)=2111156s(s+4 n) = s(s+22.7)3-11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。解

17、：(1) s3+20s2+9s+100=0勞斯表如下：s3 19s2 20 100s1 4S0 100系統(tǒng)穩(wěn)定。(3) s4+8s3+18s2+16s+5=0勞斯表如下：S4S2S1S01816216 16"518 516 5系統(tǒng)穩(wěn)定。.KQ5S+1) 3-12已知單包負(fù)反饋系綠酹%也+1麻05S2+S+1)解：s4+3s3+4s2+2s+Ks+2K=0s4S2S113b31b4142+K 2K2Kb41 =K2+10K-2010-K(K-1.7)(K+11.7)>0K<10R(Sf,暹10 C(s)3-13已知系統(tǒng)結(jié)構(gòu)如圖,試確定系s(s+1)l統(tǒng)穩(wěn)定肘tfi 1S 解

18、：G(s)= 10(1 + S)10(S+1)+S+10T S=S(S2+S+10r s)tzs3s2心心10”)(s)=S3+S2+10+1'0S+103叫+快叫110(1+10T ) 10b3110T >03-14已知系統(tǒng)結(jié)構(gòu)如圖，試確定系< _10_ C(s)ls(s+1)l '解：G(s)=10 (tS+1)S3S2S211b3110S+T) 10T 10O(s)=3 1Q fes+1)(S) s3 +S2+10r s+10. 10r-10 sb31 = " >0T >13-16已知單位反饋系統(tǒng)的開(kāi)環(huán)傳遞函數(shù),r(t)=I(t)+2t+

19、t 2R(s尸 士+酒+0_ _ R0 _ 1 ess1"iTK =21ess2=°°ess3=o°一 Kp=20解： G(s)=(0.is+2%.2s+) ”=0 J K =0一 Ka=0 G(s)=s(s+22(s+10) =10s(0.5s+1)(0.1s+1) i 10(2s+1) j (2s+1) (3) G(s)= s2(s2+4s+10) = s2(0.1s2+0.4s+1)ess=00L Kp=0°ess1=0u=1, Ku =103$52=企=卷I Ka=0ess3=Xess=°°Kp=0°ess

20、1=0Ku = "ess2=0Ka=1ess3=2ess=2解: 求系統(tǒng)的穩(wěn)態(tài)誤差：r(t)=I(t), t , 解：G(s尸？+fo = s(5s+1) u=1I R(s)=4 R(s)=-4 、R(s)=?Kp=0°ess1=0Ku =Kess2 =0.24Ka=0ess3=0°3-17已知系統(tǒng)結(jié)構(gòu)如圖。(1)單位階躍輸入：(T %=20% ts = 1.8(5%)確定 K1 和 p 值K1木/、 K1左wn=K1r1/g=02G(s)=/ c (s尸 q2+Ks+K21< e 3s+K 1 ss+K1 s+K1La n2 =K1L t = 7T3- =

21、 1.8_ &_° _ _s q3n=0.453n=1 830 453.7 K1=宙=13.7 T =0.243-18已知系統(tǒng)結(jié)構(gòu)如圖。為使 諄0.7時(shí), Ks2+(2+Kr)s+K單位斜坡輸入的穩(wěn)態(tài)誤差ess=0.25確定KK_新2s尸s2+壽徐=sgj = 2 3n=2+Kr =2*0.7<esF2=0.25K=31.6L n2=K-卡L。863-19系統(tǒng)結(jié)構(gòu)如圖。r(t尸di(t)=d2(t)=I(t)(1)求r(t)作下的穩(wěn)態(tài)誤差.Di(s)1D2(s)解：essr=Simo s1+G(s)F(s)=1+G(0)F(0)(2)求d1(trnd2同時(shí)作用下的穩(wěn)態(tài)誤

22、差.Ed(s)=-G2(s)H(s)1+G i(s)G2(s)H(s)D(s)e 干 lim s -F(s)+-11-1+F(s)essc= lsmo s1+G(s)F(s) +1+G(s)F(s) s=1+G(0)F(0) 求d1(t)作用下的穩(wěn)態(tài)誤差.G(s)=Kp + -KF(s)= Js嗎=sim。s1+G(s)F(s)=simo s1鄧鼻)71 =04-1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。4-2已知開(kāi)環(huán)傳遞函數(shù)，試用解析法繪制出系統(tǒng)的根軌G(s)= (s+1)-3+j2 j卜 0+j1*：M7-1-2-10 bK解：(s尸 s+K+Ks=-1-KrKr=0s=-1s

23、=-2+j0Kr=700 s=-°° s=0+j1 s=-3+j24-3已知系統(tǒng)的開(kāi)環(huán)傳遞函數(shù)， 試?yán)L制出根軌跡圖。解： Gs)= Kr(s+1.5)(s+5.5)用牛. G(s)=s(s+1)(s+5)1)開(kāi)環(huán)零、極點(diǎn)Pl = 0 P2=-1 P3=-5z1=-1.5 Z2=-5.52)實(shí)軸上根軌跡段 PlP2 ZP3 4-0°3)根軌跡的漸近線 n-m= 10 = + 180o4)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)'=3s 2+12s+5i B(s)

24、'=2s+7Si =-0.63 s2=-2.5s3=-3.6s4=-7.28(2、G(7 - (彘1.5)(2) G(s)= s(s+1)(s+4)1)開(kāi)環(huán)零、極點(diǎn)P1=0 P2=-1 P3=-4z1=-1.52)實(shí)軸上根軌跡段P1P2 P343)根軌跡的漸近線 n-m= 29 = +90o-1-4+1.5>2=-1.754)分離點(diǎn)和會(huì)合點(diǎn)'A(s)=s3+5s2+4s f B(s)=s+1.5lA(s)'=3s2+10s+4 Ib(s)'=1s=-0.62 G(s)=S(S+1)開(kāi)環(huán)零、極點(diǎn)P1=0 P2=-1 P3=-12)實(shí)軸上根軌跡段P1P2P3-

25、°°3)根軌跡的漸近線 n-m=3e = +60o, +180°(t=-J31-0.674)根軌跡與虛軸的交點(diǎn)s3+2s2+S+K。J Kr=0 a 1=01 Kr=23 2,3= - 15)分離點(diǎn)和會(huì)合點(diǎn)A(s)=s3+2s2+S B(s)=1A(s)'=3s2+4s+1iB(s)'=0s=-0.33(4) G(s)=Kr(s+8)s(s+3)(s+7)(s+15)1）開(kāi)環(huán)零、極點(diǎn)Pi=0 p2=-3 p3=-7 p4=-15 Zi=-82)實(shí)軸上根軌跡段P1P2 p3Zl p4-oo3)根軌跡的漸近線n-m=3 a = -3-785+8 =-5

26、.679 = +60o, +180°4)根軌跡與虛軸的交點(diǎn) s4+25s3+171s2+323s+8Kr=0Kr=0 a 1=0 Kr=638 3 2,3= ± 6.25)分離點(diǎn)和會(huì)合點(diǎn)A(s)=s4+25s3+171s2+315sA(s)'=4s3+75s2+342s+315B(s)=s+8B(s)'=2s+7s=-1.44-5已知系統(tǒng)的開(kāi)環(huán)傳遞函數(shù)。（1）試?yán)L制出根軌跡圖。（2）增益Kr為何值時(shí),復(fù)數(shù)特征根的實(shí)部為-2。Ks+2)G(s)= s(s +1)解：p1=0 p2=-1 Z1=-2p1p2 z1-°°分離點(diǎn)和會(huì)合點(diǎn)s2+4s

27、+2=0 s1=-3.41 s2=-0.59 閉環(huán)特征方程式 s2+s+Krs+2Kr=0 - s=-2+ja(-2+jw )2+(-2+j)(1+Kr)+2Kr=0；-4co +(1+Kr =0 、4-32-2(1+K)+2Kr=0；3 = ±1.41I Kr = 34-6已知系統(tǒng)的開(kāi)環(huán)傳遞函數(shù)，試確定閉環(huán)極點(diǎn)諄0.5時(shí)的Kr值(1) G(s)H (s)=s(s+1)(s+3)解：p1=0 p2=-1 p3=-3 p1p2 p3-8(y = 33=-1.3 9 =+60°, +180°根軌跡的分離點(diǎn)：A(s)B'(s)=A' (s)B(s) 3s

28、2+8s+3=0s1=-0.45s2=-2.2 舍去與虛軸交點(diǎn)s3+4s2+3s+Kr=0-a 3+% =0 f Kr=0 a 1=0 ：Kr-4a2=0 1Kr = 12 a 2,3= ±1.7(=0.5 得 s1=-0.37+j0.8 s3=-4+0.37 x2=-3.26 Kr=|s311s3+1|怕3+3|=3.26X2.26X0.26=1.9(2) G(s)H(s)=Krs(s+3)(s2+2s+2)解：Pl = 0 P2=-3 P3.4=-1 ±j P1P2 (T = z3411=-1.25 9 = +45°, +135°根軌跡的出射角。3=

29、±兀-0 1-02-04=土兀-135g90g26.6c=-71.6°與虛軸的交點(diǎn)s(s+3)(s2+2s+2)+Kr=0 s4+5s3+8s2+6s+Kr=03 )4+5(j 3 )3+8(j w)2+j6w +Kr=0j 3分離點(diǎn)和會(huì)合點(diǎn)4s3+15s2+16s+6=0解得 s=-2.3 工=0.5得 s1=-0.36+j0.75Kr=|s1|s1+3|s1+1+j|s1 + 1-j|=2.92心 4-8co 2+Kr=0 ” i-5co 3+6oo =0"Kr=03 1=0Kr=8.16 3 2,3= ± 1.14-7已知系統(tǒng)的開(kāi)環(huán)傳遞函數(shù)，(1)

30、試?yán)L制出根軌跡圖。Ys)H (s)=s(s+2)(s+4)解：P1=0 P2=-2 P3=-4 p1p2 P3-8 (r= 234=-2 9 = +60°, +180°根軌跡的分離點(diǎn)：,-3 3+(8jd =01Kr=0 a 1 =0 1Kr-632=0 1 Kr=48 3 2,3= ±2.8 L =0.5s1=-0.7+j1.2s3=-6+0.7x2=-4.6Kr=4.6 X2.6X0.6=7.2A(s)B'(s)=A' (s)B(s) s1=-0.85 %=-3.15 舍去(2)阻尼振蕩響應(yīng)的Kr值范圍s=-0.85 Kr=0.85x 1.15

31、X3.15=3.1s=± j2.8 Kr=48(3)與虛軸交點(diǎn)s3+6s2+8s+K05-1已知單位負(fù)反饋系統(tǒng)開(kāi)環(huán)傳遞函數(shù)，當(dāng)輸入信號(hào)r(t)=sin(t+30°),試求系解統(tǒng)G(s汽搟 j(s)=(哥日A(3)市與苛皚=112=0.905 巾心)=也-1濘=咆-1#=-5.2。Cs=0.9sin(t+24.8°)5-2已知單位負(fù)反饋系統(tǒng)開(kāi)環(huán)傳遞函數(shù)，試?yán)L制系統(tǒng)開(kāi)環(huán)幅相頻率特性曲線解：（1） G（s尸750s(s+5)(s+15)I型系統(tǒng)n-m=3(D =0= = 00A(a 尸 00A(oo 尸0(|)(co )=-90o(co )=-270oImCO =

32、76;°co =0解：(3) G(s)=10=00 a()=0解：G(s)=(J)(cd )=-1800o10(s+0.2)(2s+1)(8s+1)II型系統(tǒng)解：I型系統(tǒng)CD =0 A()=°°= =°° A ()=00型系統(tǒng)n-m=2 3=0 A()=10 Meo )=0oIm3 = 00 3=0解：（5） G（s尸朋）oon-m=24 (a )=-270(a )=-1l80)= =0s2(s+0.1)(s+15)A(a )=°°A(3 )= 0(J) (<jl) K-I8O)0()(co )=-270)0n-m=3

33、CO =05-2已知單位負(fù)反饋系統(tǒng)開(kāi)環(huán)傳遞函數(shù),解：(1) G(s)=G(s)=s750s(s+5)(s+15)10L(3 )dB40-20dB/decIm3 =000Re試?yán)L制系統(tǒng)開(kāi)環(huán)對(duì)數(shù)頻力冏性曲線。(3) G(s)=10(2s+1)(8s+1)L(co©s+1)*s+1)20lgK=d2tt 1=5 co2=15= =0= =°°6 心)=-9906(3 )=7O)00-20 -,$ (o )10-90- il”，-180、-270+ F1515-40dB/dec-60dB/decCD解：20lgK弓20 g=0.1252=0.53 =0(J)(CO )=(

34、0o3 =8 () (3 ) = -W0°200-200-90-180-20dB/dec20lgK -f 0.1250.5，-40dB/decCD解：(5) G(s)= s(s°1)001=1 20lgK=20= =040200d (co )=-270o -20小(a )="180o 0-90-180-270L()dB-20dB/dec-40dB/decm GG= , 10(s+0.2)=.1.33(5s+1)(7) G(s)= s2(s+0,1)(s+15)=s2(10s+1)(0.67s+1)解:20lgK=2.5dB必=0.132 = 0.2 33=153=

35、03(3 )=-180。a =oo 3 6)=-270。5-4(a)已知系統(tǒng)的開(kāi)環(huán)幅頻率特性曲線，寫(xiě)出傳遞函數(shù)并畫(huà)出對(duì)數(shù)相頻特性曲線。20lgK=20K=1020G(s)=(0：11s+1)03L( 3 )/B-20dB/dec20lgK10(b)20lgK=-20K=0.10G(0_0.1s -20G(s)=(0.05s+1)L( »dB1k20dB/dec0)(d) 20lgK=4848K=2510卜 ()/dBJ - -20dB/dec 20lgK1K=100251-40dB/dec-60dB/de10 50 100 3G(s)=(s+1)(0.1s+(K).01s+1) K=

36、100G(s尸100(e)由圖可得：20lgMr=4.58dB Mr=1.7 =。= ±0.94 (2= ±0.323r =3 nJT-22 3n = 50|L( ydB一20dB/dec0.01-40dB/dec -60dB/decG(s)=s(100s端01s+1)0s(100s+1)(0.01s+1)L9 )/dB-20dB/dec001100-40dB/dech-60dB/dec2 <1-12 t=0.3K=必=100T2=焦)2=0.0222T =0.01得:G(s尸5-7已知奈氏曲線，p為不穩(wěn)定極點(diǎn)個(gè)數(shù),系統(tǒng)不穩(wěn)定LG ydB&0dB/dec4 4.58dB100,01003 r=45.-60dB/decs(0.02s)2+0.01s+1)u為積分環(huán)節(jié)個(gè)數(shù)，試判別系統(tǒng)穩(wěn)(b)p=0“Imu =2co =00 Re*系統(tǒng)穩(wěn)定(c)p=0(d)i系統(tǒng)不穩(wěn)定系統(tǒng)穩(wěn)定Imp=0(e)(f)0=0系統(tǒng)穩(wěn)定(h)io系統(tǒng)穩(wěn)定系統(tǒng)不穩(wěn)定穩(wěn)定性將對(duì)數(shù)幅頻特性向右平移十倍頻程K=1010G(s)=0s(10s+1)(0.05s+1