自动控制原理_第五版_胡寿松_课后答案[1-8章].khdaw

1、固笞案网www.daanwa ng. tom（原廉后曇案网）最专业的课后刃题琶案月爭社区教材课后答案I练习册答案I期末考卷答案I实験报告答案答案网 (http:/www. daanwang com)部分习题答案 (仅供参考，不对之处敬请批评指正，谢谢！)22但)mxo(t) = /i戌壬。卜f2xo(t)即：mx0(t) + (f1 +f2 )xo(t) = fjxt) f(k +k2 )xo(t) + k1k2xo(t) = k2fxi(t)(c) fxo(t)+ (kt + k2)xo(t)=+ k1xi(t)2-3(a) RRCjCiio(0 + ( + R2C2 + RC? )%(t)

2、 + %(f )=R1R2C1C2念(t) + (RC + R2C2 )i(t)(b)(0 + (/li + f*2 + f2kr)x0(t)+k1k2x0(t)=(t)+(+ hh 川(Q+hk2xi( Q2-4(a) RjR2(Juo仃)+ (R + R? )u0(t) = RjR2( / ) + RqH：() (b)R2CiC2”o(f) + (RC2 + 2RC )uo(t)+uo(t)=RC1C2W/(0 + 2RC + Ui(t)答案网 (http:/www daanwang com)(2-5-2-10 题)2-5(1)运动模态：e-5t(2) 运动模态：严sin琴t(3) 运动模

3、态：(l+t)et 绘 APAF = 12.11 yS = -Edo (sina0 )(a-ao)s2 +4s + 2x(t) = t-2 + 2ex(t )=宁 e5 sin 耳 t x(t) = l-(l + t)e,2-62-72-82-92-10二 / w ；、 k(t)=畔=8(t) + 2e-2t -e-f (s + l)(s + 2)dt零初态响应 c1(t) = l-2et+e-2t 零输入响应c2(t) = e-2t-2e_t 总输出 c(f) = C(f) + c2(t) = 1 - 4el + 2e2tC(s) 100(4s+l)211 R 一 12” +23s+ 252

4、12(町职峠(rcs+1)Ui Ro(b)丛旦=-昱答案网 (http:/www daanwang com)(211题215题)E(s) 10(12“ +23s+5)RG)(b)12s2 + 23s+ 25U(s)二(RCiS+l)(RCs+10)E一RC】s(R2C2s+1)1 U(s) Ro (R1 + R2)C2s+1Uo(s)R1R22-13244U.(s) R：RCiC2s2 + RC2s + RxR2 。皿_ KGmUa(s) Tms +1245QiOo31.260i(s) Tms2 + (1 + 3k3ktkm )s + 31.26k3km答案网 (http:/www daanw

5、ang com)(217题221题)答案网 (http:/www daanwang com)(b)RG)1 + G2G4 +G3G42-17(a)j灼_ j i亠2R(s)_ 1 + G2G3(b)(C)C(s)(G1+G3)G2(d)C(s)R(s)1 + G? H + GG? H?RG)(e)C.G 十G1G2G34 wggHi+G2G3h2 (f)2-18(a)Cs) _G1G2C(s)_ (I + GJG2G4+G3G42C(S)G&2(1 + H1H2)R一 1-GiHi + HR6G2G31 + G1H1 + G2H2 + G3H5 + G1H1G3H3C(s)二(G + Gs)G

6、2R(s) l + GdC(s) G3G2 (1 + GiGqH)N(s) l + GG+GGHC(s) =G4N(s) I + G2G4+G3G4答案网 (http:/www daanwang com)答案网 (http:/www daanwang com)219与217同220与218同答案网 (http:/www daanwang com)答案网 (http:/www daanwang com)2-21(a)C(s)R(s) 1 + G1H1 + G3H2 + G1G2G3H1H2 + GiHGHE(s) (I+G3H2)-GQ3H2HR(s )1 +G1H +G3H2 +G7G2G?H

7、7H2 +G7H7G?H2C(s)GGR(s) 1 GG?3GiG?答案网 (http:/www daanwang com)答案网 (http:/www daanwang com)2 22(曲 co = g +GGG3G4G5(b)9个单独回路：Lx = -G2Hr,L2 =-G4H2,L3 =-G6H3,L4 = -G3G4G5H4,L5 =-G/GsGaGsGsHL& G7G3G4G5G6H5,L7 = -G1G8G6H5 ,L8 = G7H1G8G6H5 ,L9 = G8H4H16对两两互不接触回路：三个互不接触回路1组：4条前向通路及其余子式：:HuL1L2 L1L3 L2L3 L7L

8、2L1L2L3 p1=g1g2g3g4g5g6，产 1 ; P3二G7HGG6 ,A3=1+G4H2;4 c(s)=PkAkR(S)1-f L. + f-Lc-L4L3 a=l1P2F7G3G4G5G6, a2=i ;P4=G1G8G6, A4=1+G4H2;(C) = = 15.1282 R(s) 39 仔平/、C(s) bcde + ade + (a + bc)(l + eg)(e)= 7 R(s)1 + cf + eg + bcdeh + cefg + adehR2(s) 1 + cf + eg + bcdeh + cefg + adehC(s) _ ah(l 一 fg) + aej +

9、 aegi + (bdh + bdej + b deg i) + (cfdh + cfdej + ci)1 Ms) =C(s) fdh + fdej + i + fabcd + ed(l- bg)(丿 R(s) 1 - af - bg - ch - ehgf + afchC(s)le(l + cf) 一 lehbc 一 leha1-f deg- fgC(s)h(l-fg) + ej + egi答案网 (http:/www daanwang com)R2(s) 1-fdeg-fgR3G) 1-fdeg-fg答案网 (http:/www daanwang com)答案网 (http:/www da

10、anwang com)答案网 (http:/www daanwang com)3-1 h(t) = l-e_l3-2 (2) k(t ) = e3t sin 4t2 2 小序/001253-3 (1)(s) =s +1.25 =2 弘=13-63-8g = 06 r = 1.0066 tr = 1.45 sg = 1.43(a)(b)3-2 (1) k(t) =10 h(t) = lOth(t) = l-丄严 sin(4t + 53.13 )4(2) (S) = + 2、(3)(叫=，丄八s s +16s( 3 s 1)a% = 9.478%tp = 1.96 st .= 3.156 s(Dn

11、 24.5 轴=1 s + 1 = 0(S)= 2,S +S + 1(s)=-s+s + 1=仏5 Z = 2.5 0 =乡ts = 6.0133s cr% =17.99%3-7 k = 1.44系统临界稳定g = 0.5a)n = 1ts= 2.917 s屮= _1.686k2 =0.311a% = 29.8%a% = 16.3%ts =7.5 Ist_ = 8.08 s3-9 (1)G( s) = -s(0.5s + l) G(s)=1(1S)s(s + l)g = 0.5 cdu =(b)比(c)多一个零点，附加零点有削弱阻尼的作用。k = 5 g = Ia)n = y107% = 35

12、.09% ts = 3.5 s ess = 0.2yj 10k = 10 g = JoZ = 10 rIF屮=7% = 37.06% ts = 3s e ss = 0.10d = 1.249答案网 (http:/www daanwang com)341劳斯表变号两次，有两个特征根在s右半平面，系统不稳定。312 (1)有一对纯虚根：=j2系统不稳定。(2) s】2=jVs34 =1 s5 = 1 s6 = 5系统不稳定。3-13 0k0x#0答案网 (http:/www daanwang com)答案网 (http:/www daanwang com)3-20答案网 (http:/www da

13、anwang com)答案网 (http:/www daanwang com)答案网 (http:/www daanwang com)(411题412题)KOOt LOCUS-44-11System: sysGain: 6.9Pole: -9.98Damping: 1Overshoot (%): 0Frequency (rad/sec): 9.98Root LocusSi= -9.98$2,3= 2.46 士 j3.354-10-8-6-410&B4204-12(2) iIk* = 22.75【 (o = j255、J.4-8-100 Vk決V22.75时稳定11A16420答案网 (http

14、:/www daanwang com)4-14(2)B； ,30bG (s) =s(s + 40) 1答案网 (http:/www daanwang com)电4320110Root LocusJ1I答案网 (http:/www daanwang com)答案网 (http:/www daanwang com)-2Root LocusRoot Locus4-20(1)-2G3 =10SG + 20 入 j1G + 233)G + 324/ 6?-481(111Q1-30-25-20-15-10502-4-64-20(3)主导极点离虚轴太 近！10如G；G)=2 灭 2( + 233)( + 3

15、24)- + 6寸-30-25 -20 -15-10 -S 0Q i-40-35-25-20-15 W4-20(2)主导极点离虚轴太近!& = 10化心 + 20/ 12( + 233)6 + 324) + 6訂一I-25-20-15404-210Root Locus0kl系统稳定，点)振荡收敛，； 5(8)= 0.521I11：i-2.5-2-1.5-1-0.500.5oSZ-78rO = (Zf)DorfSI-76-I = (0f)3=(s)o(1 + SSO 0-)011T时系统稳定， Tx时系统不稳定。5-6 : v=l时系统稳定，其余不穗定。答案网 (http:/ww

16、w. daanwang com)(514题523题)5-14(1) Z=0-2(-l)=2 不稳定(2) z=0 稳定(3) z=0-2(-l)=2 不稳定(4) z=0 稳定(5) Z=0-2(-l)=2 不稳定(6) z=0-2(l-l)=0 稳定 z=0 稳定(8) z=0稳定(9) Z=1不稳定(10) z=2不稳定 5-15 z=0-2(-l)=2 不稳定5-16(1) k1.5 (2) Tl/9 (3) k-l(3i)=(p(a)2)= -18(r(右图)原系统z=0-2(-l)=2不稳定。改变k值，使3cV3时系统稳定其中 9(!)= 1805-19 K0时应有 0k2.64 ;

17、 kvO 时应有-lk05-20 0t 1.36865-21 a = = 0,845-22 dc = 1.94455 y = 65.1565-23 g = 0.517 con = 2.24 y = 53.175-24 Yi = 90y2 = 151.8= 60 y4 = 33.56答案网 (http:/www. daanwang com)答案网 (http:/www. daanwang com)6-1(1)(2)6-3(61题67题)丫 = 4.0521=3.1623Y = 29.7673v=7.835I = 2；11 27Q = 2.9243 = 3.8473(2) -= 2.2673s q

18、(s)= (s + 3)(s +0.091)k=57G + o.oi) =8A:, =28.2226 Z =0.7sVs1 + 0.1956-5 (1) G (s) =八 1 +0.004sG(s)=26s + 1八180s+ 16-6 o； =1.969/ = 14.25验算得：y = 30.r验算得：y = 504h = 1.1667方=3.1249b%=32%验算得:曲=793卅=62h99 = 16 JdBh = 30.3JBY” = 74.5131 = 0.796267 取匕=5 co： = 0.5 GciG)=芳+1 验算得：= 0.45,y = 53.4。 200s +1/、 2

19、0iS + 16$ + X rA ,t=r fl C AU ff 殳。(2) G(s) = 验算得： =2.05, 丫 =51.3c 200s +1 0.3s+1答案网 (http:/www. daanwang com)(68题-6-15题)s +16-8 (1) (a)= r系统不稳定lUs + 1(b) gc(s) =0.1s+ 1001s+ 1co： = 36.6 f = 34.5h” = 4(c) Gc(s) =(0.5s +1)2(10s + l)(0025s + l)Oc，= 10= 48.2(D： = 58 8,h = 21 7dB6-(2) 釆用(c)6-14 0k0.045时

20、系统为过阻尼；Gr(s)=金时e$s= 0，其中0k0 klko1 + TkiGc2(s) = , 0ko0, k20 sfk2(616题617题)6-16 Gf() = -(s2 + 1.3747s) GcG) = 0.466s 处=2.95w / 02$3 + 1.4s? +50九 50九(0.25+1)5(5+ 2) + 100 入2 = 0.028九 i = 0.04(Ts+1)寄生因子不影响系统稳定性，且因为它为非主导极点, 所以也不太影响动态性能，但附加极点有增大阻尼的作用。61 祀 G)=21答案网 (http:/www daanwang com)7-3(73题710题)(1)

21、e(nT)=10(27)(2) e*(t) = -(2n + 3)8(t一nT) = -38(t)一58(t-T)-78(t-2T)-n=0(1) E(z) = |z-1(l-z-1)S z-2n,an =3n -an.x (2) e(n) = L5 n + (3n3n=0 32.25(1) e(oo) = oo(2) e(oo) = 0c(2) = 4c(3) = 15c(4) = 567-(1) c(l) = 0 c * (t) = E (丄一丄 2*+l4b(t_nT)go 326T(2) c(nT)=扣 一 1)P - (一“ (3) c(n) = 1 l(-l)n + 7(-2)n + 5(-3)n10z2c/、1/ zz7-9(a) G(Z)=z2-(e-2T+e-5T)z + e-7TG血)(b)G(z) = *(冇)3 z-e z-e7-10(a)O)(z) =ns 山)c(z)= RGiQIGhGsGS + RGzG/z)1 + G1G2(z) + G1(z)G3(z)l + GhG3G4(z)l + D1(z)GhG1G2(z)(c) eq)二 RQ)D|d) + D2(z)GhG|G2(z) + NG2(z)7-117-13(2)(3)7-157-167-177-187-197-217-22(7-11题721题