自动控制原理第四版习题答案

部分习题答案2-2(a))()]()([)(21txftxtxftxmooio--即：)t(xf)t(x)ff()t(xmi1o21o(b))t(xfk)t(xkk)t(x)kk(fi2o21o21(c))()()()()(121txktxftxkktxfiioo+=++2-3(a))t(u)t(u)CRCRCR()t(uCCRRooo++++2122112121(b))t(xkk)t(x)kfkfkf()t(xffooo2112211121++++2-4(a))t(uR)t(uCRR)t(u)RR()t(uCRRi2i21o21o21(b))t(u)t(u)RCRC()t(uCCRooo+++122122（2-2题~2-4题）（仅供参考，不对之处敬请批评指正，谢谢！）)()(2)(121tutuRCtuCRCiii)t(xkk)t(x)kfkf()t(xffiii21122121+++=)t(u)CRCR()t(uCCRRii22112121++=2-5(1)运动模态：t5.0etettx5.022)(-(2)tsine23t5.0tsine)t(x23t5.0332(3)(1+t)e-tte)t1(1)t(x2-6PQoQ22k2-7y11.12F2-8))((sinEeooodd2-9)2s)(1s(2s4s(s)2tt2ee2)t(dt)t(dc)t(k2-10零初态响应t2ee21)t(ct1零输入响应tt22e2e)t(c总输出tteetctctc221241)()()(（2-5题~2-10题）运动模态：运动模态：（2-11题~2-15题）2-1125s23s12)1s4(100)s(R)s(C225s23s12)5s23s12(10)s(R)s(E222-12(a)sCR)10sCR)(1sCR()s(U)s(U1ooo11io(b))1sCR(RR)s(U)s(Uooo1io(b)1sC)RR()1sCR(RR)s(U)s(U22122o1io2-132123o22113o21ioRRsCRsCCRRRR)s(U)s(U2-141sTK)s(U)s(m1am1sTK)s(M)s(m2am2-1511132k3)1sT(skmm3skt111ioiuouu1u2uautum3mt32miokk26.31s)kkk31(sT26.31)s()s(（2-17题~2-21题）2-17(a)3221GG1GG)s(R)s(C(b)21112121HHHG1)HH1(GG)s(R)s(C(c)22112231HGGHG1G)GG()s(R)s(C(d)3311332211321HGHGHGHGHG1GGG)s(R)s(C(e)232121123214HGGHGGHG1GGGG)s(R)s(C(f)121231HGG1G)GG()s(R)s(C2-18(a)1212121HGGGG1GG)s(R)s(C1212112123HGGGG1)HGG1(GG)s(N)s(C(b)434243421GGGG1GGGG)G1()s(R)s(C43424GGGG1G)s(N)s(C2-19与2-17同2-20与2-18同2-21(a)23112132123111134321HGHGHHGGGHGHG1)HG1(GGGGG)s(R)s(C2311213212311123423HGHGHHGGGHGHG1HHGG)HG1()s(R)s(E(b)21212121GG3GG1GG2GG)s(R)s(C212121GG3GG1GG1)s(R)s(C（2-22题）2-22(a)23234313543216HGGHGGHG1GGGGGG)s(R)s(C(b)9个单独回路：5654321545434363242121HGGGGGGL,HGGGL,HGL,HGL,HGL1489568178568175654376HHGL,HGGHGL,HGGGL,HGGGGGL6对两两互不接触回路：L1L2L1L3L2L3L7L2L8L2L9L2三个互不接触回路1组：L1L2L34条前向通路及其余子式：P1=G1G2G3G4G5G6,Δ1=1;P2=G7G3G4G5G6,Δ2=1;P3=-G7H1G8G6,Δ3=1+G4H2;P4=G1G8G6,Δ4=1+G4H2;91a61321cba41kkkLLLLLL1P)s(R)s(C(c)128.1539590)s(R)s(C(d)afchehgfchbgaf1)bg1(edabcd)s(R)s(C(e)adehcefgbcdehegcf1)eg1)(bca(adebcde)s(R)s(C1adehcefgbcdehegcf1lehalehbc)cf1(le)s(R)s(C2(f)fgdegf1)cicfdejcfdh()idegbbdejbdh(]aegiaej)fg1(ah[)s(R)s(C1fgdegf1fjifdejfdh)s(R)s(C2fgdegf1egiej)fg1(h)s(R)s(C3（3-1题~3-9题）3-1Tte1)t(hTT3-2(1)10)t(kt10)t(h3-2(2)t4sine425)t(kt3)13.53t4sin(e451)t(hot33-3(1)25.1s0125.0)s((2)16s)4s(50s5)s(22(3))1s3(s1.0)s(3-4s917.2ts96.1t%478.9%26.0spn3-5686.15.2z5.010066.1r2dn%99.17%s0133.6ts156.3ts45.1tspr3-65.2443.1n3-7311.0k44.1k213-8(a)10n系统临界稳定(b)s51.7t%8.29%15.01ss1s)s(sn2(c)s08.8t%3.16%15.01ss1)s(sn2(b)比(c)多一个零点,附加零点有削弱阻尼的作用。3-9(1)2.0es5.3t%09.35%101015k)1s5.0(s5)s(Gsssn(2)249.11r10z1010110k)1s(s)1s1.0(10)s(Gdn1.0es3t%06.37%2sss（3-11题~3-20题）3-11劳斯表变号两次，有两个特征根在s右半平面,系统不稳定。3-12(1)有一对纯虚根：2js2,1系统不稳定。(2)5s1s1s2js654,32,1系统不稳定。(3)有一对纯虚根：5js2,1系统不稳定。3-137.1k03-14003-15(1)ssssee20k(2)sssse2.0e10k(3)20e0e1.0kssss3-16(1)(3)(2)0k0k50kavp0k200kkkavp1kkkavp3-18(1)(2)(3)0essr0e1ssn0e2ssn3-20k1)1sT(sk121sT12RuBCE(s)=R(s)-C(s)由题意得：2122112o2TTkTTk)T(k1（4-4题~4-5题）-0.8810j7=k-0.293-1.707-0.9-4.236op135±=op0=（4-6题~4-10题）4-6(1)11k=(2)30*k=63.630199z==-0.4042.73*k=036.1=op73.92±=6j+-2-210j±260*k=-3.2921j96*k=oop135,45±±=150kc=62.9ko=-21.137.70=（4-11题~4-12题）s1=-9.98s2,3=-2.46354.3j±-4系统始终不稳定！55.2j=75.22*k=75.22*k0时稳定（4-13题~4-14题）)3s2s(s)2s(k)s(G2+++=′)2s)(1s(s)10s(k)s(G+++=′-0.434j1.6976k=-8.47224)2s()4s(b)s(G+++=′)40s(sb30)s(G+=′（4-15题~4-19题）2j2*k=-0.732536.0*k=2.732464.7*k=整条实轴为根轨迹212*k0k*12时系统稳定系统始终不稳定（4-20题~4-21题）主导极点离虚轴太近！主导极点离虚轴太近！]6)24.3)[(3.23()20(10)(221ssssksGt]6)24.3)[(3.23()20(10)(2222ssssksGt]6)24.3)[(3.23(10)(2223sssksGt21k0k1系统不稳定，c(t)振荡发散K=1系统临界稳定，c(t)等幅振荡k1系统稳定，c(t)振荡收敛，5.0)(cn（5-2题~5-8题）5-213j)36(36)j()9s)(4s(36)s(25-3)57.26t2cos(79.0)4.48tsin(632.0)t(eooss或：)90t2cos(707.0)4.3tsin(447.0)t(cooss)t(c)t(r)t(essss5-4848.1653.0n5-50j0jTT5-60j0j0j0j1v2v3v4v5-7)1s20(s)1s05.0(10)s(G5-8o4.1539.17)5.0j(Go53.327383.0)2j(G61（5-9题~5-13题）j-10220o90)0j(Go4.153)2j(Go4.333)2j(Go3600)j(G5-95-11(1)02.6]20[125.025.0]40[dB)(LdB05-11(2)]40[]60[]80[861.02611.2dB0dB)(L5-11(3)06.3802.261.01243.5]20[]40[]60[707.0rdB25.1LmdB0]20[]40[]60[]20[4006.78dB011.020dB)(L5-11(4)5-135-5：τT时系统稳定，5-6：ν=1时系统稳定，其余不稳定。Tτ时系统不稳定。（5-14题~5-23题）5-14(1)Z=0-2(-1)=2不稳定(2)z=0稳定(3)z=0-2(-1)=2不稳定(4)z=0稳定(5)Z=0-2(-1)=2不稳定(6)z=0-2(1-1)=0稳定(7)z=0稳定(8)z=0稳定(9)Z=1不稳定(10)z=2不稳定5-15z=0-2(-1)=2不稳定5-16(1)k1.5(2)T1/9(3)k-11/T5-17z=0-0=0稳定5-18（左图）原系统稳定，改变k值。使ωcω1或ωcω2时系统稳定,（右图）原系统z=0-2(-1)=2不稳定。改变k值，使ωcω1时系统稳定5-19K0时应有0k2.64;k0时应有-1k05-203686.105-2184.021a45-22oc156.6594455.15-23on17.5324.2517.0o