自控原理习题解答(第二章)

自控原理习题解答（第二章）。cos12te)t(x)3();35t.0(sin)t(x)2();cos2t1(05.0)t(x)1(4t.02-1设tO时，x(t)=0，试求下列函数的拉氏变换：)4s(s2.04sss10.05cos2t)]-L[0.05(12225.0s25.0866s.014s1s3225.0ss2325.0s5.021]3L[sin(0.5t222216.1448s.0s4.0s144)4.0s(4.0s]cos12te[L224t.0[答2-1⑴][答2-1⑵][答2-1⑶])3s()1s(s2s)s(x)3()42ss)(2s(s82s3s)s(x)2()2s)(1s(s)s(x(1)2222-2试求下列函数的拉氏反变换：2)2s(2)1)(s(ssB1)1s(2)1)(s(ssA2ee2s21s1L2sB1sAL)2s)(1s(sL)t(x)]1(22[2s1s2tt11-1-答.1d;1d.3jd3j;0dd)42ss()42ss)(2s(s82s3sdsd2)2s()42ss)(2s(s82s3sB1428s)42ss)(2s(s82s3sAt3cose2e13)1s(1s2s2s1L42ssdsd2sBsAL)42ss)(2s(s82s3sL)t(2(2)]x-2[211122223j1s212s220s22t2t21221112213j1s解得：答121)3s()3s()1s(s2sB32s)3s()1s(s2sAe43te21e121321s43)1s(213s121s32L1sA)1s(A3SBsAL)3s()1s(s2sLt)(x2(3)-23s20s2tt3t21-221121答43)3ss(64ss1)s()3s()1s(s2sdsdA211)s()3s()1s(s2sA1s2221s222-1s2212-3试用拉氏变换求解下列微分方程(设为零初始条件)：)t(1)t(x)t(x2)t(x)3();t()t(x)t(x)t(x)2();t(1t)t(1)t()t(r),t(r)t(x)t(xT)1(和、分别为Tt11eT1T1sT1L)s(xL)t(xT1sT11Ts1)s(x1)s(x)s(Tsx)t()t(x)t(xT]1132[））（答Tt11-T1-s0se1T1s1s1L)s(xLx(t)1)T1s()T1s(sT1B1;s)T1s(sT1AT1sBsA)T1s(sT11)Ts(s1)s(xs1)s(x)s(Tsx);t(1)t(x)t(xT:]2132[））（答Tt211-0s2220s221T1-s2221222TeTtT1sTsTs1L)s(xLx(t)Ts)T1s(sT1dsdA1s)T1s(sT1AT;T1s()T1s(sT1AsAsAT1sA)T1s(sT11)Ts(s1)s(xs1)s(x)s(Tsx);t(1t)t(x)t(xT:]3132[）））（答t23sine332)t(x2321s233322321s322343)21(s11)ss1)s(x1)s(x)s(sx)s(xs);t()t(x)t(x)t(x:]232[2t2222222）（答)1t(e1ete11s1)1s(1s1L)s(xL)t(x1)1s()1s(s1dsdA1)1s()1s(s1A;1s)1s(s1A)1s(A)1s(AsA)1s(s11)2ss(s1)s(xs1)s(x)s(2sx)s(xs);t(1)t(x)t(x2)t(x:]332[ttt211-1s222-1s2210s2221222）（答•2-4试求图2-62中各电路的传递函数Ey(s)/Ex(s)，并说明其是什么环节(环节类型)。exeyRexeyR1C1R2C2exR1C1R2C2ey为实际微分环节）（答1TsT1sRLsRLLsRLsLsR)s(ILs)s(I)s(e)s(ea42ddxyexeyR二阶环节为比例环节（答1s)RCCRCR(sCCRRsRC11s)RCCRCR(sCCRRsRC-sRC1s)CRCR(sCCRR1s)RCCRCR(sCCRR1s)CRCR(sCCRRsC1RsC1RsC1R)s(I)sC1sR)s(I)s(e)s(e)b(4212221122121121222112212112122211221211222112212122112212122111122xyexeyR1C1R2C2惯性环节为实际微分环节（答1TsK1sTsTK1CCsC)CR(RCCC1CCsC)CR(RsCCCCRCCsC)CR(RCCCsC)CR(RsCCRCCsC)CR(RCsCCRsCsCsC)CR(R1sCsCCRsC1RsC1R)s(I)sC1R)s(I)s(e)s(e)c(42ddd1221211211221211221212212111221212121221211212122212112212221122xyexR1C1R2C2ey•2-5设控制系统的方框图如图2-63所示，试用框图简化的方法求系统的传递函数Y(s)/X(s)。X(s)G1G2G3H1H2Y(s)G1G2G3H2G1G2H11G1G2G3)s(x)s(Y•2-6试用框图简化的方法求图2-64所示控制系统的传递函数Y(s)/X(s)。X(s)G1G2G3H3H2Y(s)G1H1----X(s)G2G3G4H3H2Y(s)G1H3----G4G4X(s)G2G4H3H2-H1G4Y(s)G1--433GG1GX(s)G4H2-H1G4Y(s)G1-433323324332GG1HGG1HGG1GG1GGX(s)G4H2-H1G4Y(s)G1-3324332HGGGG1GGX(s)G4Y(s)3324341232133243321HGGGG1)GH(HGGG1HGGGG1GGGX(s)Y(s)143212321332434321HGGGGHGGGHGGGG1GGGG143212321332434321HGGGGHGGGHGGGG1GGGG)s(X)s(Y2-7试求图2-65所求系统的输出拉氏变换Y(s)。X(s)G2H2H1Y(s)G1---D1(s)D2(s)D3(s)121223121221221331212212132121222211212221121222103210HGGHG1)s(DHGG)s(DG)s(DG)s(XGG)s(Y)s(DHGGHG1HGG)s(Y)s(DHGGHG1G)s(Y)s(DHGGHG1G)s(Y)s(XHGGHG1GG)s(Y)s(Y)s(Y)s(Y)s(Y)s(Y•2-8已知反馈放大器的信号流图如图2-66所示，试用梅森公式求传递函数G(s)=Eo(s)/Ei(s)/。jeiae0bcdefghiebjk)hiefgjeghijfkjckhcfbe(1abijdagjdaghcdabcd)s(e)s(ep1PLLLLLL-1ebjkLL:(3)hiL;efghL;eghL;ijfL;kjL;ckhL;cfL;beL(2)1,abijdp;1,aghcdp;1,agjdp;1,abcdp)]1(82[i0n1kkkcbacba418765432144332211益乘积每两个互不接触回路增不同回路增益前向通路增益答•2-9应用梅森公式试求图2-67信号流图的传递函数G(s)=Y(s)/X(s)。jbycdefgahixbijf)bijbcdjihgfhdgc(1)f-1abie(abcde)s(e)s(ep1PLLLLLL-1bijfLL(3)bijL;bcdjL;ihgL;fL;hdL;gcL(2);f-1,abiep;1,abcdep)1(9]2[i0n1kkkcbacba636543212211乘积：每两个不接触回路增益不同回路增益前向通路增益答•2-10已知系统由如下方程组组成：)s(x)s(G)s(Y)s(G)s(G)s(Y)s(x)s(x)s(x)s(G)s(x)s(G)s(x)s(Y)s(G)s(G)s(G)s(x)s(G)s(x3435233612287111试画出该系统的方框图，并求出闭环传递函数Y(s)/X(s)。X(s)G2G3G4G5G7Y(s)G1G8---G6-G1x1x2x3答2-10X(s)G3G4G5G7-G8Y(s)G1--G6-G1x1x3答2-10G22G18743215436324321GGGGGGGGGGGG1GGGG)s(X)s(Y)t(ydt)t(dyT)t(xK)t(xKdt)t(dx)t(yK)t(x)t(x)t(x)t(xK)t(x)t(xKdt)t(dx)t(x)t(y)t(x)t(x544355534223111212-11设一系统由下列代数方程和微分方程组成：54321K,K,K,K,K,T,式中，x(t)和y(t)分别为系统的输入信号和输出信号；均为常系数。试建立系统的方框图，并求出系统的传递函数。2-12设系统由下列代数方程和微分方程组成：dt)t(dydt)t(yd)t(xK)t(nK)t(x)t(x)t(xdt)t(dxT)t(x)t(x)t(x)t(xK)t(x)t(n)t(y)t(x)t(x225022433452311211式中，x(t)、n1(t)、n2(t)为系统的三个输入信号；y(t)为输出信号；K0、K1、K2、T均为常系数。试画出系统的方框图，并求出传递函数Y(s)/X(s)、Y(s)/N1(s)、Y(s)/N2(s)。•2-13已知一控制系统的框图如图2-68所示，试求当x(