2-10设系统传递函数为2C(s)R(s)2+3s+2且初始条件c(0)=-1,c(0)=0。试求阶跃输入r（t)=1(t)时，系统的输出响应c（t)。解：由系统的传递函数得：d°c( +3 dc(0) +2c(t) =2r()(1)dt?dt对式（1）取拉氏变换得：s2C(s) - Sc(0) -c(0) +3sC(s) -3c(0)+2C(s)=2R(s)(2)将初始条件代入（2）式得(s* +3$ +2)(s) +$ +3 =2)s122-s2-3s22s +64即: C(s)=ss2+3s+2S+2s(s + 3s +2)$+1s所以：c(t)=2-4e-+2e-2r2-12求图2-61所示有源网络的传递函数RU1CR,(a)RUU-K93-KU(b)(c)图2-61有源网络1RoCosR.1解：（a）Z。=R/lT= RCo1Tos+1CasRo+Cos
U()--R--R(Tos+1)RoU,(s)Zo1RoCosRo1(b)Z= R IlT= R.Co1CasTos+1Ro+CosT,s +11Z, = R +T, = R,CC,sCis1ZU.(s)(Ts +I)(Tos +I)-ZoR.C,sU,(s)/15+11=R1Z/2 = R, (R, +C,sC,sT,s+1(c)C.sR,(T,s+ I)T, = R,C2二R+15+1T,s+R, +1C2sU.(s)Z12R,T,s+1U,(s)RoR T,s+R +12-17已知控制系统结构图如图2-65所示。试通过结构图等效变换求系统传递函数C(s)/R(s)。HG,()GsGAS(s)G()H)G,(s)H)(a)(d)G(8)HHOGGA)G(s)H()(b)GF(e)HOH(0H(O)G(s)G,(s)(c)GAs)(f)图2-65题2-17系统结构图
解：(a)R(s)C(s)G;(s)R(s)C(s)GsG(sG2(s)G(SG;(s)GalsC(s)R(s)1G(s)+ G(s)1+G2(s) G(s)C(s)G, +G,所以：R(s)1+G,G,(b)C(s)R(s)G(s)Gi(s)H(s)1+ Hi(s)H2(s)C(s)R(s)G (1+ H,H2)G(s)1+ H,H2- G,HiC(s)G,G,(I+H,H,)所以：R(s)I+H,H,-GH(c)R(s)C(s)G21+G2Hi
R(s)C(s)G21+G2H1R(s)C(s)G2G;+G31+G2H,+G,G2H2C(s)G(G,+G,)所以：R(s)1+G,H, +G,G,H2(d)H2/GR(s)(SGHH2H2/ G,G3R(s)G3C(s)G1+GH3H2H2/ G,G3R(s)G3C(s)GiG21+ G,H31+ G,H,C(s)G,G,G所以：R(s)(I+G,H)I+G,H)+G,H
(e)R(s)C(s)H2+H/G3H/G334R(s)C(s)GG2G31+ G2G3H2+ H,G2H,/G3G4R(s)C(s)G,GG1+ G2G;H2+ H,G2- G,G2H1GaG,G,G3C(s)所以：=G.+R(s)1+G,G,H,+H,G,-G,G,H,(f)HiR(s)C(s)GTR(s)G2C(s)G,+G1+G,G2H;C(s)(G, +G,)G,所以：R(s)1+G,G,H