自动控制理论第三版课后习题答案（夏德钤翁贻方版）第2～5章

1、自动控制理论第三版课后习题答案（夏德钤翁贻方版）第25章 自动控制理论 第3版习题参考答案 第二章 2-1 (a) u2?s?r1r2cs?r2r2r1cs?1? rru1?s?r1r2cs?r1?r2r1?r212cs?1r1?r2(b) u1?s?1 ?u2?s?r1r2c1c2s2?(r1c1?r1c2?r2c2)s?12-2 (a) u?s?ru2?s?rcs?1u?s?r?r1?1? (b) 2 (c) 2?1?1cs?1? u1?s?rru1?s?rcsu1?s?r?4?cs?142-3 设激磁磁通?kfif恒定 cm?s? 60ua?s?s?lajs2?laf?raj?s?raf

2、?ce?cm?2?2-4 c?s?r?s?kacm? 60?ilajs3?i?laf?raj?s2?i?raf?ce?cm?s?kacm?2?2-5 id?2.19?10?3?0.084?ud?0.2? 2-8 (a) ?g1?g2?g3g1?g2g3?g4?c?s?c?s? (b) ?r?s?1?g1?h1?g3r?s?1?g1g2h1?g2g3?g4?h2?g1h3?r(s) + 2-9 框图化简中间结果如图a-2-1所示。 0.7_ + + 1s2?0.3s?10.161.2?2sc(s) ks 图a-2-1 题2-9框图化简中间结果 c?s?0.7s?0.42 ?3r?s?s?0.9?

3、0.7k?s2?1.18?0.42k?s?0.522-10 g1g2g3c?s?g4 r?s?1?g2h1?g1g2h1?g2g3h22-11 系统信号流程图如图a-2-2所示。 图a-2-2 题2-11系统信号流程图 c1?s?g1g2g3r?s?1?g1g2?g4?g1g2g4g5h1h2c 2?s?g?1g2g4g5g6h2rs?1?g1g2?g4?g1g2g4g5h1h22-12 (a) c?s?1r?s?1?cdh?abcdef?agdef?abcdi?adgi?c?s?r?s?r2r 1c1r2c2s2?r1c1?r2c1?r2c2?s?12-13 由选加原理,可得 c?s?11?

4、hg?g1g2r?s?g2d1?s?g2d2?s?g1g2h1d3?s? 11?h2?g2 第三章 3-1 分三种情况讨论 (a) 当?1时 s1?2?1?2n,s2?1?n?2?1?2?1?ntc?t?t?2?1?nt?e?e? 21?2?2n2?2n?2?1?1?(b) 当0?1时 s1?j1?2?n,s2?j1?2?nc?t?t?2?2?nt21?2?2?nt?n?e?cos1?nt?n1?2?e?sin1?2?nt n?t?2?1?sin?nt?2?1?2?1?2?nt?arctg?2n1?en?1?2?2?(c) 当?1时 s1,2?n c?t?t?22?e?nt?1?n?n?t?n

5、?2?(b) 设系统为单位反馈系统,有 er?s?r?s?c?s?r?s?s?s?2?n? 22s?2?n?ns?s?2?n?12? ?s2s2?2?ns?n2?n系统对单位斜坡输入的稳态误差为 esr?ims?s?03-2 (1) kp?50,kv?0,ka?0 (2) kp?,kv?k,ka?0 (3) kp?,kv?,ka?kk,ka?0 (4) kp?,kv?102003-3 首先求系统的给定误差传递函数 ?e?s?误差系数可求得如下 e(s)1s(0.1s?1)? r(s)1?g(s)0.1s2?s?10s(0.1s?1)?02s?0s?00.1s?s?10d10(0.2s?1)c1

6、?lim?e?s?lim?0.122s?0s?0ds(0.1s?s?10)c0?lim?e?s?limd22(0.1s2?s?10)?20(0.2s?1)2c2?lim2?e?s?lim?0s?0s?0ds(0.1s2?s?10)3(1) r(t)?r0，此时有rs(t)?r0, ?s(t)?rr?(t)?0，于是稳态误差级数为 sesr?t?c0rs(t)?0，t?0 (2) r(t)?r0?r1t，此时有rs(t)?r0?r1t,?s(t)?r1,?rr(t)?0，于是稳态误差级数为 s?s(t)?0.1r1，t?0 esr?t?c0rs(t)?c1r(3) r(t)?r0?r1t?为 1

7、1?s(t)?r1?r2t，?r2t2，此时有rs(t)?r0?r1t?r2t2,r，于是稳态误差级数r?s(t)?r222?s(t)?esr?t?c0rs(t)?c1r3-4 首先求系统的给定误差传递函数 c2?t?0 r?s(t)?0.1(r1?r2t)，2!?e?s?误差系数可求得如下 e(s)1s(0.1s?1)? r(s)1?g(s)0.1s2?s?500s(0.1s?1)?02s?0s?00.1s?s?500d500(0.2s?1)1?c1?lim?s?lim?es?0s?0ds(0.1s2?s?500)2500c0?lim?e?s?limd2100(0.1s2?s?500)?10

8、00(0.2s?1)298?c2?lim?s?lim?es?0s?0ds2(0.1s2?s?500)35002? rs(t)?sin5t?s(t)?5cos5tr?r?s(t)?25sin5t稳态误差级数为 c?esr?t?c0?2?25?sin5t?c1?5?cos5t 2?4.9?10?4?sin5t?1?102?cos5t3-6 系统在单位斜坡输入下的稳态误差为 esr?加入比例微分环节后 2?n c?s?r?s?1?as?c?s?g?s?1?as?g?s?r?s?1?as?c?s?r?s?1?g?s?s?2?s?s?2?a?se?s?r?s?c?s?r?s? 2n22nn2ns?2?n

9、s?n2n2r?s?1s22?a?nesr?imse?s?s?0?n可见取a?2?n，可使esr?0 3-7 ?0.598,?n?19.588 4 ss2?4s?63-8 g?s?3-9 按照条件（2）可写出系统的特征方程 (s?1?j)(s?1?j)(s?a)?(s2?2s?2)(s?a)?s?(2?a)s?(2?2a)s?2a?0将上式与1?g(s)?0比较，可得系统的开环传递函数 32 g(s)?2a s?s2?(2?a)s?(2?2a)?根据条件（1），可得 kv?解得a?1，于是由系统的开环传递函数为 12a ?0.5?esr2?2ag(s)?2 s?s2?3s?4? ?1?m3-1

10、0 ?2?mpp?46.6%,ts?7.99s?2%?,(?n?2.12rad/s,?0.24)?16.3%,ts?8s?2%?,(?n?1rad/s,?0.5)?3?ts?15s,(?n?0.4rad/s,?1.25)，过阻尼系统，无超调。 3-11 （1）当a = 0时，?0.354,?n?22。 （2）?n不变，要求?0.7，求得a = 0.25 3-12 1. 单位脉冲响应 (a) 无零点时 c?t?n1?2e?ntsin1?2?nt,?t?0? （b）有零点z?1时 2?1?n?2?,?t?0? c?t?esin1?t?arctgn?1?n?1?2?比较上述两种情况，可见有z?1零点时，单位脉冲响应的振幅较无零点时小，而且产生相移，相移角为 1?2?n?n?n2?nt1?2?n。 arctg1?n2单位阶跃响应 (a) 无零点时 c?t?1?11?2e?nt2?1?2sin?1?nt?arctg?,?t?0? ?（b）有零点z?1时 c?t?1?1?2?n?n1?222?1?e?ntsin?1?2?nt?arctg?n?,?t?0? ?加了z?1的零点之后，超调量mp和超调时间tp都