10s?1s(s?1)10(s?1)??2解：(a). G(s)?
2s?10s1?s(s?21)s(s?1)
D(s)= s2(s?21)?10(s?1)?s3?21s2?10s?1 Routh.s3 1 10 s2 21 1
210?1 s >0 21 s0 11
系统稳定。
1010s(s?2)(b). ?(s)??2
10(10s?1)s?102s?101?s(s?2)
D(s)= s2?102s?10
满足必要条件，故系统稳定。
3-15. 已知单位反馈系统的开环传递函数为
G(s)?K
s(0.01s2?0.2?s?1)试求系统稳定时，参数K和?的取值关系。 解：D(s)?s(0.01s?0.2?s?1)?k?0
2D(s)?s3?20?s2?100s?100k?0
Routh: s3 1 100 s2 20??0 100k
s1 2000??100k?020?
s0 100k?0
???0k?由Routh表第一列系数大于0得?k?0，即?20(??0,k?0)
??k?20??设系统结构图如图3-55所示，已知系统的无阻尼振荡频率?n?3rads。试确定
3-16.
系统作等幅振荡时的K和a值(K、a均为大于零的常数)。
解：??1?1KK1K???? s?2s(s?a)(s?2)(s?a)s?2s(s?a)
D(s)?s(s?2)(s?a)?s(s?a)?K(s?a)?Ks?K
?s2?(3?a)s2?3as?3K?0
32D(j?n)??j?n?(3?a)?n?j3a?n?3K?0 2??Re[D(j?n)]??(3?a)?n?3K?0 ?3??Im[D(j?n)]???n?3a?n?0
解得：??a?3
?K?1823-17.
已知单位反馈控制系统开环传递函数如下，试分别求出当输入信号为1(t)、t和t时系统的稳态误差。 1． G(s)?10
(0.1s?1)(0.5s?1) 2． G(s)?7(s?3)
s(s?4)(s2?2s?2) 3． G(s)?8(0.5s?1) 2s(0.1s?1)解： 1. G(s)??K?1010 ?(0.1s?1)(0.5s?1)?v?0
D(s)?(0.1s?1)(0.5s?1)?10?0经判断系统稳定
r(t)?1(t)A1ess?
1?K11ess?r(t)?tessr(t)?t2?
7?321?K??7(s?3)?2. G(s)? 4?28 ?2s(s?4)(s?2s?2)??v?1
D(s)?s(s?4)(s2?2s?2)?7(s?3)?0
经判断：系统不稳定。
3. G(s)??K?88(0.5s?1)经判断系统稳定 ?2s(0.1s?1)?v?2essr(t)?t0
essr(t)?1(t)
r(t)?t2A21ess??
K843-18. 设单位反馈系统的开环传递函数
G(s)?100
s(0.1s?1)试求当输入信号r(t)?1?2t时，系统的稳态误差。
解： G(s)??K?100100 ?s(0.1s?1)?v?1
D(s)?s(0.1s?1)?100?0.1s2?s?100?0满足必要条件，系统稳定。 ess1?0
ess2?A21?? K100501ess?ess1?ess2?
503-19. 控制系统的误差还有一种定义，这就是无论对于单位反馈系统还是非单位反馈系统，误差均定义为系统输入量与输出量之差，即
E(s)?R(s)?C(s)
现在设闭环系统的传递函数为
bmsm?bm?1sm?1???b1s?b0 n?m ?(s)?nn?1s?an?1s???a1s?a0试证：系统在单位斜坡函数作用下，不存在稳态误差的条件是a0?b0和a1?b1。 证明：E(s)?R(s)?C(s)
E(s)R(s)C(s)???1??(s)??e(s) R(s)R(s)R(s)
bmsm?bm?1sm?1??b1s?b0 ??e(s)?1??(s)?1?ns?an?1sn?1??a1s?a0
?sn?an?1sn?1??bmsm?bm?1sm?1??(a1?b1)s?a0?b0 nn?1s?an?1s??a1s?a01 s2
r(t)?t R(s)?s?0s?0ess?lims?E(s)?lims??e(s)?R(s)
1sn?an?1sn?1??bmsm?bm?1sm?1??(a1?b1)s?a0?b0 ?lim?nn?1s?0ss?an?1s??a1s?a0
要使ess?0，只有让a1?b1?0,a0?b0?0，即a1?b1,a0?b0
3-20.
1(t)具有扰动输入n(t)的控制系统如图3-56所示。试计算阶跃扰动输入n(t)?N2
时系统的稳态误差。
K2T2s?1K2(T1s?1)E(s)???解：?en(s)?
KN(s)1?(T2s?1)(T1s?1?K1)1T1s?s?
n(t)?N0?1(t) N(s)?ess?lims??en(s)?N(s)
s?0N0 s
?lims?s?0N?K2N0?K2(T1s?1) ?0?(T1s?1?K1)(T2s?1)sK1?1
3-21.
试求图3-57所示系统总的稳态误差。
解：(a). ?e(s)?E(s)?R(s)1?1s(0.5s?1)?
2000.5s2?s?200s(0.5s?1)1s(0.5s?1)?
2000.5s2?s?200s(0.5s?1)s?0
?en(s)?E(s)?N(s)1?
ess?ess1?ess2?lims??e(s)?R(s)?lims??en(s)?N(s)
s?0
?lims?s?0s(0.5s?1)1s(0.5s?1)0.1??lims???0 22s?00.5s?s?200s0.5s?s?200s
s?1s(s?1)2 (b). ?e(s)? ?21s?s?11?s(s?1)
?en(s)?1?11s(s?1)?s(s?1) 2s?s?1