自控原理第六章习题参考答案

2n6 G(s)222s(s4s6)s(s2nsn)2n6 n6=2.45, 2n=4 42n20.816 6

K1

所以，c1 20lgK0

(c)90arctg2c/n2*0.816*1/2.4590arctg 22211/2.451c/n2*0.816*1/2.450.66690arctg90arctg90arctg0.7995211/2.450.8339038.64128.64

180(c)180128.6451.36

L()(dB)504030201000.01-10-20-30-40-20dB/decn0.112.4510-60dB/dec

(2) 11, 2=1/0.2=5

(c)90arctg2c/nccarctgarctg 221/cn1211128.64arctgarctg128.644511.3194.95

15180(c)18094.9585.05

1

L(dB)504030201000.01-10-20-30-40-20dB/dec20dB/decGc0.11n2.45-40dB/dec-60dB/dec510-60dB/dec 6-5

(1)

G(s)10

s(0.5s1)(0.1s1)

1, 20lgK=20lg10=20dB

11/0.52, 21/0.110

12 时，L(1)2020(lg2lg1)20lg1020lg220lg514dB

210 时，L(2)1440(lg10lg2)13.96dB

所以，1c2

L(1)40(lgclg2)40(lgc/2)14dB

c4.48

(c)90arctg0.5carctg0.1c90arctg2.24arctg0.448

9065.9424.13180.07

180(c)180180.070.07

L()(dB)50403020100-10-20-30-40-20dB/dec-40dB/dec0.11210100c-60dB/dec

2

(2)

G(s)Gc(s)10(0.33s1)

s(0.5s1)(0.1s1)(0.033s1)1, 20lgK=20lg10=20dB

11/0.52, 21/0.333, 31/0.110, 41/0.03330

23 时，L(1)L(2)40(lg2lg1) 14L(2)40(lg4.35lg2)

L(2)7dB

L(310)L(23)20(lg3lg2)3.37dB

所以2c23

L(2)20(lgc2lg2)20(lgc2/3)7dB

c26.72

(c)90arctg0.5c2arctg0.1c2arctg0.33c2arctg0.033c2

90arctg3.36arctg0.672arctg2.22arctg0.222 9073.4333.9065.7512.52144.1

2180(c2)180144.135.9

L()(dB)50403020100-10-20-30-40-20dB/dec-40dB/dec20dB/decGcc20.11231030GcG100c1-40dB/dec-20dB/dec-60dB/dec-60dB/dec 校正环节为相位超前校正，校正后系统的相角裕量增加，系统又不稳定变为稳定，且有一定

的稳定裕度，降低系统响应的超调量；剪切频率增加，系统快速性提高；但是高频段增益提高，系统抑制噪声能力下降。

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