34试用劳斯判据确定具有下列特征方程是的系统稳定性
3-4试用劳斯判据确定具有下列特征方程是的系统稳定性。
32(1)2092000S,S,S,,
解:
319S
2 20200S
11S,
0S200
劳斯表第一列元的符合变化两次,系统有两实部根。不稳定。
432(2)8181650S,S,S,S,,
解:
4185S
3816S
2155S
113.5S
05S
劳斯表第一列元的符合没有变化两次,系统没有有正部实根。所以系统稳定。
5432(3)S,6S,3S,2S,S,1,0
解:
5S131
4S621
853S 36
12S18
,1231S6
0S1
劳斯表第一列元的符合变化两次,系统有两实部根。不稳定。3-5设单位负反馈系统的开环传递
为
KG(S), S(0.1S,1)(0.2,1)试确定系统稳定时k的取值范围。
解:又开环传递函数的系统的闭环特征方程为:
32D(S),0.02S,0.3S,S,K,03S0.021
2 S0.3K
0.3,0.02K1SK0.3
0SK
0.3,0.02K,0,于是系统稳定,则有,0,K,15,K,0,
3-6已知系统的闭环特征方程为
S,1)(S,1.5)(S,2),K,0
试由劳斯判据确定使得系统闭环特征根的实部均小于-1的最大k值。
解:根据题意可作线性变换
,,1ZS
此时系统的闭环特征方程为:
32(),(,0.5)(,1),,,1.5,0.5,,0DSZZZKZZZK1S,Z,
310.5Z
21.5ZK
0.75,K1Z1.5
0ZK
0.75,,0K,依题意得,0,K,0.75,K,0,
?K的最大值为0.75(临界稳定)。3-7 设单位负反馈系统的开环传递函数如下,
10G(S),(1) (0.1S,1)(0.2,1)
50s,4,,G(S),(2) 2S(S,1)(s,2s,2)
20s,1,,G(S),(,) 2S(0.5,1)
(1)解:根据误差系数
有:
10()10K,GS,,Plimlim(0.1,1)(0.2,1)SS,,00SS
10,(),,,0KSGSSVlimlim(0.11)(0.21)S,S,,,S0S0
1022,(),,,0KaSGSSlimlim(0.11)(0.21)S,S,,,00SS
?
11,,1t0.09当输入为时,e,,,ss1,1,10Kp
1,,t1t当输入为时,e,,,ssKv
12,,1t当输入为时,,,,essKa
(2)解:
50(S,4),,,,KG(S)Plimlim2S(S,1)(S,2S,2),,S0S0
50(S,4)K,SG(S),S,,100Vlimlim2,,,S(S1)(S2S2),,00SS
50(S,4)22Ka,SG(S),S,,0limlim2S(S,1)(S,2S,2),,S0S0
?
11,,当输入为1t时,e,,,0ss1,Kp1,,
11,,当输入为t1t时,e,,,0.01ssKv100
112,,当输入为时,,,,,t1tessKa0
(3)解:
20(,1)SK,G(S),,,Plimlim2S(0.5S,1),,S0S0
20(S,1)K,SG(S),S,,,Vlimlim2S(0.5S,1),,S0S0
20(S,1)22Ka,SG(S),S,,20limlim2S(0.5S,1),,S0S0
?
11,,当输入为1t时,e,,,0ss1,Kp1,,
11,,当输入为t1t时,e,,,0ssKv,
211,,t1t当输入为时,,,,0.05ess2Ka20
3-8 设控制系统如题图所示。是否可以选择一个合适的值,使系统在单位阶K1
21,,tt跃扰动下的稳态误差,和, ,,,,1tt1t2
解:由题意得:特征方程为:
D(S),(0.1S,1)(0.2S,1)(0.5S,1),10K,01
3232?D(S),0.01S,0.17S,0.8S,1,10K,S,17S,80S,100,1000K,011?劳斯表为:
3S180
2S17100,1000K1
100,1000K11S80,17
0S100,1000K1
100,1000K,180,,0,17,若系统稳,则100,1000K,0,0,K,1.26,1
,K,01,
,
10,?(S),,en(0.1S,1)(0.2S,1)(0.5S,1),10K1
11010?e,S,,(S),,,,ssnenlimlimS(0.1S,1)(0.2S,1)(0.5S,1),10K1,10K,,SS0011
10e,,0.009时,有,0.009ssn1,10K1
解得K,111,1.261
当K,111时,系统不稳定1
9G,,s,3-9 设单位负反馈控制系统的开环传递函数为。试计算当输入,,10s,1
信号分别为,,,,,,rt,1t和rt,2sin2t时,系统的稳态误差。 解:(1)
9rttKpGS(),1(),(),,9当时,limlimS10,1SS,0,0 11e?,,,0.1ssKp1,1,9
当r(t),2sin2t时(2)
,,由(),2sin2,A,2 ,2 ,0rtt
,,,,,,e,jAsint,,,j,,,,,,sse
2,,1,25,,其中,,j,,,122216,4,,,,0,,
,0,,,,,,j,90,arctan,arctane2,,
0,180
0,,即 e,2sin2t,180ss