发布时间 : 星期一 文章自动控制原理课程设计-超前校正更新完毕开始阅读99bff9cf6137ee06eff918bc
(2)采用串联超前矫正,根据?Lo(?c)?Lc(?m)?10lga,试在Bode图上取剪切频率为6.86的点,此时等式=10,求出a=10,根据T?根据Gc(s)=
1?aTs1?Ts1求出T=0.046097,再
?ma,所以系统矫正之后
求出超前传递函数为Gc(s)=
1?0.46097s1?0.046097s的开环传递函数为
Gc(s)G0(s)=
调用MATLAB函数:
G=tf(conv([0.46097 1],6),conv(conv(conv([1 0],[0.1 1]),[0.3 1]),[0.046097 1])); [Gm,Pm,Wcg,Wcp]=margin(G) 求出:
Gm=4.0670 新的幅值裕度Pm=46.4901 新的相角裕度 Wcg=15.9019 新的相位穿越频率 Wcp=6.8431 新的幅值穿越频率 矫正之后的相角裕度为46.4901度,满足要求。
6(1?0.46097s)s(1?0.1s)(1?0.3s)(1?0.046097s)
①使用simulink绘制校正后单位阶跃响应曲线
0.46097s+10.046097s+1StepTransfer Fcn16320.03s +0.4s +sTransfer FcnScope
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②绘制校正后开环Bode图
G=tf(conv([0.46097 1],6),conv(conv(conv([1 0],[0.1 1]),[0.3 1]),[0.046097 1])); bode(G); grid on
Bode Diagram50Magnitude (dB)Phase (deg)0-50-100-150-45-90-135-180-225-27010-1100101102103Frequency (rad/sec)
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③绘制校正后闭环Nyquist图 num=[2.766 6];
den=[0.001383 0.04844 0.4461 3.766 6]; G=tf(num,den); nyquist(G); grid on
Nyquist Diagram1.52 dB0 dB-2 dB-4 dB14 dB6 dB0.510 dB20 dB0-10 dB-20 dB-6 dBImaginary Axis-0.5-1-1.5-1-0.8-0.6-0.4-0.20Real Axis0.20.40.60.81
(3)综合 figure(1); num=[6];
den=[0.03 0.4 1 6]; G=tf(num,den); nyquist(G); hold on;
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nu=[2.766 6];
de=[0.001383 0.04844 0.4461 3.766 6]; H=tf(nu,de); nyquist(H); grid on; figure(2);
J=tf(6,conv(conv([1 0],[0.1 1]),[0.3 1])); bode(J); hold on;
K=tf(conv([0.46097 1],6),conv(conv(conv([1 0],[0.1 1]),[0.3 1]),[0.046097 1])); bode(K); grid on; figure(3); nun=[6];
dem=[0.03 0.4 1 6]; L=tf(nun,dem); step(L); hold on; nug=[2.766 6];
deh=[0.001383 0.04844 0.4461 3.766 6]; F=tf(nug,deh); step(F)
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