自动控制原理课程设计-超前校正

(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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