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1、以下示例仅供参考，大家不要完全照抄例：设单位反馈系统的开环传递函数为G(S)二KS(0.2S 1)(0.0625S 1)(1) 若要求校正后系统的相位裕量为(2) 若要求校正后系统的相位裕量为250,幅值裕量为10 12 dB。试设计串联超前校正装置。500,幅值裕量为30 40 dB。试设计串联滞后校正装置。(1)解：K = 10;G1 = tf ( K,1 0)；G2 = tf ( 1,0.21)；G3 = tf ( 1,0.06251);GS = G1*G2*G3;Bode DisgrsmGm = 6.44 dB 8.94 rad/sec) , Pin -19.3 deg (at 5.9
2、9rad/sec)margi n( GS)500-50-WO-150-90135-130-225-270-ia12310 10 10 10 10Frequency (rfld/sec超前校正装置最大超前角phi = 25- 19 + 6 = 12 alpha=(1-sin(phi *pi/180)”(1+sin( 32*pi/180) alpha =0.517710*log10(alpha) ans =-2.8589|GS(jw)| = 20*log10(K) - 20*log10(w)-20*log10( sqrt( (0.2*w)A2+1 ) - 20*log10( sqrt( (0.06
3、25*w)A2+1 )使用 Symbolic Math Toolbox Solving Algebraic Equations 求解 wsyms wsqrt( (0.2*w)A2+1)-s =20*log10(K)-20*log10(w)-20*log10(20*log10( sqrt( (0.0625*w)A2+1 ) - 2.8589;solve(s)ans =matrix(4.9087390022664795121494014132089437)验证x,y=bode(GS,4.9);20*log10(x)ans =2.8833正确所以 w = 4.9 是校正后的剪切频率T = 1/(sq
4、rt(alpha)*w)T = 0.2836T1=T*alphaans =0.1468超前校正环节传递函数GCS= tf (T 1, T1 1) GCS= tf (0.2836 1, 0.1468 1)Transfer function:0.2153 s + 1 0.0662 s + 1总传递函数GS1 = GCS*GSTransfer function:4.306 s + 200.0331 sA3 + 0.5662 sA2 + s margin(GS)hold onmargi n(GCS)margi n( GS1)(2)串联滞后校正装置，解:K = 10;G1 = tf ( K,1 0)；G
5、2 = tf ( 1,0.21)；G3 = tf ( 1,0.06251);GS = G1*G2*G3;bode(GS)相位裕度 +补偿角=50 deg + 5 deg = 55 -180 deg + 55 deg = - 125 deg可以直接从图中读出，2.45计算mag,phase,w = bode(GS,2.4, 2.45, 2.5)mag(:,:,2)=3.6230phase(:,:,2)=-124.8106magdb = 20*log10(mag(:,:,2)magdb =11.1815可知大约在 w = 2.45处将成为新的穿越频率。校正环节 Wt = w/10 = 0.245由 20lgb = -11.1815得 b = 10A(-11.1815/20) = 0.276T = 1/Wt/b = 14.788 bt = 4.0816滞后校正环节传递函数GCS= tf( 4.0816 1,14.788 1)总传递函数GS1 = GCS*GSmargi n( GS1)101010IS1O3Frequency (rad-wcjPlot type :hanged.
