超前滞后校正设计示例

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 例：设单位反馈系统的开环传递函数为G (S)=S(0.2 S + 1)(0.0625S +1)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。若要求校正后系统的相位裕量为500，幅值裕量为3040 dB。试设计串联滞后校正装置。解：K = 10;G1 = tf ( K,1 0)；G2 = tf ( 1,0.2 1);G3 = tf ( 1,0.0625 1);GS = G1*G2*G3;margin(GS)Sode Diagram彻n = 6.44 dB (at 8.94 rad

2、/fiec) , Pm = 19.3 deg (at 5.99Sode Diagram彻n = 6.44 dB (at 8.94 rad/fiec) , Pm = 19.3 deg (at 5.99七|101WD10110210JFrequency GCS= tf (0.2836 1, 0.1468 1)Transfer function:0.2153 s + 10.0662 s + 1总传递函数GS1 = GCS*GSTransfer function:4.306 s + 200.0331 sA3 + 0.5662 sA2 + smargin(GS)hold onmargin(GCS)mar

3、gin(GSl)(2)串联滞后校正装置，解：K = 10;G1 = tf ( K,1 0);G2 = tf ( 1,0.2 1);G3 = tf ( 1,0.0625 1);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，2.45Bode Diagramm9IDpnllu罢1011010Bode Diagramm9IDpnllu罢10110101102103Frequency (rad/sec)计算mag,phase,w = bode(GS,2.4, 2.45, 2

4、.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 = 10(-11.1815/20) = 0.276T = 1/Wt/b = 14.788bt = 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*GS margin(GS1)Bode Diagram100System: GS1Gain Margin (dB): 17.2At frequency (rad/ssc): B.73System: GS1Frequency (rad/sec): 70.4Magnitude (dB): -64.2System: GSFr&quency