超前滯后校正設計示例

1、以下示例僅供參考，大家不要完全照抄例：設單位反饋系統(tǒng)的開環(huán)傳遞函數(shù)為G(S)二KS(0.2S 1)(0.0625S 1)(1) 若要求校正后系統(tǒng)的相位裕量為(2) 若要求校正后系統(tǒng)的相位裕量為250,幅值裕量為10 12 dB。試設計串聯(lián)超前校正裝置。500,幅值裕量為30 40 dB。試設計串聯(lián)滯后校正裝置。(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超前校正裝置最大超前角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)驗證x,y=bode(GS,4.9);20*log10(x)ans =2.8833正確所以 w = 4.9 是校正后的剪切頻率T = 1/(sq

4、rt(alpha)*w)T = 0.2836T1=T*alphaans =0.1468超前校正環(huán)節(jié)傳遞函數(shù)GCS= tf (T 1, T1 1)>> GCS= tf (0.2836 1, 0.1468 1)Transfer function:0.2153 s + 1 0.0662 s + 1總傳遞函數(shù)GS1 = GCS*GSTransfer function:4.306 s + 200.0331 sA3 + 0.5662 sA2 + s margin(GS)hold onmargi n(GCS)margi n( GS1)(2)串聯(lián)滯后校正裝置，解:K = 10;G1 = tf (

5、K,1 0)；G2 = tf ( 1,0.21)；G3 = tf ( 1,0.06251);GS = G1*G2*G3;bode(GS)相位裕度 +補償角=50 deg + 5 deg = 55 -180 deg + 55 deg = - 125 deg可以直接從圖中讀出，2.45計算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可知大約在 w = 2.45處將成為新的穿越頻率。校正環(huán)節(jié) 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滯后校正環(huán)節(jié)傳遞函數(shù)GCS= tf( 4.0816 1,14.788 1)總傳遞