考虑动力学系统：G(s)=25s(O.ls + l)(s + 5)设计一个控制器，使得系统满足下列性能：（1）单位速度输入下的稳态误差等于 0.1（2）调节时间小于 1s；（3）超调量小于 5%。解：化成标准形式：5s (0.1s +1)(0.2 s +1)G (s) H (s)=1 = ss5K1二 0.110 A( s)s(0.1s + 1)(0.2s +1) B (s)得k = 2K = 10s=tf(s)G=5/(s*(0.1*s+1)*(0.2*s+1)figure(1)margin(G)gridBode DiagramGm = 9.54dB (at 7.07 rad/s, Pm = 32.6 deg (at S.75 rad/s)10101102Frequency frad/s)10Jmag,phase,w=bode(G); Gm,Pm=margin(G); QWPm=95; FIm=QWPm-Pm+5 FIm=FIm*pi/180; alfa=(1+sin(FIm)/(1-sin(FIm); adb=20*log10(mag); am=10*log10(alfa); wc2=spline(adb,w,-am); T=1/(wc2*sqrt(alfa); alfat=alfa*T;Gc=tf(alfa*T 1,T 1); Gh=Gc*G; figure(2) margin(Gh) grid pause ysb=feedback(Gh,1,-1); figure(3) step(ysb) gridBode DiagramGm= 15 dB 25 rad/s), Pm = 54.4 deg 9.01 rad/s)Frequency (rad/s)(8P) 2pnnu6es(6ap) QseLo:Qpnp_dStep Response0.511.52Time (sec on ds)2.53