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1、第六章的题。例 已知原系统的开环传递函数，试设计相位超前校正装置，使)20030(400)(2ssssG静态速度误差系数，相角裕度。10vk040解：由速度静态误差系数的要求，可知校正环节的开环增益应为；5ck理论上，求解的步骤为：（1）确定未校正前系统的截止频率和相位裕度；c相位裕度此时为 32 度，不合要求；（2）根据要求，需把相位裕度增加到 40 度，确定需增加的量+修正量；3240m利用相位超前作用，可确定校正装置中的参数满足：（3）由于超前校正装置对幅值的提升作用，可知最大的相位提升量为1log20)(mL于是从原 bode 幅值曲线中，去找幅值为的对应点的频率，作为新的截止频)(m
2、L率，再由确定参数 T。c1cT 程序为：ng=400;dg=1 30 200 0;G0=tf(ng,dg);kc=5;dpm=40+10;mag,phase,w=bode(G0*kc);Mag=20*log10(mag);Gm,Pm,Wcg,Wcp=margin(G0*kc)phi=(dpm-Pm)*pi/180;alpha=(1-sin(phi)/(1+sin(phi);Mn=-10*log10(1/alpha);Wcgn=spline(Mag,w,Mn);T=1/Wcgn/sqrt(alpha);Gc=tf(T,1,alpha*T,1);figure(1)bode(G0*kc,G0*kc
3、*Gc);F0=feedback(G0*kc,1);mm sin1sin11 F=feedback(G0*kc*Gc,1);figure(2)step(F0,F); 运行结果为： 0.1489 s + 1 - 0.0804 s + 1-150-100-50050100Magnitude (dB)10-1100101102103-270-225-180-135-90Phase (deg)Bode DiagramFrequency (rad/sec)00.511.522.500.20.40.60.811.21.4Step ResponseTime (sec)Amplitude第五章的题 p129
4、作业5.5 1：-10-505101520Magnitude (dB)10-210-1100101-90-450Phase (deg)Bode Diagram Gm = Inf , Pm = 95.8 deg (at 4.97 rad/sec)Frequency (rad/sec)2：-60-40-20020Magnitude (dB)10-310-210-1100101-180-135-90-450Phase (deg)Bode Diagram Gm = 75.2 dB (at 26.8 rad/sec) , Pm = 101 deg (at 0.195 rad/sec)Frequency
5、(rad/sec)4：-150-100-50050100150Magnitude (dB)10-210-1100101102-360-315-270-225-180Phase (deg)Bode Diagram Gm = -123 dB (at 0.0038 rad/sec) , Pm = -147 deg (at 1.68 rad/sec)Frequency (rad/sec)6：-100-50050100150200Magnitude (dB)10-310-210-1100101102-270-240-210-180Phase (deg)Bode Diagram Gm = -178 dB (at 0.0005 rad/sec) , Pm = -28.8 deg (at 3.41 rad/sec)Frequency (rad/sec)8：-200-150-100-50050100Magnitude (dB)10-310-210-1100101102-450-360-270-180-90Phase (deg)Bode Diagram Gm = -102 dB (at 0.0005 rad/sec) , Pm = 48.4 deg (at 0.355 rad/sec)Frequency (rad/sec)
