自动控制原理课程设计

(jw)=5/jw(jw+1)(j0.1w+1)Р,(w)=-90-arctanw-arctan0.1wР=-5.5Р与实轴交点:令Q(w)=0,得w=,P(w)=-5/11,故交点为(-5/11,j0)。在(-1,j0)右侧。Р如图,可知系统稳定。Р20lg5/(lg1-lg)=-40,得=,Р=,GM=20lgh=-20lgA()=6.85Р可知,系统稳定。РMATLAB分析:РMatlab代码绘制图形如下Р>> num=50;Рden=[1 11 10 0];Рfigure(1);Рnyquist(num,den)Р频域法设计Р采用滞后校正:Р=1/K<0.02Р故K>50,取K=50,Р20lg50/(lg1-lg)=-40,得=5,=-90°-arctan-arctan0.1=-207.2°Р=-27.2°Р取=60°,180°-90°-arctan-arctan0.1=60°Р=0.5,Р[20lg-20lg50]/[lg-lg1]=-20,得=50/0.5=100Р=0.5/5=0.1Р=0.1/100=0.001Р=(1/100)(s+0.1)/(s+0.001)=10(10s+1)/(100s+1)РG(s)’H(s)=G(s)H(s)=500(10s+1)/s(100s+1)(s+1)(0.1s+1)Р校正后,=180°-90°-arctan100-arctan-arctan0.1-arctan10=49.04°Р>45°,故达到了校正效果。РMATLAB分析:РMatlab代码及绘制伯德图如下Р>> num=[5]Рnum = Р5Р>> den=[0.1 1.1 1 0]Рden = Р 0.1000 1.1000 1.0000 0Р>> bode(num,den)Р校正前的伯德图如下:Р校正装置的伯德图为:(s)=100s+10)/(100s+1)