如何用matlab画出反正切函数的曲线

xiaofang123dd

请各位大神指点呀：syms  u
c=0.2;u=1;
x=0:0.01:5;
y=arctan((2*c)/(4*u*(c^2)*x.^2-2*c-x.^2+2*c*x.^2));
axis([0,5,0,4]);
plot(x,y,'b');
xlabel('固有频率比');
ylabel('相位角');
axis([0,5,0,4]);
legend('u=3');
set(gca,'Fontsize',12)
legend('u=1','u=2','u=3','u=4');
这个程序错在哪了？总是提示说y=arctan((2*c)/(4*u*(c^2)*x.^2-2*c-x.^2+2*c*x.^2));出错

fzwxjtu

clc;clear;
c=0.2;
x=0:0.01:5;
for u=1:4
    a=(4*u*(c^2)*x.^2-2*c-x.^2+2*c*x.^2);
    b=2*c;
    z=a+i*b;
    y=unwrap(angle(z));
    plot(x,y,'color',[u*0.2 u*0.1 u*0.1]);
    hold on
end
legend('u=1','u=2','u=3','u=4','location','northeastoutside');
axis tight
xlabel('固有频率比');
ylabel('相位角');

fzwxjtu

颜色区分度不是很好，另外求相位角一般涉及到相位解缠，用atan不是很理想