matlab 极坐标曲线拟合方程怎么求？

Question

θ=[0,15,30,45,60,75,90,105,120,135,150,165,180,195,210,225,240,255,270,285,300,315,330,345,360]; θ的值的单位为度
R=[320,321,341,375,420,450,460,466,474,468,466,466,466,466,466,466,466,461,443,420,400,383,360,337,320];
各点之间由光滑曲线连接（闭合的样条曲线），怎么求θ，R之间的极坐标方程?
很着急啊！那位大侠帮帮忙！？

Answer 1

theta = [0,15,30,45,60,75,90,105,120,135,150,165,180,195,210,225,240,255,270,285,300,315,330,345,360]; 

R = [320,321,341,375,420,450,460,466,474,468,466,466,466,466,466,466,466,461,443,420,400,383,360,337,320];

plot(theta,R,'o');

%将数据标准化

theta1 = (theta-mean(theta))./std(theta);

p=polyfit(theta1,R,8);

pval=polyval(p,theta1);

hold on;

plot(theta,pval);

Answer 2

theta1=linspace(theta(1),theta(end),5000);

yy=spline(theta,R,theta1);

polar(theta1,yy)

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