搜狗做题网matlab拟合离散点曲线
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matlab拟合离散点曲线
有X=（ 0.48 ,1.58 ,2.428,3.032,3.36,3.368,2.92,2.172 ,1.368 ,0.512 ,-0.436 ,0.876,-0.48,-1.58,-2.428,-3.032,-3.36,-3.368,-2.92,-2.172,-1.368,-0.512,0.436,0.48 ） Y=（0.752,1.8572,2.39,2.24,1.766,0.8828,0.0572,-0.3136,-0.472,-0.6064,-0.7864,-0.2804,-0.752,-1.8572 ,-2.39,-2.24,-1.766,-0.8828,-0.0572 ,0.3136,0.472,0.6064,0.7864,0.752 ）两组数据,画出组成的曲线,怎样用MATLAB将这条曲线拟合成椭圆方程,然后求出这个椭圆的面积,得到这个椭圆的主轴和X轴的夹角.并且得到这个椭圆的X最大值对应的X,Y坐标.
有X=（ 0.48 ,1.58 ,2.428,3.032,3.36,3.368,2.92,2.172 ,1.368 ,0.512 ,-0.436 ,0.876,-0.48,-1.58,-2.428,-3.032,-3.36,-3.368,-2.92,-2.172,-1.368,-0.512,0.436,0.48 ） Y=（0.752,1.8572,2.39,2.24,1.766,0.8828,0.0572,-0.3136,-0.472,-0.6064,-0.7864,-0.2804,-0.752,-1.8572 ,-2.39,-2.24,-1.766,-0.8828,-0.0572 ,0.3136,0.472,0.6064,0.7864,0.752 ）两组数据,画出组成的曲线,怎样用MATLAB将这条曲线拟合成椭圆方程,然后求出这个椭圆的面积,得到这个椭圆的主轴和X轴的夹角.并且得到这个椭圆的X最大值对应的X,Y坐标.
问题分析
1、从数据看,与椭圆的形状吻合不是很好,尤其是短轴部分明显凹入.
2、拟合的基本思路是,使用长半轴a、短半轴b和倾角Q作为参数描述椭圆方程（进一步还可以考虑圆心不在原点,再增加两个参数）,然后通过优化,目标是使得所有已知点到椭圆的总距离最短.
3、关于点到椭圆的最短距离,有很多讨论,都非常复杂,并不实用.这里采用在椭圆上取足够多的点然后直接计算的距离的方式.
4、拟合得到椭圆参数后,面积S=πab直接计算,主轴和X轴的夹角为拟合得到的数据,而关于椭圆的X最大值对应的X,Y坐标的计算,可以把点取密集一些,直接求最大值,也可以对X求导数,然后解X'=0得到对应的t,再计算X、Y.
参考代码
function zd
X=[0.48, 1.58,2.428,3.032,3.36,3.368,2.92,2.172,1.368,0.512,-0.436,0.876,-0.48,-1.58,-2.428,-3.032,-3.36,-3.368,-2.92,-2.172,-1.368,-0.512,0.436,0.48];
Y=[0.752,1.8572,2.39,2.24,1.766,0.8828,0.0572,-0.3136,-0.472,-0.6064,-0.7864,-0.2804,-0.752,-1.8572,-2.39,-2.24,-1.766,-0.8828,-0.0572,0.3136,0.472,0.6064,0.7864,0.752];
clf
plot(X,Y,'x')
hold on
c = lsqcurvefit(@curve,[3.5 1.5 30*pi/180 0 0],[X;Y],X*0)
a = c(1);
b = c(2);
Q = c(3);
T=linspace(0,2*pi,100);
x1=a*cos(T);
y1=b*sin(T);
x=x1*cos(Q)-y1*sin(Q);
y=x1*sin(Q)+y1*cos(Q);
plot(x,y,'r--')
% 椭圆的面积
S = pi*a*b
% 椭圆的X最大值对应的X,Y坐标
t=fzero(@(t)-a*cos(Q)*sin(t)-b*sin(Q)*cos(t),0);
x=a*cos(t)*cos(Q)-b*sin(t)*sin(Q)
y=a*cos(t)*sin(Q)+b*sin(t)*cos(Q)
plot(x,y,'mo')
legend('原始数据','拟合曲线','X最大值',4)
% 椭圆的主轴和X轴的夹角
Q = Q * 180/pi
function f = curve(c,xy)
a = c(1);
b = c(2);
Q = c(3);
t=linspace(0,2*pi,500);
x1=a*cos(t);
y1=b*sin(t);
x=x1*cos(Q)-y1*sin(Q) + c(4);
y=x1*sin(Q)+y1*cos(Q) + c(5);
f = xy(1,:)*0;
for i=1:length(xy)
f(i) = min( sqrt((xy(1,i)-x).^2 + (xy(2,i)-x).^2 ) );
end
结果
程序输出：
c =
3.4794 1.4809 0.5155 -0.0066 0
S =
16.1875
x =
3.1141
y =
1.3654
Q =
29.5357
也就是说a=3.48,b=1.48,Q=29.5度,S=16.2,如图所示.
1、从数据看,与椭圆的形状吻合不是很好,尤其是短轴部分明显凹入.
2、拟合的基本思路是,使用长半轴a、短半轴b和倾角Q作为参数描述椭圆方程（进一步还可以考虑圆心不在原点,再增加两个参数）,然后通过优化,目标是使得所有已知点到椭圆的总距离最短.
3、关于点到椭圆的最短距离,有很多讨论,都非常复杂,并不实用.这里采用在椭圆上取足够多的点然后直接计算的距离的方式.
4、拟合得到椭圆参数后,面积S=πab直接计算,主轴和X轴的夹角为拟合得到的数据,而关于椭圆的X最大值对应的X,Y坐标的计算,可以把点取密集一些,直接求最大值,也可以对X求导数,然后解X'=0得到对应的t,再计算X、Y.
参考代码
function zd
X=[0.48, 1.58,2.428,3.032,3.36,3.368,2.92,2.172,1.368,0.512,-0.436,0.876,-0.48,-1.58,-2.428,-3.032,-3.36,-3.368,-2.92,-2.172,-1.368,-0.512,0.436,0.48];
Y=[0.752,1.8572,2.39,2.24,1.766,0.8828,0.0572,-0.3136,-0.472,-0.6064,-0.7864,-0.2804,-0.752,-1.8572,-2.39,-2.24,-1.766,-0.8828,-0.0572,0.3136,0.472,0.6064,0.7864,0.752];
clf
plot(X,Y,'x')
hold on
c = lsqcurvefit(@curve,[3.5 1.5 30*pi/180 0 0],[X;Y],X*0)
a = c(1);
b = c(2);
Q = c(3);
T=linspace(0,2*pi,100);
x1=a*cos(T);
y1=b*sin(T);
x=x1*cos(Q)-y1*sin(Q);
y=x1*sin(Q)+y1*cos(Q);
plot(x,y,'r--')
% 椭圆的面积
S = pi*a*b
% 椭圆的X最大值对应的X,Y坐标
t=fzero(@(t)-a*cos(Q)*sin(t)-b*sin(Q)*cos(t),0);
x=a*cos(t)*cos(Q)-b*sin(t)*sin(Q)
y=a*cos(t)*sin(Q)+b*sin(t)*cos(Q)
plot(x,y,'mo')
legend('原始数据','拟合曲线','X最大值',4)
% 椭圆的主轴和X轴的夹角
Q = Q * 180/pi
function f = curve(c,xy)
a = c(1);
b = c(2);
Q = c(3);
t=linspace(0,2*pi,500);
x1=a*cos(t);
y1=b*sin(t);
x=x1*cos(Q)-y1*sin(Q) + c(4);
y=x1*sin(Q)+y1*cos(Q) + c(5);
f = xy(1,:)*0;
for i=1:length(xy)
f(i) = min( sqrt((xy(1,i)-x).^2 + (xy(2,i)-x).^2 ) );
end
结果
程序输出：
c =
3.4794 1.4809 0.5155 -0.0066 0
S =
16.1875
x =
3.1141
y =
1.3654
Q =
29.5357
也就是说a=3.48,b=1.48,Q=29.5度,S=16.2,如图所示.