How to get the intersection points of a line and a curve which was fit to data?

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Khanh 2014-10-24

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/159912-how-to-get-the-intersection-points-of-a-line-and-a-curve-which-was-fit-to-data

编辑： Hussein Qenawy 2019-4-13

采纳的回答： Andrei Bobrov

在 MATLAB Online 中打开

Hi,

I have a line and a curve that was fit to a data. I also get Coefficients of Equation of the Curve, but don't know how to find its equation to make two equations equal to find the points of the tangency. Could someone give me some recommends?

Here is my code:

clc
array=[515 525 561 600 632 700 761 800 900 1000 1014 1750;
    0 150 300 394 450 540 600 631 696 745 750 865];
x=linspace(array(1,1),array(1,end),101)
y=interp1(array(1,:),array(2,:),x,'pchip')
x=transpose(x)
y=transpose(y)
%
f=fit(y,x,'pchip')
a=coeffvalues(f)
plot(f,y,x)
hold on
% Equation of line that pass through origin
x1=0:1000;
slope=tan(51.5*pi/180);
y1=slope*x1
plot(x1,y1)

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采纳的回答

Andrei Bobrov 2014-10-24

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链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/159912-how-to-get-the-intersection-points-of-a-line-and-a-curve-which-was-fit-to-data#answer_156434

编辑：Andrei Bobrov 2014-10-25

在 MATLAB Online 中打开

one way with Curve Fitting Toolbox

array=[515 525 561 600 632 700 761 800 900 1000 1014 1750;
         0 150 300 394 450 540 600 631 696  745  750  865];
x = array([2 1],:)';
f = fit(x(:,1),x(:,2),'cubicinterp');
df = fit(x(:,1),differentiate(f,x(:,1)),'cubicinterp');
xx = fzero(@(x)f(x)/x - df(x),[1 750]);
x1 = linspace(x(1,1),x(end,1),300);
plot(x1,f(x1),x1,x1*df(xx),xx,f(xx),'ro');

well, more

f = fit(x(:,1),x(:,2),'cubicinterp');
df = fit(x(:,1),differentiate(f,x(:,1)),'cubicinterp');
k = tand(10);
xx = fzero(@(x)df(x) - k,[1 x(end,1)]);

you line: y = k*x + b

b = f(xx) - k*xx;
x1 = linspace(x(1,1),x(end,1),300);
plot(x1,f(x1),x1,k*x1 + b,xx,f(xx),'ro');

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Khanh 2014-10-27

Wow. Perfect. Thank you so much.

Hussein Qenawy 2019-4-13

编辑：Hussein Qenawy 2019-4-13

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