필터 지우기
필터 지우기

Need to fit a curve to some data points

조회 수: 25 (최근 30일)
ISh
ISh 2024년 2월 26일
편집: Sam Chak 2024년 2월 26일
I am new to matlab, and I have a problem.
I have the data points:
x = [1 2 100]
y = [55 22 0]
I need to generate a curve that goes through these values. I thought some version of polyfit would work, but I also can't have the y values go below 0. I am looking for something like the upper half of the function 1/x. What should I use
  댓글 수: 2
the cyclist
the cyclist 2024년 2월 26일
Here is a plot of your data:
x = [1 2 100];
y = [55 22 0];
scatter(x,y,64)
Are you saying you want a smooth curve that passes through these points and never goes below zero? Doesn't go below zero for any x, or just x in [1,100]?
Out of curiosity, what's the purpose?
ISh
ISh 2024년 2월 26일
I would like a smooth curve that goes through or at least close to these points. I can't have it ever go below 0 for any positive x values. I cannot go into details of the purpose.

댓글을 달려면 로그인하십시오.

채택된 답변

Sam Chak
Sam Chak 2024년 2월 26일
Here's another candidate. An exponential decay function model should align with the description you provided.
Try learning how to fit it into the data.
%% Data
x = [ 1 2 100];
y = [55 22 0];
%% Exponential decay function
xx = linspace(x(1), x(end));
f = 137.5*exp(-0.9163*xx);
%% Plot results
plot(x, y, 'o', 'markersize', 10, 'linewidth', 2), hold on
plot(xx, f, 'linewidth', 1.5), grid on
xlabel x, ylabel y
  댓글 수: 2
Alexander
Alexander 2024년 2월 26일
Where do you get 137.5 and -0.91629 from?
Sam Chak
Sam Chak 2024년 2월 26일
From the fitting app. But I didn't use lsqcurvefit().
dXdata = [1 2 100];
dYdata = [55 22 0];
x = linspace(0,100,1000);
% y = a*exp(-b*x);
fun2 = @(w,xdata)(w(1)*exp(dXdata*(w(2))));
x02 = [0,0];
xFit2 = lsqcurvefit(fun2,x02,dXdata,dYdata)
Local minimum found. Optimization completed because the size of the gradient is less than the value of the optimality tolerance.
xFit2 = 1×2
137.5000 -0.9163
yAsy2 = xFit2(1).*exp(x*(xFit2(2)));
plot(x,yAsy2,dXdata,dYdata,'o'); grid;

댓글을 달려면 로그인하십시오.

추가 답변 (4개)

Matt J
Matt J 2024년 2월 26일
편집: Matt J 2024년 2월 26일
x = [1 2 100];
y = [55 22 0];
flist={@(p,x)1./(x-p(1)).^p(2)};
[p,A]=fminspleas(flist,[0,1], x,y, [-inf,0])
p = 1x2
0.0005 1.3219
A = 54.9679
f=@(x) A./(x-p(1)).^p(2); %The fitted function
xs=linspace(x(1),x(end));
plot(x,y,'or',xs,f(xs),'b--');

Alexander
Alexander 2024년 2월 26일
편집: Alexander 2024년 2월 26일
Seems to be an exponetial behavior. Use lsqcurvefit to approximate a curve according your needs. My code:
dXdata = [1 2 100]
dYdata = [55 22 0]
x = linspace(0,100,1000);
% y = a*exp(-b*x);
fun2 = @(w,xdata)(w(1)*exp(dXdata*(w(2))));
x02 = [0,0];
xFit2 = lsqcurvefit(fun2,x02,dXdata,dYdata);
yAsy2 = xFit2(1).*exp(x*(xFit2(2)));
plot(x,yAsy2,dXdata,dYdata,'o'); grid;

Matt J
Matt J 2024년 2월 26일
편집: Matt J 2024년 2월 26일
x = [1 2 100];
y = [55 22 0];
ylog=log(y+eps);
p=polyfit(x,ylog,2);
f=@(x) exp(polyval(p,x)); %The fitted function
fitError=abs(f(x)-y)
fitError = 1x3
1.0e-12 * 0.8669 0.0355 0.0002
xs=linspace(x(1),x(end));
plot(x,y,'or',xs,f(xs),'b--');

Sam Chak
Sam Chak 2024년 2월 26일
편집: Sam Chak 2024년 2월 26일
Hi @ISh
This Rational function model (Rat11) precisely fits the three data points.
format long g
%% Data
xdat = [ 1 2 100];
ydat = [55 22 0];
%% A Rational function model (Rat11) with coefficients {p1, p2, p3} is proposed
yfit = @(p, xdat) (p(1)*xdat + 1)./(p(2)*xdat + p(3));
%% Initial guess of coefficients {p1, p2, p3}
p0 = [1, 2, 3];
%% Call lsqcurvefit to fit the model
[psol, resnorm] = lsqcurvefit(yfit, p0, xdat, ydat)
Local minimum found. Optimization completed because the size of the gradient is less than the value of the optimality tolerance.
psol = 1×3
-0.00999999999999777 0.0265454545454659 -0.00854545454549746
resnorm =
9.40049033638573e-21
%% Plot fitting result
xq = linspace(xdat(1), xdat(end), 1000);
plot(xdat, ydat, 'o', 'markersize', 10, 'linewidth', 2), hold on
plot(xq, yfit(psol, xq), 'linewidth', 1.5), grid on
legend('Data points', 'Fitted curve', 'location', 'best', 'fontsize', 12)
xlabel('x'), ylabel('y')
title({'$y(t) = \frac{-0.01 x + 1}{\frac{73}{2750} x - \frac{47}{5500}}$'}, 'interpreter', 'latex', 'fontsize', 16)
%% Test
p = [-0.01, 73/2750, -47/5500]; % <-- True values of the coefficients
ytest = yfit(p, xdat)
ytest = 1×3
55 22 0

카테고리

Help CenterFile Exchange에서 Get Started with Curve Fitting Toolbox에 대해 자세히 알아보기

제품


릴리스

R2023b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by