How to fit a smooth curve on discrete data.
40 ビュー (過去 30 日間)
古いコメントを表示
Hello all, I want to produce an equation that can develop a continous smooth curve (does not matter whether it follow any distribution or any plot) which connect the data given below. Using that equation I can interpolate data in between but I want a smoooth curve not a discrete curve. Can anyone please help me with this.
x = [3, 2.5, 2, 1.5, 1, 0.5]
y = [1.8, 1.75, 1.71, 1.55, 0.8, 1.25]
plot(x,y)
0 件のコメント
採用された回答
Torsten
2022 年 10 月 19 日
Maybe something like this:
x = [3, 2.5, 2, 1.5, 1, 0.5];
y = [1.8, 1.75, 1.71, 1.55, 0.8, 1.25];
plot(x,y)
xx = linspace(x(1),x(end),100);
yy = interp1(x,y,xx,'cubic');
hold on
plot(xx,yy)
hold off
4 件のコメント
Torsten
2022 年 10 月 20 日
編集済み: Torsten
2022 年 10 月 20 日
Alex's idea is to approximate your data by a curve. This curve does not pass exactly through each of your data points, but gives a good approximation over the complete interval with a small number of coefficients.
x = [3, 2.5, 2, 1.5, 1, 0.5];
y = [1.8, 1.75, 1.71, 1.55, 0.8, 1.25];
fun = @(p,x) p(1)./(p(2)+(x-p(3)).^2)+p(4) ;
fun1 = @(p) fun(p,x)-y;
p0 = [-0.1;0.1;0.9;2] ;
sol = lsqnonlin(fun1,p0)
hold on
plot(x,y,'o')
xx = x(1):-0.01:x(end);
plot(xx,fun(sol,xx))
hold off
その他の回答 (0 件)
参考
カテゴリ
Help Center および File Exchange で Linear and Nonlinear Regression についてさらに検索
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!