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样条插值及其(精确)导数

derivative interpolation matlab

Alex · 技术社区 · 7 年前

假设我有以下数据和命令:

clc;clear;
t = [0:0.1:1];
t_new = [0:0.01:1];
y = [1,2,1,3,2,2,4,5,6,1,0];
p = interp1(t,y,t_new,'spline');
plot(t,y,'o',t_new,p)

可以看到,它们工作得很好,从某种意义上说,插值函数与节点处的数据点匹配得很好。但我的问题是,我需要计算y(即p函数)w.r.t.时间的精确导数,并将其与t向量作图。怎么做?我将不使用diff命令,因为我需要确保导数函数的长度与t向量相同。谢谢。

1 回复 | 直到 7 年前

Leander Moesinger 7 年前

方法A:使用导数

% calculate the polynominal
pp = interp1(t,y,'spline','pp')
% take the first order derivative of it
pp_der=fnder(pp,1);
% evaluate the derivative at points t (or any other points you wish)
slopes=ppval(pp_der,t);

如果没有曲线拟合工具箱,可以替换 fnder 符合:

% piece-wise polynomial
[breaks,coefs,l,k,d] = unmkpp(pp);
% get its derivative
pp_der = mkpp(breaks,repmat(k-1:-1:1,d*l,1).*coefs(:,1:k-1),d);

This mathworks question . 感谢m7913d的链接。

附录:

p = interp1(t,y,t_new,'spline');

是的快捷方式

% get the polynomial 
pp = interp1(t,y,'spline','pp');
% get the height of the polynomial at query points t_new
p=ppval(pp,t_new);

t = [0:0.1:1];
t_new = [0:0.01:1];
y = [1,2,1,3,2,2,4,5,6,1,0];

% fit a polynomial 
pp = interp1(t,y,'spline','pp');
% get the height of the polynomial at query points t_new
p=ppval(pp,t_new);
% plot the new interpolated curve
plot(t,y,'o',t_new,p)

% piece-wise polynomial
[breaks,coefs,l,k,d] = unmkpp(pp);
% get its derivative
pp_der = mkpp(breaks,repmat(k-1:-1:1,d*l,1).*coefs(:,1:k-1),d);
% evaluate the derivative at points t (or any other points you wish)
slopes=ppval(pp_der,t);

方法B:使用有限差分法

一个连续函数的导数在其基部就是f(x)到f(x+无穷小差)的差除以所述无穷小差。

在matlab中, eps 是具有双精度的最小差异。因此在每次 t_new 我们添加第二点,即放大并插值 y 对于新点。然后,每个点与其+eps对之间的差除以eps得到导数。

% how many floating points the derivatives can have
precision = 10;
% add after each t_new a second point with +eps difference
t_eps=[t_new; t_new+eps*precision];
t_eps=t_eps(:).';
% interpolate with those points and get the differences between them
differences = diff(interp1(t,y,t_eps,'spline'));
% delete all differences wich are not between t_new and t_new + eps
differences(2:2:end)=[];
% get the derivatives of each point
slopes = differences./(eps*precision);

t\u新 具有 t (或任何其他你想得到的微分)如果你想得到旧点的导数。