在Matlab中进行样条插值以预测值 [英] Spline interpolation in matlab in order to predict value

问题描述

我的情况如下图所示:

此图是两个向量的结果:

This plot is the result of two vectors:

fi = [41.309180589278, 41.8087915220215, 42.8081880760916, ...
      43.8078181874395, 44.8076823745539, 45.8077808710707, 46.3079179803177]
m = [1.00047608139868, 1.00013712198767, 0.999680989440986, ...
     0.999524195487826, 0.999671686649694, 1.00012913666266, 1.00047608139868]

我需要获取fi的值，其中m等于1.所以大约是42.2和42.5.

I need to get the values of fi where m is equal to 1. So approximately that will be 42.2 and 42.5.

我试图进行样条插值:

xq = [fi(1):0.25:fi(7)];
vq1 = interp1(fi,m,xq);
[fi1, fi2] = interp1(m, xq, 1)

但这不起作用.有人可以帮我吗?

But that is not working. Can someone help me with this?

推荐答案

查找过零的一种方法是使图形侧向翻转"，使fi是m的函数，然后进行插值以找到m=0.但是interp1要求m输入必须是单调的，但不是.实际上，对于每个m，此函数都有两个不同的值.

One way to find a zero crossing is to "turn the graph sideways", having fi be a function of m, and interpolate to find m=0. But interp1 requires the m input to be monotonic, which this is not. In fact, this function has two different values for each m.

MATLAB知道 fzeros函数，该函数可以找到一个函数的数字过零.它需要一个函数作为输入.我们可以使用interp1定义匿名函数，该函数为x的任何值返回m-1.在这里，x由fi定义，而f(x)由m定义:

MATLAB knows the fzeros function, which finds a zero crossing of a function numerically. It requires a function as input. We can define an anonymous function using interp1, which returns m-1 for any value of x. Here, x is defined by fi and f(x) by m:

fi = [41.309180589278, 41.8087915220215, 42.8081880760916, ...
      43.8078181874395, 44.8076823745539, 45.8077808710707, 46.3079179803177];
m = [1.00047608139868, 1.00013712198767, 0.999680989440986, ...
     0.999524195487826, 0.999671686649694, 1.00012913666266, 1.00047608139868];
fun = @(x)interp1(fi,m,x)-1;
x1 = fzero(fun,42)
x2 = fzero(fun,46)

这给了我

x1 =  42.109
x2 =  45.525

请注意，我们需要知道这两个零的大概位置.据我所知，没有简单的方法可以解决此问题.如果知道两个零交叉以及函数的一般形状，则可以找到局部最小值:

Note that we needed to know the approximate locations for these two zeros. There is no easy way around this that I know of. If one knows that there are two zero crossings, and the general shape of the function, one can find the local minimum:

[~,fimin] = min(m);
fimin = fi(fimin);

，然后找到每个端点和局部最小值之间的零交叉点:

and then find the zero crossings between each of the end points and the local minimum:

x1 = fzero(fun,[fi(1),fimin])
x2 = fzero(fun,[fimin,fi(end)])

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