Find x for known y from fit

Hi all

I would like to find values of x when y values are known using fzero. Assuming I have the following data:

x = [1      3      6       8      14    19   20  22]; 

y = [0.3   0.5     0.8    0.85    1    1.05  1.5  1.9];

xf=linspace(0,22,300)

yf=interp1(x,y,xf,'spline'); 

plot(x,y,'LineStyle','none','Color','k', 'MarkerSize',4 ,'Marker','square');
hold on
plot(xf,yf,'-r')

To find for example the value of x at yy=1.5 I used:

xdatax=fzero(@(xi)interp1(x,y,xi,'spline')-yy,5)

it works and it gives 3. But changing the value of yy (e.g.yy=1.5) gives xdata=-5.04 which is worong!!!

Does anyone know why it gives wrong results for some yy values? I appreciate your help and thanks..

Adam

[Adam]

Correction:

xdatax=fzero(@(xi)interp1(x,y,xi,'spline')-yy,5)

must be

xdatax=fzero(@(xf)interp1(x,y,xf,'spline')-yy,5)

Answer 1

[Walter Roberson]

-5.04 is a valid answer considering that you did not constrain the search range. Perhaps you want

xdatax = fzero(@(xf)interp1(x,y,xf,'spline')-yy,[min(x) max(x)])

Note that this will give an error if y(1)-yy is the same sign as y(end)-yy

Also, there are values such as 0.9 that occur multiple times; your code does not define which of the values will be located.

With spline fit, you are going to get "overcorrections". If, for example, you have a line that angles up to the right and it has a peak, then the spline will typically have its peak a little higher, because splines do not have sharp angles. This can result in false matches.

I suggest you reconsider your algorithm. The false matches you can get cannot be justified unless you know that the underlying physical process happens to have a spline response (for example your measurements happen to be along the edge of some bent wood.)