Fit of groups of data

Hi,

I have two groups of data. Both of them have the same x-axis but differ in y-axis. I want to to solve an equation that depend on these two groups of data.

The x-axis is representing time x=[1; 2; 3; 4; 5; 10; 15; 30; 45; 60; 90]

and y-axis for the first group is: y1=[56.88; 62.75; 87.07; 114.54; 139.14; 177.24; 216.39; 285.77; 399.99; 357.28; 359.53]

and y-axis for the second group is: y2=[353.64; 343.60; 303.67; 312.73; 328.91; 306.76; 299.97; 259.01; 222.10; 190.04; 228.87]

Here is my equation: y1/(y1+y2)=[k1/(k1+k2)]*{1-exp[-(k1+k2)*t]}

Note: all the equation parameters are known except k1 and k2.

How can I find them by fitting the data please?

[the cyclist]

In your equation, you wrote t. I assume that should be x.

[the cyclist]

I suggest you simplify by defining

z = y2./y1

Then you just have 1/(1+z) on the left.

[Hisham]

sorry sir, I did not understand the last comment. Why did you do z = y2./y1 ?

[the cyclist]

An equation with one variable is simpler than an equation with two variables. If you use z = y2./y1, then you can write 1./(1+z) instead of y1./(y1+y2).

[Matt J]

You could even define

 z = y1/(y1+y2)

and perform a fit to the equation

 z=A*{1-exp[-B*t]}

There aren't truly 2 data sets being fitted, here.

[Walter Roberson]

Hisham commented "yes, t should be x"

Answer 1

[the cyclist]

If you have the Statistics Toolbox, I think the nlinfit() function will do what you need: http://www.mathworks.com/help/stats/nlinfit.html