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From: Greg Heath <heath@alumni.brown.edu>
Newsgroups: comp.soft-sys.matlab
Subject: Re: simple parameter estimation
Date: Fri, 10 Oct 2008 20:25:35 -0700 (PDT)
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On Oct 9, 9:53 am, "Marcio " <marciobarba...@yahoo.com> wrote:
> please, dear friends,
>
> how'd you solve a system like this:
>
> for a given y1, y2 and y3:
>
> dy1/dt = a*y1 +     b*y2 - c*y3
> dy2/dt = b*y1 - (a-1)*y2 - c*y1*y3
> dy3/dt = a*y1 - b
>
> How to find a, b and c?
>
> Many thanks

If y1,y2,y3 are given, dy1/dt,dy2/dt and dy3/dt
are readily estimated.

You then have 3 systems of N linear equations for
A = [a b c]'.

Define

Y1 = [y1 y2 -y3];
Y2 = [-y2 y1 -y1.*y3];
Y3 = [y1 -ones(N,1) zeros(N,1)];

Then

dy1/dt    = Y1*A
dy2 - y2  = Y2*A
dy3/dt    = Y3*A

Note that Y3 is rank deficient. Therefore, the LS
solution of the 3rd system may not be unique.
Although Y1 and Y2 are full rank, the 1st two equations
yield 2 unique, but very different LS solutions
for A. In general, all of the three LS solutions will
be different.

A more satisfactory approach is to use a combined LS
solution. If there is evidence that points or equations
have unequal importance, weighting is easily introduced.

The ordinary LS solution results from minimizing


f = ||dy1/dt-Y1*A||^2 + ||dy2/dt-y2-Y2*A||^2
     + ||dy3/dt-Y3*A||^2.


Hope this helps.

Greg