nonlinear headache

Hi

could some expert recommend a Liapunov function that might help
control this plant represented by a nonlinear differential equation:

y(t) = x(t) * ( 1 - F(s) * (|x(t)|^2) )

here F(s) = K/(1+sb), represents a 1st order low pass filter
(convolution with Kexp(-bt)) with time constant 1/b, and K is a
"small" positive number, so y is nearly x but not exactly; both b and
K are unknown although I have a good idea of their value within a
factor of 2.

( By F(s) * (|x|^2) is meant the square of x(t), x(t)^2, is low pass
filtered by F(s).)

I need a control law x = C[u, y],for positive u>0, such that the
tracking error |y-u| is small.

Thanks

robert egri wrote:

> Hi
>
> could some expert recommend a Liapunov function that might help
> control this plant represented by a nonlinear differential equation:
>
> y(t) = x(t) * ( 1 - F(s) * (|x(t)|^2) )
>
> here F(s) = K/(1+sb), represents a 1st order low pass filter
> (convolution with Kexp(-bt)) with time constant 1/b, and K is a
> "small" positive number, so y is nearly x but not exactly; both b and
> K are unknown although I have a good idea of their value within a
> factor of 2.
>
> ( By F(s) * (|x|^2) is meant the square of x(t), x(t)^2, is low pass
> filtered by F(s).)
>
> I need a control law x = C[u, y],for positive u>0, such that the
> tracking error |y-u| is small.
>
> Thanks

You are mixing Laplace notation with time notation.

Tom

rge11x@netscape.net (robert egri) wrote:

>Hi
>
>could some expert recommend a Liapunov function that might help
>control this plant represented by a nonlinear differential equation:
>
>y(t) = x(t) * ( 1 - F(s) * (|x(t)|^2) )
>
>here F(s) = K/(1+sb), represents a 1st order low pass filter
>(convolution with Kexp(-bt)) with time constant 1/b, and K is a
>"small" positive number, so y is nearly x but not exactly; both b and
>K are unknown although I have a good idea of their value within a
>factor of 2.
>
>( By F(s) * (|x|^2) is meant the square of x(t), x(t)^2, is low pass
>filtered by F(s).)
>
>I need a control law x = C[u, y],for positive u>0, such that the
>tracking error |y-u| is small.
>
>Thanks

I think what you said was 
dx/dt = x - F x^3 
Is this what you mean?  If so, you get 
dz/dt = (-z + F )/2
with z = 1/x^2
This equation should be much easier to work with.