about approximate linearization

On Sun, 13 Dec 2009 06:23:21 -0800, melda wrote:


Explaining it by Taylor's expansion is one way.

What you're really trying to do is find the derivative around the
operating point for the whole state vector.  So your

x1dot = x2 * x3

would linearize as

x1dot = x2(0) * x3 + x3(0) * x2,

where the x2(0) and x3(0) are the values of those two states at the
nominal operating point.

More formally (but still from memory, so I may not be quoting textbooks
exactly), if x is a vector, and if your system is described as

dx
-- = f(x, u, t)
dt

then your linearization would be

dx                     del              |
-- = f(x(0), u, t) +  ----- f(x, u, t)  |           ,
dt                    del x             | x = x(0)

where the "del / del x" operation takes the Jordanian of the system
function.

This still isn't a perfectly linear system because you're adding in the
value of the system function at the operating point, but it's affine, and
pretty easy to turn into a linear system by replacing that expression
with some extra integrators, appropriately initialized.

--
www.wescottdesign.com

Small correction: designating the derivative/Jacobian of f(x,u,t) with
respect to x as f'(x0,u,t)
the linearization would be
f(x0,u,t)+f'(x0,u,t)(x-x0)
Where the second term yields a linear correction; the Jacobian matrix
times a variation vector x0.
Ray

On Mon, 14 Dec 2009 08:46:06 -0800, RRogers wrote:


+  -----

        | x =

Thank you, yes.

--
www.wescottdesign.com