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about approximate linearization

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melda

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Dec 13, 2009, 9:23:21 AM12/13/09
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hello all,

I'm a little bit unexperienced in the area and wanna learn probably a
basic key issue about approximate linearization of nonlinear models
for control.

For an example if a process model is like "x1dot=square(x2) +..", then
I know that approximate linearization of the model is extracted via
taylor expansion,
but if the nonlinear model is like "x1dot=x2.x3 +..", how should I
realize the linearization?

Thank you in advance...

Tim Wescott

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Dec 13, 2009, 1:39:03 PM12/13/09
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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

RRogers

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Dec 14, 2009, 11:46:06 AM12/14/09
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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

Tim Wescott

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Dec 14, 2009, 1:11:02 PM12/14/09
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Thank you, yes.

--
www.wescottdesign.com

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