Nonlinear Control Sy.. - Free

10.3. EXAMPLES 271
provided that
which results in
0. Thus we can choose
T1(x) = xi
T=[e-x21 1J
Notice that this coordinate transformation maps the equilibrium point at the origin in the x
plane into the origin in the z plane.
The functions 0 and w can be obtained as follows:
w = 29(x) = ex2
It is easy to verify that, in the z-coordinates
which is of the form
with
{ zl = z2
(aT2/ax)f(x) _
(0T2/8x)9(x)
f z1 = x1
ll z2=eX2-1
z2 = az1 z2 + azl + (z2 + 1)u
z = Az + Bw(z) [u - O(z)]
A A,- [ 0
O I, B=B,
= 1 0 J
w(z) = z2 + 1, O(z) _ -azi.
Example 10.11 (Magnetic Suspension System)
Consider the magnetic suspension system of Chapter 1. The dynamic equations are
x2 0
xl
aµa2
k
x2 = 9 mX2 2m 1+µa1 + 0
xg 1+µa1 (-Rx + l µ x )
a 3 1+µa1) 2x3 A
(
11