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 ential equation:
             dy
diy  d" y+..-+aid aoy=u(t)  (5.222)
dty + an_i dt                   l1 d~ + ao
Let us select a set of state vanables such that each subsequent state variable is
defined as the derivative of the previous state variable. That is,
Then,
XI =y
XI - X2
X2 -. X3
     dn-iy
X2 - y - Xl, - -.,Xn =  d~y = k" _1             (5.223)
(5.224)
(5.225)
Xn - -aoxi - alx2 - a2X3 - - - - - an -lXn + u(t)             (5.226)
Or, in matrix form,
x -. Ax + Bu
(5.227)
where
LINEAR SYSTEMS, THEORY, AND DESIGN: A BRIEF REVIEW       515
The output is given by
B-
                     Xl
                X2
y=Cx=[l O ... O] X3
                   xl
(5.228)
(5.229)
(5.230)
(5.231)
Thus, by choosing each successive state variable to be the derivative ofthe previous
one, we are able to express the given differential equation in the state-space, phase-
variable form.
5.10.7  Conversion of General State-Space Representation to
         Phase-Variable Form
  A system given in a general state-space form can be expressed iri the phase-
variable form if the system is controllable. Let
x - Ax + Bu
   y = Cx
(5.232)
(5.233)

 oioo.o
 ooio.o
A= 0 0 O 1  O
            ..,,
    ....-.
     -ao -ai  -a2  '  ' -an_i

516            PERFORMANCE, STABILITY, DYNAMICS, AND CONTROL
be the given plant, which is notin phase-variable form. We assume that this system
is controllable. Furthermore, let us assume that there exists a matrix P, which is
defined as
z - Px                              (5.234)
which transforms the given system into phase-variable form,

1
0
O
00
10
01
0
O
0
0
0
-ao  -ai  '  '  '  -an_i

The transformation matrix P has the form
                   p\\  P12  '  '  '
 p21p22"
                   p31  P32  .  ' -
                   P -.
             .-...
              .....
                   Pnl  pn2 '  ' -
   We have
                                      Zi - Pix
so that
u    (5.235)
(5.23 6)
(5.237)
                                             Zi = Pi Ax + Pi Bu                               (5.238)
Because this transformed equation is suppased to be in phase-variable form, we
must have zi - 22. This gives 22 - PiAx and PiB - 0. From Eq. (5.234),
we have 22 - P2X. Therefore, P2 - PiA. Continuing this further, we find that
22 - Pl A2X, Pi AB  - 0, P3  - Pl A2,...,2n_l  = Pi An-lX, PiAl-2B = O,
and Pn-I = PiAn-2.Finally,we have
　
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