Page 760 - The Mechatronics Handbook

0066_Frame_C24 Page 8 Thursday, January 10, 2002 3:43 PM

At
where the transition matrix e satisfies
∞
At 1 k k
e = I + ∑ ---- A t (24.45)
k=1 k!

The interested reader can check that (24.44) satisfies (24.43). To do that he/she should use the Leibnitz’s
rule for the derivative of an integral.
With the above result, the solution for (24.43) is given by

(
(
y t() = Ce A t−t ) o x o + C ∫ t e A t−t) Bu t() t + Du t() (24.46)
d
o t
System Dynamics
The state of the system has two components: the unforced component, x u (t), and the forced component,
x f (t), where

(
A t−t )
x u t() = e o x o (24.47)
(
x f t() = ∫ t t o e A t−t ) Bu t() t (24.48)
d
To gain insight into the state space model and its solution, consider the case when t o = 0 and u(t) = 0
∀t ≥ 0, i.e., the state has only the unforced part. Then

x t() = e x o (24.49)
At
n
Further assume that A ∈ and that, for simplicity, it has distinct eigenvalues λ 1 , λ 2 ,…,λ n with n
(linearly independent) eigenvectors v 1 , v 2 ,…, v n . Then there always exists a set of constants α 1 , α 2 ,…,α n
such that

n
x o ∑ a v , a ∈ (24.50)
=
=1
k k k k
,
A well-known result from linear algebra tells us that the eigenvalues of A are l 1 l 2 ,…,l n with
corresponding eigenvectors v 1 , v 2 ,…, v n . The application of this result yields

∞
n
n
1
l t
x t() = e x o = I + ∑ a ∑ ---- A v t = ∑ a e v (24.51)
At

k
k
=1 k=1 k!  k   =1
l v

This equation shows that the unforced component of the state is a linear combination of natural modes,
l t

{e }, each of which is associated with an eigenvalue of A. Hence the matrix A determines:
• the structure of the unforced response
• the stability (or otherwise) of the system
• the speed of response
When the matrix A does not have a set of n independent eigenvectors, Jordan forms can be used (see,
e.g., [9,10]).
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