Page 190 - Numerical Analysis Using MATLAB and Excel

The State Transition Matrix

(
At – t ) 0 At t – Aτ
xt() = e x + e ∫ e bu τ() τ (5.123)
d
0
t 0
Proof:
Let be any n × n matrix whose elements are constants. Then, another n × n matrix denoted as
A
ϕ t() , is said to be the state transition matrix of (5.34), if it is related to the matrix as the matrix
A
power series
1
1
1
2 2
3 3
n n
ϕ t() ≡ e At = I + At + -----A t + ----A t + … + -----A t (5.124)
-
2! 3! n!
where is the n × n identity matrix.
I
From (5.124), we find that
A0
ϕ 0() = e = I + A0 + … = I (5.125)
Differentiation of (5.124) with respect to yields
t
d At 2 2
⋅
ϕ't() = -----e = 0 + A 1 + A t + … = A + A t + … (5.126)
dt
and by comparison with (5.124) we obtain

d At At
-
----e = Ae (5.127)
dt
To prove that (5.123) is the solution of the first equation of (5.120), we must prove that it satisfies
both the initial condition and the matrix differential equation. The initial condition is satisfied
from the relation

(
At – t ) 0 At 0 t 0 – Aτ A0
0
d
xt () = e x + e ∫ e bu τ() τ = e x + 0 = Ix = x 0 (5.128)
0
0
0
0
t 0
where we have used (5.125) for the initial condition. The integral is zero since the upper and lower
limits of integration are the same.
To prove that the first equation of (5.120) is also satisfied, we differentiate the assumed solution
(
At – t ) 0 At t – Aτ
d
xt() = e x + e ∫ e bu τ() τ
0
t 0
with respect to and we use (5.127), that is,
t
d At At
-
----e = Ae
dt

Numerical Analysis Using MATLAB® and Excel®, Third Edition 5−29

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