Page 768 - The Mechatronics Handbook
P. 768
0066_Frame_C24 Page 16 Thursday, January 10, 2002 3:44 PM
If we consider the continuous, time-invariant, and linear state space model defined by equations (24.42)
and (24.43), with initial state x(k 0 ∆) = x 0 , we can use Eq. (24.44) to calculate the next value of the state:
k ∆+∆
(
(
0
(
x k 0 ∆ + ∆) = e A k ∆+∆−k ∆) x k 0 ∆) + ∫ k ∆ e A k ∆+∆−t) B u t() t (24.90)
(
0
0
0
d
0
Furthermore, if a zero order hold is used, i.e., u(t) = u(k 0 ∆) for k 0 ∆ ≤ t < k 0 ∆ + ∆, we obtain
A∆
(
Ah
x k 0 ∆ +( ∆) = e x k 0 ∆) + ∫ ∆ e dhB u k 0 ∆) (24.91)
(
0
And, if we know the state and the input at time k 0 ∆, the output is defined by Eq. (24.43):
y k 0 ∆( ) = Cx k 0 ∆) + D u k 0 ∆) (24.92)
(
(
We can now conclude that given a continuous-time model with state space matrices {A, B, C, D}, and
we sample inputs and outputs every ∆ seconds then, the equivalent sampled data systems will be described
by the discrete-time state space model:
x k∆ + ∆) = A d x k∆( ) + B d u k∆) (24.93)
(
(
(
(
(
y k∆) = C d x k∆) + D d u k∆) (24.94)
where
Ah
A d = e , B d = ∫ ∆ e dhB, C d = C, D d = D (24.95)
A∆
0
There are different methods to obtain A d defined in (24.95), but a simple way to calculate this matrix
is to use Laplace transformation. This yields
A d = e A∆ = L { ( sIA) } t =∆ (24.96)
1
1
–
–
–
Example 24.6
Consider the mechanical system of Example 24.1 on the page 4, that was described by the state space
model:
x ˙ 1 t() 0 1 x 1 t() 0
= + ft() (24.97)
K
1
x ˙ 2 t() – ----- – D x 2 t() – -----
-----
M
M
M
where f(t) is the external force, and where we can choose either the mass position, x 1 (t), or the mass
velocity, x 2 (t), of the mass, as the system output.
For the purpose of a numerical illustration, we set M = 1 kg, D = 1.2 N s/m, and K = 0.32 N/m.
The matrix A d is obtained from (24.96), applying inverse Laplace transformation
1 – 1 – 0.4∆ – 0.8∆ ( – 0.4∆ – 0.8∆ )
A d = L s – 1 = 2e – e 2.5 e – e (24.98)
–
0.32 s + 1.2 ( – 0.4∆ – 0.8∆ – 0.4∆ + – 0.8∆
–
t =∆ 0.8 e – e ) e 2e
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