Page 837 - The Mechatronics Handbook

066_Frame_C26 Page 22 Wednesday, January 9, 2002 1:59 PM

The th order Padé approximation is defined as follows:

k
k k
N h s() = ∑ – ( 1) c kh s
k=0

k k
D h s() = ∑ c k h s
k=0
where coefficients c k ’s are computed from
( 2 k)! !
–
c k = --------------------------------, k = 0, 1,…,
(
2 !k! k)!
–
First- and second-order approximations are in the form
 1 hs/2
–
 -------------------, = 1
N h s()  1 + hs/2
------------- = 
D h s()  ----------------------------------------------, =
1 hs/2 +
2
(
hs) /12
–
 1 + hs/2 + ( 2 2
 hs) /12
Given the problem data {h, K max , N(s), D(s)}, how do we find the smallest degree, , of the Padé
approximation, so that ∆ h ≤ d (or ∆ h /K max ≤ d′ ) for a specified error d, or a specified relative error d¢ ?
The answer lies in the following result [7]: for a given degree of approximation we have
 2 +1
ehw

 2 ----------  , w ≤ 4
------
(
N h jw)   4  eh
−jhw
e – ------------------ ≤ 
D h jw)  4
(
 2, w ≥ ------
 eh
In light of this result, we can solve the approximation order selection problem by using the following
procedure:
1. Determine the frequency w x such that
K max Njw) d for all w ≥
(
--------------------------- ≤
--,
(
Djw) 2 w x
and initialize = 1.
2. For each ≥ 1 define
 4 

w = max w x , ------ 
 eh 
and plot the function

 2 +1
K max Njw(
 2 --------------------------- )  ehw  , for w ≤ 4
----------
------
(
Φ w() : =  Djw)  4  eh

 K max Njw( ) 4
 2 --------------------------- , for w ≥ w ≥ ------
(
 Djw) eh
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