Page 78 - MODUL DARING SISTEM PENGATURAN_Neat

CONTOH 4.3 :

Jika dari persamaan (4-8) diketahui :
− 2 0 1  1 
   
A = 1 − 2 0 , B = 0 , C = 2 1 −  1 , gunakan algoritma Leverrier
   
 1 1 −1  1 




untuk menentukan fungsi alihnya.
JAWAB :
Dalam hal ini n = 3, yaitu matriks A berdimensi 3 x 3
* P = I
2
* a = t − r A= -(-2-2-1) = 5
2
* P = P A + a I = I A + a I = A + 5 I
2
2
2
1
− 2 0 1  1 0  0 3 0  1






= 1 − 2 0 + 5 0 1 0 = 1 3 0






 1 1 −1  0 0  1  1 1  4 




3 0  1 − 2 0 1  
1 1     
* a = − t (P ) A = − t r  1 3 0 1 − 2 0
1
2 r 1 2    
1 1 4   1 1 −1    
 

− 5 1 2 
1  
= − t 1 − 6 1 

r
2
 3 2 −  3 

1
= − (-5-6-3)
2
= 7
3 0  1 − 2 0 1  1 0  0
 




* P = P A + a I = 1 3 0 1 − 2 0 + 7 0 1 0

 



0
1
1
1 1  4   1 −1  0 0  1 
 1



2 1  2


= 1 1 1


3 2  4 

Metode Ruang-Keadaan
77