Page 79 - MODUL DARING SISTEM PENGATURAN_Neat
P. 79

2   1    2  − 2  0  1   
                                       1             1                           
                              *  a =  −   ( APt  ) = −  t r  1  1  1     1  − 2  0   
                                  0
                                       3  r   0      3
                                                         3   2  4    1  1   −1     
                                                         
                                                                      
                                           −1    0    0  
                                       1               
                                     = −  t   0  −1    0
                                       3  r            
                                            0    0   −1 
                                           
                                                        
                                      1
                                    = − (-1-1-1)
                                      3

                                    = 1
                              maka :

                                                2
                              adj (s I – A) =  P 2  s +  P 1  s +  P
                                                          0
                                               1  0    0   3  0     1  2  1    2
                                             2                                
                                                  =  s   0  1  0  + s  1  3  0  +  1  1  1
                                                       
                                                                                   
                                                                          
                                                                      
                                                             
                                               0  0    1   1  1    4   3  2   4 
                                               
                                                                          
                                                             
                                            s 2  + 3s  + 2  1          s + 2  
                                                                              
                                                  =    s  +1  s 2  + 3s  +1  1  
                                               s + 3       s  + 2   s 2  + 4s  + 4 
                                                                              
                                                                          2
                                                    2
                                             3
                                                                    3
                                                                             7
                              det (s I – A) =  s +  a 2  s + a 1 s +  a =  s + 5s + s + 1
                                                               0
                              maka :
                                      Y (s )                    adj (sI −  ) A
                                                       -1
                              G(s) =       = C (s I – A)  B = C               B
                                      U (s )                    det(sI −  ) A
                                                       s
                                                 s 2  + 3 + 2    1          s + 2  
                                                              2                    
                                                                    s
                                                    s  + 1   s  + 3 + 1       1    
                                                    s + 3       s  + 2   s 2  +  4 +  4   1 
                                                                                s
                                                                                        0
                                     = 2  1 −   1                                   
                                                                                         
                                                                                       
                                                                       s
                                                           s 3  + 5s  2  + 7 + 1
                                                                                       1 
                                                                                         
                                                                                       
                                                         s
                                                  s 2  +  4 +  4
                                                             
                                                     s  +  2  
                                                   s 2  +  5 + 7 
                                                         s
                                      = 2  1 −   1         
                                                 s 3  + 5s 2  +  7 + 1
                                                            s
                                        s  2  + 4 + 3
                                              s
                                      =
                                                 s
                                     s 3  + 5s 2  +  7 + 1
                        Metode Ruang-Keadaan
                                                                                                     78
   74   75   76   77   78   79   80   81   82   83   84