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336 ———  MATLAB: An Introduction with Applications


                   6.5.4  Park Stiffly Stable Method
                   The Park Stiffly Stable method is an accurate method for low frequency ranges and stable for all higher-
                   frequency components. Using a linear combination of the following two difference formulas derives the
                   velocity in the Park Stiffly Stable method:
                                         1
                                  X    t+ ∆  =  6 t ∆  [11X t+ ∆t  − 18X +  9X t− ∆ t  t  −  2X t− 2 t ∆  ]  ...(6.76)
                                                       t
                                         1
                                  X    t+ ∆  =  2 t ∆  ( 2X t+ ∆t  − 4X +  X t− ∆ t  t )            ...(6.77)
                                                      t
                   The linear combination of (6.76) and (6.77) gives
                                         1
                                  X    t+ ∆  =  4 t ∆   (3X t+ ∆t  − 4X +  X t− ∆ t  t )
                                                      t
                                           1
                                               +  (11X  − 18X +  9X  −  2X  )                       ...(6.78)
                                         12 t ∆   t+ ∆    t   t− ∆t  t  t− 2 t ∆
                                         1
                   or             X    t+ ∆  =  6 t ∆  (10X t+ ∆t  − 15X +  6X t− ∆ t  t  −  X t− 2 t ∆  )  ...(6.79)
                                                       t
                   Similarly, for the acceleration, we obtain
                                         1


                                  X t+ ∆  =  6 t ∆  (10X   t+ ∆t  t  − 15X +  6X   t−∆ t –X   t –2 t ∆  )  ...(6.80)
                                                       t
                   The difference formulas in the Park Stiffly method are given by
                                                                             }
                                         1
                                                                      −
                                                     −
                                { X    t+ ∆  } =  6 t ∆    10 { X   t+ ∆t  t } 15 X    t  + 6    t− ∆  t } { X   t− 2 t ∆    ...(6.81)
                                                        { } { X
                                         1
                                                     −
                                                                             }
                                                                      −
                                { X   t+ ∆  } =  6 t ∆    10 {X t+ ∆t  t  } 15 X t  +  {  t− ∆  t } {X t− 2 t ∆   ...(6.82)
                                                        { } 6 X
                   We consider Eq. (6.13) to obtain solution for the displacements, velocities and accelerations at time  t + t∆ .
                                                         {

                   By substituting the expressions for { X t+ ∆  }  and  X   t+ ∆t  t }  from (6.81) and (6.82), respectively, into (6.13), we get
                                      =
                              | M  { | X t+∆ t  } { t+∆ t }                                         ...(6.83)
                                         F
                                              
                                                                             F
                                                 
                                              
                   where the effective mass matrix  M   and the effective force vector { t+∆ t } are given by
                                                    10
                                          100
                                   
                                     M  =  36 t ∆  2 [ ] − 6 t ∆  [] [ ]                           ...(6.84)
                                                        C +
                                                            K
                                               M
                                      
                                   
                                          15           1              1           }
                                                                         M
                                                          M
                                                  X
                                  F
                                              M
                                 { t+∆ t } =  6 t ∆  [ ]{ } −  ∆t  [ ]{ X t− ∆t } + 6 t ∆  [ ]{ X t− 2 t
                                                                                ∆
                                                   t
                                            150      15            10      1    
                                                +   36 t ∆  2  [ ] +  6 t ∆  [] { } −   6 t ∆  2  [ ] +  ∆t [] {X t − ∆t }  ...(6.85)
                                                 M
                                                              X
                                                                         M
                                                                                 C
                                                          C
                                                                                   
                                                            
                                                            
                                                               t
                                                                                   
                                                            
                                                       1
                                             1
                                                +    36 t ∆  2  [ ] +  6 t ∆  [] {X t−∆  }
                                                 M
                                                          C
                                                            
                                                                 2 t
                                                            
                   The solution of Eq.(6.83) gives  {X t+ ∆t } , which is then substituted in Eq.(6.82) to obtain velocities. The
                   values of { X    t+ ∆t }  are then obtained by the use of Eq.(6.81). Note that in the Park Stiffly stable method, the
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