Page 911 - The Mechatronics Handbook
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0066_Frame_C30  Page 22  Thursday, January 10, 2002  4:44 PM









                       We will see below that this will be the approximate open loop transfer function PK opt  for the final design.
                       In this sense, G KF  represents our target open loop transfer function.
                         Control Gain Matrix G c . Since  D 12 C 1 =  0  , the associated CARE is given by
                                                  T
                                       A X +  XA +  C 1 C 1 –  XB 2 R B 2 X =  X +  X +  1 –  1  2  0  (30.117)
                                                                              ---X =
                                                   T
                                         T
                                                             1
                                                            –
                                                                              r
                       or
                                                        2
                                                       X –  2rX r =  0                         (30.118)
                                                               –
                       Application of the quadratic formula and selecting the positive (stabilizing) root yields:
                                                                 2
                                                       X =  r +  r +  r                        (30.119)
                       This yields the following control gain matrix:


                                                   G c =  R B 2 X =  1 +  1 +  1               (30.120)
                                                         −1
                                                            T
                                                                        ---
                                                                        r
                       This results in a closed loop (regulator) pole at
                                                                   --- =
                                              AB 2 G c =  11 –  1 +  1  –  1 +  1              (30.121)
                                                                             ---
                                                –
                                                          –
                                                                   r         r
                       Note that for large r (referred to as expensive control in LQR problems) we have a closed loop pole at
                       s = −1, at the left half plane reflection of the plant pole at s = 1. We will select the design parameter r to
                       be small (referred to as cheap control in LQR problems) so that this closed loop (regulator) pole s ≈ −1/ r
                       is fast and the closed loop filter pole at s = −5 at is the dominant closed loop pole.
                          2
                                                                                 2
                         H  Optimal Output Feedback Model-Based Compensator. The resulting H  optimal output feedback
                       model-based compensator is given by
                                                       AB 2 G c –  H f C 2  H f
                                                         –
                                                K opt =                                        (30.122)
                                                            – G c     0 n ×  n
                                                                        u  y
                       where

                                                                                1
                                                                      1
                                                                         1
                                             –
                                                             –
                                           AB 2 G c – H f C 2 =  11 –  1 +  --- – –  1 +  ---  (30.123)
                                                                      r         m
                                                         =  11 –  1 +  1  1  1 +  24           (30.124)
                                                                      --- – –
                                                             –
                                                                      r
                                                                    1
                                                         =  –  6 –  1 +  ---                   (30.125)
                                                                    r
                                                                   1
                                                      H f =  1 +  1 +  ---                     (30.126)
                                                                   m
                                                         =  1 +  1 +  24                       (30.127)
                                                         =  6                                  (30.128)
                                                      G c =  1 +  1 +  1                       (30.129)
                                                                   ---
                                                                   r

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