Page 304 - Dynamics and Control of Nuclear Reactors
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306    APPENDIX F State variable models and transient analysis




                         or
                                                    dx
                                                      ¼ AxtðÞ + BftðÞ
                                                     dt
                         This is the original set of differential equations.
                            The state transition matrix φ(t)¼exp(At) also has the following properties.
                                                    ðÞφ t 2 ¼ φ t 1 + t 2 Þ
                                                   φ t 1  ðÞ  ð                         (F.31)
                                                         1
                                                             ð
                                                     φ tðÞ  ¼ φ  tÞ                     (F.32)
                                                             ð t
                                            xtðÞ ¼ φ t t 0 Þxt 0 +  φ t τÞBf τðÞdτ      (F.33)
                                                               ð
                                                  ð
                                                        ðÞ
                                                             t 0

                           Example F.3
                            For the system defined in Example F.2, determine the state transition matrix.



                           Solution


                                                          h      i
                                                   φ tðÞ ¼ L  1  ð sI  AÞ  1

                                                  2  3          s +2  3
                                             A ¼       , ð sI  AÞ ¼
                                                  1  4            1  s +4
                                                          1      s +4  3
                                                    1
                                              ð sI  AÞ  ¼
                                                       ð s +1Þ s +5Þ  1 s +2
                                                           ð
                                                  2                 3
                                                      s +4      3
                                                                ð
                                                            ð
                                                   ð s +1Þ s +5Þ s +1Þ s +5Þ
                                                       ð
                                                  6                 7
                                                ¼ 6    1       s +2  7
                                                  4                 5
                                                            ð
                                                       ð
                                                                ð
                                                   ð s +1Þ s +5Þ s +1Þ s +5Þ
                                             2                           3
                                              3  1   1  1    3  1   3  1
                                                    +              +
                                                              ð
                                                                     ð
                                              4 s +1Þ  4 s +5Þ  4 s +1Þ  4 s +5Þ
                                               ð
                                                      ð
                                             6                           7
                                           ¼ 6                           7
                                                1  1   1  1  1  1   3  1
                                             4                           5
                                                     +            +
                                                              ð
                                                                     ð
                                                 ð
                                                        ð
                                                4 s +1Þ  4 s +5Þ 4 s +1Þ  4 s +5Þ
                              Taking Laplace inverse transform gives the state transition matrix.
                                                  2                    3
                                                    3    1     3    3
                                                                  t
                                                       t
                                                     e + e  5t    e + e  5t
                                                    4    4     4    4
                                                  6                    7
                                              φ tðÞ ¼ 4  1  1  1   3   5
                                                       t
                                                                 t
                                                     e + e  5t  e + 3  5t
                                                     4   4    4    4
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