Page 303 - Dynamics and Control of Nuclear Reactors
P. 303

APPENDIX F State variable models and transient analysis  305





                       Using partial fractions (or the method of residues) gives.
                                               1    1   3  1
                                        YsðÞ ¼    +
                                              5s  2 s +1Þ  10 2+ 5Þ
                                                          ð
                                                  ð
                                         1  t
                                      1
                                             3
                       Time response ytðÞ ¼   + e   e  5t
                                      5  2   10
                  F.3.4 The state transition matrix
                  The matrix φ(t) in Eq. (F.20) is often called the State Transition Matrix (STM)
                  because it provides the solution x(t) at time t from an initial time t¼0 (or t¼t 0 ).
                  In general Eq. (F.20) may be written in the form.
                                                     ð t
                                      xtðÞ ¼ φ t, t 0 Þxt 0 +  φ t, τÞBf τðÞdτ  (F.25)
                                                ðÞ
                                                       ð
                                            ð
                                                     t 0
                  We showed that for the special case of time-invariant system.
                                                   h      1  i
                                           φ tðÞ ¼ L  1  ð sI  AÞ              (F.26a)
                  and
                                              φ tðÞ ¼ exp AtðÞ                 (F.26b)
                  For time-invariant systems the state transition matrix φ(t) satisfies the following
                  properties [3].
                                               dφ
                                                  ¼ Aφ tðÞ                      (F.27)
                                               dt
                                            φ t 0 , t 0 Þ ¼ I,φ 0 ðÞ ¼ I        (F.28)
                                             ð
                                              ð
                                                     ð
                                             φ t, t 0 Þ ¼ φ t t 0 Þ             (F.29)
                  These properties satisfy the solution.
                                                   ð t
                                       xtðÞ ¼ φ tðÞx 0ðÞ + φ t τÞBf τðÞdτ       (F.30)
                                                      ð
                                                   0
                  Differentiate Eq. (F.30) with respect to t.
                                                       ð t
                                  dx  dφ tðÞ
                                                           ð
                                    ¼    x 0ðÞ + φ 0ðÞBftðÞ + Aφ t τÞBf τðÞdτ
                                  dt   dt
                                                       0
                  Using Eqs. (F.27) and (F.28) in the above gives.
                                                       ð t
                                   dx
                                     ¼ Aφ tðÞx 0ðÞ + BftðÞ + A φ t τð  ÞBf τðÞdτ
                                   dt
                                                       0
                                                ð t
                                       2                    3
                                    ¼ A φ tðÞx 0ðÞ + φ t τð  ÞBf τðÞdτ + BftðÞ
                                       4
                                                            5
                                                0
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