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Section 3.3  The State Differential Equation                        167
                      where y is the set of output signals expressed in column vector form. The state-space
                      representation  (or  state-variable  representation)  comprises  the  state  differential
                      equation  and the output  equation.
                          We use Equations (3.8) and (3.9) to obtain the state variable differential  equation
                      for the RLC  of Figure 3.4 as

                                                       - 1
                                                   0    —
                                                       C
                                             x  =          x +   C                       (3.18)
                                                  J_   -R        0   «(0
                                                   L   L
                      and the output as
                                                    y  =  [0  R]x.                       (3.19)
                      When R  =  3, L  =  1, and C  =  1/2,  we have

                                                    "o  -2~     V
                                               x  = 1       x +
                                                   L    -3 J  L°J
                      and

                                                    y  =  [0  3]x.
                          The solution  of the state differential  equation  (Equation  3.16) can be obtained
                      in  a manner  similar  to  the  method  for  solving  a  first-order  differential  equation.
                      Consider the first-order  differential  equation
                                                    x  = ax  + bu,                       (3.20)
                      where x(t)  and u(t) are scalar functions  of time. We expect  an exponential solution of
                               at
                      the form e . Taking the Laplace transform  of Equation (3.20), we have

                                            sX{s)  -  x(0)  =  aX{s)  + bU(s);
                      therefore,
                                                     x(0)      b
                                             X(s)  =  T ^ :  +  T^-tf(s).                (3.21)
                                                     s  —  a  s  —  a
                      The inverse Laplace transform  of Equation  (3.21) can be shown to be

                                                 at          +a( T)
                                          x{t)  =  e x(0)  +  [  e '- bu(T)  dr.         (3.22)
                                                         Jo
                      We  expect  the  solution  of  the  general  state  differential  equation  to  be  similar  to
                      Equation  (3.22)  and  to be  of  exponential  form. The  matrix exponential  function is
                      defined  as
                                                                        k k
                                                             2,2        \ t
                                    ,A/  _                  Mi        +  ——  +•
                                       =  exp(Ar)  =  I  +  At  +  +                     (3.23)
                                                             2!          k\
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