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Section 3.5  Alternative Signal-Flow  Graph and Block Diagram  Models  185

                                           l
                          " . O   • -







                          " 2 O — *



                                              (a)


                          4          1      x

                          -      — •         ' .   a
                       u x           S

                                                       1 r +
                                                        > —   s      X-i
                                                        i
                                 7—
       FIGURE 3.20     u 2  -        1
       Model for the                 5      X,   y
      spread of an        -   ^
      epidemic disease.       T
      (a) Signal-flow                r
      graph, (b) Block
      diagram model.                          (b)

                       By examining Equation (3.63) and the models depicted in Figure 3.20, we find that the
                       state variable * 3 is dependent  on x-[  and x 2  and does not  affect  the variables Xi and x 2-
                           Let  us consider  a closed  population,  so that  u\(t)  =  ^2(/)  =  0. The  equilibri-
                       um  point  in the  state  space  for  this  system  is obtained  by setting  dx/dt  — 0. The
                       equilibrium  point  in the state space  is the point  at which the system settles in  the
                       equilibrium, or rest, condition. Examining Equation  (3.63), we find  that  the  equi-
                                                        =  0. Thus, to determine whether the  epidemic
                       librium point for this system is x v  — x 2
                       disease is eliminated  from  the population, we must obtain the characteristic  equation
                       of  the  system. From  the  signal-flow  graph  shown  in Figure  3.20, we  obtain  the  flow
                       graph  determinant
                                                                    2
                                                                             2
                                      A(s)  =  1  -  (-as~ l  -  ys' 1  -  jSV )  +  (ays' ),  (3.64)
                       where  there  are  three  loops, two  of  which  are  nontouching. Thus, the  characteristic
                       equation  is
                                                                           2
                                             2
                                     q{s)  =  s A(s)  =  s 2  +  (a  +  y)s  +  (ay  +  jS )  =  0.  (3.65)
                           The  roots  of  this  characteristic  equation  will  lie  in  the  left-hand  s-plane  when
                       a  +  y  >  0 and  ay  +  ft 2  >  0. When roots  are  in the  left-hand  plane, we expect  the
                       unforced  response  to decay to zero as t —*• 00.  •
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