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CLOSED LOOP CONTROL  71
                                            x(t)



                                         k
                                                        m            f(t)
                                          c








                              f(t)                                            x(t)
                             Input               Mass-spring-damper          Output

                             FIGURE 2.21: Mass-spring system dynamics and its position control.


                             Further, we will consider the step response when there is an additional pole and zero
                                                        (    )
                                                         s
                                                          + 1          2
                                                         a           n
                                                        (    )
                                                         s  + 1 s +      s +    2
                                                               2
                                                         b           n    n
                             The utility of this is that given transient response specifications for a control system
                             (PO%, t ) we can determine where the dominant poles should be in order to meet the
                                   s
                             specifications. Even though our system may not be second order, second-order system
                             pole-zero locations can provide a good starting point in design, especially if the higher
                             order system can be made to have a dominant second-order dynamics.
                                  Consider the step response of a second-order system (Figure 2.21);
                                                          m̈ x + c ̇ x + kx = f

                             let f(t) = kr(t)
                                                             c    k    k
                                                         ̈ x +  ̇ x +  x =  r
                                                             m    m    m
                                 c         k   2
                             Let   = 2     ,  =    and take the Laplace transform of the differential equation with
                                 m      n  m   n
                             zero initial conditions,
                                                        x(s)          2 n
                                                           =
                                                              2
                                                        r(s)  s + 2     s +    2
                                                                     n    n
                                                    1
                             If r(t) is a step input, r(s) = , the response x(s) is given by
                                                    s
                                                                    2 n    1
                                                      x(s) =              ⋅
                                                             2
                                                            s + 2     s +    2  s
                                                                   n    n
                             Using the partial fraction expansions and taking the inverse Laplace transform, the response
                             can be found as
                                            (                                    )
                                                 √                     √
                                                                             2
                                                       2
                              x(t) = 1 − e −     n t  cos  1 −       t + √  sin  1 −       t  for 0 ≤   < 1 range
                                                                               n
                                                         n
                                                               1 −    2
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