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Section  2.10  Sequential  Design  Example: Disk Drive  Read  System  129

                       Table 2.10  Typical Parameters for Disk Drive Reader
                       Parameter                 Symbol          Typical Value
                       Inertia of arm and
                                                                       2
                        read head                J               1 N m s /rad
                       Friction                  b               20 N m s/rad
                       Amplifier                 K a             10-1000
                       Armature resistance       R               i n
                       Motor constant            K m             5Nm/A
                       Armature  inductance      L               lmH


                      Figure 2.20 with Kb =  0. The model shown in Figure 2.76(b) assumes that the  flex-
                      ure is entirely rigid and does not significantly flex. In Chapter 4, we will consider the
                      model when the flexure cannot be assumed to be completely rigid.
                          Typical parameters for the disk drive system are given in Table 2.10.Thus, we have
                                                           K„
                                             G(s)  =
                                                     s(Js  +  b)(Ls  +  R)
                                                           5000
                                                                                        (2.138)
                                                     s(s  + 20)(5  +  1000)'
                      We can also write

                                                          K m/(bR)
                                              G(s)  =                                   (2.139)
                                                     S(T LS  +  1)(TS  +  1)'

                      where T L  =  J/b  =  50 ms and  T =  L/R  =  1 ms. Since  T  <*C T L, we often  neglect  T.
                      Then, we would have
                                                   KJ(bR)         0.25
                                           G(s)
                                                  S(T LS  +  1)  5(0.055  +  1)'

                      or


                                                  G(s)  =
                                                         s(s  + 20)
                      The  block  diagram  of  the  closed-loop  system  is  shown  in  Figure  2.77. Using  the
                      block diagram transformation  of Table 2.6, we have

                                                 Y(s)     K aG(s)
                                                                                        (2.140)
                                                 R(s)   1 +  K aG(s)'



                      /?(.*)  HQ       ,   *«               G(s)           •  Yis)
                               • ^
     FIGURE 2.77               J      *
     Block diagram of
     closed-loop
     system.
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