Page 14 - Servo Motors and Industrial Control Theory -
P. 14

6                                             1  Feedback Control Theory

            Example 3  Speed of DC motors assuming negligible inductance.
              In this example, only a brief discussion on modeling of DC motors will be given.
            Full details of analysis will be discussed in a different chapter.
              The voltage equation can be written as
                                       v =  Ri C ω+  m  m                (1.16)
                                        i

            the torque current relation and the equation of motion become
                                              dω
                                        T =  J  m                        (1.17)
                                         m
                                               dt

                                         T =  K i                        (1.18)
                                          m    t

            The parameters K ,C  are the torque and voltage constants of the motor respectively.
                            m
                          t
            Eliminating T ,i from the three equations gives
                       m
                                        RJ  d  ω
                                    v =       m  + C ω                   (1.19)
                                     i
                                        K t  dt    m  m
            Taking Laplace Transform from both sides of Eq. (1.19) and referring to Table 1.1
            yields
                                       RJ
                                   () =
                                 Vs       sω m () s + C ω m  () s        (1.20)
                                                   m
                                  i
                                       K
                                         t
            rearranging gives
                                              1
                                   ω  () s =       Vs
                                                    ()
                                     m     RJ  sC+  i                    (1.21)
                                           K t   m

            now Eq. (1.21) can be solved for any input function, for a step input of V, where
                  V
            LV() =  , using partial fraction method, we have
                   s
                                           1   τ −  1
                                  ω m ()s =      +   V                 (1.22)
                                              s
                                          C τ + 1  s 
                                            m
            Taking the inverse of Laplace Transform of Eq. (1.22) gives
                                            V
                                    ω  ( )t =  (1 e−  −   tτ  )          (1.23)
                                      m
                                           C m
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