Page 94 - Mechatronics with Experiments
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80   MECHATRONICS
                              The most general form of a transfer function and its frequency domain representation,
                                                                  2
                                                  Π(s∕z + 1)Π[(s∕w ) + 2   (s∕w ) + 1]
                                                                 zi
                                                                       zi
                                                      i
                                                                            zi
                                       G(s) = K 0                                                (2.47)
                                                                    2
                                                s ±N Π(s∕p + 1)Π[(s∕w ) + 2   (s∕w ) + 1]
                                                                  pi
                                                        i
                                                                              pi
                                                                         pi
                                                                     2
                                                  Π(jw∕z + 1)Π[(jw∕w ) + 2   (jw∕w ) + 1]
                                                        i
                                                                   zi
                                                                                zi
                                                                          zi
                                      G(jw) = K 0                                                (2.48)
                                                                       2
                                                jw ±N Π(jw∕p + 1)Π[(jw∕w ) + 2   (jw∕w ) + 1]
                                                                                  pi
                                                                     pi
                                                                            pi
                                                          i
                                                               2 1∕2 j tan
                                                                                      2 2
                                                                                   2
                                                    Π[1 + (w∕z ) ]  e  −1 (w∕z i ) Π[(1 − w ∕w )
                                                             i                        z i
                                                                         −1 (2   zi w∕w zi )
                                                                      j tan  (   )
                                                                               2
                                                                 2 1∕2
                                                      + (2   w∕w ) ]  e    (1−w 2 ∕w )
                                                                               zi
                                                          z i  z i
                                      G(jw) = K 0                         −1                     (2.49)
                                                                                          2 2
                                                                                       2
                                                                   2 1∕2 j tan
                                                   e
                                                w ±N ±jN90 Π[1 + (w∕p ) ]  e  (w∕p i ) Π[(1 − w ∕w )
                                                                  i                       pi
                                                                  j tan −1 (  (2   pi w∕w pi ) )
                                                                           2
                                                             2 1∕2
                                                                           pi
                                                  + (2   w∕w ) ]  e   (1−w 2 ∕w )
                                                           pi
                                                      pi
                                                    jψ(w)
                                           = |G(jw)|e                                            (2.50)
                              Now, let us express the magnitude and phase information separately, and take the logarithm
                              of the magnitude information. Further, let us multiply the logarithm of the magnitude by
                              20 in order to express the magnitude information in dB (decibel) units.
                                                            ∑                 2 1∕2
                                  20 log |G(jw)| = 20 log K +  20 log [1 + (w∕z ) ]              (2.51)
                                                                    10
                                      10
                                                                             i
                                                      10
                                                         0
                                                   ∑              2  2 2           2 1∕2
                                                 +   20 log [(1 − w ∕w ) + (2   w∕w ) ]          (2.52)
                                                          10         zi      zi  zi
                                                                  ∑               2 1∕2
                                                 − 20 (±N) log w −   log [1 + (w∕p ) ]           (2.53)
                                                                       10
                                                                                 i
                                                            10
                                                   ∑           2   2 2           2 1∕2
                                                 −   log [(1 − w ∕w ) + (2   w∕w ) ]             (2.54)
                                                        10         pi     pi   pi
                                                         ∑              ∑       (2   w∕w )
                                                                                   zi
                                                               −1
                                                                                        zi
                                            (w) =∠G(jw) =   tan (w∕z ) +   tan −1                (2.55)
                                                                     i               2  2
                                                                                (1 − w ∕w )
                                                                                        zi
                                                                                    (2   w∕w )
                                                   ∑        ∑     −1       ∑     −1   pi   pi
                                                 −   ±N90 −    tan (w∕p ) −   tan                (2.56)
                                                                        i               2   2
                                                                                   (1 − w ∕w )
                                                                                            pi
                              A Bode plot of the frequency response G(jw) is the two plots of the above two equations
                              versus the log w.
                                         10
                                   The implication of taking the logarithm of the magnitude information is that the
                              contribution of gain, zeros, and poles to the overall gain plot becomes additive. The phase
                              information is already additive. When designing compensators, the additive nature of
                              frequency response in Bode plots is very helpful. As we try different controllers, we do not
                              have to replot the open loop system frequency response. Logarithmic scale in frequency
                              allows us to capture the behavior of the system at very low frequencies as well as very high
                              frequencies while using a reasonable size for the x-axis.
                                   A Bode plot of any frequency response which can be expressed as a rational polyno-
                              mial can be drawn as a linear summation of magnitude and phase contribution of (i) gain,
                              (ii) zero/pole at origin, (iii) first-order zero/pole, (iv) second-order zero/pole. Quite often,
                              the asymptotic sketches of the contribution of each of these dynamic components are more
                              useful than their exact plots due to the fact that the asymptotic approximate sketches can
                              be plotted rather quickly by hand.
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