Page 133 - Mechatronics with Experiments
P. 133

CLOSED LOOP CONTROL  119
                                r (s)                                                   y (s)
                                                 K             e –t  s      1/(s+1)
                                                                 d
                                      –



                                                                                         t d = 0.5;
                                                                                   Im (s)
                                                                                         e –tds  1
                                As K: 0  to  infinity                                        t d  s+1
                                                Im (s)                                       1  1
                                                                                         1+K      = 0
                                  t d = 0;                                                 t d  s+1  s+1
                                    1
                                1+K   =0
                                   s+1

                                                           Re (s)                               Re (s)
                                         -1                                –2    –1





                                                (a)                               (b)

                                                                              Im (s)
                                                                                      1+ K  e -td s  1  =0
                                                                                            s+1
                                                                             .               .
                                 t d = 0.5;                                  .               .
                               e –tds  1  2     Im (s)                             j3π  t d
                                     s+1)
                                   (t d                                            j2π  t d
                                    1   1
                               1+K    2  s+1  = 0
                                 (t d  s+1)                                        jπ  t d
                                                          Re (s)                                Re (s)
                                   –2     –1                                  –1
                                                                                   –jπ  t d

                                                                                   –j2π  t d
                                                                                   –j3π  t d
                                                                             .
                                                                             .
                                                (c)                          .    (d)
                             FIGURE 2.54: Root locus of the closed loop system poles for (a) no time delay (t = 0.0), and
                                                                                           d
                             (b) and (c) two different approximations to the pure time delay (t d  ≠ 0.0): one with a first-order
                             filter and one with a second-order filter, and (d) pure time delay accurately taken into account
                             without approximation.



                                  A more detailed analysis of the root locus without approximation to the time delay
                             shows that the closed loop root locus would have infinite number of branches and all of
                             them would eventually go to the right hand plane for large values of K. The nature of the
                             infinite number of branches of the root locus can be observed from the fact that
                                                                     1
                                                              −t d s
                                                       1 + K ⋅ e  ⋅      = 0                   (2.184)
                                                                   (s + a)
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