Page 530 - Mechanical Engineers' Handbook (Volume 2)
P. 530

5 Root Locus   521

                           Rule 9. The angle condition is made use of to determine the angle by which a branch would
                              depart from a pole or would arrive at a zero as K →  .
                                  A point s is considered very near the pole (zero) and the angle G(s )H(s )is
                                                                                               0
                                                                                           0
                                          0
                              computed. The fact that s is very near the pole (zero) makes all but one angle fixed.
                                                   0
                              Thus by employing the angle condition, the unknown angle of departure (arrival) can
                              be computed.
                              An example is given next to illustrate the various rules for constructing a root locus.
                           Example 3 Consider CLCE
                                                                    K(s   6)
                                                1   KG(s)H(s)   1
                                                                  s(s   1)(s   4)
                           R1: n   3 ⇒ 3 branches originating at 0,  1,  4at K   0.
                           R2: m   1 ⇒ 1 branch terminates at  6, at K  .
                           R3: n   m   2 branches approach   along asymptotes.
                           R4: Hub 
   (0   1   4   ( 6))/2   0.5.
                                                                 180  N    180  N
                                             Asymptote angles:
                                                                 n   m       2
                                                         Set N   1 ⇒   90
                           R1–R4: Yield the sketch of Fig. 18b.
                           R5: Sections on the real line are 0 to  l and  4to  6.
                           R6: There must be a breakaway point between 0 and  1.
                           R7: Break points dK/ds   0.
                                                                         2
                                                                    3
                                                     2
                                           ⇒ (s   6)(3s   10s   4)   (s   5s   4s)   0
                                                             3
                                                                    2
                                                            s   11.5s   30s   12   0
                                                      (s   0.49)(s   7.89)(s   3.12)   0
                           Values for s of  7.89 and  3.12 are unacceptable from R5. Therefore the only breakaway
                              point is at  0.49. A sketch of the root loci is given in Fig. 18c.
                           R8: Imaginary axis crossings:
                                                       3
                                                            2
                                               CLCE   s   5s   (4   K)s   6K   0
                           Now lets s   j :
                                                   3
                                               ( j )   5( j )   (4   K)j    6K   0
                                                          2
                                                 (6k   5  )   j [(4   K)     ]   0
                                                                         2
                                                         2
                           Therefore
                                                                6K   5  2
                                                                 2
                                                       [4   K     ]   0
                           yielding
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