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0066_frame_C27  Page 17  Wednesday, January 9, 2002  7:10 PM









                            Let c = 0, and solving for K, we get the critical gains

                                                  K cr =  – 3 ±  21  0.79, 3.79
                                                       ---------------------- =
                                                                      –
                                                          2
                            Substituting the values of K cr  in the auxiliary equation
                                                     ( K cr +  3)s + 2K cr =  0
                                                             2
                            we obtain the critical frequencies

                                                            0.65,  when K cr =  0.79
                                           w cr =  ---------------- =  
                                                   2K cr
                                                  K cr +  3   3.10,  when K cr =  – 3.79

                            At the critical frequency, we have the characteristic equation

                                                             (
                                                      1 + K cr Ljw cr ) =  0
                            Hence the points of the L(jω) locus that cross the real-axis are

                                                             1
                                                                        1
                                                                    1
                                                   (
                                                  Ljw cr ) =  – ------- =  – ----------, ----------
                                                            K cr   0.79 3.79
                         The complete Nyquist plot is shown not to scale in Fig. 27.18(b). The range of the gain K for which
                       the system is stable can be determined using Nyquist criterion. For different values of K, the Nyquist
                       diagram needs to be redrawn in order to count the number of encirclement of the −1 point. We can
                       avoid this by counting the number of encirclement of −1/K point instead. From the Nyquist criterion,
                       Z = N + P, where P = 0. It can be seen from Fig. 27.18(b) that there are four cases of the encirclements
                       of the −1/K point.
                         1. K > 0 and −1/K < −1/0.79 ⇒ 0 < K <  0.79,  and N = 0. We have Z = 0 and the system is stable.
                         2. K > 0 and −1/K > −1/0.79 ⇒  K >  0.79,  and N = 2. We have Z = 2 and the system has two unstable
                            poles.
                         3. K < 0 and −1/K < 1/3.79 ⇒  K <  – 3.79,  and N = 3. We have Z = 3 and the system has three unstable
                            poles.
                         4. K < 0 and −1/K > 1/3.79 ⇒  – 3.79 <  K <  0  and N = 1. We have Z = 1 and the system has one
                            unstable pole.

                       The root locus of system (27.13) is also shown in Fig. 27.18(c) for comparison.

                       27.7 Relative Stability

                       In designing a control system, it is required that the system be stable. In addition to stability, there are
                       important concerns such as acceptable transient response and capability to deal with model uncertainty.
                       Since the model used in the design and analysis of a control system is never exact, it may suggest a stable
                       system; but the physical system turns out to be unstable. Therefore, it is required that the system not
                       only be stable but also have some stability margin or adequate relative stability.
                         Suppose that the sinusoidal loop transfer function L(jω) locus intersects the −1 point for some critical
                       frequency ω cr . Then

                                                                  (
                                                           (
                                                 (
                                                Ljw cr ) =  Gjw cr )Hjw cr ) =  −1
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