Nonlinear Control Sy.. - Free

1.7. PHASE-PLANE ANALYSIS OF NONLINEAR SYSTEMS 19
Figure 1.11: Stable limit cycle: (a) vector field diagram; (b) the closed orbit.
which is linear time-invariant. Moreover, the eigenvalues of the A matrix are A1,2 = ±3,
which implies that the equilibrium point [0, 0] is a center. The term µ(1- x1)x2 in equation
(1.26) provides additional dynamics that, as we will see, contribute to maintain the oscillations.
Figure 1.11(a) shows the vector field diagram for the system (1.25)-(1.26) assuming
µ = 1. Notice the difference between the Van der Pol oscillator of this example and the
center of Example 1.11. In Example 1.11 there is a continuum of closed orbits. A trajectory
initiating at an initial condition xo at t = 0 is confined to the trajectory passing through x0
for all future time. In the Van der Pol oscillator of this example there is only one isolated
orbit. All trajectories converge to this trajectory as t -> oo. An isolated orbit such as this
is called a limit cycle. Figure 1.11(b) shows a clearer picture of the limit cycle.
We point out that the Van der Pol oscillator discussed here is not a theoretical
example. These equations derive from simple electric circuits encountered in the first radios.
Figure 1.12 shows a schematic of such a circuit, where R in represents a nonlinear
resistance. See Reference [84] for a detailed analysis of the circuit.
As mentioned, the Van der Pol oscillator of this example has the property that all
trajectories converge toward the limit cycle. An orbit with this property is said to be a
stable limit cycle. There are three types of limit cycles, depending on the behavior of the
trajectories in the vicinity of the orbit: (1) stable, (2) unstable, and (3) semi stable. A
limit cycle is said to be unstable if all trajectories in the vicinity of the orbit diverge from
it as t -> oo. It is said to be semi stable if the trajectories either inside or outside the orbit
converge to it and diverge on the other side. An example of an unstable limit cycle can be
obtained by modifying the previous example as follows:
±1 = -x2
±2 = xl - µ(1 - xi)x2.