Question 13.5: FINDING THE TRANSIENT PLOT OF THE VAN DER POL EQUATION The w...

The Van der Pol equation can be written as

\frac{d^{3} y}{dt^{2} } =\mu (1-y^{2} ) \frac{dy}{dt} -y                 (13.4)

This is identical to that for the LC circuit of Figure 13.20. Let us assume that L = 1 H and C = 1 F. The equation describing the LC circuit of Figure 13.20 is

v=L\frac{di}{dt} +\frac{1}{C} \int{i  dt} =\frac{di}{dt} +\int{i  dt}

Differentiating both sides yields

\frac{dv}{dt} =\frac{d^{2}i }{dt^{2} } +i

Solving for d²i/dt², we get

\frac{d^{2}i }{dt^{2} } =\frac{dv}{dt} -i                                        (13.5)

Equation 13.4 will be identical to Equation 13.5 if the current i represents y and dv/dt equals

\frac{dv}{dt} =\mu (1-y^{2} )\frac{dy}{dt} =\mu(1-i^{2} )\frac{di}{dt}       (13.6)

which can be integrated to give

v=\int{\mu (1-i^{2} )di} =μ( i-\frac{i^{3}}{3})         (13.7)

Equation 13.7 is a polynomial of the form

v=P_{0}+P_{1}i +P_{2}i^{2}+P_{3}i^{3}

where P_{o} = P_{2} = 0, P_{1} = 2,  and  P_{3} = −μ/3.

The voltage across the inductor v_{L} = di/dt represents dy/dt.

The PSpice schematic is shown in Figure 13.21.

The listing of the circuit file is as follows:

Example 13.5 Van der Pol’s equation

The plots of the transient response and the phase plane for the Van der Pol equation are shown in Figure 13.22 and Figure 13.23, respectively. The system is unstable and oscillates between limit cycles. A smoother curve can be obtained by reducing the printing time of 0.01 sec.