Computer Algebra Recipes

1.1. PHASE-PLANE PORTRAITS 29
The battery voltage VB is adjusted to coincide with the in°ection point VS of
the iD vs. VD curve, i.e., VB = VS. Near this operating point, one may write
i = ¡av+ bv3 ,wherei = iD ¡ iS and v = VD ¡ VS, andaand b are positive.
The governing Van der Pol (VdP) ODE, which will presently be derived, is
Äx ¡ ² (1 ¡ x 2 )_x + x =0; ² > 0; (1.8)
with x proportional to v. Equation (1.8) is just the simple harmonic oscillator
equation for unit frequency and mass with an amplitude-dependent damping
term. For x1 the damping is positive, tending to reduce the
oscillations. The negative damping is responsible for the growth of any small
spontaneous circuit \noise" into stable oscillations, i.e., into a stable limit cycle.
Let's now derive the VdP equation and demonstrate the growth of a small
input signal into a limit cycle for a typical tunnel diode, 1N3719, for which
a =0:05 and b =1:0 in SI units.
The DEtools and PDEtools packages are loaded. The former contains the
DEplot3d command, which is a three-dimensional generalization of the DEplot
command. The PDEtools package contains the dchange command, which will
allow us to easily make a somewhat complicated variable transformation.
> restart: with(DEtools): with(PDEtools):
The time-dependent tunnel diode current and voltage expressions are entered.
> i[D]:=i[S]-a*v(t)+b*v(t)^3; V[D]:=V[S]+v(t);
iD := iS ¡ a v(t)+b v(t) 3 VD := VS + v(t)
The voltage drop across both the resistor R and the capacitor C is the same as
across the diode D, i.e., VR = VD and VC = VD. The voltage drop across the
inductor L is VL = VB ¡ VD = VS ¡ VD, the latter form being entered.
> V[R]:=V[D]: V[C]:=V[D]: V[L]:=V[S]-V[D];
VL := ¡v(t)
By Ohm's law, the current through the resistor is iR = VR=R. The current
through the capacitor is iC = C (dVC=dt).
> i[R]:=V[R]/R; i[C]:=C*diff(V[C],t);
iR := VS
μ
+ v(t)
d
iC := C
R
dt v(t)
Using Kirchho®'s current rule, eq1 states that the current leaving L must be
equal to the sum of the currents entering R, C, andD.
> eq1:=-i[L](t)+i[R]+i[C]+i[D]=0; #Kirchhoff's current rule
eq1 := ¡iL(t)+ VS
μ
+ v(t) d
+ C
R dt v(t)
+ iS ¡ a v(t)+b v(t) 3 =0
Di®erentiating eq1 with respect to t eliminates the in°ection point current iS.
> eq2:=expand(diff(eq1,t)/C);
d
eq2 := dt v(t)
CR ¡
d
dt iL(t)
μ
d
μ 2 a
d dt
+ v(t) ¡
C dt2 v(t)
3 b v(t)
+
C
2
μ
d
dt v(t)
=0
C