Oscillations in parallel-RLC network with a limiting VCCS

Question

Here is a parallel-RLC network (where G = 1/R is the admittance) with a limiting VCCS. The characteristic of the VCCS is shown in Fig (b), where the slope is G1. The problem asked is as follows:

Find the condition on slope of the VCCS for oscillations to start.
In Fig (c), assuming Vo = A sin(wt), determine and plot the amplitude of the fundamental component of i as A is varied from 0 to Infinity.
What will be amplitude of oscillation of the circuit in part (a) of the figure, in steady state?

I've found that the slope of the VCCS should be greater than or equal to G in order for oscillations to start. I also obtained the expression for the first fourier coefficient of i (assuming that's the "fundamental component" asked) in terms of A for the circuit in Fig (c).

The part I'm struggling with is how to find the amplitude of oscillations of the circuit in Fig (a). My best guess would be Imax/G for the amplitude of Vo, but I don't know how to prove it analytically.

Try proving two separate things. 1) if \$A > I_{max} / G\$, then the amplitude gradually reduces. 2)if \$A < I_{max} / G\$, then the amplitude gradually increases. Rather than working in terms of voltage and current, try working in terms of energy. For case 1, at peak voltage A, the current through the admittance is \$A \cdot G\$ out of which \$I_{max}\$ is coming from the source. The rest is coming from the inductor or capacitor resulting in energy stored in them being dissipated in G resulting in reduction of amplitude. Similar argument for case 2. — AJN, Jul 27 '20 at 07:26
The energy (or rate of change of energy) technique is popular in the area of non linear (control) systems to predict stability of oscillations and [limit cycles](https://en.wikipedia.org/wiki/Limit_cycle). The *energy like* function is called a [Lyapunov function](https://en.wikipedia.org/wiki/Lyapunov_function). [Lyapunov stability](https://en.wikipedia.org/wiki/Lyapunov_stability). This comment is just additional info and not required to solve the problem at hand. — AJN, Jul 27 '20 at 07:50

AJN · Answer 1 · 2020-07-27T10:19:16.880

Energy method

Assume oscillation is \$v_o = A \sin (\omega t)\$. Current and instantaneous power going into each element is (respectively)

Capacitor

\$i_C(t) = C \frac{d v_o}{dt} = C\omega A \cos(\cdot)\$

\$P_C(t) = C A^2 \omega\sin(\cdot) \cos(\cdot)\$

Inductor

\$i_L(t) = \frac{-A \cos(\cdot)}{\omega L}\$

\$P_I(t) = \frac{-A^2 \sin(\cdot) \cos(\cdot)}{\omega L}\$

Note that

the capacitor and inductor powers are exactly opposite phase; i.e. they are exchanging energy with each other.
At perfect sinusoidal oscillation and steady state, \$P_C = -P_I\$, i.e; \$\omega^2 = \frac{1}{LC}\$

Admittance

\$i_G(t) = G \cdot A \sin (\omega t)\$

\$P_G(t) = G A^2 \sin^2(\cdot)\$

Source

\$i_S(t) = -H \cdot A \sin (\omega t)\$ when \$v(t) \cdot H < I_{max}\$

\$P_S(t) = -H A^2 \sin^2(\cdot)\$ when \$v(t) \cdot H < I_{max}\$

\$P_S(t) = -I_{max} A \sin(\cdot)sgn(\sin(\cdot))\$ when \$v(t) \cdot H > I_{max}\$

-ve sign shows that power is coming out of the device.

Analysis of energy

During the oscillation \$v(t)\$, the circuit goes through points 1&2 or 3&4 explained below.

When \$H = G\$ and \$H \cdot v(t) < I_{max}\$, The dissipation in the admittance is exactly supplied by the source and we have steady oscillation. This may be what the OP has mentioned in their Question.
When \$H = G\$ and \$v(t) > I_{max}\$ We can see that there is a short fall in the power supplied by the source when compared to the power consumed by the admittance. So the perfect balance between \$P_C\$ and \$P_I\$ is disturbed. Energy is taken from them. This energy is not supplied back to them during the time instants where the instantaneous voltage is low since the condition \$H = G\$ ensures that the source supplied just enough power which is required by the admittance. Hence through out the cycle energy decreases until the amplitude is small enough for the circuit to remain in point 1 above through out the cycle.
When \$H > G\$ and \$G \cdot v(t) < I_{max}\$. The source supplies more power than the admittance can dissipate. This gets stored in the inductance and capacitance.
When \$H > G\$ and \$G \cdot v(t) > I_{max}\$. The admittance consumes more power than the source can provide. This gets taken from the inductance and capacitance. further analysis is need to see if this shortfall can be taken from point 3 above or if the energy stored in point 3 got wasted in the harmonics of the non-sinusoidal waveform.

Note

It can be noted that some assumptions are made in the answer above (perfect sinusoidal oscillation, etc.) and the answer is not mathematically rigorous. It should be taken only as a starting point for an analysis involving energy. There are established methods for limit cycle amplitude prediction in literature as I have mentioned in the comment to OP's question.

So the amplitude of Vo in steady-state is indeed Imax/G? – SacredMechanic Jul 30 '20 at 18:26 — SacredMechanic, Jul 30 '20 at 18:26

score 1 · Answer 2 · answered Jul 27 '20 at 08:26

1

Assuming the slope of your current source is higher than the LC tank conductance and steady state amplitude is much higher than the linear range of your current source, the fundamental component of current will be: $$I_{f_o} = \frac{4I_{max}}{\pi}$$ This current will flow only through the conductance since LC impedance is infinite at resonance, so the voltage amplitude will be: $$A = \frac{4I_{max}}{G\pi}$$

answered Jul 27 '20 at 08:26

sarthak

3,646
4
19
31

1

Linear range of current source is \$v = I_{max}/G\$. The assumption you have used is \$A >> I_{max}/G\$. But \$A = \frac{4}{\pi} \cdot \frac{I_{max}}{G}\$ is only 28% higher. So I think the amplitude will be even lower as OP has guessed (\$I_{max}/G\$). – AJN Jul 27 '20 at 09:36
@AJN No the linear range is \$v = I_{max}/G_m\$, where \$G_m > G\$ which is required for oscillations. This is assumed to be very small compared to amplitude. – sarthak Jul 27 '20 at 11:21
Understood. ____ – AJN Jul 27 '20 at 11:28
What is that current that you have mentioned? Is it the current in the LC tank or the one flowing out of the LC tank? – SacredMechanic Jul 30 '20 at 18:33
Current source outside the LC tank – sarthak Jul 31 '20 at 07:06

score 0 · Accepted Answer · answered Jul 31 '20 at 10:57

I have found out that the method to get the amplitude is actually quite simple. It has to do with damping due to the resistance in the circuit.

- Case 1: If G1 > G:

Here, the Amplitude is Imax/G, since if it's greater than that, there is a positive resistance (1/G) which will dampen it back to Imax/G and if it's less than that, there is a negative resistance 1/(G-G1) which will have the opposite effect of damping and bring it back to Imax/G.

- Case 2: If G1 = G:

Here, similar logic holds, but for A < Imax/G, it will stay that way, since there is no damping (because G = G1). And for A > Imax/G, the positive resistance (1/G) causes damping and brings it back to Imax/G.