To get a feel for how to generate the MLN equations for a circuit that has both RLC and non-linear components, consider the circuit of fig. 1.

fig. 1. An RLC circuit with a diode.

The KCL equations for this circuit are

1. $$0 = i_s – i_d$$
2. $$i_L + \frac{v_2 – v_3}{R} = i_d$$
3. $$\frac{v_3 – v_2}{R} + C \frac{dv_3}{dt} = 0$$
4. $$-v_2 + L \frac{d i_L}{dt} = 0$$
5. $$i_d = I_0 \lr{ e^{(v_1 – v_2)/v_T} – 1}$$

FIXME: for the diode, is that the right voltage sign with respect to the current direction?

With $$Z = 1/R$$, these can be put into the standard MLN matrix form as

\label{eqn:diodeRLCSample:20}
\begin{bmatrix}
0 & 0 & 0 & 0 \\
0 & Z & -Z & 1 \\
0 & -Z & Z & 0 \\
0 & -1 & 0 & 0 \\
\end{bmatrix}
\begin{bmatrix}
v_1 \\
v_2 \\
v_3 \\
i_L \\
\end{bmatrix}
+
\begin{bmatrix}
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & C & 0 \\
0 & 0 & 0 & L \\
\end{bmatrix}
{\begin{bmatrix}
v_1 \\
v_2 \\
v_3 \\
i_L \\
\end{bmatrix}}’
=
\begin{bmatrix}
I_0 & 1 \\
-I_0 & 0 \\
0 & 0 \\
0 & 0 \\
\end{bmatrix}
\begin{bmatrix}
1 \\
i_s(t) \\
\end{bmatrix}
+
\begin{bmatrix}
-I_0 \\
I_0 \\
0 \\
0 \\
\end{bmatrix}
\begin{bmatrix}
e^{(v_2 – v_3)/v_T}
\end{bmatrix}

Let’s write this as

\label{eqn:diodeRLCSample:40}
\BG \BX(t) + \BC \dot{\BX}(t) = \BB \Bu(t) + \BD \Bw(t).

Here $$\Bu(t)$$ collects up all the unique signature sources (for example sources with each different frequency in the system), and $$\Bw(t)$$ is a vector of all the unique non-linear (time dependent) terms.

Assuming a bandwidth limited periodic source we know how to express any of the time dependent variables $$v_1, …$$ in terms of their (discrete) Fourier transforms. Suppose that

the Fourier coefficients for $$v_a(t), u_b(t), w_c(t)$$ are given by

\label{eqn:diodeRLCSample:60}
\begin{aligned}
v_a(t) &= \sum_{n = -N}^N V_n^{(a)} e^{j \omega_0 n t} \\
u_b(t) &= \sum_{n = -N}^N U_n^{(b)} e^{j \omega_0 n t} \\
w_c(t) &= \sum_{n = -N}^N W_n^{(c)} e^{j \omega_0 n t}.
\end{aligned}

For example, in this circuit, if the source was zero phase signal at the fundamental frequency of our Fourier basis ($$i_s(t) = e^{j \omega_0 t}$$), the only non-zero Fourier components $$U_n^{(a)}$$ would be $$U_0^{(1)} = 1, U_1^{(2)} = 1$$.

Equation \ref{eqn:diodeRLCSample:40} then becomes

\label{eqn:diodeRLCSample:80}
0 = \sum_{n=-N}^N
e^{j n \omega_0 t}
\lr{
\lr{
\BG + j \omega_0 n \BC
}
\begin{bmatrix}
V_n^{(1)} \\
V_n^{(2)} \\
\vdots
\end{bmatrix}
– \BB
\begin{bmatrix}
U_n^{(1)} \\
U_n^{(2)} \\
\end{bmatrix}
– \BD
\begin{bmatrix}
W_n^{(1)} \\
\end{bmatrix}
}

The time dependence in the linear terms is nicely taken of by this transformation to the frequency domain. However, we have a fairly messy structure with sums of Fourier components instead of the nice Fourier component vectors that we see in \S A.4 of [1]. That reference does consider multivariable problems like this one, so it looks like fully digesting that methodology is the next step.

# References

Giannini and Giorgio Leuzzi. Nonlinear Microwave Circuit Design. Wiley Online Library, 2004.