## A sample diode RLC circuit

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.

## Matrix form for discrete time Fourier transform

December 4, 2014 ece1254 , ,

## Matrix form

The discrete time Fourier transform has been seen to have the form

\label{eqn:discreteFourierMatrixForm:200}
x_k = \sum_{n = -N}^N X_n e^{ 2 \pi j n k /(2 N + 1)}

\label{eqn:discreteFourierMatrixForm:220}
X_n = \inv{2 N + 1} \sum_{k = -N}^N x_k e^{- 2 \pi j n k /(2 N + 1)}.

A matrix representation of this form is desired. Let

\label{eqndiscreteFourierMatrixForm:240}
\Bx =
\begin{bmatrix}
x_N \\
\vdots \\
x_0 \\
\vdots \\
x_{-N}
\end{bmatrix}

\label{eqndiscreteFourierMatrixForm:260}
\BX =
\begin{bmatrix}
X_N \\
\vdots \\
X_0 \\
\vdots \\
X_{-N}
\end{bmatrix}

\ref{eqn:discreteFourierMatrixForm:200} written out in full is
\label{eqn:discreteFourierMatrixForm:280}
\begin{aligned}
x_k
&= X_N e^{ 2 \pi j N k /(2 N + 1)} \\
&+ X_{N-1} e^{ 2 \pi j \lr{N-1} k /(2 N + 1)} \\
&+ \cdots \\
&+ X_0 \\
&+ \cdots \\
&+ X_{1-N} e^{ -2 \pi j \lr{N-1} k /(2 N + 1)} \\
&+ X_{-N} e^{ -2 \pi j N k /(2 N + 1)}.
\end{aligned}

With $$\alpha = e^{ 2 \pi j /(2 N + 1) }$$ the matrix form is

\label{eqn:discreteFourierMatrixForm:300}
\Bx =
\begin{bmatrix}
\alpha^{ N N } & \alpha^{ \lr{N-1} N } & \cdots & 1 & \cdots & \alpha^{ -\lr{N-1} N } & \alpha^{ -N N } \\
\alpha^{ N \lr{N-1} } & \alpha^{ \lr{N-1} \lr{N-1} } & \cdots & 1 & \cdots & \alpha^{ -\lr{N-1} \lr{N-1} } & \alpha^{ -N \lr{N-1} } \\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
1 & 1 & 1 & 1 & 1 & 1 & 1 \\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
\alpha^{ -N \lr{N-1} } & \alpha^{ -\lr{N-1} \lr{N-1} } & \cdots & 1 & \cdots & \alpha^{ {N-1} \lr{N-1} } & \alpha^{ N \lr{N-1} } \\
\alpha^{ -N N } & \alpha^{ -N N } & \cdots & 1 & \cdots & \alpha^{ \lr{N-1} N } & \alpha^{ N N } \\
\end{bmatrix}
\BX

Similarily, from \ref{eqn:discreteFourierMatrixForm:220}, the inverse relation expands out to

\label{eqn:discreteFourierMatrixForm:320}
\begin{aligned}
( 2 N + 1 ) X_n
&= x_N e^{- 2 \pi j n N /(2 N + 1)} \\
&+ x_{N-1} e^{- 2 \pi j n \lr{N-1}/(2 N + 1)} \\
&\cdots \\
&+ x_0 \\
&\cdots \\
&+ x_{1-N} e^{ 2 \pi j n \lr{ N-1 }/(2 N + 1)} \\
&+ x_{-N} e^{ 2 \pi j n N/(2 N + 1)},
\end{aligned}

with a matrix form of

\label{eqn:discreteFourierMatrixForm:340}
( 2 N + 1 ) \BX =
\begin{bmatrix}
\alpha^{- N N } & \alpha^{- N \lr{N-1}} &\cdots & 1 &\cdots & \alpha^{ N \lr{ N-1 }} & \alpha^{ N N} \\
\alpha^{- \lr{N-1} N } & \alpha^{- \lr{N-1} \lr{N-1}} &\cdots & 1 &\cdots & \alpha^{ \lr{N-1} \lr{ N-1 }} & \alpha^{ \lr{N-1} N} \\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
1 & 1 & 1 & 1 & 1 & 1 & 1 \\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
\alpha^{\lr{N-1} N } & \alpha^{ \lr{N-1} \lr{N-1}} &\cdots & 1 &\cdots & \alpha^{ -\lr{N-1} \lr{ N-1 }} & \alpha^{ -\lr{N-1} N} \\
\alpha^{ N N } & \alpha^{ N \lr{N-1}} &\cdots & 1 &\cdots & \alpha^{ -N \lr{ N-1 }} & \alpha^{ -N N} \\
\end{bmatrix}

Lettting

\label{eqn:discreteFourierMatrixForm:360}
\BF =
\begin{bmatrix}
\alpha^{ N N } & \alpha^{ \lr{N-1} N } & \cdots & 1 & \cdots & \alpha^{ -\lr{N-1} N } & \alpha^{ -N N } \\
\alpha^{ N \lr{N-1} } & \alpha^{ \lr{N-1} \lr{N-1} } & \cdots & 1 & \cdots & \alpha^{ -\lr{N-1} \lr{N-1} } & \alpha^{ -N \lr{N-1} } \\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
1 & 1 & 1 & 1 & 1 & 1 & 1 \\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
\alpha^{ -N \lr{N-1} } & \alpha^{ -\lr{N-1} \lr{N-1} } & \cdots & 1 & \cdots & \alpha^{ {N-1} \lr{N-1} } & \alpha^{ N \lr{N-1} } \\
\alpha^{ -N N } & \alpha^{ -N N } & \cdots & 1 & \cdots & \alpha^{ \lr{N-1} N } & \alpha^{ N N } \\
\end{bmatrix},

the discrete transform pair has the following compactly matrix representation

\label{eqn:discreteFourierMatrixForm:380}
\Bx = \BF \BX

\label{eqn:discreteFourierMatrixForm:400}
\BX = \inv{2 N + 1} \overline{{\BF}} \Bx,

where $$\overline{{\BF}}$$ is the complex conjugate of $$\BF$$.

# References

## Discrete Fourier Transform

In [1] a verification of the discrete Fourier transform pairs was performed. A much different looking discrete Fourier transform pair is given in [2] \$ A.4. This transform pair samples the points at what are called the Nykvist time instants given by

\label{eqn:discreteFourier:20}
t_k = \frac{T k}{2 N + 1}, \qquad k \in [-N, \cdots N]

Note that the endpoints of these sampling points are not $$\pm T/2$$, but are instead at

\label{eqn:discreteFourier:40}
\pm \frac{T}{2} \inv{1 + 1/N},

which are slightly within the interior of the $$[-T/2, T/2]$$ range of interest. The reason for this slightly odd seeming selection of sampling times becomes clear if one calculate the inversion relations.

Given a periodic ($$\omega_0 T = 2 \pi$$) bandwith limited signal evaluated only at the Nykvist times $$t_k$$,

\label{eqn:discreteFourier:60}
\boxed{
x(t_k) = \sum_{n = -N}^N X_n e^{ j n \omega_0 t_k},
}

assume that an inversion relation can be found. To find $$X_n$$ evaluate the sum

\label{eqn:discreteFourier:80}
\begin{aligned}
&\sum_{k = -N}^N x(t_k) e^{-j m \omega_0 t_k} \\
\sum_{k = -N}^N
\lr{
\sum_{n = -N}^N X_n e^{ j n \omega_0 t_k}
}
e^{-j m \omega_0 t_k} \\
\sum_{n = -N}^N X_n
\sum_{k = -N}^N
e^{ j (n -m )\omega_0 t_k}
\end{aligned}

This interior sum has the value $$2 N + 1$$ when $$n = m$$. For $$n \ne m$$, and
$$a = e^{j (n -m ) \frac{2 \pi}{2 N + 1}}$$, this is

\label{eqn:discreteFourier:100}
\begin{aligned}
\sum_{k = -N}^N
e^{ j (n -m )\omega_0 t_k}
&=
\sum_{k = -N}^N
e^{ j (n -m )\omega_0 \frac{T k}{2 N + 1}} \\
&=
\sum_{k = -N}^N a^k \\
&=
a^{-N} \sum_{k = -N}^N a^{k+ N} \\
&=
a^{-N} \sum_{r = 0}^{2 N} a^{r} \\
&=
a^{-N} \frac{a^{2 N + 1} – 1}{a – 1}.
\end{aligned}

Since $$a^{2 N + 1} = e^{2 \pi j (n – m)} = 1$$, this sum is zero when $$n \ne m$$. This means that

\label{eqn:discreteFourier:120}
\sum_{k = -N}^N
e^{ j (n -m )\omega_0 t_k} = (2 N + 1) \delta_{n,m},

which provides the desired Fourier inversion relation

\label{eqn:discreteFourier:140}
\boxed{
X_m = \inv{2 N + 1} \sum_{k = -N}^N x(t_k) e^{-j m \omega_0 t_k}.
}

# References

[1] Peeter Joot. Condensed matter physics., appendix: Discrete Fourier transform. 2013. URL https://peeterjoot.com/archives/math2013/phy487.pdf. [Online; accessed 02-December-2014].

[2] Giannini and Leuzzi Nonlinear Microwave Circuit Design. Wiley Online Library, 2004.