Fourier series

Geometric algebra notes collection split into two volumes

November 10, 2015 math and physics play No comments , , , , , , , , , , , , ,

I’ve now split my (way too big) Exploring physics with Geometric Algebra into two volumes:

Each of these is now a much more manageable size, which should facilitate removing the redundancies in these notes, and making them more properly book like.

Also note I’ve also previously moved “Exploring Geometric Algebra” content related to:

  • Lagrangian’s
  • Hamiltonian’s
  • Noether’s theorem

into my classical mechanics collection (449 pages).

Development of Harmonic balance equations (considering a sample diode RLC circuit)

December 9, 2014 ece1254 No comments , , , , , ,

[Click here for a PDF of this post with nicer formatting]

Previously, the time domain MNA equations and first steps at producing the Harmonic Balance equations were performed. That was a frequency domain analysis with an assumed Fourier solution associated with discrete time sampling.

The next goal is to put this in block matrix form. First introducing discrete time sampling vectors

\begin{equation}\label{eqn:diodeRLCSample:100}
\Bv_a =
\begin{bmatrix}
v_a(t_{-N}) \\
v_a(t_{1-N}) \\
\vdots \\
v_a(t_{N-1}) \\
v_a(t_{N}) \\
\end{bmatrix}, \qquad
\Bu_a =
\begin{bmatrix}
u_b(t_{-N}) \\
u_b(t_{1-N}) \\
\vdots \\
u_b(t_{N-1}) \\
u_b(t_{N}) \\
\end{bmatrix}, \qquad
\Bw_a =
\begin{bmatrix}
w_c(t_{-N}) \\
w_c(t_{1-N}) \\
\vdots \\
w_c(t_{N-1}) \\
w_c(t_{N}) \\
\end{bmatrix},
\end{equation}

and Fourier component vectors

\begin{equation}\label{eqn:diodeRLCSample:120}
\BV_a =
\begin{bmatrix}
V^{(a)}_{-N} \\
V^{(a)}_{1-N} \\
\vdots \\
V^{(a)}_{N-1} \\
V^{(a)}_{N} \\
\end{bmatrix}, \qquad
\BU_b =
\begin{bmatrix}
U^{(b)}_{-N} \\
U^{(b)}_{1-N} \\
\vdots \\
U^{(b)}_{N-1} \\
U^{(b)}_{N} \\
\end{bmatrix}, \qquad
\BW_c =
\begin{bmatrix}
W^{(c)}_{-N} \\
W^{(c)}_{1-N} \\
\vdots \\
W^{(c)}_{N-1} \\
W^{(c)}_{N} \\
\end{bmatrix}.
\end{equation}

With \( \alpha = e^{ 2 \pi j /(2 N + 1) } \), and

\begin{equation}\label{eqn:diodeRLCSample:140}
\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},
\end{equation}

in each case the time domain sampling vectors are related to the Fourier components by relations of the form

\begin{equation}\label{eqn:diodeRLCSample:160}
\Bx = \BF \BX.
\end{equation}

Block matrix form, with physical parameter ordering

To understand how to put \ref{eqn:diodeRLCSample:240} in block matrix form, it is helpful to consider a specific example. Consider again the specific example of the RLC circuit above, which has the form

\begin{equation}\label{eqn:diodeRLCSample:260}
\begin{bmatrix}
0 & 0 & 0 & 0 \\
0 & Z & -Z & 1 \\
0 & -Z & Z + j \omega_0 n C & 0 \\
0 & -1 & 0 & + j \omega_0 n L \\
\end{bmatrix}
\begin{bmatrix}
V_n^{(1)} \\
V_n^{(2)} \\
V_n^{(3)} \\
I_n^{(L)} \\
\end{bmatrix}
=
\begin{bmatrix}
I_n^{(1)} \\
I_n^{(2)} \\
I_n^{(3)} \\
I_n^{(4)} \\
\end{bmatrix}
\end{equation}

Here the \( I^{(i)} \) terms are the DFT representations of both the linear and non-linear sources.

Suppose for example that \( N = 1 \). One way to write \ref{eqn:diodeRLCSample:260} would be

\begin{equation}\label{eqn:diodeRLCSample:320}
\begin{aligned}
&
\begin{bmatrix}
I_{-1}^{(1)} \\
I_0^{(1)} \\
I_{1}^{(1)} \\
I_{-1}^{(2)} \\
I_0^{(2)} \\
I_{1}^{(2)} \\
I_{-1}^{(3)} \\
I_0^{(3)} \\
I_{1}^{(3)} \\
I_{-1}^{(4)} \\
I_0^{(4)} \\
I_{1}^{(4)} \\
\end{bmatrix}
=
\left[
\begin{array}{c|c|c|c}
\begin{matrix}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{matrix} &
\begin{matrix}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{matrix} &
\begin{matrix}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{matrix} &
\begin{matrix}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{matrix} \\ \hline
\begin{matrix}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{matrix} &
\begin{matrix}
Z & 0 & 0 \\
0 & Z & 0 \\
0 & 0 & Z \\
\end{matrix} &
\begin{matrix}
-Z & 0 & 0 \\
0 & -Z & 0 \\
0 & 0 & -Z \\
\end{matrix} &
\begin{matrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{matrix} \\ \hline
\begin{matrix}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{matrix} &
\begin{matrix}
-Z & 0 & 0 \\
0 & -Z & 0 \\
0 & 0 & -Z \\
\end{matrix} &
\begin{matrix}
Z & 0 & 0 \\
0 & Z & 0 \\
0 & 0 & Z \\
\end{matrix} &
\begin{matrix}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{matrix} \\ \hline
\begin{matrix}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{matrix} &
\begin{matrix}
-1 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & -1 \\
\end{matrix} &
\begin{matrix}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{matrix} &
\begin{matrix}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{matrix} \\
\end{array}
\right]
\begin{bmatrix}
V_{-1}^{(1)} \\
V_{0}^{(1)} \\
V_{1}^{(1)} \\
V_{-1}^{(2)} \\
V_{0}^{(2)} \\
V_{1}^{(2)} \\
V_{-1}^{(3)} \\
V_{0}^{(3)} \\
V_{1}^{(3)} \\
I_{-1}^{(L)} \\
I_{0}^{(L)} \\
I_{1}^{(L)} \\
\end{bmatrix} \\
&+
\left[
\begin{array}{c|c|c|c}
\begin{matrix}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{matrix} &
\begin{matrix}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{matrix} &
\begin{matrix}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{matrix} &
\begin{matrix}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{matrix} \\ \hline
\begin{matrix}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{matrix} &
\begin{matrix}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{matrix} &
\begin{matrix}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{matrix} &
\begin{matrix}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{matrix} \\ \hline
\begin{matrix}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{matrix} &
\begin{matrix}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{matrix} &
\begin{matrix}
j \omega_0 (-1) C & 0 & 0 \\
0 & j \omega_0 (0) C & 0 \\
0 & 0 & j \omega_0 (1) C \\
\end{matrix} &
\begin{matrix}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{matrix} \\ \hline
\begin{matrix}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{matrix} &
\begin{matrix}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{matrix} &
\begin{matrix}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{matrix} &
\begin{matrix}
j \omega_0 (-1) L & 0 & 0 \\
0 & j \omega_0 (0) L & 0 \\
0 & 0 & j \omega_0 (1) L \\
\end{matrix} \\
\end{array}
\right]
\begin{bmatrix}
V_{-1}^{(1)} \\
V_{0}^{(1)} \\
V_{1}^{(1)} \\
V_{-1}^{(2)} \\
V_{0}^{(2)} \\
V_{1}^{(2)} \\
V_{-1}^{(3)} \\
V_{0}^{(3)} \\
V_{1}^{(3)} \\
I_{-1}^{(L)} \\
I_{0}^{(L)} \\
I_{1}^{(L)} \\
\end{bmatrix} \\
\end{aligned}
\end{equation}

With a vector of fourier coeffient vectors

\begin{equation}\label{eqn:diodeRLCSample:280}
\BV =
\begin{bmatrix}
\BV^{(1)} \\
\BV^{(2)} \\
\BV^{(3)} \\
\BI^{(L)}
\end{bmatrix}, \qquad
\BI =
\begin{bmatrix}
\BI^{(1)} \\
\BI^{(2)} \\
\BI^{(3)} \\
\BI^{(4)}
\end{bmatrix}.
\end{equation}

and a \( (2 N + 1) \times (2 N + 1) \) matrix of indexes

\begin{equation}\label{eqn:diodeRLCSample:220}
\BN =
\begin{bmatrix}
-N & & & & \\
& 1-N & & & \\
& & \ddots & & \\
& & & N-1 & \\
& & & & N \\
\end{bmatrix},
\end{equation}

the complete block diagonalization is

\begin{equation}\label{eqn:diodeRLCSample:300}
\boxed{
{\begin{bmatrix}
g_{rs} \BI_{2 N + 1} +
j \omega_0 c_{rs} \BN
\end{bmatrix}
}_{rs}
\BV
=
\BI.
}
\end{equation}

Block matrix form, with frequency ordering

It turns out that a better way of ordering the vector of Fourier components is using a frequency ordering that interleaves the physical parameters. With such an ordering the DFT MNA equations are

\begin{equation}\label{eqn:diodeRLCSample:340}
\begin{aligned}
\BI =
&\begin{bmatrix}
I_{-1}^{(1)} \\
I_{-1}^{(2)} \\
I_{-1}^{(3)} \\
I_{-1}^{(4)} \\
I_0^{(1)} \\
I_0^{(2)} \\
I_0^{(3)} \\
I_0^{(4)} \\
I_{1}^{(1)} \\
I_{1}^{(2)} \\
I_{1}^{(3)} \\
I_{1}^{(4)} \\
\end{bmatrix}
+
\left[
\begin{array}{c|c|c}
\begin{matrix}
0 & 0 & 0 & 0 \\
0 & Z & -Z & 1 \\
0 & -Z & Z & 0 \\
0 & -1 & 0 & 0 \\
\end{matrix} & 0 & 0 \\ \hline
0 &
\begin{matrix}
0 & 0 & 0 & 0 \\
0 & Z & -Z & 1 \\
0 & -Z & Z & 0 \\
0 & -1 & 0 & 0 \\
\end{matrix} & 0 \\ \hline
0 & 0 &
\begin{matrix}
0 & 0 & 0 & 0 \\
0 & Z & -Z & 1 \\
0 & -Z & Z & 0 \\
0 & -1 & 0 & 0 \\
\end{matrix} \\
\end{array}
\right]
\begin{bmatrix}
V_{-1}^{(1)} \\
V_{-1}^{(2)} \\
V_{-1}^{(3)} \\
I_{-1}^{(L)} \\
V_{0}^{(1)} \\
V_{0}^{(2)} \\
V_{0}^{(3)} \\
I_{0}^{(L)} \\
V_{1}^{(1)} \\
V_{1}^{(2)} \\
V_{1}^{(3)} \\
I_{1}^{(L)} \\
\end{bmatrix} \\
&+
j \omega_0
\left[
\begin{array}{c|c|c}
(-1)
\begin{bmatrix}
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & C & 0 \\
0 & 0 & 0 & L \\
\end{bmatrix} & 0 & 0 \\ \hline
0 &
(0)
\begin{bmatrix}
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & C & 0 \\
0 & 0 & 0 & L \\
\end{bmatrix} & 0 \\ \hline
0 & 0 &
(1)
\begin{bmatrix}
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & C & 0 \\
0 & 0 & 0 & L \\
\end{bmatrix} \\
\end{array}
\right]
\begin{bmatrix}
V_{-1}^{(1)} \\
V_{-1}^{(2)} \\
V_{-1}^{(3)} \\
I_{-1}^{(L)} \\
V_{0}^{(1)} \\
V_{0}^{(2)} \\
V_{0}^{(3)} \\
I_{0}^{(L)} \\
V_{1}^{(1)} \\
V_{1}^{(2)} \\
V_{1}^{(3)} \\
I_{1}^{(L)} \\
\end{bmatrix} \\
\end{aligned}
\end{equation}

This ordering matches that of [1].

Representing the linear sources

Assuming real sources with frequencies that are only multiples of the fundamental harmonic, a reasonable way to represent them in storage is with a pair of matrices

\begin{equation}\label{eqn:diodeRLCSample:360}
\begin{bmatrix}
\BI \sim \BB \Bomega
\end{bmatrix}.
\end{equation}

If \( R \) is the dimension of \(\BG\) and \( \BC \), then \( \BB \) is a \( R \times S \) dimension matrix, where \( S \) is the sum of

  • 1, if there are any DC sources, plus
  • 2 times the number of unique frequency sources.

For example, if there is a DC source and one AC source with frequency \( \nu \), then for column vectors \( \Bb_i \) this pair is of the form

\begin{equation}\label{eqn:diodeRLCSample:380}
\BU \Bomega =
\begin{bmatrix}
\Bb_{-1} & \Bb_0 & \Bb_1
\end{bmatrix}
\begin{bmatrix}
– 2 \pi \nu \\
0 \\
2 \pi \nu
\end{bmatrix}.
\end{equation}

This representation produces the time domain representation exactly when there are only DC sources, and can be used to construct the Fourier coefficients by inspection when there are AC sources. For example, for \( N = 1 \) in the example above, the Fourier coefficent vector is

\begin{equation}\label{eqn:diodeRLCSample:400}
\BI
=
\begin{bmatrix}
\Bb_{-1} \\
\Bb_{0} \\
\Bb_{1} \\
\end{bmatrix}.
\end{equation}

If \( N = 2 \) was used, then we would have instead

\begin{equation}\label{eqn:diodeRLCSample:420}
\BI
=
\begin{bmatrix}
\Bzero \\
\Bb_{-1} \\
\Bb_{0} \\
\Bb_{1} \\
\Bzero \\
\end{bmatrix}.
\end{equation}

Representing non-linear sources

The time domain MNA fig. 1.

diodeWithResistorFig1

fig. 1. Simple diode circuit

 

 

With \( Z = 1/R, Z_g = 1/R_g \), the KCL equations are

  1. \( \lr{ v_1 – v_2 } Z_s = i_s – i_d \)
  2. \( \lr{ v_2 – v_1 } Z_s + v_2 Z_g = -i_s + i_d \)

Using the model \( i_d = I^{(0)} \lr{ e^{ (v_1 – v_2)/V_T } – 1 } \), with source \( i_s = I^{(s)} \cos( \omega_0 t ) \),
this has the block matrix form

\begin{equation}\label{eqn:diodeRLCSample:580}
\BG =
\begin{bmatrix}
Z_s & -Z_s \\
-Z_s & Z_s + Z_g \\
\end{bmatrix}, \qquad
\Bx =
\begin{bmatrix}
v_1(t) \\
v_2(t) \\
\end{bmatrix}
\end{equation}
\begin{equation}\label{eqn:diodeRLCSample:600}
\BB =
\begin{bmatrix}
I^{(s)}/2 & -I^{(0)} & I^{(s)}/2 \\
-I^{(s)}/2 & I^{(0)} & -I^{(s)}/2
\end{bmatrix}, \qquad
\Bu(t) =
\begin{bmatrix}
e^{-j \omega_0 t} \\
1 \\
e^{j \omega_0 t}
\end{bmatrix}
\end{equation}
\begin{equation}\label{eqn:diodeRLCSample:620}
\BD
=
\begin{bmatrix}
I^{(0)} \\
-I^{(0)}
\end{bmatrix}, \qquad
\Bw(t) = e^{(v_1(t) – v_2(t))/V_T}.
\end{equation}

If \( E_n \) is the nth DFT coefficient for \( e(t) = e^{(v_1(t) – v_2(t))/V_T} \), then the DFT equations for the \( N = 1 \) DFT is

\begin{equation}\label{eqn:diodeRLCSample:640}
\begin{aligned}
\lr{ V_{-1}^{(1)} – V_{-1}^{(2)} } Z_s &= I^{(s)}/2 – I^{(0)} E_{-1} \\
\lr{ V_{-1}^{(2)} – V_{-1}^{(1)} } Z_s + V_{-1}^{(2)} Z_g &= -I^{(s)}/2 + I^{(0)} E_{-1} \\
\lr{ V_{0}^{(1)} – V_{0}^{(2)} } Z_s &= I^{(0)} – I^{(0)} E_{0} \\
\lr{ V_{0}^{(2)} – V_{0}^{(1)} } Z_s + V_{0}^{(2)} Z_g &= -I^{(0)} + I^{(0)} E_{0} \\
\lr{ V_{1}^{(1)} – V_{1}^{(2)} } Z_s &= I^{(s)}/2 – I^{(0)} E_{1} \\
\lr{ V_{1}^{(2)} – V_{1}^{(1)} } Z_s + V_{1}^{(2)} Z_g &= -I^{(s)}/2 + I^{(0)} E_{1}
\end{aligned}
\end{equation}

Let \( \Bb = [ \Bb_{-1}\, \Bb_0\, \Bb_1 ] \), and \( \BD = [ \Bd_1 ] \). The block matrix equivalent form, by inspection, is

\begin{equation}\label{eqn:diodeRLCSample:660}
\begin{bmatrix}
\BG & 0 & 0 \\
0 & \BG & 0 \\
0 & 0 & \BG
\end{bmatrix}
\begin{bmatrix}
V_{-1}^{(1)} \\
V_{-1}^{(2)} \\
V_{0}^{(1)} \\
V_{0}^{(2)} \\
V_{1}^{(1)} \\
V_{1}^{(2)}
\end{bmatrix}
=
\begin{bmatrix}
\Bb_{-1} \\
\Bb_{0} \\
\Bb_{1} \\
\end{bmatrix}
+
\begin{bmatrix}
\Bd_1 E_{-1} \\
\Bd_1 E_{0} \\
\Bd_1 E_{1}
\end{bmatrix}.
\end{equation}

This shows the stamping pattern required to form the non-linear portion of the Harmonic balance equations. The general pattern can be written as

\begin{equation}\label{eqn:diodeRLCSample:820}
\boxed{
\mathcal{G} \BV = \BI + \mathcal{I}(\BV),
}
\end{equation}

Here \( \mathcal{G} \) is block diagonal, and in general has blocks of \( \BG + j \omega_0 n \BC \). The matrix \( \BI \) was generated from the Fourier coeffients of all the linear sources, and \( \mathcal{I}(\BV) \) encodes all the non-linear contributions to the system.

More general non-linear structure, for multiple diodes

For the diode exponentials, these non-linear term will be of the form

\begin{equation}\label{eqn:diodeRLCSample:680}
\BD \Bw(t)
=
\begin{bmatrix}
\Bd_1 & \Bd_2 & \cdots & \Bd_S
\end{bmatrix}
\begin{bmatrix}
w_1(t) \\
w_2(t) \\
\vdots \\
w_S(t) \\
\end{bmatrix},
\end{equation}

where \( w_i(t) = \exp( (v_{i,1}(t) – v_{i,2}(t))/V_{T,i} ) \). If the DFT coordinates of these functions are \( E_n^{(i)} \), then the frequency domain representation is

\begin{equation}\label{eqn:diodeRLCSample:700}
\mathcal{I}(\BV)
=
\sum_{i = 1}^S
\begin{bmatrix}
\Bd_i E_{-N}^{(i)} \\
\Bd_i E_{1-N}^{(i)} \\
\vdots \\
\Bd_i E_{N-1}^{(i)} \\
\Bd_i E_{N}^{(i)} \\
\end{bmatrix}.
\end{equation}

This is a \( R (2 N + 1) \times 1 \) matrix, as expected.

The computation of these DFT coordinates is a bit messy since they are time dependent, and thus dependent on the (unknown) values of \( V_n^{(1)} \). Consider the above circuit as an example where we have

\begin{equation}\label{eqn:diodeRLCSample:720}
w(t) = \exp\lr{ (v_1(t) – v_2(t))/V_T }.
\end{equation}

The DFT inverse is

\begin{equation}\label{eqn:diodeRLCSample:740}
E_n = \sum_{k=-N}^N \exp\lr{ (v_1(t_k) – v_2(t_k))/V_T } e^{-j \omega_0 n t_k }.
\end{equation}

With \( \BE = ( E_{-N}, E_{1-N}, \cdots, E_{N-1}, E_N ) \), this is

\begin{equation}\label{eqn:diodeRLCSample:760}
\BE =
\inv{2 N + 1} \overline{{\BF}}
\begin{bmatrix}
\exp\lr{ (v_{-N}^{(1)} – v_{-N}^{(2)})/v_T } \\
\exp\lr{ (v_{1-N}^{(1)} – v_{1-N}^{(2)})/v_T } \\
\vdots \\
\exp\lr{ (v_{N-1}^{(1)} – v_{N-1}^{(2)})/v_T } \\
\exp\lr{ (v_{N}^{(1)} – v_{N}^{(2)})/v_T } \\
\end{bmatrix}
=
\inv{2 N + 1} \overline{{\BF}}
\begin{bmatrix}
\exp\lr{ {[\BF (\BV^{(1)} – \BV^{(2)})/v_T]}_{-N} } \\
\exp\lr{ {[\BF (\BV^{(1)} – \BV^{(2)})/v_T]}_{1-N} } \\
\vdots \\
\exp\lr{ {[\BF (\BV^{(1)} – \BV^{(2)})/v_T]}_{N-1} } \\
\exp\lr{ {[\BF (\BV^{(1)} – \BV^{(2)})/v_T]}_{N} }
\end{bmatrix}.
\end{equation}

With the introduction a term by term exponentiation operator

\begin{equation}\label{eqn:diodeRLCSample:780}
\exp[ \Bx ] =
\begin{bmatrix}
\exp(x_1) \\
\exp(x_2) \\
\vdots \\
\end{bmatrix},
\end{equation}

this one Fourier coefficient vector is

\begin{equation}\label{eqn:diodeRLCSample:800}
\BE =
\inv{2 N + 1} \overline{{\BF}}
\exp[ \BF (\BV^{(1)} – \BV^{(2)})/v_T ].
\end{equation}

This is now completely expressed in terms of the unknown Fourier component vectors, each a subset of the aggreggated “voltage”, range selectable with the Matlab operation \( \BV^{(i)} = \BV(i:R:end) \).

Newton’s method

In order to solve the system, Newton’s method on the Fourier coeffients is required. Solutions to \( \mathcal{F}(\BV) = 0 \) are sought, where

\begin{equation}\label{eqn:diodeRLCSample:840}
\mathcal{F}(\BV) = \mathcal{G} \BV – \BI – \mathcal{I}(\BV).
\end{equation}

Here the sources current vector DFT coordinates have been split into the linear contributions \( \BI \) and nonlinear contributions \( \mathcal{I} \) defined by \ref{eqn:diodeRLCSample:700}.

Working with ones-indexed coordinates of \( \BV = [V_k]_k \), where \( k \in [1, R(2 N + 1) ] \), the Jacobian is

\begin{equation}\label{eqn:diodeRLCSample:860}
\BJ = \mathcal{G} – {[ \PDi{V_s}{\mathcal{I}_r} ]}_{rs}.
\end{equation}

To calculate these partials we need the partials of the coordinates of \( \BE \) of
\ref{eqn:diodeRLCSample:800}. The kth coordinate of \( \BV^{(1)}, \BV^{(2)} \) in terms of the coordinates of the \( R(2 N + 1) \) vector of unknowns \( \BV \) are

\begin{equation}\label{eqn:diodeRLCSample:880}
\begin{aligned}
[\BV^{(1)}]_k &= V_{1 + (k-1) R} \\
[\BV^{(2)}]_k &= V_{2 + (k-1) R}
\end{aligned}
\end{equation}

Using summation convention, with sums over any repeated indexes implied, those coordinates are

\begin{equation}\label{eqn:diodeRLCSample:900}
E_r =
\inv{2 N + 1} \overline{{F}}_{r a}
\exp\lr{ F_{ab}
(
V_{1 + (b-1) R} –
V_{2 + (b-1) R}
)/v_T }.
\end{equation}

The partials now follow immediately

\begin{equation}\label{eqn:diodeRLCSample:920}
\PD{V_s}{E_r} =
\inv{2 N + 1} \inv{v_T} \overline{{F}}_{r a} F_{ab}
\exp\lr{ F_{ab}
(
V_{1 + (b-1) R} –
V_{2 + (b-1) R}
)/v_T }
\lr{
\delta_{s,1 + (b-1) R} –
\delta_{s,2 + (b-1) R}
}.
\end{equation}

Generalization

To generalize this, suppose that the diode exponential was associated with voltages spanning nodes \( m, n\) so that

\begin{equation}\label{eqn:diodeRLCSample:980}
\BE =
\inv{2 N + 1} \overline{{\BF}}
\exp[ \BF (\BV^{(m)} – \BV^{(n)})/v_T ].
\end{equation}

In this case, the coordinates of the physical “voltages” are

\begin{equation}\label{eqn:diodeRLCSample:1020}
\begin{aligned}
[\BV^{(m)}]_k &= V_{m + (k-1) R} \\
[\BV^{(n)}]_k &= V_{n + (k-1) R}
\end{aligned},
\end{equation}

so

\begin{equation}\label{eqn:diodeRLCSample:1040}
E_r =
\inv{2 N + 1} \overline{{F}}_{r a}
\exp\lr{ F_{ab}
(
V_{m + (b-1) R} –
V_{n + (b-1) R}
)/v_T }.
\end{equation}

The Jacobian partials are

\begin{equation}\label{eqn:diodeRLCSample:1060}
\PD{V_s}{E_r} =
\inv{2 N + 1} \inv{v_T} \overline{{F}}_{r a} F_{ab}
\exp\lr{ F_{ab}
(
V_{m + (b-1) R} –
V_{n + (b-1) R}
)/v_T }
\lr{
\delta_{s,m + (b-1) R} –
\delta_{s,n + (b-1) R}
}.
\end{equation}

Note that this Jacobian

\begin{equation}\label{eqn:diodeRLCSample:1080}
\BJ_\BE =
{
\begin{bmatrix}
\PD{V_s}{E_r}
\end{bmatrix}}_{rs}
\end{equation}

is a \( (2 N + 1) \times R(2N + 1) \) matrix.

The full Jacobian of \( \mathcal{I}(\BV) \) is

\begin{equation}\label{eqn:diodeRLCSample:1120}
\BJ_{\mathcal{I}}(\BV)
=
\sum_{i = 1}^S
\begin{bmatrix}
\Bd_i \PD{\BV}{E_{-N}^{(i)}} \\
\Bd_i \PD{\BV}{E_{1-N}^{(i)}} \\
\vdots \\
\Bd_i \PD{\BV}{E_{N-1}^{(i)}} \\
\Bd_i \PD{\BV}{E_{N}^{(i)}} \\
\end{bmatrix}.
\end{equation}

Where \( \PDi{\BV}{E_{k}^{(i)}} \) is the kth row of \( \BJ_{\BE^{(i)}} \). The complete Jacobian is

\begin{equation}\label{eqn:diodeRLCSample:1100}
\BJ = \mathcal{G} – \BJ_{\mathcal{I}}(\BV).
\end{equation}

References

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

Discrete Fourier Transform

December 2, 2014 ece1254 No comments , , , ,

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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

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

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

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

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 \( t_k \),

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

assume that an inversion relation can be found. To find \( X_n \) evaluate the sum

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

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

\begin{equation}\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}
\end{equation}

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

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

which provides the desired Fourier inversion relation

\begin{equation}\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}.
}
\end{equation}

References

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

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