math and physics play

Impedance transformation

May 16, 2015 ece1229 , , , ,

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In our final problem set we used the impedance transformation for calculations related to a microslot antenna. This transformation wasn’t familiar to me, and is apparently covered in the third year ECE fields class. I found a derivation of this in [1], but the idea is really simple and follows from the reflection coefficient calculation for a normal reflection configuration.

Consider a normal field reflection between two interfaces, as sketched in fig. 1.

normalTransmissionFig1

fig. 1. Normal reflection and transmission between two media.

The fields are

\begin{equation}\label{eqn:impedanceTransformation:40}
\BE^\textrm{i} = \xcap E_0 e^{-j k_1 z}
\end{equation}
\begin{equation}\label{eqn:impedanceTransformation:60}
\BH^\textrm{i} = \ycap \frac{E_0}{\eta_1} e^{-j k_1 z}
\end{equation}
\begin{equation}\label{eqn:impedanceTransformation:80}
\BE^\textrm{r} = \xcap \Gamma E_0 e^{j k_1 z}
\end{equation}
\begin{equation}\label{eqn:impedanceTransformation:100}
\BH^\textrm{r} = -\ycap \Gamma \frac{E_0}{\eta_1} e^{j k_1 z}
\end{equation}
\begin{equation}\label{eqn:impedanceTransformation:120}
\BE^\textrm{t} = \xcap E_0 T e^{-j k_2 z}
\end{equation}
\begin{equation}\label{eqn:impedanceTransformation:140}
\BH^\textrm{t} = \ycap \frac{E_0}{\eta_1} T e^{-j k_2 z}.
\end{equation}

The field orientations have been picked so that the tangential component of the electric field is \( \xcap \) oriented for all of the incident, reflected, and transmitted components. Requiring equality of the tangential field components at the interface gives

\begin{equation}\label{eqn:impedanceTransformation:180}
1 + \Gamma = T
\end{equation}
\begin{equation}\label{eqn:impedanceTransformation:200}
\inv{\eta_1} – \frac{\Gamma}{\eta_1} = \frac{T}{\eta_2}.
\end{equation}

Solving for the transmission coefficient gives

\begin{equation}\label{eqn:impedanceTransformation:220}
\begin{aligned}
T
&= \frac{2}{ 1 + \frac{\eta_1}{\eta_2} } \\
&= \frac{2 \eta_2}{ \eta_2 + \eta_1 },
\end{aligned}
\end{equation}

and for the reflection coefficient

\begin{equation}\label{eqn:impedanceTransformation:240}
\begin{aligned}
\Gamma
&= T – 1 \\
&= \frac{2 \eta_2 – \eta_1 – \eta_2}{ \eta_2 + \eta_1 } \\
&= \frac{\eta_2 – \eta_1 }{ \eta_2 + \eta_1 }.
\end{aligned}
\end{equation}

The total fields in medium 1 at the point \( z = -l \) are

\begin{equation}\label{eqn:impedanceTransformation:280}
\BE^\textrm{i} + \BE^\textrm{r}
=
\xcap E_0 \lr{ e^{ -j k_1 (-l)} + \Gamma e^{ j k_1 (-l) } }
\end{equation}
\begin{equation}\label{eqn:impedanceTransformation:300}
\BH^\textrm{i} + \BH^\textrm{r}
=
\ycap \frac{E_0}{\eta_1} \lr{ e^{ -j k_1 (-l)} – \Gamma e^{ j k_1 (-l) }}.
\end{equation}

The ratio of the electric field strength to the magnetic field strength is defined as the input impedance

\begin{equation}\label{eqn:impedanceTransformation:320}
Z_{\textrm{in}} \equiv \evalbar{\frac{E^\textrm{i} + E^\textrm{r}}{H^\textrm{i} + H^\textrm{r}}}{ z = -l}.
\end{equation}

That is

\begin{equation}\label{eqn:impedanceTransformation:340}
\begin{aligned}
Z_{\textrm{in}}
&=
\eta_1 \frac{
e^{ j k_1 l} + \Gamma e^{ -j k_1 l }
}{
e^{ j k_1 l} – \Gamma e^{ -j k_1 l }
} \\
&=
\eta_1 \frac{
\lr{ \eta_1 + \eta_2} e^{ j k_1 l} + \lr{ \eta_2 – \eta_1} e^{ -j k_1 l }
}{
\lr{ \eta_1 + \eta_2} e^{ j k_1 l} – \lr{ \eta_2 – \eta_1} e^{ -j k_1 l }
} \\
&=
\eta_1 \frac{
\eta_2 \cos( k_1 l ) + \eta_1 j \sin( k_1 l)
}{
\eta_2 j \sin( k_1 l ) + \eta_1 \cos( k_1 l)
},
\end{aligned}
\end{equation}

or
\begin{equation}\label{eqn:impedanceTransformation:360}
\boxed{
Z_{\textrm{in}}
=
\eta_1 \frac{
\eta_2 + j \eta_1 \tan( k_1 l)
}{
\eta_1 + j \eta_2 \tan( k_1 l )
}.
}
\end{equation}

References

[1] Constantine A Balanis. Advanced engineering electromagnetics, chapter {Reflection and transmission}. Wiley New York, 1989.

Plane wave solution directly from Maxwell’s equations

May 6, 2015 math and physics play , , , ,

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Here’s a problem that I thought was fun, an exercise for the reader to show that the plane wave solution to Maxwell’s equations can be found with ease directly from Maxwell’s equations. This is in contrast to the what seems like the usual method of first showing that Maxwell’s equations imply wave equations for the fields, and then solving those wave equations.

Problem. \( \xcap \) oriented plane wave electric field ([1] ex. 4.1)

A uniform plane wave having only an \( x \) component of the electric field is traveling in the \( + z \) direction in an unbounded lossless, source-0free region. Using Maxwell’s equations write expressions for the electric and corresponding magnetic field intensities.

Answer

The phasor form of Maxwell’s equations for a source free region are

\begin{equation}\label{eqn:ExPlaneWave:40}
\spacegrad \cross \BE = -j \omega \BB
\end{equation}
\begin{equation}\label{eqn:ExPlaneWave:60}
\spacegrad \cross \BH = j \omega \BD
\end{equation}
\begin{equation}\label{eqn:ExPlaneWave:80}
\spacegrad \cdot \BD = 0
\end{equation}
\begin{equation}\label{eqn:ExPlaneWave:100}
\spacegrad \cdot \BB = 0.
\end{equation}

Since \( \BE = \xcap E(z) \), the magnetic field follows from \ref{eqn:ExPlaneWave:40}

\begin{equation}\label{eqn:ExPlaneWave:120}
-j \omega \BB
= \spacegrad \cross \BE
=
\begin{vmatrix}
\xcap & \ycap & \zcap \\
\partial_x & \partial_y & \partial_z \\
E & 0 & 0
\end{vmatrix}
=
\ycap \partial_z E(z)
– \zcap \partial_y E(z),
\end{equation}

or

\begin{equation}\label{eqn:ExPlaneWave:140}
\BB =
-\inv{j \omega} \partial_z E.
\end{equation}

This is constrained by \ref{eqn:ExPlaneWave:60}

\begin{equation}\label{eqn:ExPlaneWave:160}
j \omega \epsilon \xcap E
=
\inv{\mu} \spacegrad \cross \BB
=
-\inv{\mu j \omega}
\begin{vmatrix}
\xcap & \ycap & \zcap \\
\partial_x & \partial_y & \partial_z \\
0 & \partial_z E & 0
\end{vmatrix}
=
-\inv{\mu j \omega}
\lr{
-\xcap \partial_{z z} E
+ \zcap \partial_x \partial_z E
}
\end{equation}

Since \( \partial_x \partial_z E = \partial_z \lr{ \partial_x E } = \partial_z \inv{\epsilon} \spacegrad \cdot \BD = \partial_z 0 \), this means

\begin{equation}\label{eqn:ExPlaneWave:180}
\partial_{zz} E = -\omega^2 \epsilon\mu E = -k^2 E.
\end{equation}

This is the usual starting place that we use to show that the plane wave has an exponential form

\begin{equation}\label{eqn:ExPlaneWave:200}
\BE(z) =
\xcap
\lr{
E_{+} e^{-j k z}
+
E_{-} e^{j k z}
}.
\end{equation}

The magnetic field from \ref{eqn:ExPlaneWave:140} is

\begin{equation}\label{eqn:ExPlaneWave:220}
\BB
= \frac{j}{\omega} \lr{ -j k E_{+} e^{-j k z} + j k E_{-} e^{j k z} }
= \inv{c} \lr{ E_{+} e^{-j k z} – E_{-} e^{j k z} },
\end{equation}

or

\begin{equation}\label{eqn:ExPlaneWave:240}
\BH
= \inv{\mu c} \lr{ E_{+} e^{-j k z} – E_{-} e^{j k z} }
= \inv{\eta} \lr{ E_{+} e^{-j k z} – E_{-} e^{j k z} }.
\end{equation}

A solution requires zero divergence for the magnetic field, but that can be seen to be the case by inspection.

References

[1] Constantine A Balanis. Advanced engineering electromagnetics. Wiley New York, 1989.

Coupled wave equation in cylindrical coordinates

May 5, 2015 math and physics play , ,

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In [1], for a sourceless configuration, it is noted that the electric field equations \( \spacegrad^2 \BE = -\beta^2 \BE \) have the form

\begin{equation}\label{eqn:cylindricalFieldSolution:20}
\spacegrad^2 E_\rho – \frac{E_\rho}{\rho^2} – \frac{2}{\rho^2} \PD{\phi}{E_\phi} = -\beta^2 E_\rho
\end{equation}
\begin{equation}\label{eqn:cylindricalFieldSolution:60}
\spacegrad^2 E_\phi – \frac{E_\phi}{\rho^2} + \frac{2}{\rho^2} \PD{\phi}{E_\rho} = -\beta^2 E_\phi
\end{equation}
\begin{equation}\label{eqn:cylindricalFieldSolution:80}
\spacegrad^2 E_z = -\beta^2 E_z,
\end{equation}

where

\begin{equation}\label{eqn:cylindricalFieldSolution:100}
\spacegrad^2 \psi =
\inv{\rho} \PD{\rho}{} \lr{ \rho \PD{\rho}{\psi}} + \inv{\rho^2}\PDSq{\phi}{\psi} + \PDSq{z}{\psi}.
\end{equation}

He applies separation of variables to the last equation, ending up with the usual Bessel function solution, but the first two coupled equations are dismissed as coupled and difficult. It looks like separation of variables works for this too, but we have to prep the system slightly by writing \( \psi = E_\rho + j E_\phi \), which gives

\begin{equation}\label{eqn:cylindricalFieldSolution:120}
\spacegrad^2 \psi – \frac{\psi}{\rho^2} + \frac{2 j}{\rho^2} \PD{\phi}{\psi} = -\beta^2 \psi,
\end{equation}

or

\begin{equation}\label{eqn:cylindricalFieldSolution:140}
\inv{\rho} \PD{\rho}{} \lr{ \rho \PD{\rho}{\psi}} + \inv{\rho^2}\PDSq{\phi}{\psi} + \PDSq{z}{\psi}
– \frac{\psi}{\rho^2} + \frac{2 j}{\rho^2} \PD{\phi}{\psi} = -\beta^2 \psi.
\end{equation}

With a separation of variables substitution \( \psi = f(\rho) g(\phi) h(z) \) this gives

\begin{equation}\label{eqn:cylindricalFieldSolution:160}
\inv{\rho f} \PD{\rho}{} \lr{ \rho \PD{\rho}{f}}
+ \inv{\rho^2 g}\PDSq{\phi}{g}
+ \inv{z} \PDSq{z}{h}
– \frac{1}{\rho^2} + \frac{2 j}{\rho^2 g} \PD{\phi}{g} = -\beta^2.
\end{equation}

Assuming a solution for the function \( h \) of

\begin{equation}\label{eqn:cylindricalFieldSolution:180}
\inv{z} \PDSq{z}{h} = -\alpha^2,
\end{equation}

the PDE is reduced to an equation in two functions

\begin{equation}\label{eqn:cylindricalFieldSolution:200}
\inv{\rho f} \PD{\rho}{} \lr{ \rho \PD{\rho}{f}}
+ \inv{\rho^2 g}\PD{\phi}{} \lr{ g + 2 j g}
+ \beta^2 -\alpha^2
– \frac{1}{\rho^2}
= 0,
\end{equation}

or

\begin{equation}\label{eqn:cylindricalFieldSolution:220}
\frac{\rho}{f} \PD{\rho}{} \lr{ \rho \PD{\rho}{f}}
+ \inv{g}\PD{\phi}{} \lr{ g + 2 j g}
+ \lr{ \beta^2 -\alpha^2 }\rho^2
= 1.
\end{equation}

With the term in \( g \) having only \( \phi \) dependence, we can assume

\begin{equation}\label{eqn:cylindricalFieldSolution:240}
\inv{g}\PD{\phi}{} \lr{ g + 2 j g} = 1 – \gamma^2,
\end{equation}

for

\begin{equation}\label{eqn:cylindricalFieldSolution:260}
\frac{\rho}{f} \PD{\rho}{} \lr{ \rho \PD{\rho}{f}}
– \gamma^2
+ \lr{ \beta^2 -\alpha^2 }\rho^2
= 0.
\end{equation}

I’m not sure off hand if these can be solved in known special functions, especially since the constants in the mix are complex.

References

[1] Constantine A Balanis. Advanced engineering electromagnetics, volume 20. Wiley New York, 1989.

Tangential and normal field components

May 4, 2015 ece1229 , , , , , , , , , , , , ,

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The integral forms of Maxwell’s equations can be used to derive relations for the tangential and normal field components to the sources. These relations were mentioned in class. It’s a little late, but lets go over the derivation. This isn’t all review from first year electromagnetism since we are now using a magnetic source modifications of Maxwell’s equations.

The derivation below follows that of [1] closely, but I am trying it myself to ensure that I understand the assumptions.

The two infinitesimally thin pillboxes of fig. 1, and fig. 2 are used in the argument.

pillboxForTangentialFieldsFig1

fig. 2: Pillboxes for tangential and normal field relations

pillboxForNormalFieldsFig2

fig. 1: Pillboxes for tangential and normal field relations

Maxwell’s equations with both magnetic and electric sources are

\begin{equation}\label{eqn:normalAndTangentialFields:20}
\spacegrad \cross \boldsymbol{\mathcal{E}} = -\PD{t}{\boldsymbol{\mathcal{B}}} -\boldsymbol{\mathcal{M}}
\end{equation}
\begin{equation}\label{eqn:normalAndTangentialFields:40}
\spacegrad \cross \boldsymbol{\mathcal{H}} = \boldsymbol{\mathcal{J}} + \PD{t}{\boldsymbol{\mathcal{D}}}
\end{equation}
\begin{equation}\label{eqn:normalAndTangentialFields:60}
\spacegrad \cdot \boldsymbol{\mathcal{D}} = \rho_\textrm{e}
\end{equation}
\begin{equation}\label{eqn:normalAndTangentialFields:80}
\spacegrad \cdot \boldsymbol{\mathcal{B}} = \rho_\textrm{m}.
\end{equation}

After application of Stokes’ and the divergence theorems Maxwell’s equations have the integral form

\begin{equation}\label{eqn:normalAndTangentialFields:100}
\oint \boldsymbol{\mathcal{E}} \cdot d\Bl = -\int d\BA \cdot \lr{ \PD{t}{\boldsymbol{\mathcal{B}}} + \boldsymbol{\mathcal{M}} }
\end{equation}
\begin{equation}\label{eqn:normalAndTangentialFields:120}
\oint \boldsymbol{\mathcal{H}} \cdot d\Bl = \int d\BA \cdot \lr{ \PD{t}{\boldsymbol{\mathcal{D}}} + \boldsymbol{\mathcal{J}} }
\end{equation}
\begin{equation}\label{eqn:normalAndTangentialFields:140}
\int_{\partial V} \boldsymbol{\mathcal{D}} \cdot d\BA
=
\int_V \rho_\textrm{e}\,dV
\end{equation}
\begin{equation}\label{eqn:normalAndTangentialFields:160}
\int_{\partial V} \boldsymbol{\mathcal{B}} \cdot d\BA
=
\int_V \rho_\textrm{m}\,dV.
\end{equation}

Maxwell-Faraday equation

First consider one of the loop integrals, like \ref{eqn:normalAndTangentialFields:100}. For an infinestismal loop, that integral is

\begin{equation}\label{eqn:normalAndTangentialFields:180}
\begin{aligned}
\oint \boldsymbol{\mathcal{E}} \cdot d\Bl
&\approx
\mathcal{E}^{(1)}_x \Delta x
+ \mathcal{E}^{(1)} \frac{\Delta y}{2}
+ \mathcal{E}^{(2)} \frac{\Delta y}{2}
-\mathcal{E}^{(2)}_x \Delta x
– \mathcal{E}^{(2)} \frac{\Delta y}{2}
– \mathcal{E}^{(1)} \frac{\Delta y}{2} \\
&\approx
\lr{ \mathcal{E}^{(1)}_x
-\mathcal{E}^{(2)}_x } \Delta x
+ \inv{2} \PD{x}{\mathcal{E}^{(2)}} \Delta x \Delta y
+ \inv{2} \PD{x}{\mathcal{E}^{(1)}} \Delta x \Delta y.
\end{aligned}
\end{equation}

We let \( \Delta y \rightarrow 0 \) which kills off all but the first difference term.

The RHS of \ref{eqn:normalAndTangentialFields:180} is approximately

\begin{equation}\label{eqn:normalAndTangentialFields:200}
-\int d\BA \cdot \lr{ \PD{t}{\boldsymbol{\mathcal{B}}} + \boldsymbol{\mathcal{M}} }
\approx
– \Delta x \Delta y \lr{ \PD{t}{\mathcal{B}_z} + \mathcal{M}_z }.
\end{equation}

If the magnetic field contribution is assumed to be small in comparison to the magnetic current (i.e. infinite magnetic conductance), and if a linear magnetic current source of the form is also assumed

\begin{equation}\label{eqn:normalAndTangentialFields:220}
\boldsymbol{\mathcal{M}}_s = \lim_{\Delta y \rightarrow 0} \lr{\boldsymbol{\mathcal{M}} \cdot \zcap} \zcap \Delta y,
\end{equation}

then the Maxwell-Faraday equation takes the form

\begin{equation}\label{eqn:normalAndTangentialFields:240}
\lr{ \mathcal{E}^{(1)}_x
-\mathcal{E}^{(2)}_x } \Delta x
\approx
– \Delta x \boldsymbol{\mathcal{M}}_s \cdot \zcap.
\end{equation}

While \( \boldsymbol{\mathcal{M}} \) may have components that are not normal to the interface, the surface current need only have a normal component, since only that component contributes to the surface integral.

The coordinate expression of \ref{eqn:normalAndTangentialFields:240} can be written as

\begin{equation}\label{eqn:normalAndTangentialFields:260}
– \boldsymbol{\mathcal{M}}_s \cdot \zcap
=
\lr{ \boldsymbol{\mathcal{E}}^{(1)} -\boldsymbol{\mathcal{E}}^{(2)} } \cdot \lr{ \ycap \cross \zcap }
=
\lr{ \lr{ \boldsymbol{\mathcal{E}}^{(1)} -\boldsymbol{\mathcal{E}}^{(2)} } \cross \ycap } \cdot \zcap.
\end{equation}

This is satisfied when

\begin{equation}\label{eqn:normalAndTangentialFields:280}
\boxed{
\lr{ \boldsymbol{\mathcal{E}}^{(1)} -\boldsymbol{\mathcal{E}}^{(2)} } \cross \ncap = – \boldsymbol{\mathcal{M}}_s,
}
\end{equation}

where \( \ncap \) is the normal between the interfaces. I’d failed to understand when reading this derivation initially, how the \( \boldsymbol{\mathcal{B}} \) contribution was killed off. i.e. If the vanishing area in the surface integral kills off the \( \boldsymbol{\mathcal{B}} \) contribution, why do we have a \( \boldsymbol{\mathcal{M}} \) contribution left. The key to this is understanding that this magnetic current is considered to be confined very closely to the surface getting larger as \( \Delta y \) gets smaller.

Also note that the units of \( \boldsymbol{\mathcal{M}}_s \) are volts/meter like the electric field (not volts/squared-meter like \( \boldsymbol{\mathcal{M}} \).)

Ampere’s law

As above, assume a linear electric surface current density of the form

\begin{equation}\label{eqn:normalAndTangentialFields:300}
\boldsymbol{\mathcal{J}}_s = \lim_{\Delta y \rightarrow 0} \lr{\boldsymbol{\mathcal{J}} \cdot \ncap} \ncap \Delta y,
\end{equation}

in units of amperes/meter (not amperes/meter-squared like \( \boldsymbol{\mathcal{J}} \).)

To apply the arguments above to Ampere’s law, only the sign needs to be adjusted

\begin{equation}\label{eqn:normalAndTangentialFields:290}
\boxed{
\lr{ \boldsymbol{\mathcal{H}}^{(1)} -\boldsymbol{\mathcal{H}}^{(2)} } \cross \ncap = \boldsymbol{\mathcal{J}}_s.
}
\end{equation}

Gauss’s law

Using the cylindrical pillbox surface with radius \( \Delta r \), height \( \Delta y \), and top and bottom surface areas \( \Delta A = \pi \lr{\Delta r}^2 \), the LHS of Gauss’s law \ref{eqn:normalAndTangentialFields:140} expands to

\begin{equation}\label{eqn:normalAndTangentialFields:320}
\begin{aligned}
\int_{\partial V} \boldsymbol{\mathcal{D}} \cdot d\BA
&\approx
\mathcal{D}^{(2)}_y \Delta A
+ \mathcal{D}^{(2)}_\rho 2 \pi \Delta r \frac{\Delta y}{2}
+ \mathcal{D}^{(1)}_\rho 2 \pi \Delta r \frac{\Delta y}{2}
-\mathcal{D}^{(1)}_y \Delta A \\
&\approx
\lr{ \mathcal{D}^{(2)}_y
-\mathcal{D}^{(1)}_y } \Delta A.
\end{aligned}
\end{equation}

As with the Stokes integrals above it is assumed that the height is infinestimal with respect to the radial dimension. Letting that height \( \Delta y \rightarrow 0 \) kills off the radially directed contributions of the flux through the sidewalls.

The RHS expands to approximately

\begin{equation}\label{eqn:normalAndTangentialFields:340}
\int_V \rho_\textrm{e}\,dV
\approx
\Delta A \Delta y \rho_\textrm{e}.
\end{equation}

Define a highly localized surface current density (coulombs/meter-squared) as

\begin{equation}\label{eqn:normalAndTangentialFields:360}
\sigma_\textrm{e} = \lim_{\Delta y \rightarrow 0} \Delta y \rho_\textrm{e}.
\end{equation}

Equating \ref{eqn:normalAndTangentialFields:340} with \ref{eqn:normalAndTangentialFields:320} gives

\begin{equation}\label{eqn:normalAndTangentialFields:380}
\lr{ \mathcal{D}^{(2)}_y
-\mathcal{D}^{(1)}_y } \Delta A
=
\Delta A \sigma_\textrm{e},
\end{equation}

or

\begin{equation}\label{eqn:normalAndTangentialFields:400}
\boxed{
\lr{ \boldsymbol{\mathcal{D}}^{(2)} – \boldsymbol{\mathcal{D}}^{(1)} } \cdot \ncap = \sigma_\textrm{e}.
}
\end{equation}

Gauss’s law for magnetism

The same argument can be applied to the magnetic flux. Define a highly localized magnetic surface current density (webers/meter-squared) as

\begin{equation}\label{eqn:normalAndTangentialFields:440}
\sigma_\textrm{m} = \lim_{\Delta y \rightarrow 0} \Delta y \rho_\textrm{m},
\end{equation}

yielding the boundary relation

\begin{equation}\label{eqn:normalAndTangentialFields:420}
\boxed{
\lr{ \boldsymbol{\mathcal{B}}^{(2)} – \boldsymbol{\mathcal{B}}^{(1)} } \cdot \ncap = \sigma_\textrm{m}.
}
\end{equation}

References

[1] Constantine A Balanis. Advanced engineering electromagnetics, volume 20, chapter Time-varying and time-harmonic electromagnetic fields. Wiley New York, 1989.

Updated notes for ece1229 antenna theory

April 10, 2015 ece1229 , , ,

I’ve now posted a second update of my notes for the antenna theory course that I am taking this term at UofT.

Unlike most of the other classes I have taken, I am not attempting to take comprehensive notes for this class. The class is taught on slides which go by faster than I can easily take notes for (and some of which match the textbook closely). In class I have annotated my copy of textbook with little details instead. This set of notes contains musings of details that were unclear, or in some cases, details that were provided in class, but are not in the text (and too long to pencil into my book), as well as some notes Geometric Algebra formalism for Maxwell’s equations with magnetic sources (something I’ve encountered for the first time in any real detail in this class).

This new update includes the following new content:

March 29, 2015 Antenna array design (problem)

March 23, 2015 Antenna array design with Chebychev polynomials (problem)

March 22, 2015 Chebychev antenna design (problem)