## E and H plane directivities

In [2] directivities associated with the half power beamwidths are given as

\label{eqn:taiAndPereira:20}
D_1 = \frac{\Abs{E_\theta}^2_{\textrm{max}}}{\inv{2} \int_0^\pi \Abs{E_\theta(\theta, 0)}^2 \sin\theta d\theta}

\label{eqn:taiAndPereira:40}
D_2 = \frac{\Abs{E_\phi}^2_{\textrm{max}}}{\inv{2} \int_0^\pi \Abs{E_\phi(\theta, \pi/2)}^2 \sin\theta d\theta},

whereas [1] lists these as

\label{eqn:taiAndPereira:60}
\inv{D_1} = \inv{2 \ln 2} \int_0^{\Theta_{1 r}/2} \sin\theta d\theta

\label{eqn:taiAndPereira:80}
\inv{D_2} = \inv{2 \ln 2} \int_0^{\Theta_{2 r}/2} \sin\theta d\theta.

where the total directivity is given by the associated arithmetic mean formula

\label{eqn:taiAndPereira:160}
\inv{D_0} = \inv{2}\lr{\inv{D_1} + \inv{D_2}}.

This should follow from the far field approximation formula for $$U$$. I intended to derive that result, but haven’t gotten to it. What follows instead are a few associated notes from a read of the paper, which I may revisit later to complete.

## Short horizontal electrical dipole

### Problem

In [2] a field for which directivities can be calculated exactly was used in comparisons of some directivity approximations

\label{eqn:taiAndPereira:140}
\BE = E_0 \lr{ \cos\theta \cos\phi \thetacap – \sin\phi \phicap }.

(Observe that an inverse radial dependence in $$E_0$$ must be implied here for this to be a valid far-field representation of the field.)

Show that Tai & Pereira’s formula gives $$D_1 = 3$$, and $$D_2 = 1$$ respectively for this field.

Calculate the exact directivity for this field.

The field components are

\label{eqn:taiAndPereira:180}
E_\theta = E_0 \cos\theta \cos\phi

\label{eqn:taiAndPereira:200}
E_\phi = -E_0 \sin\phi

Using \ref{eqn:taiAndPereira:10} from the paper, the directivities are

\label{eqn:taiAndPereira:220}
D_1 = \frac{2}{\int_0^\pi \cos^2 \theta \sin\theta d\theta}
= \frac{2}{\evalrange{-\inv{3}\cos^3\theta}{0}{\pi}}
= 3,

and

\label{eqn:taiAndPereira:240}
D_2
= \frac{2}{\int_0^\pi \sin\theta d\theta}
= \frac{2}{\evalrange{-\cos\theta}{0}{\pi}}
= 1.

To find the exact directivity, first the Poynting vector is required. That is

\label{eqn:taiAndPereira:260}
\begin{aligned}
\BP
&= \frac{
\Abs{E_0}^2
}{2 c \mu_0}
\lr{ \cos\theta \cos\phi \thetacap – \sin\phi \phicap }
\cross
\lr{ \rcap \cross \lr{ \cos\theta \cos\phi \thetacap – \sin\phi \phicap } } \\
&= \frac{
\Abs{E_0}^2
}{ 2 c \mu_0}
\lr{ \cos\theta \cos\phi \thetacap – \sin\phi \phicap }
\cross
\lr{ \cos\theta \cos\phi \phicap + \sin\phi \thetacap } \\
&= \frac{
\Abs{E_0}^2 \rcap
}{2 c \mu_0}
\lr{ \cos^2\theta \cos^2\phi + \sin^2\phi },
\end{aligned}

\label{eqn:taiAndPereira:280}
U(\theta, \phi) \propto \cos^2\theta \cos^2\phi + \sin^2\phi.

The $$\thetacap$$, and $$\phicap$$ contributions to this intensity, and the total intensity are all plotted in fig. 1, fig. 2, and fig. 3 respectively.

fig 1. The theta direction contribution to the radiation intensity.

fig 2. The phi direction contribution to the radiation intensity.

fig 3. Radiation intensity (both theta and phi direction contributions).

Given this the total radiated power is

\label{eqn:taiAndPereira:300}
\lr{ \cos^2\theta \cos^2\phi + \sin^2\phi } \sin\theta d\theta d\phi
= \frac{8 \pi}{3}.

Observe that the radiation intensity $$U$$ can also be decomposed into two components, one for each component of the original $$\BE$$ phasor.

\label{eqn:taiAndPereira:320}
U_\theta = \cos^2 \theta \cos^2 \phi

\label{eqn:taiAndPereira:340}
U_\phi = \sin^2 \phi

This decomposition allows for expression of the partial directivities in these respective (orthogonal) directions

\label{eqn:taiAndPereira:360}
D_\theta = \frac{4 \pi U_\theta}{P_{\textrm{rad}}} = \frac{3}{2} \cos^2 \theta \cos^2 \phi

\label{eqn:taiAndPereira:380}
D_\phi = \frac{4 \pi U_\phi}{P_{\textrm{rad}}} = \frac{3}{2} \sin^2 \phi

The maximum of each of these partial directivities is both $$3/2$$, giving a maximum directivity of

\label{eqn:taiAndPereira:400}
D_0 =
\evalbar{D_\theta}{{\textrm{max}}}
+\evalbar{D_\phi}{{\textrm{max}}} = 3,

the exact value from the paper.

# References

[1] Constantine A Balanis. Antenna theory: analysis and design. John Wiley & Sons, 3rd edition, 2005.

[2] C-T Tai and CS Pereira. An approximate formula for calculating the directivity of an antenna. IEEE Transactions on Antennas and Propagation, 24:235, 1976.

## Maxwell’s equations review (plus magnetic sources and currents)

These are notes for the UofT course ECE1229, Advanced Antenna Theory, taught by Prof. Eleftheriades, covering ch. 3 [1] content.

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 that match the textbook so closely, there is little value to me taking notes that just replicate the text. Instead, I am annotating my copy of textbook with little details instead. My usual notes collection for the class will contain 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.)

## Maxwell’s equation review

For reasons that are yet to be seen (and justified), we work with a generalization of Maxwell’s equations to include
electric AND magnetic charge densities.

\label{eqn:chapter3Notes:20}
\spacegrad \cross \boldsymbol{\mathcal{E}} = – \boldsymbol{\mathcal{M}} – \PD{t}{\boldsymbol{\mathcal{B}}}

\label{eqn:chapter3Notes:40}
\spacegrad \cross \boldsymbol{\mathcal{H}} = \boldsymbol{\mathcal{J}} + \PD{t}{\boldsymbol{\mathcal{D}}}

\label{eqn:chapter3Notes:60}

\label{eqn:chapter3Notes:80}

Assuming a phasor relationships of the form $$\boldsymbol{\mathcal{E}} = \text{Real} \lr{ \BE(\Br) e^{j \omega t}}$$ for the fields and the currents, these reduce to

\label{eqn:chapter3Notes:100}
\spacegrad \cross \BE = – \BM – j \omega \BB

\label{eqn:chapter3Notes:120}
\spacegrad \cross \BH = \BJ + j \omega \BD

\label{eqn:chapter3Notes:140}

\label{eqn:chapter3Notes:160}

In engineering the fields

• $$\BE$$ : Electric field intensity (V/m, Volts/meter).
• $$\BH$$ : Magnetic field intensity (A/m, Amperes/meter).

are designated primary fields, whereas

• $$\BD$$ : Electric flux density (or displacement vector) (C/m, {Coulombs/meter).
• $$\BB$$ : Magnetic flux density (W/m, Webers/meter).

are designated the induced fields. The currents and charges are

• $$\BJ$$ : Electric current density (A/m).
• $$\BM$$ : Magnetic current density (V/m).
• $$\rho$$ : Electric charge density (C/m^3).
• $$\rho_m$$ : Magnetic charge density (W/m^3).

Because $$\spacegrad \cdot \lr{ \spacegrad \cross \Bf } = 0$$ for any
(sufficiently continuous) vector $$\Bf$$, divergence relations between the
currents and the charges follow from \ref{eqn:chapter3Notes:100}…

\label{eqn:chapter3Notes:180}
0
= -\spacegrad \cdot \BM – j \omega \rho_m,

and

\label{eqn:chapter3Notes:200}
0
= \spacegrad \cdot \BJ + j \omega \rho,

These are the phasor forms of the continuity equations

\label{eqn:chapter3Notes:220}
\spacegrad \cdot \BM = – j \omega \rho_m

\label{eqn:chapter3Notes:240}
\spacegrad \cdot \BJ = -j \omega \rho.

### Integral forms

The integral forms of Maxwell’s equations follow from Stokes’ theorem and the divergence theorems. Stokes’ theorem is a relation between the integral of the curl and the outwards normal differential area element of a surface, to the boundary of that surface, and applies to any surface with that boundary

\label{eqn:chapter3Notes:260}
\iint
= \oint \Bf \cdot d\Bl.

The divergence theorem, a special case of the general Stokes’ theorem is

\label{eqn:chapter3Notes:280}
= \iint_{\partial V} \Bf \cdot d\BA,

where the integral is over the surface of the volume, and the area element of the bounding integral has an outwards normal orientation.

See [5] for a derivation of this and various generalizations.

Applying these to Maxwell’s equations gives

\label{eqn:chapter3Notes:320}
\oint d\Bl \cdot \BE = –
\iint d\BA \cdot \lr{
\BM + j \omega \BB
}

\label{eqn:chapter3Notes:340}
\oint d\Bl \cdot \BH =
\iint d\BA \cdot \lr{
\BJ + j \omega \BD
}

\label{eqn:chapter3Notes:360}
\iint_{\partial V} d\BA \cdot \BD = \iiint \rho dV

\label{eqn:chapter3Notes:380}
\iint_{\partial V} d\BA \cdot \BB = \iiint \rho_m dV

## Constitutive relations

For linear isotropic homogeneous materials, the following constitutive relations apply

• $$\BD = \epsilon \BE$$
• $$\BB = \mu \BH$$
• $$\BJ = \sigma \BE$$, Ohm’s law.

where

• $$\epsilon = \epsilon_r \epsilon_0$$, is the permutivity (F/m, Farads/meter ).
• $$\mu = \mu_r \mu_0$$, is the permeability (H/m, Henries/meter), $$\mu_0 = 4 \pi \times 10^{-7}$$.
• $$\sigma$$, is the conductivity ($$\inv{\Omega m}$$, where $$1/\Omega$$ is a Siemens.)

In AM radio, will see ferrite cores with the inductors, which introduces non-unit $$\mu_r$$. This is to increase the radiation resistance.

## Boundary conditions

For good electric conductor $$\BE = 0$$.
For good magnetic conductor $$\BB = 0$$.

(more on class slides)

## Linear time invariant

Linear time invariant meant that the impulse response $$h(t,t’)$$ was a function of just the difference in times $$h(t,t’) = h(t-t’)$$.

## Green’s functions

For electromagnetic problems the impulse function sources $$\delta(\Br – \Br’)$$ also has a direction, and can yield any of $$E_x, E_y, E_z$$. A tensor impulse response is required.

Some overview of an approach that uses such tensor Green’s functions is outlined on the slides. It gets really messy since we require four tensor Green’s functions to handle electric and magnetic current and charges. Because of this complexity, we don’t go down this path, and use potentials instead.

In \S 3.5 [1] and the class notes, a verification of the spherical wave form for the Helmholtz Green’s function was developed. This was much simpler than the same verification I did in [4]. Part of the reason for that was that I worked in Cartesian coordinates, which made things much messier. The other part of the reason, for treating a neighbourhood of $$\Abs{\Br – \Br’} \sim 0$$, I verified the convolution, whereas Prof. Eleftheriades argues that a verification that $$\int \lr{\spacegrad^2 + k^2} G(\Br, \Br’) dV’ = 1$$ is sufficient. Balanis, on the other hand, argues that knowing the solution for $$k \ne 0$$ must just be the solution for $$k = 0$$ (i.e. the Poisson solution) provided it is multiplied by the $$e^{-j k r}$$ factor.

Note that back when I did that derivation, I used a different sign convention for the Green’s function, and in QM we used a positive sign instead of the negative in $$e^{-j k r }$$.

## Notation

• Phasor frequency terms are written as $$e^{j \omega t}$$, not $$e^{-j \omega t}$$, as done in physics. I didn’t recall that this was always the case in physics, and wouldn’t have assumed it. This is the case in both [3] and [2]. The latter however, also uses $$\cos(\omega t – k r)$$ for spherical waves possibly implying an alternate phasor sign convention in that content, so I’d be wary about trusting any absolute “engineering” vs. physics sign convention without checking carefully.
• In Green’s functions $$G(\Br, \Br’)$$, $$\Br$$ is the point of observation, and $$\Br’$$ is the point in the convolution integration space.
• Both $$\BM$$ and $$\BJ_m$$ are used for magnetic current sources in the class notes.

# References

[1] Constantine A Balanis. Antenna theory: analysis and design. John Wiley \& Sons, 3rd edition, 2005.

[2] David Jeffrey Griffiths and Reed College. Introduction to electrodynamics, chapter {Electromagnetic Waves}. Prentice hall Upper Saddle River, NJ, 3rd edition, 1999.

[3] JD Jackson. Classical Electrodynamics, chapter {Simple Radiating Systems, Scattering, and Diffraction}. John Wiley and Sons, 2nd edition, 1975.

[4] Peeter Joot. Quantum Mechanics II., chapter {Verifying the Helmholtz Green’s function.} peeterjoot.com, 2011. URL http://peeterjoot.com/archives/math2011/phy456.pdf. [Online; accessed 28-January-2015].

[5] Peeter Joot. Exploring physics with Geometric Algebra, chapter {Stokes theorem}. peeterjoot.com, 2014. URL http://peeterjoot.com/archives/math2009/gabook.pdf. [Online; accessed 28-January-2015].

## Polarization review

It seems worthwhile to review how a generally polarized field phasor leads to linear, circular, and elliptic geometries.

The most general field polarized in the $$x, y$$ plane has the form

\label{eqn:polarizationReview:20}
\BE
= \lr{ \xcap a e^{j \alpha} + \ycap b e^{j \beta} } e^{j \lr{ \omega t -k z }}
= \lr{ \xcap a e^{j \lr{\alpha – \beta}/2} + \ycap b e^{j \lr{ \beta – \alpha}/2} } e^{j \lr{ \omega t -k z + \lr{\alpha + \beta}/2 }}.

Knowing to factor out the average phase angle above is only because I tried initially without that and things got ugly and messy. I guessed this would help (it does).

Let $$\boldsymbol{\mathcal{E}} = \text{Re} \BE = \xcap x + \ycap y$$, $$\theta = \omega t + (\alpha + \beta)/2$$, and $$\phi = (\alpha – \beta)/2$$, so that

\label{eqn:polarizationReview:40}
\BE
= \lr{ \xcap a e^{j \phi} + \ycap b e^{-j \phi} } e^{j \theta }.

The coordinates can now be read off

\label{eqn:polarizationReview:60}
\frac{x}{a} = \cos\phi \cos\theta – \sin\phi \sin\theta

\label{eqn:polarizationReview:80}
\frac{y}{b} = \cos\phi \cos\theta + \sin\phi \sin\theta,

or in matrix form

\label{eqn:polarizationReview:100}
\begin{bmatrix}
x/a \\
y/b \\
\end{bmatrix}
=
\begin{bmatrix}
\cos\phi & – \sin\phi \\
\cos\phi & \sin\phi
\end{bmatrix}
\begin{bmatrix}
\cos\theta \\
\sin\theta
\end{bmatrix}.

The goal is to eliminate all the $$\theta$$ (i.e. time dependence), converting the parametric relationship into a conic form.
Assuming that neither $$\cos\theta$$, nor $$\sin\theta$$ are zero for now (those are special cases and lead to linear polarization), inverting the matrix will allow the $$\theta$$ dependence to be eliminated

\label{eqn:polarizationReview:120}
\inv{\sin\lr{ 2\phi }}
\begin{bmatrix}
\sin\phi & \sin\phi \\
– \cos\phi & \cos\phi
\end{bmatrix}
\begin{bmatrix}
x/a \\
y/b \\
\end{bmatrix}
=
\begin{bmatrix}
\cos\theta \\
\sin\theta
\end{bmatrix}.

Squaring and summing both rows of these equation gives
\label{eqn:polarizationReview:140}
\begin{aligned}
1
&=
\inv{\sin^2 \lr{ 2\phi}}
\lr{
\sin^2\phi
\lr{
\frac{x}{a}
+\frac{y}{b}
}^2
+
\cos^2\phi
\lr{
-\frac{x}{a}
+\frac{y}{b}
}^2
} \\
&=
\inv{\sin^2 \lr{ 2\phi}}
\lr{
\frac{x^2}{a^2}
+\frac{y^2}{b^2}
+2 \frac{x y}{a b} \lr{ \sin^2\phi – \cos^2\phi }
} \\
&=
\inv{\sin^2 \lr{ 2\phi}}
\lr{
\frac{x^2}{a^2}
+\frac{y^2}{b^2}
-2 \frac{x y}{a b} \cos \lr{2\phi}
}
\end{aligned}

Time to summarize and handle the special cases.

1. To have $$\cos\phi = 0$$, the phase angles must satisfy $$\alpha – \beta = \lr{ 1 + 2 k } \pi, \, k \in \mathbb{Z}$$.

For this case \ref{eqn:polarizationReview:50} reduces to

\label{eqn:polarizationReview:160}
-\frac{x}{a} = \frac{y}{b},

which is just a line.

### Example.

Let $$\alpha = 0, \beta = -\pi$$, so that the phasor has the value

\label{eqn:polarizationReview:260}
\BE = \lr{ \xcap a – \ycap b } e^{j \omega t}

2. For have $$\sin\phi = 0$$, the phase angles must satisfy $$\alpha – \beta = 2 \pi k, \, k \in \mathbb{Z}$$.

For this case \ref{eqn:polarizationReview:60} and \ref{eqn:polarizationReview:80} reduce to

\label{eqn:polarizationReview:180}
\frac{x}{a} = \frac{y}{b},

also just a line.

### Example.

Let $$\alpha = \beta = 0$$, so that the phasor has the value

\label{eqn:polarizationReview:280}
\BE = \lr{ \xcap a + \ycap b } e^{j \omega t}

3. Last is the circular and elliptically polarized case. The system is clearly elliptically polarized if $$\cos(2 \phi) = 0$$, or $$\alpha – \beta = (\pi/2)( 1 + 2 k ), k \in \mathbb{Z}$$. When that is the case and $$a = b$$ also holds, the ellipse is a circle.

When the $$\cos( 2 \phi) = 0$$ condition does not hold, a rotation of coordinates

\label{eqn:polarizationReview:200}
\begin{bmatrix}
x \\
y
\end{bmatrix}
=
\begin{bmatrix}
\cos\mu & \sin\mu \\
-\sin\mu & \cos\mu
\end{bmatrix}
\begin{bmatrix}
u \\
v
\end{bmatrix}

where

\label{eqn:polarizationReview:220}
\mu = \inv{2} \tan^{-1} \lr{ \frac{ 2 \cos (\alpha – \beta)}{b – a}}

puts the trajectory into a standard (but messy) conic form

\label{eqn:polarizationReview:240}
1 = \frac{u^2}{ab} \lr{
\frac{b}{a} \cos^2 \mu
+ \frac{a}{b} \sin^2 \mu
+ \inv{2} \sin\lr{2 \mu + \alpha – \beta}
}
+
\frac{v^2}{ab} \lr{
\frac{b}{a} \sin^2 \mu
+ \frac{a}{b} \cos^2 \mu
– \inv{2} \sin\lr{2 \mu + \alpha – \beta}
}

It isn’t obvious to me that the factors of the $$u^2, v^2$$ terms are necessarily positive, which is required for the conic to be an ellipse and not a hyperbola.

### Circular polarization example.

With $$a = b = E_0$$, $$\alpha = 0$$, $$\beta = \pm \pi/2$$, all the circular polarization conditions are met, leaving the phasor with values

\label{eqn:polarizationReview:300}
\BE = E_0 \lr{ \xcap \pm j \ycap } e^{j \omega t}

## Fundamental parameters of antennas

This is my first set of notes for the UofT course ECE1229, Advanced Antenna Theory, taught by Prof. Eleftheriades, covering ch. 2 [1] content.

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 that match the textbook so closely, there is little value to me taking notes that just replicate the text. Instead, I am annotating my copy of textbook with little details instead. My usual notes collection for the class will contain 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.)

## Poynting vector

The Poynting vector was written in an unfamiliar form

\label{eqn:chapter2Notes:560}
\boldsymbol{\mathcal{W}} = \boldsymbol{\mathcal{E}} \cross \boldsymbol{\mathcal{H}}.

I can roll with the use of a different symbol (i.e. not $$\BS$$) for the Poynting vector, but I’m used to seeing a $$\frac{c}{4\pi}$$ factor ([6] and [5]). I remembered something like that in SI units too, so was slightly confused not to see it here.

Per [3] that something is a $$\mu_0$$, as in

\label{eqn:chapter2Notes:580}
\boldsymbol{\mathcal{W}} = \inv{\mu_0} \boldsymbol{\mathcal{E}} \cross \boldsymbol{\mathcal{B}}.

Note that the use of $$\boldsymbol{\mathcal{H}}$$ instead of $$\boldsymbol{\mathcal{B}}$$ is what wipes out the requirement for the $$\frac{1}{\mu_0}$$ term since $$\boldsymbol{\mathcal{H}} = \boldsymbol{\mathcal{B}}/\mu_0$$, assuming linear media, and no magnetization.

It was mentioned that

U(\theta, \phi)
=
\frac{r^2}{2 \eta_0} \Abs{ \BE( r, \theta, \phi) }^2
=
\frac{1}{2 \eta_0} \lr{ \Abs{ E_\theta(\theta, \phi) }^2 + \Abs{ E_\phi(\theta, \phi) }^2},

where the intrinsic impedance of free space is

\eta_0 = \sqrt{\frac{\mu_0}{\epsilon_0}} = 377 \Omega.

(this is also eq. 2-19 in the text.)

To get an understanding where this comes from, consider the far field radial solutions to the electric and magnetic dipole problems, which have the respective forms (from [3]) of

\label{eqn:chapter2Notes:740}
\begin{aligned}
\boldsymbol{\mathcal{E}} &= -\frac{\mu_0 p_0 \omega^2 }{4 \pi } \frac{\sin\theta}{r} \cos\lr{w t – k r} \thetacap \\
\boldsymbol{\mathcal{B}} &= -\frac{\mu_0 p_0 \omega^2 }{4 \pi c} \frac{\sin\theta}{r} \cos\lr{w t – k r} \phicap \\
\end{aligned}

\label{eqn:chapter2Notes:760}
\begin{aligned}
\boldsymbol{\mathcal{E}} &= \frac{\mu_0 m_0 \omega^2 }{4 \pi c} \frac{\sin\theta}{r} \cos\lr{w t – k r} \phicap \\
\boldsymbol{\mathcal{B}} &= -\frac{\mu_0 m_0 \omega^2 }{4 \pi c^2} \frac{\sin\theta}{r} \cos\lr{w t – k r} \thetacap \\
\end{aligned}

In neither case is there a component in the direction of propagation, and in both cases (using $$\mu_0 \epsilon_0 = 1/c^2$$)

\label{eqn:chapter2Notes:780}
\Abs{\boldsymbol{\mathcal{H}}}
= \frac{\Abs{\boldsymbol{\mathcal{E}}}}{\mu_0 c}
= \Abs{\boldsymbol{\mathcal{E}}} \sqrt{\frac{\epsilon_0}{\mu_0}}
= \inv{\eta_0}\Abs{\boldsymbol{\mathcal{E}}} .

A superposition of the phasors for such dipole fields, in the far field, will have the form

\label{eqn:chapter2Notes:800}
\begin{aligned}
\BE &= \inv{r} \lr{ E_\theta(\theta, \phi) \thetacap + E_\phi(\theta, \phi) \phicap } \\
\BB &= \inv{r c} \lr{ E_\theta(\theta, \phi) \thetacap – E_\phi(\theta, \phi) \phicap },
\end{aligned}

with a corresponding time averaged Poynting vector

\label{eqn:chapter2Notes:820}
\begin{aligned}
\BW_{\textrm{av}}
&= \inv{2 \mu_0} \BE \cross \BB^\conj \\
&=
\inv{2 \mu_0 c r^2}
\lr{ E_\theta \thetacap + E_\phi \phicap } \cross
\lr{ E_\theta^\conj \thetacap – E_\phi^\conj \phicap } \\
&=
\frac{\thetacap \cross \phicap}{2 \mu_0 c r^2}
\lr{ \Abs{E_\theta}^2 + \Abs{E_\phi}^2 } \\
&=
\frac{\rcap}{2 \eta_0 r^2}
\lr{ \Abs{E_\theta}^2 + \Abs{E_\phi}^2 },
\end{aligned}

verifying \ref{eqn:advancedantennaL1:20} for a superposition of electric and magnetic dipole fields. This can likely be shown for more general fields too.

## Field plots

We can plot the fields, or intensity (or log plots in dB of these).
It is pointed out in [3] that when there is $$r$$ dependence these plots are done by considering the values of at fixed $$r$$.

The field plots are conceptually the simplest, since that vector parameterizes
a surface. Any such radial field with magnitude $$f(r, \theta, \phi)$$ can
be plotted in Mathematica in the $$\phi = 0$$ plane at $$r = r_0$$, or in
3D (respectively, but also at $$r = r_0$$) with code like that of the
following listing

Intensity plots can use the same code, with the only difference being the interpretation. The surface doesn’t represent the value of a vector valued radial function, but is the magnitude of a scalar valued function evaluated at $$f( r_0, \theta, \phi)$$.

The surfaces for $$U = \sin\theta, \sin^2\theta$$ in the plane are parametrically plotted in fig. 2, and for cosines in fig. 1 to compare with textbook figures.

Visualizations of $$U = \sin^2 \theta$$ and $$U = \cos^2 \theta$$ can be found in fig. 3 and fig. 4 respectively. Even for such simple functions these look pretty cool.

fig 3. Square sinusoidal radiation intensity

fig 4. Square cosinusoidal radiation intensity

## dB vs dBi

Note that dBi is used to indicate that the gain is with respect to an “isotropic” radiator.
This is detailed more in [2].

## Trig integrals

Tables 1.1 and 1.2 produced with tableOfTrigIntegrals.nb have some of the sine and cosine integrals that are pervasive in this chapter.

## Polarization vectors

The text introduces polarization vectors $$\rhocap$$ , but doesn’t spell out their form. Consider a plane wave field of the form

\label{eqn:chapter2Notes:840}
\BE
=
E_x e^{j \phi_x} e^{j \lr{ \omega t – k z }} \xcap
+
E_y e^{j \phi_y} e^{j \lr{ \omega t – k z }} \ycap.

The $$x, y$$ plane directionality of this phasor can be written

\label{eqn:chapter2Notes:860}
\Brho =
E_x e^{j \phi_x} \xcap
+
E_y e^{j \phi_y} \ycap,

so that

\label{eqn:chapter2Notes:880}
\BE = \Brho e^{j \lr{ \omega t – k z }}.

Separating this direction and magnitude into factors

\label{eqn:chapter2Notes:900}
\Brho = \Abs{\BE} \rhocap,

allows the phasor to be expressed as

\label{eqn:chapter2Notes:920}
\BE = \rhocap \Abs{\BE} e^{j \lr{ \omega t – k z }}.

As an example, suppose that $$E_x = E_y$$, and set $$\phi_x = 0$$. Then

\label{eqn:chapter2Notes:940}
\rhocap = \xcap + \ycap e^{j \phi_y}.

## Phasor power

In section 2.13 the phasor power is written as

\label{eqn:chapter2Notes:620}
I^2 R/2,

where $$I, R$$ are the magnitudes of phasors in the circuit.

I vaguely recall this relation, but had to refer back to [4] for the details.
This relation expresses average power over a period associated with the frequency of the phasor

\label{eqn:chapter2Notes:640}
\begin{aligned}
P
&= \inv{T} \int_{t_0}^{t_0 + T} p(t) dt \\
&= \inv{T} \int_{t_0}^{t_0 + T} \Abs{\BV} \cos\lr{ \omega t + \phi_V }
\Abs{\BI} \cos\lr{ \omega t + \phi_I} dt \\
&= \inv{T} \int_{t_0}^{t_0 + T} \Abs{\BV} \Abs{\BI}
\lr{
\cos\lr{ \phi_V – \phi_I } + \cos\lr{ 2 \omega t + \phi_V + \phi_I}
}
dt \\
&= \inv{2} \Abs{\BV} \Abs{\BI} \cos\lr{ \phi_V – \phi_I }.
\end{aligned}

Introducing the impedance for this circuit element

\label{eqn:chapter2Notes:660}
\BZ = \frac{ \Abs{\BV} e^{j\phi_V} }{ \Abs{\BI} e^{j\phi_I} } = \frac{\Abs{\BV}}{\Abs{\BI}} e^{j\lr{\phi_V – \phi_I}},

this average power can be written in phasor form

\label{eqn:chapter2Notes:680}
\BP = \inv{2} \Abs{\BI}^2 \BZ,

with
\label{eqn:chapter2Notes:700}
P = \textrm{Re} \BP.

Observe that we have to be careful to use the absolute value of the current phasor $$\BI$$, since $$\BI^2$$ differs in phase from $$\Abs{\BI}^2$$. This explains the conjugation in the [4] definition of complex power, which had the form

\label{eqn:chapter2Notes:720}
\BS = \BV_{\textrm{rms}} \BI^\conj_{\textrm{rms}}.

### Flat plate.

\label{eqn:chapter2Notes:960}
\sigma_{\textrm{max}} = \frac{4 \pi \lr{L W}^2}{\lambda^2}

fig. 6. Square geometry for RCS example.

### Sphere.

In the optical limit the radar cross section for a sphere

fig. 7. Sphere geometry for RCS example.

\label{eqn:chapter2Notes:980}
\sigma_{\textrm{max}} = \pi r^2

Note that this is smaller than the physical area $$4 \pi r^2$$.

### Cylinder.

fig. 8. Cylinder geometry for RCS example.

\label{eqn:chapter2Notes:1000}
\sigma_{\textrm{max}} = \frac{ 2 \pi r h^2}{\lambda}

### Tridedral corner reflector

fig. 9. Trihedral corner reflector geometry for RCS example.

\label{eqn:chapter2Notes:1020}
\sigma_{\textrm{max}} = \frac{ 4 \pi L^4}{3 \lambda^2}

## Scattering from a sphere vs frequency

Frequency dependence of spherical scattering is sketched in fig. 10.

• Low frequency (or small particles): Rayleigh\label{eqn:chapter2Notes:1040}
\sigma = \lr{\pi r^2} 7.11 \lr{\kappa r}^4, \qquad \kappa = 2 \pi/\lambda.
• Mie scattering (resonance),\label{eqn:chapter2Notes:1060}
\sigma_{\textrm{max}}(A) = 4 \pi r^2

\label{eqn:chapter2Notes:1080}
\sigma_{\textrm{max}}(B) = 0.26 \pi r^2.
• optical limit ( $$r \gg \lambda$$ )\label{eqn:chapter2Notes:1100}
\sigma = \pi r^2.

fig 10. Scattering from a sphere vs frequency (from Prof. Eleftheriades’ class notes).

FIXME: Do I have a derivation of this in my optics notes?

## Notation

• Time average.
and the text [1] use square brackets $$[\cdots]$$ for time averages, not $$<\cdots>$$. Was that an engineering convention?
writes $$\Omega$$ as a circle floating above a face up square bracket, as in fig. 1, and $$\sigma$$ like a number 6, as in fig. 1.
• Bold vectors are usually phasors, with (bold) calligraphic script used for the time domain fields. Example: $$\BE(x,y,z,t) = \ecap E(x,y) e^{j \lr{\omega t – k z}}, \boldsymbol{\mathcal{E}}(x, y, z, t) = \textrm{Re} \BE$$.

fig. 11. Prof. handwriting decoder ring: Omega

fig 12. Prof. handwriting decoder ring: sigma

# References

[1] Constantine A Balanis. Antenna theory: analysis and design. John Wiley \& Sons, 3rd edition, 2005.

[2] digi.com. Antenna Gain: dBi vs. dBd Decibel Detail, 2015. URL http://www.digi.com/support/kbase/kbaseresultdetl?id=2146. [Online; accessed 15-Jan-2015].

[3] David Jeffrey Griffiths and Reed College. Introduction to electrodynamics. Prentice hall Upper Saddle River, NJ, 3rd edition, 1999.

[4] J.D. Irwin. Basic Engineering Circuit Analysis. MacMillian, 1993.

[5] JD Jackson. Classical Electrodynamics. John Wiley and Sons, 2nd edition, 1975.

[6] L.D. Landau and E.M. Lifshitz. The classical theory of fields. Butterworth-Heinemann, 1980. ISBN 0750627689.

## Cancellation billing correspondence with Bell Canada

Jan 3, 2015.

From:

Account # xxxyyyzzz

Customer ID Number: xxxyyyzzzwww

Re: Disputing billing past cancellation.

To Whom it may concern,

I am writing to dispute the continued billing that I have received from Bell Canada, for services cancelled in August 2014.

During that second call I was told that not only could you not cancel the service until an effective date of September 30th, you would also bill me for this additional service, whether I wanted it or not. I was and amazed at the meaningless charges that were imposed on me, but by this time felt that I did not have much choice. You have a corporate policy to extort your customers should they desire to keep using your services, so the fact that you are also willing to impose the same type of extortion on customers that no longer wish to use your services is not surpising.

Most recent payment history:

I have records of having paid the $39.49 home phone bill for August 2014 on Sept 9, 2014, and have records of having paid the ($39.49) bill for the services that should have ended Sept 30, 2014 on October 10, 2014.

Failure to cancel:

The house with the undesired Bell service was due to be sold in early November, and to ensure the new owners did not arrive in the house with a functioning phone service in my name, I contacted Bell through your online contact service on October 28, to ensure that the services had been properly cancelled. I talked to Marianne at that point who stated “I was asking because your home phone service is still showing active on my end. If the people you spoke with are not the Loyalty team then it will be the reason why the cancellation was not process. What needs to be done now is to call 310-BELL (2355) 1-800-668-6878 form 10 am – 9 pm and look for the Loyalty team to process the cancellation. I will note the account regarding what happened so you won’t have to render the 30 days notice anymore. “

It appears that your company lost the cancellation request that I had made. This is the same cancellation request where I was told explicitly by a Bell representative that the service would be cancelled effective September 30 2014.

Your representative Oszman (6028990), a nasty fellow to deal with, full of scorn and impatience, did manage to cancelled the service for real on November 3, effective November 4th. I was not surprised to find that your company’s claim that services could not be cancelled immediately was a fabrication.

I do not know if the lost cancellation request was made with Bell “Loyalty” or some other Bell customer service division. I understand from others in customer service roles that there are sometimes penalties for representatives that are not able to retain customers by offering incentives. Perhaps my cancellation request was “lost” to avoid such a penalty?

I have no intention of paying any bills from Bell Canada past the bill paid on October 10 2014 for the September service that I did not desire. For the record, I also consider that September bill to be invalid, but have paid it assuming that in some fine print somewhere I had agreed to be victimized by your company should I wish to cancel the service.

Sincerely,

Peeter Joot

A copy of this correspondence will be made publicly available on the internet.