## Antenna array design with Chebychev polynomials

Prof. Eleftheriades desribed a Chebychev antenna array design method that looks different than the one of the text .

Portions of that procedure are like that of the text. For example, if a side lobe level of $$20 \log_{10} R$$ is desired, a scaling factor

\begin{equation}\label{eqn:chebychevSecondMethod:20}
x_0 = \cosh\lr{ \inv{m} \cosh^{-1} R },
\end{equation}

is used. Given $$N$$ elements in the array, a Chebychev polynomial of degree $$m = N – 1$$ is used. That is

\begin{equation}\label{eqn:chebychevSecondMethod:40}
T_m(x) = \cos\lr{ m \cos^{-1} x }.
\end{equation}

Observe that the roots $$x_n’$$ of this polynomial lie where

\begin{equation}\label{eqn:chebychevSecondMethod:60}
m \cos^{-1} x_n’ = \frac{\pi}{2} \pm \pi n,
\end{equation}

or

\begin{equation}\label{eqn:chebychevSecondMethod:80}
x_n’ = \cos\lr{ \frac{\pi}{2 m} \lr{ 2 n \pm 1 } },
\end{equation}

The class notes use the negative sign, and number $$n = 1,2, \cdots, m$$. It is noted that the roots are symmetric with $$x_1′ = – x_m’$$, which can be seen by direct expansion

\begin{equation}\label{eqn:chebychevSecondMethod:100}
\begin{aligned}
x_{m-r}’
&= \cos\lr{ \frac{\pi}{2 m} \lr{ 2 (m – r) – 1 } } \\
&= \cos\lr{ \pi – \frac{\pi}{2 m} \lr{ 2 r + 1 } } \\
&= -\cos\lr{ \frac{\pi}{2 m} \lr{ 2 r + 1 } } \\
&= -\cos\lr{ \frac{\pi}{2 m} \lr{ 2 ( r + 1 ) – 1 } } \\
&= -x_{r+1}’.
\end{aligned}
\end{equation}

The next step in the procedure is the identification

\begin{equation}\label{eqn:chebychevSecondMethod:120}
\begin{aligned}
u_n’ &= 2 \cos^{-1} \lr{ \frac{x_n’}{x_0} } \\
z_n &= e^{j u_n’}.
\end{aligned}
\end{equation}

This has a factor of two that does not appear in the Balanis design method. It seems plausible that this factor of two was introduced so that the roots of the array factor $$z_n$$ are conjugate pairs. Since $$\cos^{-1} (-z) = \pi – \cos^{-1} z$$, this choice leads to such conjugate pairs

\begin{equation}\label{eqn:chebychevSecondMethod:140}
\begin{aligned}
\exp\lr{j u_{m-r}’}
&=
\exp\lr{j 2 \cos^{-1} \lr{ \frac{x_{m-r}’}{x_0} } } \\
&=
\exp\lr{j 2 \cos^{-1} \lr{ -\frac{x_{r+1}’}{x_0} } } \\
&=
\exp\lr{j 2 \lr{ \pi – \cos^{-1} \lr{ \frac{x_{r+1}’}{x_0} } } } \\
&=
\exp\lr{-j u_{r+1}}.
\end{aligned}
\end{equation}

Because of this, the array factor can be written

\begin{equation}\label{eqn:chebychevSecondMethod:180}
\begin{aligned}
\textrm{AF}
&= ( z – z_1 )( z – z_2 ) \cdots ( z – z_{m-1} ) ( z – z_m ) \\
&=
( z – z_1 )( z – z_1^\conj )
( z – z_2 )( z – z_2^\conj )
\cdots \\
&=
\lr{ z^2 – z ( z_1 + z_1^\conj ) + 1 }
\lr{ z^2 – z ( z_2 + z_2^\conj ) + 1 }
\cdots \\
&=
\lr{ z^2 – 2 z \cos\lr{ 2 \cos^{-1} \lr{ \frac{x_1′}{x_0} } } + 1 }
\lr{ z^2 – 2 z \cos\lr{ 2 \cos^{-1} \lr{ \frac{x_2′}{x_0} } } + 1 }
\cdots \\
&=
\lr{ z^2 – 2 z \lr{ 2 \lr{ \frac{x_1′}{x_0} }^2 – 1 } + 1 }
\lr{ z^2 – 2 z \lr{ 2 \lr{ \frac{x_2′}{x_0} }^2 – 1 } + 1 }
\cdots
\end{aligned}
\end{equation}

When $$m$$ is even, there will only be such conjugate pairs of roots. When $$m$$ is odd, the remainding factor will be

\begin{equation}\label{eqn:chebychevSecondMethod:160}
\begin{aligned}
z – e^{2 j \cos^{-1} \lr{ 0/x_0 } }
&=
z – e^{2 j \pi/2} \\
&=
z – e^{j \pi} \\
&=
z + 1.
\end{aligned}
\end{equation}

However, with this factor of two included, the connection between the final array factor polynomial \ref{eqn:chebychevSecondMethod:180}, and the Chebychev polynomial $$T_m$$ is not clear to me. How does this scaling impact the roots?

### Example: Expand $$\textrm{AF}$$ for $$N = 4$$.

The roots of $$T_3(x)$$ are

\begin{equation}\label{eqn:chebychevSecondMethod:200}
x_n’ \in \setlr{0, \pm \frac{\sqrt{3}}{2} },
\end{equation}

so the array factor is

\begin{equation}\label{eqn:chebychevSecondMethod:220}
\begin{aligned}
\textrm{AF}
&=
\lr{ z^2 + z \lr{ 2 – \frac{3}{x_0^2} } + 1 }\lr{ z + 1 } \\
&=
z^3
+ 3 z^2 \lr{ 1 – \frac{1}{x_0^2} }
+ 3 z \lr{ 1 – \frac{1}{x_0^2} }
+ 1.
\end{aligned}
\end{equation}

With $$20 \log_{10} R = 30 \textrm{dB}$$, $$x_0 = 2.1$$, so this is

\begin{equation}\label{eqn:chebychevSecondMethod:240}
\textrm{AF} = z^3 + 2.33089 z^2 + 2.33089 z + 1.
\end{equation}

With

\begin{equation}\label{eqn:chebychevSecondMethod:260}
z = e^{j (u + u_0) } = e^{j k d \cos\theta + j k u_0 },
\end{equation}

the array factor takes the form

\begin{equation}\label{eqn:chebychevSecondMethod:280}
\textrm{AF}
=
e^{j 3 k d \cos\theta + j 3 k u_0 }
+ 2.33089
e^{j 2 k d \cos\theta + j 2 k u_0 }
+ 2.33089
e^{j k d \cos\theta + j k u_0 }
+ 1.
\end{equation}

This array function is highly phase dependent, plotted for $$u_0 = 0$$ in fig. 1, and fig. 2.

This can be directed along a single direction (z-axis) with higher phase choices as illustrated in fig. 3, and fig. 4.

These can be explored interactively in this Mathematica Manipulate panel.

# References

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

## Chebychev antenna array design

In our text  is a design procedure that applies Chebychev polynomials to the selection of current magnitudes for an evenly spaced array of identical antennas placed along the z-axis.

For an even number $$2 M$$ of identical antennas placed at positions $$\Br_m = (d/2) \lr{2 m -1} \Be_3$$, the array factor is

\begin{equation}\label{eqn:chebychevDesign:20}
\textrm{AF}
=
\sum_{m=-N}^N I_m e^{-j k \rcap \cdot \Br_m }.
\end{equation}

Assuming the currents are symmetric $$I_{-m} = I_m$$, with $$\rcap = (\sin\theta \cos\phi, \sin\theta \sin\phi, \cos\theta )$$, and $$u = \frac{\pi d}{\lambda} \cos\theta$$, this is

\begin{equation}\label{eqn:chebychevDesign:40}
\begin{aligned}
\textrm{AF}
&=
\sum_{m=-N}^N I_m e^{-j k (d/2) ( 2 m -1 )\cos\theta } \\
&=
2 \sum_{m=1}^N I_m \cos\lr{ k (d/2) ( 2 m -1)\cos\theta } \\
&=
2 \sum_{m=1}^N I_m \cos\lr{ (2 m -1) u }.
\end{aligned}
\end{equation}

This is a sum of only odd cosines, and can be expanded as a sum that includes all the odd powers of $$\cos u$$. Suppose for example that this is a four element array with $$N = 2$$. In this case the array factor has the form

\begin{equation}\label{eqn:chebychevDesign:60}
\begin{aligned}
\textrm{AF}
&=
2 \lr{ I_1 \cos u + I_2 \lr{ 4 \cos^3 u – 3 \cos u } } \\
&=
2 \lr{ \lr{ I_1 – 3 I_2 } \cos u + 4 I_2 \cos^3 u }.
\end{aligned}
\end{equation}

The design procedure in the text sets $$\cos u = z/z_0$$, and then equates this to $$T_3(z) = 4 z^3 – 3 z$$ to determine the current amplitudes $$I_m$$. That is

\begin{equation}\label{eqn:chebychevDesign:80}
\frac{ 2 I_1 – 6 I_2 }{z_0} z + \frac{8 I_2}{z_0^3} z^3 = -3 z + 4 z^3,
\end{equation}

or

\begin{equation}\label{eqn:chebychevDesign:100}
\begin{aligned}
\begin{bmatrix}
I_1 \\
I_2
\end{bmatrix}
&=
{\begin{bmatrix}
2/z_0 & -6/z_0 \\
0 & 8/z_0^3
\end{bmatrix}}^{-1}
\begin{bmatrix}
-3 \\
4
\end{bmatrix} \\
&=
\frac{z_0}{2}
\begin{bmatrix}
3 (z_0^2 -1) \\
z_0^2
\end{bmatrix}.
\end{aligned}
\end{equation}

The currents in the array factor are fully determined up to a scale factor, reducing the array factor to

\begin{equation}\label{eqn:chebychevDesign:140}
\textrm{AF} = 4 z_0^3 \cos^3 u – 3 z_0 \cos u.
\end{equation}

The zeros of this array factor are located at the zeros of

\begin{equation}\label{eqn:chebychevDesign:120}
T_3( z_0 \cos u ) = \cos( 3 \cos^{-1} \lr{ z_0 \cos u } ),
\end{equation}

which are at $$3 \cos^{-1} \lr{ z_0 \cos u } = \pi/2 + m \pi = \pi \lr{ m + \inv{2} }$$

\begin{equation}\label{eqn:chebychevDesign:160}
\cos u = \inv{z_0} \cos\lr{ \frac{\pi}{3} \lr{ m + \inv{2} } } = \setlr{ 0, \pm \frac{\sqrt{3}}{2 z_0} }.
\end{equation}

showing that the scaling factor $$z_0$$ effects the locations of the zeros. It also allows the values at the extremes $$\cos u = \pm 1$$, to increase past the $$\pm 1$$ non-scaled limit values. These effects can be explored in this Mathematica notebook, but can also be seen in fig. 1.

The scale factor can be fixed for a desired maximum power gain. For $$R \textrm{dB}$$, that will be when

\begin{equation}\label{eqn:chebychevDesign:180}
20 \log_{10} \cosh( 3 \cosh^{-1} z_0 ) = R \textrm{dB},
\end{equation}

or

\begin{equation}\label{eqn:chebychevDesign:200}
z_0 = \cosh \lr{ \inv{3} \cosh^{-1} \lr{ 10^{\frac{R}{20}} } }.
\end{equation}

For $$R = 30$$ dB (say), we have $$z_0 = 2.1$$, and

\begin{equation}\label{eqn:chebychevDesign:220}
\textrm{AF}
= 40 \cos^3 \lr{ \frac{\pi d}{\lambda} \cos\theta } – 6.4 \cos \lr{ \frac{\pi d}{\lambda} \cos\theta }.
\end{equation}

These are plotted in fig. 2 (linear scale), and fig. 3 (dB scale) for a couple values of $$d/\lambda$$.

To explore the $$d/\lambda$$ dependence try this Mathematica notebook.

# References

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

## Questions about bill C-51 to my “representative” John McCallum in parliment.

Hi John,

As a new home owner in the Markham-Unionville riding it appears you are my representative in parliament. I have some questions about the Canadian Bill C-51, which appears is being carried along with the spree of fear porn that the media is pushing after the recent shooting at the capital.

I would like to know the following:

– Who are the specific authors of this bill? If they were bureaucrats and lawyers that were not members of parliament themselves, who were the members of parliament that backed their work?

– Who are the primary financial backers of the members of parliament that either wrote or supported the writing of this bill?

– Are there any known copies of this bill that precede the capital shooting? If so, the same questions apply to the authors or supporters of those bills.

Sincerely,

Peeter Joot

A copy of this letter and any associated correspondence will be made publicly available.

## Updated notes for ece1229 antenna theory I’ve now posted a first 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).

The notes compilation linked above includes all of the following separate notes, some of which have been posted separately on this blog:

## Image theorem

In the last problem set we examined the array factor for a corner cube configuration, shown in fig. 1.

### Motivation

This is a horizontal dipole antenna placed next to a metallic corner. The radiation at points in the interior of the cube have contributions due to the line of sight field from the antenna as well as reflections. We looked at an approximation of ground reflections using the \underlineAndIndex{Image Theorem}, modeling the ground as a perfectly conducting surface. I completely misunderstood that theorem and how it should be applied. As presented it seemed like a simple way to figure out the reflection characteristics. This confused me since it did not seem consistent with Fresnel reflection theory. I did try to reconcile to the two, but that reconciliation only appeared to work for certain dipole orientations, and that orientation dependence remained an open question.

It turns out that the idea of the Image Theorem is to find a source configuration that contains the specified source, but contains enough other sources that the tangential component of the electric field superposition is zero on the conducting surface, as required by Maxwell’s equations. This allows the boundary to be completely removed from the problem.

Thinking of the corner cube configuration as a reflection problem, I positioned sources as in fig. 2.

Because of the horizontal orientation of the dipole, I argued that the reflection coefficient should be -1. The reflection point is a bit messy to calculate, and it turns out to zeroth order in $$h/r$$ the $$\sin\theta$$ magnitude scaling of the reflected (far-field) field is present for both reflected rays. I though that this was probably because the observation point lays at the same altitude for both the line of sight ray and the reflected ray.

Attempting this problem as a reflection problem makes it much more difficult than it needs to be. It turns out that the correct image source placement for this problem is that of fig. 3.

This wasn’t at all obvious to me. The key is understanding that the goal of the image source placement isn’t to figure out how the reflection will occur, but to manufacture a source configuration for which the tangential component of the electric field is zero on the conducting surface.

### Image placement for infinite conducting plane.

Before thinking about the corner cube configuration, consider a horizontal dipole next to an infinite conducting plane. This, and the correct image source placement is illustrated in fig. 4.

I’ll now verify that this is the correct image source. This is basically a calculation that the tangential components of the electric fields from both sources sum to zero.

Let,

\begin{equation}\label{eqn:imageTheorem:20}
r = \Abs{\Bs – \Br_0},
\end{equation}

so that the magnetic vector potential for the first quadrant dipole has the form

\begin{equation}\label{eqn:imageTheorem:40}
\BA = \frac{A_0}{4 \pi r} e^{-j k r} \zcap.
\end{equation}

With

\begin{equation}\label{eqn:imageTheorem:60}
\begin{aligned}
\kcap &= \frac{\Bs – \Br_0}{s} \\
\tilde{\BE} &= \zcap – \lr{\zcap \cdot \kcap} \kcap,
\end{aligned}
\end{equation}

the far-field electric field at the point $$\Bs$$ on the plane is

\begin{equation}\label{eqn:imageTheorem:80}
\BE = -j \omega \frac{A_0}{4 \pi r} e^{-j k r} \tilde{\BE}.
\end{equation}

If the normal to the plane is $$\ncap$$ the tangential component of this field is the projection of $$\BE$$ on the direction

\begin{equation}\label{eqn:imageTheorem:100}
\pcap = \frac{\kcap \cross \ncap}{\Abs{\kcap \cross \ncap}}.
\end{equation}

That tangential component is directed along

\begin{equation}\label{eqn:imageTheorem:120}
\lr{\tilde{\BE} \cdot \pcap } \pcap
=
\lr{\lr{\zcap – \lr{\zcap \cdot \kcap} \kcap} \cdot \lr{\kcap \cross \ncap}} \frac{\kcap \cross \ncap}{\Abs{\kcap \cross \ncap}^2}.
\end{equation}

Because the triple product $$\kcap \cdot \lr{\kcap \cross \ncap} = 0$$, the tangential component of the electric field, provided $$\kcap \cdot \ncap \ne 0$$, is

\begin{equation}\label{eqn:imageTheorem:140}
\BE_\parallel
=
-j \omega \frac{A_0}{4 \pi r} e^{-j k r} \zcap \cdot \lr{\kcap \cross \ncap} \frac{\kcap \cross \ncap}{ 1 – \lr{ \ncap \cdot \kcap }^2 }.
\end{equation}

Now the wave vector direction for the second quadrant ray on the plane is required. Both $$\kcap’$$ and $$\Bs’$$ are reflections across the plane. Any such reflection has the value

\begin{equation}\label{eqn:imageTheorem:160}
\begin{aligned}
\Bx’
&= \lr{ \Bx \wedge \ncap} \ncap – \lr{ \Bx \cdot \ncap } \ncap \\
&= – \lr{ \ncap \wedge \Bx + \ncap \cdot \Bx } \ncap \\
&= – \ncap \Bx \ncap.
\end{aligned}
\end{equation}

This multivector product nicely encapsulates the reflection operation. Consider a reflection against the y-z plane with normal $$\Be_1$$ to verify that this works

\begin{equation}\label{eqn:imageTheorem:180}
\begin{aligned}
-\Be_1 \Bx \Be_1
&=
-\Be_1 \lr{ x \Be_1 + y \Be_2 + z \Be_3 } \Be_1 \\
&=
-\lr{ x – y \Be_2 \Be_1 + z \Be_3 \Be_1 } \Be_1 \\
&=
-\lr{ x \Be_1 – y \Be_2 + z \Be_3 } \\
&=
– x \Be_1 + y \Be_2 + z \Be_3.
\end{aligned}
\end{equation}

This has the x component flipped in sign and the rest left untouched as desired for a reflection in the y-z plane.

The second quadrant field will have $$\kcap’ \cross \ncap$$ terms in place of all the $$\kcap \cross \ncap$$ terms of \ref{eqn:imageTheorem:140}. We want to know how the two compare. This calculation is simply done using the dual form of the cross product temporarily

\begin{equation}\label{eqn:imageTheorem:200}
\begin{aligned}
\kcap’ \cross \ncap
&=
-I \lr{ \kcap’ \wedge \ncap} \\
&=
&=
-I \gpgradetwo{ {-\ncap \kcap \ncap} \ncap} \\
&=
I \gpgradetwo{ \ncap \kcap } \\
&=
I \ncap \wedge \kcap \\
&=
-\ncap \cross \kcap \\
&=
\kcap \cross \ncap.
\end{aligned}
\end{equation}

So, provided the image source in the second quadrant is oppositely oriented (sign inversion), the tangential components of the two will sum to zero on that surface.

Thinking back to the corner cube, it is clear that an image source opposite to the source across from one of the walls will result in a zero tangential electric field along this boundary as is the case here (say the y-z plane). A second pair of sources opposite from each other anywhere else also about the y-z plane will not change that zero tangential electric field on this surface, but if the signs of the sources is alternated as in fig. 3 it will also result in zero tangential electric field on the z-x plane, which has the desired boundary value effects for both surfaces of the corner cube.

## Tschebyscheff polynomials

In ancient times (i.e. 2nd year undergrad) I recall being very impressed with Tschebyscheff polynomials for designing lowpass filters. I’d used Tschebyscheff filters for the hardware we used for a speech recognition system our group built in the design lab. One of the benefits of these polynomials is that the oscillation in the $$\Abs{x} < 1$$ interval is strictly bounded. This same property, as well as the unbounded nature outside of the $$[-1,1]$$ interval turns out to have applications to antenna array design.

The Tschebyscheff polynomials are defined by

\begin{equation}\label{eqn:tschebyscheff:40}
T_m(x) = \cos\lr{ m \cos^{-1} x }, \quad \Abs{x} < 1 \end{equation} \begin{equation}\label{eqn:tschebyscheff:60} T_m(x) = \cosh\lr{ m \cosh^{-1} x }, \quad \Abs{x} > 1.
\end{equation}

### Range restrictions and hyperbolic form.

Prof. Eleftheriades’s notes made a point to point out the definition in the $$\Abs{x} > 1$$ interval, but that can also be viewed as a consequence instead of a definition if the range restriction is removed. For example, suppose $$x = 7$$, and let

\begin{equation}\label{eqn:tschebyscheff:160}
\cos^{-1} 7 = \theta,
\end{equation}

so
\begin{equation}\label{eqn:tschebyscheff:180}
\begin{aligned}
7
&= \cos\theta \\
&= \frac{e^{i\theta} + e^{-i\theta}}{2} \\
&= \cosh(i\theta),
\end{aligned}
\end{equation}

or

\begin{equation}\label{eqn:tschebyscheff:200}
-i \cosh^{-1} 7 = \theta.
\end{equation}

\begin{equation}\label{eqn:tschebyscheff:220}
\begin{aligned}
T_m(7)
&= \cos( -m i \cosh^{-1} 7 ) \\
&= \cosh( m \cosh^{-1} 7 ).
\end{aligned}
\end{equation}

The same argument clearly applies to any other value outside of the $$\Abs{x} < 1$$ range, so without any restrictions, these polynomials can be defined as just

\begin{equation}\label{eqn:tschebyscheff:260}
\boxed{
T_m(x) = \cos\lr{ m \cos^{-1} x }.
}
\end{equation}

### Polynomial nature.

Eq. \ref{eqn:tschebyscheff:260} does not obviously look like a polynomial. Let’s proceed to verify the polynomial nature for the first couple values of $$m$$.

• $$m = 0$$.\begin{equation}\label{eqn:tschebyscheff:280}
\begin{aligned}
T_0(x)
&= \cos( 0 \cos^{-1} x ) \\
&= \cos( 0 ) \\
&= 1.
\end{aligned}
\end{equation}
• $$m = 1$$.\begin{equation}\label{eqn:tschebyscheff:300}
\begin{aligned}
T_1(x)
&= \cos( 1 \cos^{-1} x ) \\
&= x.
\end{aligned}
\end{equation}
• $$m = 2$$.\begin{equation}\label{eqn:tschebyscheff:320}
\begin{aligned}
T_2(x)
&= \cos( 2 \cos^{-1} x ) \\
&= 2 \cos^2 \cos^{-1}(x) – 1 \\
&= 2 x^2 – 1.
\end{aligned}
\end{equation}

To examine the general case

\begin{equation}\label{eqn:tschebyscheff:340}
\begin{aligned}
T_m(x)
&= \cos( m \cos^{-1} x ) \\
&= \textrm{Re} e^{ j m \cos^{-1} x } \\
&= \textrm{Re} \lr{ e^{ j\cos^{-1} x } }^m \\
&= \textrm{Re} \lr{ \cos\cos^{-1} x + j \sin\cos^{-1} x }^m \\
&= \textrm{Re} \lr{ x + j \sqrt{1 – x^2} }^m \\
&=
\textrm{Re} \lr{
x^m
+ \binom{ m}{1} j x^{m-1} \lr{1 – x^2}^{1/2}
– \binom{ m}{2} x^{m-2} \lr{1 – x^2}^{2/2}
– \binom{ m}{3} j x^{m-3} \lr{1 – x^2}^{3/2}
+ \binom{ m}{4} x^{m-4} \lr{1 – x^2}^{4/2}
+ \cdots
} \\
&=
x^m
– \binom{ m}{2} x^{m-2} \lr{1 – x^2}
+ \binom{ m}{4} x^{m-4} \lr{1 – x^2}^2
– \cdots
\end{aligned}
\end{equation}

This expansion was a bit cavaliar with the signs of the $$\sin\cos^{-1} x = \sqrt{1 – x^2}$$ terms, since the negative sign should be picked for the root when $$x \in [-1,0]$$. However, that doesn’t matter in the end since the real part operation selects only powers of two of this root.

The final result of the expansion above can be written

\begin{equation}\label{eqn:tschebyscheff:360}
\boxed{
T_m(x) = \sum_{k = 0}^{\lfloor m/2 \rfloor} \binom{m}{2 k} (-1)^k x^{m – 2 k} \lr{1 – x^2}^k.
}
\end{equation}

This clearly shows the polynomial nature of these functions, and is also perfectly well defined for any value of $$x$$. The even and odd alternation with $$m$$ is also clear in this explicit expansion.

### Plots ### Properties

In  a few properties can be found for these polynomials

\begin{equation}\label{eqn:tschebyscheff:100}
T_m(x) = 2 x T_{m-1} – T_{m-2}
\end{equation}
\begin{equation}\label{eqn:tschebyscheff:420}
0 = \lr{ 1 – x^2 } \frac{d T_m(x)}{dx} + m x T_m(x) – m T_{m-1}(x)
\end{equation}
\begin{equation}\label{eqn:tschebyscheff:400}
0 = \lr{ 1 – x^2 } \frac{d^2 T_m(x)}{dx^2} – x \frac{dT_m(x)}{dx} + m^2 T_{m}(x)
\end{equation}
\begin{equation}\label{eqn:tschebyscheff:440}
\int_{-1}^1 \inv{ \sqrt{1 – x^2} } T_m(x) T_n(x) dx =
\left\{
\begin{array}{l l}
0 & \quad \mbox{if $$m \ne n$$ } \\
\pi & \quad \mbox{if $$m = n = 0$$ } \\
\pi/2 & \quad \mbox{if $$m = n, m \ne 0$$ }
\end{array}
\right.
\end{equation}

### Recurrance relation.

Prove \ref{eqn:tschebyscheff:100}.

To show this, let

\begin{equation}\label{eqn:tschebyscheff:460}
x = \cos\theta.
\end{equation}

\begin{equation}\label{eqn:tschebyscheff:580}
2 x T_{m-1} – T_{m-2}
=
2 \cos\theta \cos((m-1) \theta) – \cos((m-2)\theta).
\end{equation}

\begin{equation}\label{eqn:tschebyscheff:540}
\begin{aligned}
\cos( a + b )
&=
\textrm{Re} e^{j(a + b)} \\
&=
\textrm{Re} e^{ja} e^{jb} \\
&=
\textrm{Re}
\lr{ \cos a + j \sin a }
\lr{ \cos b + j \sin b } \\
&=
\cos a \cos b – \sin a \sin b.
\end{aligned}
\end{equation}

Applying this gives

\begin{equation}\label{eqn:tschebyscheff:600}
\begin{aligned}
2 x T_{m-1} – T_{m-2}
&=
2 \cos\theta \Biglr{ \cos(m\theta)\cos\theta +\sin(m\theta) \sin\theta }
– \Biglr{
\cos(m\theta)\cos(2\theta) + \sin(m\theta) \sin(2\theta)
} \\
&=
2 \cos\theta \Biglr{ \cos(m\theta)\cos\theta +\sin(m\theta)\sin\theta) }
– \Biglr{
\cos(m\theta)(\cos^2 \theta – \sin^2 \theta) + 2 \sin(m\theta) \sin\theta \cos\theta
} \\
&=
\cos(m\theta) \lr{ \cos^2\theta + \sin^2\theta } \\
&= T_m(x).
\end{aligned}
\end{equation}

### First order LDE relation.

Prove \ref{eqn:tschebyscheff:420}.

To show this, again, let

\begin{equation}\label{eqn:tschebyscheff:470}
x = \cos\theta.
\end{equation}

Observe that

\begin{equation}\label{eqn:tschebyscheff:480}
1 = -\sin\theta \frac{d\theta}{dx},
\end{equation}

so

\begin{equation}\label{eqn:tschebyscheff:500}
\begin{aligned}
\frac{d}{dx}
&= \frac{d\theta}{dx} \frac{d}{d\theta} \\
&= -\frac{1}{\sin\theta} \frac{d}{d\theta}.
\end{aligned}
\end{equation}

Plugging this in gives

\begin{equation}\label{eqn:tschebyscheff:520}
\begin{aligned}
\lr{ 1 – x^2} &\frac{d}{dx} T_m(x) + m x T_m(x) – m T_{m-1}(x) \\
&=
\sin^2\theta
\lr{
-\frac{1}{\sin\theta} \frac{d}{d\theta}}
\cos( m \theta ) + m \cos\theta \cos( m \theta ) – m \cos( (m-1)\theta ) \\
&=
-\sin\theta (-m \sin(m \theta)) + m \cos\theta \cos( m \theta ) – m \cos( (m-1)\theta ).
\end{aligned}
\end{equation}

Applying the cosine addition formula \ref{eqn:tschebyscheff:540} gives

\begin{equation}\label{eqn:tschebyscheff:560}
m \lr{ \sin\theta \sin(m \theta) + \cos\theta \cos( m \theta ) } – m
\lr{
\cos( m \theta) \cos\theta + \sin( m \theta ) \sin\theta
}
=0.
\end{equation}

### Second order LDE relation.

Prove \ref{eqn:tschebyscheff:400}.

This follows the same way. The first derivative was

\begin{equation}\label{eqn:tschebyscheff:640}
\begin{aligned}
\frac{d T_m(x)}{dx}
&=
-\inv{\sin\theta}
\frac{d}{d\theta} \cos(m\theta) \\
&=
-\inv{\sin\theta} (-m) \sin(m\theta) \\
&=
m \inv{\sin\theta} \sin(m\theta),
\end{aligned}
\end{equation}

so the second derivative is

\begin{equation}\label{eqn:tschebyscheff:620}
\begin{aligned}
\frac{d^2 T_m(x)}{dx^2}
&=
-m \inv{\sin\theta} \frac{d}{d\theta} \inv{\sin\theta} \sin(m\theta) \\
&=
-m \inv{\sin\theta}
\lr{
-\frac{\cos\theta}{\sin^2\theta} \sin(m\theta) + \inv{\sin\theta} m \cos(m\theta)
}.
\end{aligned}
\end{equation}

Putting all the pieces together gives

\begin{equation}\label{eqn:tschebyscheff:660}
\begin{aligned}
\lr{ 1 – x^2 } &\frac{d^2 T_m(x)}{dx^2} – x \frac{dT_m(x)}{dx} + m^2 T_{m}(x) \\
&=
m
\lr{
\frac{\cos\theta}{\sin\theta} \sin(m\theta) – m \cos(m\theta)
}
– \cos\theta m \inv{\sin\theta} \sin(m\theta)
+ m^2 \cos(m \theta) \\
&=
0.
\end{aligned}
\end{equation}

### Orthogonality relation

Prove \ref{eqn:tschebyscheff:440}.

First consider the 0,0 inner product, making an $$x = \cos\theta$$, so that $$dx = -\sin\theta d\theta$$

\begin{equation}\label{eqn:tschebyscheff:680}
\begin{aligned}
\innerprod{T_0}{T_0}
&=
\int_{-1}^1 \inv{\lr{1-x^2}^{1/2}} dx \\
&=
\int_{-\pi}^0 \lr{-\inv{\sin\theta}} -\sin\theta d\theta \\
&=
0 – (-\pi) \\
&= \pi.
\end{aligned}
\end{equation}

Note that since the $$[-\pi, 0]$$ interval was chosen, the negative root of $$\sin^2\theta = 1 – x^2$$ was chosen, since $$\sin\theta$$ is negative in that interval.

The m,m inner product with $$m \ne 0$$ is

\begin{equation}\label{eqn:tschebyscheff:700}
\begin{aligned}
\innerprod{T_m}{T_m}
&=
\int_{-1}^1 \inv{\lr{1-x^2}^{1/2}} \lr{ T_m(x)}^2 dx \\
&=
\int_{-\pi}^0 \lr{-\inv{\sin\theta}} \cos^2(m\theta) -\sin\theta d\theta \\
&=
\int_{-\pi}^0 \cos^2(m\theta) d\theta \\
&=
\inv{2} \int_{-\pi}^0 \lr{ \cos(2 m\theta) + 1 } d\theta \\
&= \frac{\pi}{2}.
\end{aligned}
\end{equation}

So far so good. For $$m \ne n$$ the inner product is

\begin{equation}\label{eqn:tschebyscheff:720}
\begin{aligned}
\innerprod{T_m}{T_m}
&=
\int_{-\pi}^0 \cos(m\theta) \cos(n\theta) d\theta \\
&=
\inv{4} \int_{-\pi}^0
\lr{
e^{j m \theta}
+ e^{-j m \theta}
}
\lr{
e^{j n \theta}
+ e^{-j n \theta}
}
d\theta \\
&=
\inv{4} \int_{-\pi}^0
\lr{
e^{j (m + n) \theta}
+e^{-j (m + n) \theta}
+e^{j (m – n) \theta}
+e^{j (-m + n) \theta}
}
d\theta \\
&=
\inv{2} \int_{-\pi}^0
\lr{
\cos( (m + n)\theta )
+\cos( (m – n)\theta )
}
d\theta \\
&=
\inv{2}
\evalrange{
\lr{
\frac{\sin( (m + n)\theta )}{ m + n }
+\frac{\sin( (m – n)\theta )}{ m – n}
}
}{-\pi}{0} \\
&= 0.
\end{aligned}
\end{equation}

# References

 M. Abramowitz and I.A. Stegun. Handbook of mathematical functions with formulas, graphs, and mathematical tables, volume 55. Dover publications, 1964.

## parallelized xargs

March 12, 2015 perl and general scripting hackery No comments

I’ve used xargs before to execute commands long enough that attempting them with backquotes fails. For example, if the list of files ‘c’ is too long, a checkin attempt like:

    cleartool checkin -c 'blah blah blah' cat c


could generate an “environment too long” error in the shell. Something like:

    cat c | xargs cleartool checkin -c 'blah blah blah'


is equivalent. This also checks in one file at a time, using all the filenames in the file ‘c’, but allows xargs to farm out the commands one at a time without building one giant execv argument. What I didn’t know was xargs has a parallelization option:

    cat c | xargs -P 8 -n 1 cleartool checkin -c 'blah blah blah'


This does the same job, with -n 1 restricting xargs to passing one file at a time (which may not be necessary in this case), but starting 8 different processes for the work!

## Motivation

Lance told me they’ve been covering the circumference of a circle in school this week. This made me think of the generalization of a circle, the ellipse, but I couldn’t recall what the circumference of an ellipse was. Sofia guessed $$\pi ( a + b )$$. Her reasoning was that this goes to $$2 \pi r$$ when the ellipse is circular, just like the area of an ellipse $$\pi a b$$, goes to $$\pi a^2$$ in the circular limit. That seemed reasonable to me, but also strange since I didn’t recall any $$\pi ( a + b )$$ formula.

It turns out that there’s no closed form expression for the circumference of an ellipse, unless you count infinite series or special functions. Here I’ll calculate one expression for this circumference.

## Geometry recap

There’s two ways that I think of ellipses. One is the shape that you get when you put a couple tacks in a paper, and use a string and pencil to trace it out, as sketched in fig. 1.
The other is the basic vector parameterization of that same path

\begin{equation}\label{eqn:elipticCircumference:20}
\Br = ( a \cos\theta, b \sin\theta ).
\end{equation}

It’s been a long time since grade 11 when I would have taken it for granted that these two representations are identical. To do so, we’d have to know where the foci of the ellipse sit. Cheating a bit I find in  that the foci are located at

\begin{equation}\label{eqn:elipticCircumference:40}
\Bf_{\pm} = \pm \sqrt{ a^2 – b^2 } (1, 0).
\end{equation}

This and the equivalence of the pencil and tack representation of the ellipse can be verified by checking that the “length of the string” equals $$2 a$$ as expected.

That string length is

\begin{equation}\label{eqn:elipticCircumference:60}
\begin{aligned}
\Abs{ \Br – \Bf_{+} } + \Abs{ \Br – \Bf_{-} }
&=
\sqrt{ (a \cos\theta – f)^2 + b^2 \sin^2\theta }
+
\sqrt{ (a \cos\theta + f)^2 + b^2 \sin^2\theta } \\
&=
\sqrt{ a^2 \cos^2 \theta + f^2 – 2 a f \cos\theta + b^2 \lr{ 1 – \cos^2\theta } } \\
\sqrt{ a^2 \cos^2 \theta + f^2 + 2 a f \cos\theta + b^2 \lr{ 1 – \cos^2\theta } }.
\end{aligned}
\end{equation}

These square roots simplify nicely

\begin{equation}\label{eqn:elipticCircumference:80}
\begin{aligned}
\sqrt{ a^2 \cos^2 \theta + f^2 \pm 2 a f \cos\theta + b^2 \lr{ 1 – \cos^2\theta } }
&=
\sqrt{ (a^2 – b^2) \cos^2 \theta + a^2 – b^2 \pm 2 a f \cos\theta + b^2 } \\
&=
\sqrt{ f^2 \cos^2 \theta + a^2 \pm 2 a f \cos\theta } \\
&=
\sqrt{ (a \pm f \cos\theta)^2 } \\
&=
a \pm f \cos\theta.
\end{aligned}
\end{equation}

So the total length from one focus to a point on the ellipse, back to the other focus, is

\begin{equation}\label{eqn:elipticCircumference:100}
\Abs{ \Br – \Bf_{+} } + \Abs{ \Br – \Bf_{-} }
=
a + f \cos\theta + a – f \cos\theta = 2 a,
\end{equation}

as expected. That verifies that the trigonometric parameterization matches with the pencil and tacks representation of an ellipse (provided the foci are placed at the points \ref{eqn:elipticCircumference:40}).

## Calculating the circumference

The circumference expression can almost be written by inspection. An element of the tangent vector along the curve is

\begin{equation}\label{eqn:elipticCircumference:340}
\frac{d\Br}{d\theta} = ( -a \sin\theta, b \cos\theta ),
\end{equation}

so the circumference is just a one liner

\begin{equation}\label{eqn:elipticCircumference:280}
C = 4 \int_0^{\pi/2} \sqrt{ a^2 \sin^2\theta + b^2 \cos^2 \theta} d\theta.
\end{equation}

The problem is that this one liner isn’t easy to evaluate. The square root can be put in a slightly simpler form in terms of the eccentricity, which is defined by

\begin{equation}\label{eqn:elipticCircumference:300}
e = \frac{f}{a} = \frac{\sqrt{a^2 – b^2}}{a} = \sqrt{ 1 – \frac{b^2}{a^2} }.
\end{equation}

Factoring out $$a$$ and writing the sine as a cosine gives

\begin{equation}\label{eqn:elipticCircumference:320}
\begin{aligned}
C
&=
4 a \int_0^{\pi/2} \sqrt{ 1 – \cos^2\theta + \frac{b^2}{a^2} \cos^2 \theta} d\theta \\
&=
4 a \int_0^{\pi/2} \sqrt{ 1 + \lr{ \frac{b^2}{a^2} -1} \cos^2 \theta} d\theta \\
&=
4 a \int_0^{\pi/2} \sqrt{ 1 – e^2 \cos^2 \theta} d\theta.
\end{aligned}
\end{equation}

For the square root, it’s not hard to show that the fractional binomial expansion is

\begin{equation}\label{eqn:elipticCircumference:360}
\sqrt{1 + a}
=
1 – \sum_{k=1}^\infty \frac{(-a)^k}{2k – 1}\frac{ (2k – 1)!!}{(2k)!!},
\end{equation}

so the circumference is

\begin{equation}\label{eqn:elipticCircumference:380}
C
=
4 a \int_0^{\pi/2} d\theta
\lr{ 1 – \sum_{k=1}^\infty \frac{(e \cos\theta)^{2k}}{2k – 1}\frac{ (2k – 1)!!}{(2k)!!}}.
\end{equation}

Using \ref{eqn:elipticCircumference:260}, this is

\begin{equation}\label{eqn:elipticCircumference:400}
C
=
2 \pi a

4 a
\sum_{k=1}^\infty \frac{e^{2k}}{2k – 1}\frac{ (2k – 1)!!}{(2k)!!}
\frac{(2k – 1)!!}{(2k)!!} \frac{\pi}{2}
\end{equation}

\begin{equation}\label{eqn:elipticCircumference:420}
\boxed{
C
=
2 \pi a \lr{ 1 –
\sum_{k=1}^\infty \frac{e^{2k}}{2k – 1} \lr{ \frac{ (2k – 1)!!}{(2k)!!} }^2
}.
}
\end{equation}

Observe that this does reduce to $$2 \pi r$$ for the circle (where $$e = 0$$), and certainly isn’t as nice as $$\pi (a + b)$$.

## Appendix. Integral of even cosine powers.

The integral

\begin{equation}\label{eqn:elipticCircumference:120}
\int_0^{\pi/2} \cos^{2k} \theta d\theta
\end{equation}

can be evaluated using integration by parts.

\begin{equation}\label{eqn:elipticCircumference:140}
\begin{aligned}
\int_0^{\pi/2} \cos^{2k} \theta d\theta
&=
\int_0^{\pi/2} \cos^{2k-1} \theta \frac{d \sin\theta}{d\theta} d\theta \\
&=
\evalrange{
\cos^{2k-1} \theta \sin\theta
}{0}{\pi/2}

(2 k -1)
\int_0^{\pi/2} \cos^{2k-2} \theta (-\sin\theta) \sin\theta d\theta \\
&=
(2 k -1)
\int_0^{\pi/2} \cos^{2k-2} \theta (1 – \cos^2\theta) d\theta \\
&=
(2 k -1)
\int_0^{\pi/2} \cos^{2k-2} \theta d\theta

(2 k -1)
\int_0^{\pi/2} \cos^{2k} \theta d\theta.
\end{aligned}
\end{equation}

Bringing the $$2k$$ power integral to the other side and solving for the original integral gives a recurrence relation

\begin{equation}\label{eqn:elipticCircumference:160}
\begin{aligned}
\int_0^{\pi/2} \cos^{2k} \theta d\theta
&= \frac{2 k – 1}{2 k}
\int_0^{\pi/2} \cos^{2k -2} \theta d\theta \\
&=
\frac{2 k – 1}{2 k}
\frac{2 k – 3}{2 k – 2}
\int_0^{\pi/2} \cos^{2k -4} \theta d\theta \\
&=
\frac{2 k – 1}{2 k}
\frac{2 k – 3}{2 k – 2}
\cdots
\frac{3}{4}
\int_0^{\pi/2} \cos^{2} \theta d\theta.
\end{aligned}
\end{equation}

This last can also be solved using integration by parts

\begin{equation}\label{eqn:elipticCircumference:180}
\begin{aligned}
\int_0^{\pi/2} \cos^{2} \theta d\theta
&=
\int_0^{\pi/2} \cos \theta \frac{d \sin\theta}{d\theta} d\theta \\
&=

\int_0^{\pi/2} (-\sin \theta) \sin\theta d\theta \\
&=
\int_0^{\pi/2} \lr{ 1 – \cos^2 \theta } d\theta,
\end{aligned}
\end{equation}

or

\begin{equation}\label{eqn:elipticCircumference:200}
\begin{aligned}
\Abs{ \Br – \Bf_{+} } + \Abs{ \Br – \Bf_{-} }
&=
\int_0^{\pi/2} \cos^{2} \theta d\theta \\
&=
\inv{2} \frac{\pi}{2}.
\end{aligned}
\end{equation}

This gives

\begin{equation}\label{eqn:elipticCircumference:220}
\int_0^{\pi/2} \cos^{2k} \theta d\theta
=
\frac{2 k – 1}{2 k}
\frac{2 k – 3}{2 k – 2}
\cdots
\frac{3}{4}
\frac{1}{2}
\frac{\pi}{2}.
\end{equation}

Using the double factorial notation (factorial that skips every other value), this is

\begin{equation}\label{eqn:elipticCircumference:260}
\boxed{
\int_0^{\pi/2} \cos^{2k} \theta d\theta
=
\frac{(2k – 1)!!}{(2k)!!} \frac{\pi}{2}
}
\end{equation}

# References

 Wikipedia. Ellipse — wikipedia, the free encyclopedia, 2015. URL http://en.wikipedia.org/w/index.php?title=Ellipse&oldid=650116160. [Online; accessed 9-March-2015].

## Parallel projection of electromagnetic fields with Geometric Algebra

When computing the components of a polarized reflecting ray that were parallel or not-parallel to the reflecting surface, it was found that the electric and magnetic fields could be written as

\begin{equation}\label{eqn:gaFieldProjection:280}
\BE = \lr{ \BE \cdot \pcap } \pcap + \lr{ \BE \cdot \qcap } \qcap = E_\parallel \pcap + E_\perp \qcap
\end{equation}
\begin{equation}\label{eqn:gaFieldProjection:300}
\BH = \lr{ \BH \cdot \pcap } \pcap + \lr{ \BH \cdot \qcap } \qcap = H_\parallel \pcap + H_\perp \qcap.
\end{equation}

where a unit vector $$\pcap$$ that lies both in the reflecting plane and in the electromagnetic plane (tangential to the wave vector direction) was

\begin{equation}\label{eqn:gaFieldProjection:340}
\pcap = \frac{\kcap \cross \ncap}{\Abs{\kcap \cross \ncap}}
\end{equation}
\begin{equation}\label{eqn:gaFieldProjection:360}
\qcap = \kcap \cross \pcap.
\end{equation}

Here $$\qcap$$ is perpendicular to $$\pcap$$ but lies in the electromagnetic plane. This logically subdivides the fields into two pairs, one with the electric field parallel to the reflection plane

\begin{equation}\label{eqn:gaFieldProjection:240}
\begin{aligned}
\BE_1 &= \lr{ \BE \cdot \pcap } \pcap = E_\parallel \pcap \\
\BH_1 &= \lr{ \BH \cdot \qcap } \qcap = H_\perp \qcap,
\end{aligned}
\end{equation}

and one with the magnetic field parallel to the reflection plane

\begin{equation}\label{eqn:gaFieldProjection:380}
\begin{aligned}
\BH_2 &= \lr{ \BH \cdot \pcap } \pcap = H_\parallel \pcap \\
\BE_2 &= \lr{ \BE \cdot \qcap } \qcap = E_\perp \qcap.
\end{aligned}
\end{equation}

Expressed in Geometric Algebra form, each of these pairs of fields should be thought of as components of a single multivector field. That is

\begin{equation}\label{eqn:gaFieldProjection:400}
F_1 = \BE_1 + c \mu_0 \BH_1 I
\end{equation}
\begin{equation}\label{eqn:gaFieldProjection:460}
F_2 = \BE_2 + c \mu_0 \BH_2 I
\end{equation}

where the original total field is

\begin{equation}\label{eqn:gaFieldProjection:420}
F = \BE + c \mu_0 \BH I.
\end{equation}

In \ref{eqn:gaFieldProjection:400} we have a composite projection operation, finding the portion of the electric field that lies in the reflection plane, and simultaneously finding the component of the magnetic field that lies perpendicular to that (while still lying in the tangential plane of the electromagnetic field). In \ref{eqn:gaFieldProjection:460} the magnetic field is projected onto the reflection plane and a component of the electric field that lies in the tangential (to the wave vector direction) plane is computed.

If we operate only on the complete multivector field, can we find these composite projection field components in a single operation, instead of working with the individual electric and magnetic fields?

Working towards this goal, it is worthwhile to point out consequences of the assumption that the fields are plane wave (or equivalently far field spherical waves). For such a wave we have

\begin{equation}\label{eqn:gaFieldProjection:480}
\begin{aligned}
\BH
&= \inv{\mu_0} \kcap \cross \BE \\
&= \inv{\mu_0} (-I)\lr{ \kcap \wedge \BE } \\
&= \inv{\mu_0} (-I)\lr{ \kcap \BE – \kcap \cdot \BE} \\
&= -\frac{I}{\mu_0} \kcap \BE,
\end{aligned}
\end{equation}

or

\begin{equation}\label{eqn:gaFieldProjection:520}
\mu_0 \BH I = \kcap \BE.
\end{equation}

This made use of the identity $$\Ba \wedge \Bb = I \lr{\Ba \cross \Bb}$$, and the fact that the electric field is perpendicular to the wave vector direction. The total multivector field is

\begin{equation}\label{eqn:gaFieldProjection:500}
\begin{aligned}
F
&= \BE + c \mu_0 \BH I \\
&= \lr{ 1 + c \kcap } \BE.
\end{aligned}
\end{equation}

Expansion of magnetic field component that is perpendicular to the reflection plane gives

\begin{equation}\label{eqn:gaFieldProjection:540}
\begin{aligned}
\mu_0 H_\perp
&= \mu_0 \BH \cdot \qcap \\
&= \gpgradezero{ \lr{-\kcap \BE I} \qcap } \\
&= -\gpgradezero{ \kcap \BE I \lr{ \kcap \cross \pcap} } \\
&= \gpgradezero{ \kcap \BE I I \lr{ \kcap \wedge \pcap} } \\
&= -\gpgradezero{ \kcap \BE \kcap \pcap } \\
&= \gpgradezero{ \kcap \kcap \BE \pcap } \\
&= \BE \cdot \pcap,
\end{aligned}
\end{equation}

so

\begin{equation}\label{eqn:gaFieldProjection:560}
F_1
= (\pcap + c I \qcap ) \BE \cdot \pcap.
\end{equation}

Since $$\qcap \kcap \pcap = I$$, the component of the complete multivector field in the $$\pcap$$ direction is

\begin{equation}\label{eqn:gaFieldProjection:580}
\begin{aligned}
F_1
&= (\pcap – c \pcap \kcap ) \BE \cdot \pcap \\
&= \pcap (1 – c \kcap ) \BE \cdot \pcap \\
&= (1 + c \kcap ) \pcap \BE \cdot \pcap.
\end{aligned}
\end{equation}

It is reasonable to expect that $$F_2$$ has a similar form, but with $$\pcap \rightarrow \qcap$$. This is verified by expansion

\begin{equation}\label{eqn:gaFieldProjection:600}
\begin{aligned}
F_2
&= E_\perp \qcap + c \lr{ \mu_0 H_\parallel } \pcap I \\
&= \lr{\BE \cdot \qcap} \qcap + c \gpgradezero{ – \kcap \BE I \kcap \qcap I } \lr{\kcap \qcap I} I \\
&= \lr{\BE \cdot \qcap} \qcap + c \gpgradezero{ \kcap \BE \kcap \qcap } \kcap \qcap (-1) \\
&= \lr{\BE \cdot \qcap} \qcap + c \gpgradezero{ \kcap \BE (-\qcap \kcap) } \kcap \qcap (-1) \\
&= \lr{\BE \cdot \qcap} \qcap + c \gpgradezero{ \kcap \kcap \BE \qcap } \kcap \qcap \\
&= \lr{ 1 + c \kcap } \qcap \lr{ \BE \cdot \qcap }
\end{aligned}
\end{equation}

This and \ref{eqn:gaFieldProjection:580} before that makes a lot of sense. The original field can be written

\begin{equation}\label{eqn:gaFieldProjection:620}
F = \lr{ \Ecap + c \lr{ \kcap \cross \Ecap } I } \BE \cdot \Ecap,
\end{equation}

where the leading multivector term contains all the directional dependence of the electric and magnetic field components, and the trailing scalar has the magnitude of the field with respect to the reference direction $$\Ecap$$.

We have the same structure after projecting $$\BE$$ onto either the $$\pcap$$, or $$\qcap$$ directions respectively

\begin{equation}\label{eqn:gaFieldProjection:660}
F_1 = \lr{ \pcap + c \lr{ \kcap \cross \pcap } I} \BE \cdot \pcap
\end{equation}
\begin{equation}\label{eqn:gaFieldProjection:680}
F_2 = \lr{ \qcap + c \lr{ \kcap \cross \qcap } I} \BE \cdot \qcap.
\end{equation}

The next question is how to achieve this projection operation directly in terms of $$F$$ and $$\pcap, \qcap$$, without resorting to expression of $$F$$ in terms of $$\BE$$, and $$\BB$$. I’ve not yet been able to determine the structure of that operation.

## Resolving fields into components parallel to the reflecting plane

In order to apply the Fresnel equations, the field components have to be resolved into components where either the electric field or the magnetic field is parallel to the plane of reflection. The geometry of this, with the wave vector direction $$\kcap$$ and the electric and magnetic field phasors perpendicular to that direction is sketched in fig. 1.

If the incident wave is a plane wave, or equivalently a far field spherical wave, it will have the form

\begin{equation}\label{eqn:resolvingFieldsIncidentOnPlane:20}
\BH = \inv{\mu_0} \kcap \cross \BE,
\end{equation}

with the field directions and wave vector directions satisfying

\begin{equation}\label{eqn:resolvingFieldsIncidentOnPlane:60}
\Ecap \cross \Hcap = \kcap
\end{equation}
\begin{equation}\label{eqn:resolvingFieldsIncidentOnPlane:80}
\Ecap \cdot \kcap = 0
\end{equation}
\begin{equation}\label{eqn:resolvingFieldsIncidentOnPlane:100}
\Hcap \cdot \kcap = 0.
\end{equation}

The key to resolving the fields into components parallel to the plane of reflection lies in the observation that the cross product of the plane normal $$\ncap$$ and the incident wave vector direction $$\kcap$$ lies in that plane. With

\begin{equation}\label{eqn:resolvingFieldsIncidentOnPlane:140}
\pcap = \frac{\kcap \cross \ncap}{\Abs{\kcap \cross \ncap}}
\end{equation}
\begin{equation}\label{eqn:resolvingFieldsIncidentOnPlane:160}
\qcap = \kcap \cross \pcap,
\end{equation}

the field directions can be resolved into components

\begin{equation}\label{eqn:resolvingFieldsIncidentOnPlane:200}
\BE = \lr{ \BE \cdot \pcap } \pcap + \lr{ \BE \cdot \qcap } \qcap = E_\parallel \pcap + E_\perp \qcap
\end{equation}
\begin{equation}\label{eqn:resolvingFieldsIncidentOnPlane:220}
\BH = \lr{ \BH \cdot \pcap } \pcap + \lr{ \BH \cdot \qcap } \qcap = H_\parallel \pcap + H_\perp \qcap.
\end{equation}

This subdivides the fields into two pairs, one with the electric field parallel to the reflection plane

\begin{equation}\label{eqn:resolvingFieldsIncidentOnPlane:240}
\begin{aligned}
\BE_1 &= \lr{ \BE \cdot \pcap } \pcap = E_\parallel \pcap \\
\BH_1 &= \lr{ \BH \cdot \qcap } \qcap = H_\perp \qcap,
\end{aligned}
\end{equation}

and one with the magnetic field parallel to the reflection plane

\begin{equation}\label{eqn:resolvingFieldsIncidentOnPlane:260}
\begin{aligned}
\BH_2 &= \lr{ \BH \cdot \pcap } \pcap = H_\parallel \pcap \\
\BE_2 &= \lr{ \BE \cdot \qcap } \qcap = E_\perp \qcap.
\end{aligned}
\end{equation}

This is most of what we need to proceed with the reflection and transmission analysis. The only task remaining is to determine the reflection angle.

Using a pencil with the tip on the table I was able to convince myself by observation that there is always a normal plane of incidence regardless of any oblique angle that the ray hits the reflecting surface. This was, for some reason, not intuitively obvious to me. Having done that, the geometry must be reduced to what is sketched in fig. 2.

Once $$\pcap$$ has been determined, regardless of it’s orientation in the reflection plane, the component of $$\kcap$$ that is normal, directed towards, the plane of reflection is

\begin{equation}\label{eqn:resolvingFieldsIncidentOnPlane:280}
\kcap – \lr{ \kcap \cdot \pcap } \pcap,
\end{equation}

with (squared) length

\begin{equation}\label{eqn:resolvingFieldsIncidentOnPlane:300}
\begin{aligned}
\lr{ \kcap – \lr{ \kcap \cdot \pcap } \pcap }^2
&=
1 + \lr{ \kcap \cdot \pcap }^2 – 2 \lr{ \kcap \cdot \pcap }^2 \\
&=
1 – \lr{ \kcap \cdot \pcap }^2.
\end{aligned}
\end{equation}

The angle of incidence, relative to the normal to the reflection plane, follows from

\begin{equation}\label{eqn:resolvingFieldsIncidentOnPlane:320}
\begin{aligned}
\cos\theta
&= \kcap \cdot \frac{
\kcap – \lr{ \kcap \cdot \pcap } \pcap }{
\sqrt{
1 – \lr{ \kcap \cdot \pcap }^2
}
} \\
&=
\sqrt{
1 – \lr{ \kcap \cdot \pcap }^2
},
\end{aligned}
\end{equation}

Expanding the dot product above gives

\begin{equation}\label{eqn:resolvingFieldsIncidentOnPlane:360}
\begin{aligned}
\kcap \cdot \pcap’
&=
\kcap \cdot \lr{ \pcap \cross \ncap } \\
&=
\frac{1}{\Abs{\kcap \cross \ncap} } \kcap \cdot \lr{ \lr{\kcap \cross \ncap} \cross \ncap },
\end{aligned}
\end{equation}

where

\begin{equation}\label{eqn:resolvingFieldsIncidentOnPlane:380}
\begin{aligned}
\kcap \cdot \lr{ \lr{\kcap \cross \ncap} \cross \ncap }
&=
k_r \epsilon_{r s t} \lr{\kcap \cross \ncap}_s n_t \\
&=
k_r \epsilon_{r s t} \epsilon_{s a b} k_a n_b n_t \\
&=
-k_r \delta_{r t}^{[a b]} k_a n_b n_t \\
&=
-k_r n_t \lr{ k_r n_t – k_t n_r } \\
&=
-1 + \lr{ \kcap \cdot \ncap}^2.
\end{aligned}
\end{equation}

That gives

\begin{equation}\label{eqn:resolvingFieldsIncidentOnPlane:400}
\begin{aligned}
\kcap \cdot \pcap’
&=
\frac{-1 + \lr{ \kcap \cdot \ncap}^2}{\sqrt{1 – \lr{ \kcap \cdot \ncap}^2} } \\
&=
-\sqrt{1 – \lr{ \kcap \cdot \ncap}^2},
\end{aligned}
\end{equation}

or

\begin{equation}\label{eqn:resolvingFieldsIncidentOnPlane:420}
\begin{aligned}
\cos\theta
&= \sqrt{ 1 – \lr{-\sqrt{1 – \lr{ \kcap \cdot \ncap}^2}}^2 } \\
&= \sqrt{ \lr{ \kcap \cdot \ncap}^2 } \\
&= \kcap \cdot \ncap.
\end{aligned}
\end{equation}

This surprisingly simple result makes so much sense, it is an awful admission of stupidity that I went through all the vector algebra to get it instead of just writing it down directly.

The end result is the reflection angle is given by

\begin{equation}\label{eqn:resolvingFieldsIncidentOnPlane:340}
\boxed{
\theta = \cos^{-1} \kcap \cdot \ncap,
}
\end{equation}

where the reflection plane normal should off the back surface to get the sign right. The only detail left is the vector direction of the reflected ray (as well as the direction for the transmitted ray if that is of interest). The reflected ray direction flips the sign of the normal component of the ray

\begin{equation}\label{eqn:resolvingFieldsIncidentOnPlane:440}
\begin{aligned}
\kcap’
&= -\lr{\kcap \cdot \ncap} \ncap + \lr{ \kcap \wedge \ncap} \ncap \\
&= -\lr{\kcap \cdot \ncap} \ncap + \kcap – \lr{ \ncap \kcap} \cdot \ncap \\
&= \kcap -2 \lr{\kcap \cdot \ncap} \ncap.
\end{aligned}
\end{equation}

Here the sign of the normal doesn’t matter since it only occurs quadratically.

This now supplies everything needed for the application of the Fresnel equations to determine the reflected ray characteristics of an arbitrarily polarized incident field.