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+%
+% Copyright © 2015 Peeter Joot.  All Rights Reserved.
+% Licenced as described in the file LICENSE under the root directory of this GIT repository.
+%
+\input{../blogpost.tex}
+\renewcommand{\basename}{reciprocityTheorem}
+\renewcommand{\dirname}{notes/ece1229/}
+%\newcommand{\dateintitle}{}
+%\newcommand{\keywords}{}
+
+\input{../peeter_prologue_print2.tex}
+\usepackage{peeters_layout_exercise}
+\usepackage{macros_bm}
+
+\beginArtNoToc
+
+\generatetitle{Far field electric field for two horizontal dipole configurations}
+
+\section{Horizontal dipole reflection coefficient}
+\index{horizontal dipole}
+
+Suppose an infinitesimal horizontal dipole is in a ``coming out of the page'' configuration.  With the page representing the z-y plane, this is a magnetic vector potential directed along the x-axis direction, as follows
+
+\begin{equation}\label{eqn:t:640}
+\BA = \xcap \frac{\mu_0 I_0 l}{4 \pi r} e^{-j k r}.
+%= \frac{A_r}{4 \pi r} e^{-j k r}
+\end{equation}
+
+We've seen in class (UofT ece1229, taught by Prof. Eleftheriades) that the far-field electric field given by
+
+\begin{dmath}\label{eqn:t:n}
+\BE = j \omega \Proj_\T \BA,
+\end{dmath}
+
+where \( \Proj_\T \BA \) represents the transverse projection of \( \BA \).  So, for a wave vector directed in the z-y plane, \( \kcap = \zcap \cos\theta + \ycap \sin\theta \), the electric far field is directed along
+
+\begin{dmath}\label{eqn:t:660}
+%\lr{ \xcap \wedge \kcap } \cdot \kcap
+%= 
+\xcap - \lr{ \xcap \cdot \kcap } \kcap
+=
+\xcap - \lr{ \cancel{\xcap \cdot 
+\lr{ \zcap \cos\theta + \ycap \sin\theta }
+} } \kcap
+= \xcap.
+\end{dmath}
+
+For such a ray, the electric far field lies completely in the plane of reflection.  From \citep{hecht1998hecht} (\eqntext 4.34), the Fresnel reflection coefficients is
+\index{Fresnel equations}
+
+\begin{dmath}\label{eqn:t:680}
+R =
+\frac{
+n_i \cos\theta_i - n_t \cos\theta_t
+}
+{
+n_i \cos\theta_i + n_t \cos\theta_t
+},
+\end{dmath}
+
+which approaches \( -1 \) in the no-transmission limit where \( v_t \rightarrow 0 \), and \( n_t = c/v_t \rightarrow \infty \).
+
+\paragraph{Azimuthal angle dependency of the reflection coefficient}
+
+Now consider a horizontal dipole directed along the y-axis.  For the same wave vector direction as above, the electric far field is now directed along
+
+\begin{dmath}\label{eqn:t:700}
+\ycap - \lr{ \ycap \cdot \kcap } \kcap
+=
+\ycap - \lr{ \ycap \cdot \lr{
+\zcap \cos\theta + \ycap \sin\theta
+} } \kcap
+=
+\ycap - \kcap \sin\theta
+= 
+\ycap - \sin\theta \lr{
+\zcap \cos\theta + \ycap \sin\theta
+}
+= 
+\ycap \cos^2 \theta - \sin\theta \cos\theta \zcap
+= \cos\theta \lr{ \ycap \cos\theta - \sin\theta \zcap }.
+\end{dmath}
+
+That is
+
+\begin{dmath}\label{eqn:t:720}
+\BE = 
+-j \omega \frac{\mu_0 I_0 l}{4 \pi r} e^{-j k r}
+\cos\theta \lr{ \ycap \cos\theta - \sin\theta \zcap }.
+\end{dmath}
+
+This far field electric field lies in the plane of incidence (a direction of \( \thetacap \) rotated by \( \pi/2 \)), not in the plane of reflection.  The corresponding magnetic field should be directed along the plane of reflection, which is easily confirmed by calculation
+
+\begin{dmath}\label{eqn:t:740}
+\kcap \cross
+\lr{ \ycap \cos\theta - \sin\theta \zcap }
+=
+\lr{ \zcap \cos\theta + \ycap \sin\theta } \cross
+\lr{ \ycap \cos\theta - \sin\theta \zcap }
+=
+-\xcap \cos^2 \theta - \xcap \sin^2\theta
+= -\xcap.
+\end{dmath}
+
+The far field magnetic field is seen to be
+
+\begin{dmath}\label{eqn:t:721}
+\BH = 
+j \omega \frac{I_0 l}{4 \pi r} e^{-j k r}
+\cos\theta \xcap.
+\end{dmath}
+
+Referring again to \citep{hecht1998hecht} (\eqntext 4.40) for the coefficient of reflection coefficient for this polarization
+
+\begin{dmath}\label{eqn:t:620}
+R
+=
+\frac{
+n_t \cos\theta_i - n_i \cos\theta_t
+}
+{
+n_i \cos\theta_i + n_t \cos\theta_t
+},
+\end{dmath}
+
+In the no-transmission limit, this tends to \( 1 \), and would be the value required to calculation the superposition associated with the ground reflection.
+
+For a more general azimuthal orientation of the horizontal dipole, I'd guess (but have not calculated), there would be components of the both the electric and magnetic far-field rays that lie in the reflection plane, so both reflection coefficients would be required in proportion to the components in question.
+
+\EndArticle
+%\EndNoBibArticle