half power beamwidth

Updated notes for ece1229 antenna theory

March 16, 2015 ece1229 No comments , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , ,

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:

Notes for ece1229 antenna theory

February 4, 2015 ece1229 No comments , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , ,

I’ve now posted a first set of 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 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.)

The notes linked above include:

  • Reading notes for chapter 2 (Fundamental Parameters of Antennas) and chapter 3 (Radiation Integrals and Auxiliary Potential Functions) of the class text.
  • Geometric Algebra musings.  How to do formulate Maxwell’s equations when magnetic sources are also included (those modeling magnetic dipoles).
  • Some problems for chapter 2 content.

On Tai and Pereira’s half power beamwidth approximations

January 31, 2015 ece1229 No comments , , , , , ,

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

E and H plane directivities

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

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

whereas [1] lists these as

\begin{equation}\label{eqn:taiAndPereira:60}
\inv{D_1} = \inv{2 \ln 2} \int_0^{\Theta_{1 r}/2} \sin\theta d\theta
\end{equation}
\begin{equation}\label{eqn:taiAndPereira:80}
\inv{D_2} = \inv{2 \ln 2} \int_0^{\Theta_{2 r}/2} \sin\theta d\theta.
\end{equation}

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

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

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

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

(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.

Answer

The field components are

\begin{equation}\label{eqn:taiAndPereira:180}
E_\theta = E_0 \cos\theta \cos\phi
\end{equation}
\begin{equation}\label{eqn:taiAndPereira:200}
E_\phi = -E_0 \sin\phi
\end{equation}

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

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

and

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

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

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

so the radiation intensity is

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

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

TaiAndPereiraSampleFieldThetaCapComponentFig1pn

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

 

TaiAndPereiraSampleFieldPhiCapComponentFig2pn

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

 

TaiAndPereiraSampleFieldAllComponentsFig3pn

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

 

Given this the total radiated power is

\begin{equation}\label{eqn:taiAndPereira:300}
P_{\textrm{rad}} = \int_0^{2 \pi} \int_0^\pi
\lr{ \cos^2\theta \cos^2\phi + \sin^2\phi } \sin\theta d\theta d\phi
= \frac{8 \pi}{3}.
\end{equation}

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

\begin{equation}\label{eqn:taiAndPereira:320}
U_\theta = \cos^2 \theta \cos^2 \phi
\end{equation}
\begin{equation}\label{eqn:taiAndPereira:340}
U_\phi = \sin^2 \phi
\end{equation}

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

\begin{equation}\label{eqn:taiAndPereira:360}
D_\theta = \frac{4 \pi U_\theta}{P_{\textrm{rad}}} = \frac{3}{2} \cos^2 \theta \cos^2 \phi
\end{equation}
\begin{equation}\label{eqn:taiAndPereira:380}
D_\phi = \frac{4 \pi U_\phi}{P_{\textrm{rad}}} = \frac{3}{2} \sin^2 \phi
\end{equation}

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

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

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.