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## 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.

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.