## E and H plane directivities

In  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  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  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.

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}

\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}
\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

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

 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.