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E and H plane directivities

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

D_1 = \frac{\Abs{E_\theta}^2_{\textrm{max}}}{\inv{2} \int_0^\pi \Abs{E_\theta(\theta, 0)}^2 \sin\theta d\theta}
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},

whereas [1] lists these as

\inv{D_1} = \inv{2 \ln 2} \int_0^{\Theta_{1 r}/2} \sin\theta d\theta
\inv{D_2} = \inv{2 \ln 2} \int_0^{\Theta_{2 r}/2} \sin\theta d\theta.

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

\inv{D_0} = \inv{2}\lr{\inv{D_1} + \inv{D_2}}.

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


In [2] a field for which directivities can be calculated exactly was used in comparisons of some directivity approximations

\BE = E_0 \lr{ \cos\theta \cos\phi \thetacap – \sin\phi \phicap }.

(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

E_\theta = E_0 \cos\theta \cos\phi
E_\phi = -E_0 \sin\phi

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

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,


= \frac{2}{\int_0^\pi \sin\theta d\theta}
= \frac{2}{\evalrange{-\cos\theta}{0}{\pi}}
= 1.

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

&= \frac{
}{2 c \mu_0}
\lr{ \cos\theta \cos\phi \thetacap – \sin\phi \phicap }
\lr{ \rcap \cross \lr{ \cos\theta \cos\phi \thetacap – \sin\phi \phicap } } \\
&= \frac{
}{ 2 c \mu_0}
\lr{ \cos\theta \cos\phi \thetacap – \sin\phi \phicap }
\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 },

so the radiation intensity is

U(\theta, \phi) \propto \cos^2\theta \cos^2\phi + \sin^2\phi.

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


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



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



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


Given this the total radiated power is

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}.

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

U_\theta = \cos^2 \theta \cos^2 \phi
U_\phi = \sin^2 \phi

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

D_\theta = \frac{4 \pi U_\theta}{P_{\textrm{rad}}} = \frac{3}{2} \cos^2 \theta \cos^2 \phi
D_\phi = \frac{4 \pi U_\phi}{P_{\textrm{rad}}} = \frac{3}{2} \sin^2 \phi

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

D_0 =
+\evalbar{D_\phi}{{\textrm{max}}} = 3,

the exact value from the paper.


[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.