Dipole antenna radiation pattern derivation

For a dipole antenna oriented along the z axis with length L, the far-field radiation pattern can be derived as follows:

The current on the antenna will be approximately sinusoidal, with zeros at the ends of the antenna, represented by

\begin{eqnarray*} I(z') = I_0 \sin \left( k \left( \frac{L}{2} - | z' | \right) \right) \overline a_z \end{eqnarray*}

where

\begin{eqnarray*} k &=& \frac {2 \pi}{\lambda}
\lambda &=& \frac{c}{f}

\end{eqnarray*}

Find the vector potential in the Z direction from the current distribution

\begin{eqnarray*} A_z = \mu \int_{\frac{-L}{2}}^{\frac{L}{2}} \! \frac{I(z')e^{-jkR}}{4\pi R} \, \mathrm{d}z' \end{eqnarray*}

where

\begin{eqnarray*} \overline R = \overline r - z' \overline a_z \end{eqnarray*}

and the magnitude of R can be approximated as

\begin{eqnarray*}

\end{eqnarray*}

Substituting yields

\begin{eqnarray*} A_z = \mu \int_{\frac{-L}{2}}^{\frac{L}{2}} \! \frac{I_0 \sin \left( k \left( \frac{L}{2} - | z' | \right) \right) e^{-jk(r - z' \cos \theta)}}{4\pi (r - z' \cos \theta)} \, \mathrm{d}z' \end{eqnarray*}

As $ r » z' \cos \theta $, $ z' \cos \theta $ can be neglected in the denominator. However, it cannot be ngeglected in the exponential as it is a phase offset.

\begin{eqnarray*} A_z = \frac{\mu I_0}{4\pi} \int_{\frac{-L}{2}}^{\frac{L}{2}} \! \frac{\sin \left( k \left( \frac{L}{2} - | z' | \right) \right) e^{-jkr} e^{jkz' \cos \theta}}{r} \, \mathrm{d}z' \end{eqnarray*}

rearranging yields

\begin{eqnarray*} A_z = \frac{\mu I_0 e^{-jkr}}{4\pi r} \int_{\frac{-L}{2}}^{\frac{L}{2}} \! \sin \left( k \left( \frac{L}{2} - | z' | \right) \right) e^{jkz' \cos \theta} \, \mathrm{d}z' \end{eqnarray*}

Elimiate the absolute value by splitting into two integrals:

\begin{eqnarray*} A_z = \frac{\mu I_0 e^{-jkr}}{4\pi r} \left [ \int_{\frac{-L}{2}}^0 \! \sin \left( k \left( \frac{L}{2} + z' \right) \right) e^{jkz' \cos \theta} \, \mathrm{d}z' + \int_0^{\frac{L}{2}} \! \sin \left( k \left( \frac{L}{2} - z' \right) \right) e^{z' \cos \theta} \, \mathrm{d}z' \right ] \end{eqnarray*}

Flip limits on the first integral and combine

\begin{eqnarray*} A_z = \frac{\mu I_0 e^{-jkr}}{4\pi r} \int_0^{\frac{L}{2}} \! \sin \left( k \left( \frac{L}{2} - z' \right) \right) \left ( e^{-jkz' \cos \theta} + e^{jkz' \cos \theta} \right ) \, \mathrm{d}z' \end{eqnarray*}

Substitute $ \frac{e^{jx} + e^{-jx}}{2} = cos(x) $

\begin{eqnarray*} A_z = \frac{\mu I_0 e^{-jkr}}{2\pi r} \int_0^{\frac{L}{2}} \! \sin \left( k \left( \frac{L}{2} - z' \right) \right) \cos(kz' \cos \theta) \, \mathrm{d}z' \end{eqnarray*}

Now, integrate it, setting aside the constants and limits temporarily:

\begin{eqnarray*} \alpha = \int \! \sin \left( k \left( \frac{L}{2} - z' \right) \right) \cos(kz' \cos \theta) \, \mathrm{d}z' \end{eqnarray*}

Integrate by parts (tan cow):

\begin{eqnarray*} \int u \mathrm{d} v &=& u v - \int v \mathrm{d} u
u &=& \sin \left( k \left( \frac{L}{2} - z' \right) \right)
du &=& -k \cos \left( k \left( \frac{L}{2} - z' \right) \right)
v &=& \frac{\sin(kz' \cos \theta)}{k \cos \theta}
dv &=& \cos(kz' \cos \theta) \end{eqnarray*}

\begin{eqnarray*} \alpha = \sin \left( k \left( \frac{L}{2} - z' \right) \right) \frac{\sin(kz' \cos \theta)}{k \cos \theta} + \int \cos \left( k \left( \frac{L}{2} - z' \right) \right) \frac{\sin(kz' \cos \theta)}{\cos \theta} \mathrm{d}z' \end{eqnarray*}

And again:

\begin{eqnarray*} u &=& \cos \left( k \left( \frac{L}{2} - z' \right) \right)
du &=& k \sin \left( k \left( \frac{L}{2} - z' \right) \right)
v &=& - \frac{\cos(kz' \cos \theta)}{k \cos^2 \theta}
dv &=& \frac{\sin(kz' \cos \theta)}{\cos \theta} \end{eqnarray*}

\begin{eqnarray*} \alpha = \overbrace{\sin \left( k \left( \frac{L}{2} - z' \right) \right) \frac{\sin(kz' \cos \theta)}{k \cos \theta} - \cos \left( k \left( \frac{L}{2} - z' \right) \right) \frac{\cos(kz' \cos \theta)}{k \cos^2 \theta}}^{\beta} +
\int \sin \left( k \left( \frac{L}{2} - z' \right) \right) \frac{\cos(kz' \cos \theta)}{\cos^2 \theta} \mathrm{d}z' \end{eqnarray*}

Time for some algebra:

\begin{eqnarray*} \alpha &=& \int \! \sin \left( k \left( \frac{L}{2} - z' \right) \right) \cos(kz' \cos \theta) \, \mathrm{d}z'
\beta &=& \sin \left( k \left( \frac{L}{2} - z' \right) \right) \frac{\sin(kz' \cos \theta)}{k \cos \theta} - \cos \left( k \left( \frac{L}{2} - z' \right) \right) \frac{\cos(kz' \cos \theta)}{k \cos^2 \theta}
\alpha &=& \beta + \frac{A}{\cos^2 \theta}
\alpha \cos^2 \theta &=& \beta \cos^2 \theta + \alpha
\alpha &=& \frac{\beta \cos^2 \theta}{\cos^2 \theta -1} \end{eqnarray*}

so

\begin{eqnarray*} \alpha = \left [ \sin \left( k \left( \frac{L}{2} - z' \right) \right) \frac{\sin(kz' \cos \theta)}{k \cos \theta} - \cos \left( k \left( \frac{L}{2} - z' \right) \right) \frac{\cos(kz' \cos \theta)}{k \cos^2 \theta} \right ] \frac{\cos^2 \theta}{\cos^2 \theta -1} \end{eqnarray*}

Now substitute $\cos^2 \theta - 1 = -\sin^2 \theta$

\begin{eqnarray*} \alpha = \frac{1}{k \sin^2 \theta} \left [ \cos \left( k \left( \frac{L}{2} - z' \right) \right) \cos(kz' \cos \theta) - \sin \left( k \left( \frac{L}{2} - z' \right) \right) \cos(kz' \cos \theta) \cos \theta \right ] \end{eqnarray*}

and plug in the limits $ z' = 0 $ and $ z' = \frac{L}{2}$ and bring back the constants

\begin{eqnarray*} A_z = \frac{\mu I_0 e^{-jkr}}{2\pi r k \sin^2 \theta} \left [ \cos \left( k \left( \frac{L}{2} - \frac{L}{2} \right) \right) \cos(k\frac{L}{2} \cos \theta) -
\sin \left( k \left( \frac{L}{2} - \frac{L}{2} \right) \right) \cos(k\frac{L}{2} \cos \theta) \cos \theta -
\cos \left( k \left( \frac{L}{2} - 0 \right) \right) \cos(k 0 \cos \theta) +
\sin \left( k \left( \frac{L}{2} - 0 \right) \right) \cos(k0 \cos \theta) \cos \theta \right ] \end{eqnarray*}

simplify

\begin{eqnarray*} A_z = \frac{\mu I_0 e^{-jkr}}{2\pi r k \sin^2 \theta} \left [ \cos \left ( k\frac{L}{2} \cos \theta \right ) - \cos \left( k \left( \frac{L}{2} \right) \right) \right ] \end{eqnarray*}

Now, convert to $ E_\theta $:

\begin{eqnarray*} E_\theta &=& -j \omega \eta A_T
A_T &=& A_z \sin \theta
E_\theta &=& \frac{-j \omega \eta \mu I_0 e^{-jkr}}{2\pi r k \sin \theta} \left [ \cos \left (k\frac{L}{2} \cos \theta \right ) - \cos \left( k \left( \frac{L}{2} \right) \right) \right ] \end{eqnarray*}