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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

$\begin{eqnarray*} k &=& \frac {2 \pi}{\lambda} \\ \lambda &=& \frac{c}{f} \\ |z'| &\leq& \frac{L}{2} \end{eqnarray*}$

Use the current distribution to find the Z component of the vector potential

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

where

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

and the magnitude of R can be approximated as

$\begin{eqnarray*} |R| \approx r - z' \cos \theta \end{eqnarray*}$

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:

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

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

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

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:

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

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

Time for some algebra:

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

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 $E_\theta$ :

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