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en:electronics:antenna-theory:dipole-derivation [2013/02/19 07:25]
alex
en:electronics:antenna-theory:dipole-derivation [2014/10/20 23:04] (current)
alex
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 ====== Dipole antenna radiation pattern derivation ====== ====== 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:+For a dipole antenna ​centered on the origin and 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 The current on the antenna will be approximately sinusoidal, with zeros at the ends of the antenna, represented by
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 \end{eqnarray*} \end{eqnarray*}
  
-Find the vector potential in the Z direction from the current distribution+Use the current distribution to find the Z component of the vector potential
  
 \begin{eqnarray*} \begin{eqnarray*}
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 \end{eqnarray*} \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.  ​+As $ r >> z' \cos \theta $, $ z' \cos \theta $ can be neglected in the denominator. ​ However, it cannot be neglected ​in the exponential as it is a phase offset.  ​
  
 \begin{eqnarray*} \begin{eqnarray*}
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 \end{eqnarray*} \end{eqnarray*}
  
-Elimiate ​the absolute value by splitting into two integrals:+Eliminate ​the absolute value by splitting into two integrals:
  
 \begin{eqnarray*} \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 ]+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' ​\right. ​\\ 
 +\left. ​\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*} \end{eqnarray*}
  
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 \end{eqnarray*} \end{eqnarray*}
  
-Substitute $ \frac{e^{jx} + e^{-jx}}{2} = cos(x) $+Substitute $ e^{jx} + e^{-jx} ​\cos(x) $
  
 \begin{eqnarray*} \begin{eqnarray*}
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 \begin{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} + \\ \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'​}_{\frac{\alpha}{\cos^2 \theta}}+\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*} \end{eqnarray*}
  
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 \alpha &=& \beta + \frac{\alpha}{\cos^2 \theta} \\ \alpha &=& \beta + \frac{\alpha}{\cos^2 \theta} \\
 \alpha \cos^2 \theta &=& \beta \cos^2 \theta + \alpha \\ \alpha \cos^2 \theta &=& \beta \cos^2 \theta + \alpha \\
-\alpha &=& \frac{\beta \cos^2 \theta}{\cos^2 \theta -1}+\alpha &​=& ​\beta \frac{\cos^2 \theta}{\cos^2 \theta -1}
 \end{eqnarray*} \end{eqnarray*}
  
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 \begin{eqnarray*} \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 ]+\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) \sin(kz' \cos \theta) \cos \theta \right ]
 \end{eqnarray*} \end{eqnarray*}
  
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 \begin{eqnarray*} \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) -\\ 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 - \\+\sin \left( k \left( \frac{L}{2} - \frac{L}{2} \right) \right) \sin(k\frac{L}{2} \cos \theta) \cos \theta - \\
 \cos \left( k \left( \frac{L}{2} - 0 \right) \right) \cos(k 0 \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 ]+\sin \left( k \left( \frac{L}{2} - 0 \right) \right) \sin(k 0 \cos \theta) \cos \theta \right ]
 \end{eqnarray*} \end{eqnarray*}
  
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 \begin{eqnarray*} \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 ]+A_z = \frac{\mu I_0 e^{-jkr}}{2\pi r k \sin^2 \theta} \left [ \cos \left ( \frac{kL}{2} \cos \theta \right ) - \cos \left( \frac{kL}{2} \right) \right ]
 \end{eqnarray*} \end{eqnarray*}
  
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 \begin{eqnarray*} \begin{eqnarray*}
-E_\theta &=& -j \omega ​\eta A_T \\+E_\theta &=& -j \omega A_T \\
 A_T &=& A_z \sin \theta \\ 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 ]+E_\theta &=& \frac{-j \omega \mu I_0 e^{-jkr}}{2\pi r k \sin \theta} \left [ \cos \left (\frac{kL}{2} \cos \theta \right ) - \cos \left( \frac{kL}{2} \right) \right ]
 \end{eqnarray*} \end{eqnarray*}
 +
 +This equation for $E_\theta$ is the general form for the theta component in spherical coordinates of the far-field E field of a dipole antenna of any length oriented along the z-axis. ​ The r component is zero due to the far-field assumption and the phi component is zero due to the electric field'​s orientation along the z-axis.  ​