PERSATUAN RADIO AMATUR MELAKA (PERAKAM): Mathematical perspective and formulas

Isnin, 17 Ogos 2009

Mathematical perspective and formulas

A phased array is an example of N-slit diffraction. It may also be viewed as the coherent addition of N line sources. Since each individual antenna acts as a slit, emitting radio waves, their diffraction pattern can be calculated by adding the phase shift φ to the fringing term.

We will begin from the N-slit diffraction pattern derived on the diffraction page.

$\psi ={{\psi }_0}\left[\frac{\sin \left(\frac{{\pi a}}{\lambda }\sin\theta \right)}{\frac{{\pi a}}{\lambda }\sin\theta}\right]\left[\frac{\sin \left(\frac{N}{2}{kd}\sin\theta\right)}{\sin \left(\frac{{kd}}{2}\sin\theta \right)}\right]$

Now, adding a φ term to the $\begin{matrix}kd\sin\theta\,\end{matrix}$ fringe effect in the second term yields:

$\psi ={{\psi }_0}\left[\frac{\sin \left(\frac{{\pi a}}{\lambda }\sin \theta\right)}{\frac{{\pi a}}{\lambda }\sin\theta}\right]\left[\frac{\sin \left(\frac{N}{2}\big(\frac{{2\pi d}}{\lambda }\sin\theta + \phi \big)\right)}{\sin \left(\frac{{\pi d}}{\lambda }\sin\theta +\phi \right)}\right]$

Taking the square of the wave function gives us the intensity of the wave.

$I = I_0{{\left[\frac{\sin \left(\frac{\pi a}{\lambda }\sin\theta\right)}{\frac{{\pi a}}{\lambda } \sin [\theta ]}\right]}^2}{{\left[\frac{\sin \left(\frac{N}{2}(\frac{2\pi d}{\lambda} \sin\theta+\phi )\right)}{\sin \left(\frac{{\pi d}}{\lambda } \sin\theta+\phi \right)}\right]}^2}$

$I =I_0{{\left[\frac{\sin \left(\frac{{\pi a}}{\lambda } \sin\theta\right)}{\frac{{\pi a}}{\lambda } \sin\theta}\right]}^2}{{\left[\frac{\sin \left(\frac{\pi }{\lambda } N d \sin\theta+\frac{N}{2} \phi \right)}{\sin \left(\frac{{\pi d}}{\lambda } \sin\theta+\phi \right)}\right]}^2}$

Now space the emitters a distance $d=\begin{matrix}\frac{\lambda}{4}\end{matrix}$ apart. This distance is chosen for simplicity of calculation but can be adjusted as any scalar fraction of the wavelength.

$I =I_0{{\left[\frac{\sin \left(\frac{\pi }{\lambda } a \theta \right)}{\frac{\pi }{\lambda } a \theta }\right]}^2}{{\left[\frac{\sin \left(\frac{\pi }{4} N \sin\theta+\frac{N}{2} \phi \right)}{\sin \left(\frac{\pi }{4} \sin\theta+ \phi \right)}\right]}^2}$

Sin achieves its maximum at $\begin{matrix}\frac{\pi}{2}\end{matrix}$ so we set the numerator of the second term = 1.

$\frac{\pi }{4} N \sin\theta+\frac{N}{2} \phi = \frac{\pi }{2}$

$\sin\theta=\Big(\frac{\pi }{2} - \frac{N}{2} \phi \Big)\frac{4}{N \pi }$

$\sin\theta=\frac{2}{N}-\frac{2\phi }{\pi }$

Thus as N gets large, the term will be dominated by the $\begin{matrix}\frac{2\phi}{\pi}\end{matrix}$ term. As sin can oscillate between −1 and 1, we can see that setting $\phi=-\begin{matrix}\frac{\pi}{2}\end{matrix}$ will send the maximum energy on an angle given by

$\theta = \sin^{-1}(1) = \begin{matrix}\frac{\pi}{2}\end{matrix} = 90^{\circ}$

Additionally, we can see that if we wish to adjust the angle at which the maximum energy is emitted, we need only to adjust the phase shift φ between successive antennas. Indeed the phase shift corresponds to the negative angle of maximum signal.

A similar calculation will show that the denominator is minimized by the same factor.

PERSATUAN RADIO AMATUR MELAKA (PERAKAM)

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Isnin, 17 Ogos 2009

Mathematical perspective and formulas

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