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SBR Method

122 bytes removed, 16:28, 2 October 2016
/* Wedge Diffraction from Edges */
== Free-Space Wave Propagation ==
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In a free-space line-of-sight (LOS) communication system, the signal propagates directly from the transmitter to the receiver without encountering any obstacles (scatterers). Electromagnetic waves propagate in the form of spherical waves with a functional dependence of e<sup>j(&omega;</sup><sup>t-k<sub>0</sub>R)</sup>/R, where R is the distance between the transmitter and receiver, <math>\omega = 2\pi f</math>, f is the signal frequency, <math>k_0 = \frac{\omega}{c} = \frac{2\pi}{\lambda}</math>, c is the speed of light, and &lambda;<sub>0</sub> is the free-space wavelength at the operational frequency. By the time the signal arrives at the location of the receiver, it undergoes two changes. It is attenuated and its power drops by a factor of 1/R<sup>2</sup>, and additionally, it experiences a phase shift of <math>\frac{2\pi R}{\lambda_0}</math>, which is equivalent to a time delay of R/c. The signal attenuation from the transmitter to the receiver is usually quantified by '''Path Loss''' defined as the ratio of the received signal power (P<sub>R</sub>) to the transmitted signal power (P<sub>T</sub>). Assuming isotropic transmitting and receiving radiators (<i>i.e.</i> radiating uniformly in all directions), the Path Loss in a free-space line-of-sight communication system is given by Friis’ formula:
:<math> \mathbf{ \hat{u}_b' = \hat{k}' \times \hat{u}_f' } </math>
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The diffraction coefficients are calculated in the following way:
:<math> D_h = \frac{-e^{-j\pi/4}}{2n \sqrt{2\pi k} \sin\beta_0'} \left\lbrace \begin{align} & \cot \left(\frac{\pi + (\phi-\phi')}{2n}\right) F[kLa^+(\phi-\phi')] + \cot \left(\frac{\pi - (\phi-\phi')}{2n}\right) F[kLa^-(\phi-\phi')] + \\ & R_{0 \|} \cot \left(\frac{\pi - (\phi+\phi')}{2n}\right) F[kLa^-(\phi+\phi')] + R_{n \|} \cot \left(\frac{\pi + (\phi+\phi')}{2n}\right) F[kLa^+(\phi+\phi')] \end{align} \right\rbrace </math>
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where ''F(x)'' is the Fresnel Transition function:
:<math> F(x) = 2j \sqrt{x} e^{jx} \int_{\sqrt{x}}^{\infty} e^{-j\tau^2} \, d\tau </math>
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In the above equations, we have
where <math>N^{\pm}</math> are the integers which most closely satisfy the equations <math> 2n\pi N^{\pm} - \nu = \pm \pi </math>.
<pbr />&nbsp; </phr
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