Editing Fresnel zone (section)

===Formulation===
Consider an arbitrary point ''P'' in the LoS, at a distance <math>d_1</math> and <math>d_2</math> with respect to each of the two antennas.
To obtain the radius <math>r_n</math> of zone <math>n</math>, note that the volume of the zone is delimited by all points for which the difference in distances, between the reflected wave (<math>\overline{AP} + \overline{PB}</math>) and the direct wave (<math>D=d_1+d_2</math>) is the constant <math>n\frac{\lambda}{2}</math> (multiples of half a [[wavelength]]). This effectively defines an ellipsoid with the major axis along <math>\overline{AB}</math> and foci at the antennas (points A and B). So:

:<math>\overline{AP} + \overline{PB} - D = n\frac{\lambda}{2}</math>

Re-writing the expression with the coordinates of point <math>P</math> and the distance between antennas <math>D</math>, it gives:

:<math>\sqrt{d_1^2+r_n^2}+\sqrt{d_2^2+r_n^2}-(d_1+d_2)=n\frac{\lambda}{2}</math>
:<math>d_1\left(\sqrt{1+r_n^2/d_1^2}-1\right)+d_2\left(\sqrt{1+r_n^2/d_2^2}-1\right)=n\frac{\lambda}{2}</math>

Assuming the distances between the antennas and the point <math>P</math> are much larger than the radius and applying the [[binomial approximation]] for the square root, <math>\sqrt{1+x} \approx 1+x/2</math> (for ''x''≪1), the expression simplifies to:

:<math>\frac{r_n^2}{2}\left(\frac{1}{d_1}+\frac{1}{d_2}\right)\approx n\frac{\lambda}{2}</math>

which can be solved for <math>r_n</math>:<ref>{{Cite book|title=Electronic Communication Systems - Fundamentals Through Advanced|last=Tomasi|first=Wayne|publisher=Pearson|pages=1023}}</ref>

:<math>r_n\approx\sqrt{n\frac{d_1\ d_2}{D}\lambda},\quad d_1, d_2 \gg n\lambda,</math>

For a satellite-to-Earth link, it further simplifies to:<ref name="Braasch 2017 pp. 443–468">{{cite book | last=Braasch | first=Michael S. | title=Springer Handbook of Global Navigation Satellite Systems | chapter=Multipath | publisher=Springer International Publishing | location=Cham | year=2017 | isbn=978-3-319-42926-7 | doi=10.1007/978-3-319-42928-1_15 | pages=443–468}}</ref>

:<math>r_n\approx \sqrt{n d_1 \lambda},\quad d_1 \gg n\lambda,\quad d_2\approx D</math>

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Note that when <math>d_1=0</math> or <math>d_2=0\implies r_n=0</math>, which implies that the foci seem to coincide with the [[vertex (curve)|vertices]] of the ellipsoid. This is not correct and it's a consequence of the approximation made. 

Setting the point <math>P</math> to one of the vertices (behind an antenna), it's possible to obtain the error <math>\epsilon</math> of this approximation:

:<math>\epsilon + \left(\epsilon+D\right) - D = n\frac{\lambda}{2}\implies\epsilon=n\frac{\lambda}{4}</math>

Since the distance between antennas is generally tens of km and <math>\lambda</math> of the order of cm, the error is negligible for a graphical representation.

On the other hand, considering the clearance at the left-hand antenna, with <math>d_1=0, d_2=D</math>, and applying the binomial approximation only at the right-hand antenna, we find:
:<math>\left(\sqrt{d_1^2+r_n^2}-d_1\right)+0.5 r_n^2/d_2=0.5 n \lambda</math>
:<math>r_n +0.5 r_n^2/D=0.5 n \lambda</math>
The quadratic polynomial roots are:
:<math>r_n=D\left(-1 \pm \sqrt{1+n\lambda/D}\right)</math>
Applying the binomial approximation one last time, we finally find:
:<math>r_n=0.5n\lambda,\quad d_1=0</math>
So, there should be at least half a wavelength of clearance at the antenna in the direction perpendicular to the line of sight.
The vertical clearance at the antenna in a [[slant range|slant direction]] inclined at an [[altitude angle]] ''a'' would be:
:<math>v_n=r_n \sec(a).</math>

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