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{{Short description|Region of space between a transmitting and receiving antenna}}
{{more footnotes|date=January 2017}}
[[File:Fresnel zone disrupted.png|thumb|300px|Several examples of how Fresnel zones can be disrupted]]▼
[[File:FresnelSVG1.svg|thumb|400px|Fresnel zone: ''D'' is the distance between the transmitter and the receiver; ''r'' is the radius of the first Fresnel zone (n=1) at point P. P is d1 away from the transmitter, and d2 away from the receiver.]]▼
A '''Fresnel zone''' ({{IPAc-en|lang|f|r|eɪ|ˈ|n|ɛ|l}} {{respell|fray|NEL|'}}), named after physicist [[Augustin-Jean Fresnel]], is one of a series of confocal [[prolate spheroid|prolate]] [[ellipsoid]]al regions of space between
==Significance==
In any wave-propagated transmission between a transmitter and receiver, some amount of the radiated wave propagates off-axis (not on the line-of-sight path between transmitter and receiver). This can then [[Reflection (physics)|deflect]] off
The dependence on the interference on clearance is the cause of the
Fresnel zones are seen in [[optics]], [[radio
==Spatial structure==
[[File:1st Fresnel Zone Avoidance.png|thumb|300px|First Fresnel zone avoidance]]
Fresnel zones are confocal [[prolate spheroid|prolate]] [[ellipsoid]]al shaped regions in space (e.g. 1, 2, 3), centered around the line of the direct transmission path (path AB on the diagram). The first region includes the ellipsoidal space which the direct line-of-sight signal passes through. If a stray component of the transmitted signal bounces off an object within this region and then arrives at the receiving antenna, the [[phase shift]] will be something less than a quarter-length wave, or less than a 90º shift (path ACB on the diagram). The effect regarding phase-shift alone will be minimal. Therefore, this bounced signal can potentially result in having a positive impact on the receiver, as it is receiving a stronger signal than it would have without the deflection, and the additional signal will potentially be mostly in-phase. However, the positive attributes of this deflection also depends on the polarization of the signal relative to the object
The 2nd region surrounds the 1st region but excludes it. If a reflective object is located in the 2nd region, the stray sine-wave which has bounced from this object and has been captured by the receiver will be shifted more than 90º but less than 270º because of the increased path length, and will potentially be received out-of-phase. Generally this is unfavorable. But again, this depends on polarization. Use of same [[Circular_polarization#Reflection|circular polarization]] (
The 3rd region surrounds the 2nd region and deflected waves captured by the receiver will have the same effect as a wave in the 1st region. That is, the sine wave will have shifted more than 270º but less than 450º (ideally it would be a 360º shift) and will therefore arrive at the receiver with the same shift as a signal might arrive from the 1st region. A wave deflected from this region has the potential to be shifted precisely one wavelength so that it is exactly in sync with the line-of-sight wave when it arrives at the receiving antenna.
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If unobstructed and in a perfect environment, radio waves will travel in a relatively straight line from the transmitter to the receiver. But if there are reflective surfaces that interact with a stray transmitted wave, such as bodies of water, smooth terrain, roof tops, sides of buildings, etc., the radio waves deflecting off those surfaces may arrive either out-of-phase or in-phase with the signals that travel directly to the receiver. Sometimes this results in the counter-intuitive finding that reducing the height of an antenna increases the [[signal-to-noise ratio]] at the receiver.
Although radio waves generally travel in a relative straight line, fog and even humidity can cause some of the signal in certain frequencies to scatter or bend before reaching the receiver. This means
In the early 19th century, French scientist [[Augustin-Jean Fresnel]]
==Clearance calculation==
▲[[File:FresnelSVG1.svg|thumb|400px|Fresnel zone: ''D'' is the distance between the transmitter and the receiver; ''r'' is the radius of the first Fresnel zone (n=1) at point P. P is d1 away from the transmitter, and d2 away from the receiver.]]
▲[[File:Fresnel zone disrupted.png|thumb|300px|Several examples of how Fresnel zones can be disrupted]]
The concept of Fresnel zone clearance may be used to analyze [[Interference (communication)|interference]] by obstacles near the path of a radio beam. The first zone must be kept largely free from obstructions to avoid interfering with the radio reception. However, some obstruction of the Fresnel zones can often be tolerated. As a [[rule of thumb]] the maximum obstruction allowable is 40%, but the recommended obstruction is 20% or less.<ref>{{Cite book|title=Certified Wireless Network Administrator Official Study Guide|last=Coleman, Westcott|first=David, David|publisher=John Wiley & Sons, Inc|year=2012|isbn=978-1-118-26295-5|location=111 River St. Hoboken, NJ 07030|pages=126}}</ref>
For establishing Fresnel zones, first determine the RF line of sight (RF LoS), which in simple terms is a straight line between the transmitting and receiving antennas. Now the zone surrounding the RF LoS is said to be the Fresnel zone.<ref>{{cite web|title=Fresnel Zone Clearance|url=http://www.softwright.com/faq/engineering/Fresnel%20Zone%20Clearance.html|publisher=softwright.com|access-date=2008-02-21}}</ref>
The cross sectional radius of each Fresnel zone is the longest at the midpoint of the RF LoS, shrinking to a point at each vertex
===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
:<math>\overline{AP} + \overline{PB} - D = n\frac{\lambda}{2}</math>
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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}-
:<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
:<math>\frac{r_n^2}{2}\left(\frac{1}{d_1}+\frac{1}{d_2}\right)
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
For a satellite-to-Earth link,
:<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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==See also==
{{div col|colwidth=
* [[Beam diameter]]
* [[Diversity scheme]]
* [[Ellipse#Elliptical reflectors and acoustics]]
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*[http://www.digi.com/support/kbase/kbaseresultdetl?id=2090 More Fresnel zone details]
*[http://www.searchanddiscovery.net/documents/geophysical/sheriff/index.htm R.E. Sherriff, Understanding the Fresnel zone]
*[http://www.tapr.org/ve3jf.dcc97.html VHF/UHF/Microwave Radio Propagation: A Primer for Digital Experimenters] {{Webarchive|url=https://web.archive.org/web/20180216110440/http://www.tapr.org/ve3jf.dcc97.html |date=2018-02-16 }}
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