Talk:Fresnel zone

message: I haven't even checked the accuracy of those numerical prefactors, but people should be aware that the concept of a Fresnel zone applies as well to *any* wave phenomenon (my own familiarity with it is in acoustics). I'll change the article if and when I have time, but (1)it should be mentioned that those numerical values which people have been quibbling over are specific to optics in free-space---other values will show up for acoustics problems---still others for other wave-types in different media; or, (2) there should be *no* numerical values given, and simply stress that the formula yields different results based on wave- and media-type.
--Smoo222 03:20, 21 March 2007 (UTC)
thread_title:  General formula is fine, but numerical values only for optics in free-space 

message: Should this page make any distinction between "Fresnel zone" and "Fresnel region" (commonly used in antenna theory)?  
In antenna theory, the Fresnel region generally refers to a radial range of distances between the reactive near field (~2*lambda?) of an aperture, and the Fraunhofer region (2*D^2/lambda, where D is the largest dimension of the aperture).  Within the Fresnel region, the radiation pattern of the aperture varies significantly with radial distance since the multitude of sources that constitute a given aperture cannot yet accurately be approximated as having a common phase center.
69.137.170.154 23:42, 21 March 2007 (UTC)
thread_title:  zone/region ?

message: Currently there's a spurious '15m' above the '20m' on the bottom set of images. It seems to have been left in there by mistake. Xrobau 12:50, 10 September 2007 (UTC)
thread_title:  Error in Attached Image 

message: The article, while apparently sufficient for radio transmitter/receiver problems, mentions almost nothing of other wave types, which of course the entire concept applies to. A specific example is that of mentioning "radio frequency line-of-sight": There is absolutely NO restriction to radio frequencies. 
There is also a minor correction which should be made (I have no time to get it totally right), in that it is mentioned that "If unobstructed, radio waves will travel in a straight line from the transmitter to the receiver." This is true at some level of abstraction (that of infinitely high frequency -- the domain of "ray optics"), but *false in the domain where Fresnel zone concepts apply*. In other words, the Fresnel zone is exactly one of several concepts designed to deal with the fact that waves do not travel in straight lines, even when there are no obstructions!
Another suggestion (again, I'll fix it if/when I have time) is to mention the source/receiver reciprocity which is implied by 
the figures included in the article. For an arbitrary receiver location (imagine the _field_ from the source), the Fresnel zone does not pinch down as is shown in the figures. The "pinching" is a result of the _receiver's_ own Fresnel zone -- that is, its response to an oriented incoming plane wave because the receiver also has a finite width aperture, and an orientation. The cigar-shaped zone in the figures is the *product* of the individual Fresnel zones of the source and receiver, taken individually.
Smoo222 (talk) 13:30, 31 March 2009 (UTC)
thread_title:  Article is written to refer only [or predominantly] to radio waves 

message: Why is Fresnel pronounced "/frɛnɛl/ fre-NELL" in this article, but "(pronounced /freɪˈnɛl/ fray-NELL)" in the Fresnel lens article? Both refer to the same person. I'd fix it, except that I've always pronounced it /frɛz'nɛl/ (frez-nell). My attempt is probably incorrect, but it makes me unaware of which is correct.
118.208.45.140 (talk) 06:05, 21 October 2010 (UTC)
thread_title:  Pronunciation discontinuity 

message: I don't understand the assertion that an obstacle in the first Fresnel zone will result in an interfering signal 0-90 degrees out of phase, 90-270 degrees if in the second zone, etc.
In the first place, surely the phase change associated with a path that goes from one antenna to point P on the surface of the first Fresnel zone to the second antenna is 180 degrees (one half-wavelength path difference).
In the second place, there will likely be a phase change associated with the object causing the interference. For a specular reflection (e.g., a plane surface many wavelengths in extent, such as a flat roof), the phase change will depend on the angle of incidence, polarization and nature of the surface (e.g., dielectric, highly conducting, etc.) and can vary easily from 0-180 degrees.
108.60.121.98 (talk) 22:25, 2 October 2012 (UTC)
thread_title:  Question about phase information 

message: Some of this might already be found in the article or in other comments but I think the article needs substantial clarifications to avoid confusions and misunderstandings.
"Fresnel zones result from diffraction by the circular aperture". No, diffraction by the circular aperture will cause sidelobes in the radiation pattern of circular antennas, but the Fresnel zones are independent of the antennas used (even though both could be important for the end result).
As I understand it the Fresnel zones describes areas of different path differences between the direct beam and an indirect beam that is scattered (e.g. reflected) from some position outside the line of sight. The article may actually say this but not in a very clear way. Since interference depends on phase rather than path differences the path difference is recalculated into a phase difference or difference in wavelengths, lambda. It needs to be clarified whether the zones refers to volumes (the introductory part describes the cross section of the first as circular and the subsequent as annular, which would require the zones to be volumes and a zone in 3D space would normally be interpreted as volume) or if they refer to the limiting surfaces, the distinction does not seem to be well made.
If it is the limiting surface Fn is the zone radius, but if it is a volume it is the outer radius of this volume.
If it is a surface the path difference will be an integer number of wavelengths for the even zones but 0.5*n*lambda for the odd zones. Below I will sometimes refer to the limiting surface as the surface to distinguish it from some zone volume.
If it is a volume the phase difference due to path difference will vary from (n-1)pi to n*pi in the n:th zone, which means that it could either be in-phase or out of phase or have some intermediate phase relation with the direct signal.
In addition to the path difference the scattering process can give rise to phase differences. Often reflexions will lead to a 180 degree phase shift making rays reflected from the n:th surface out of phase with the direct beam. Thus reflexions from the first surface could enhance rather than cancel the signal. Above the Brewster angle reflexions (for beams of vertical polarisation) will not change phase. Scattering may also lead to intermediate phase shifts (typically for lossy media near the Brewster angle) and the end result may also depend on phase shifts of the antenna outside the boresight direction (but since the Fresnel zones are often reasonably narrow compared to the main lobe of the antenna this may rarely be important).
The formula for the Fresnel zone radius appears to be an approximation. Although it is probably a very good approximation in most cases it is good to indicate this with a "curly equal sign" and to clearly state that the radius corresponds to a path difference of 0.5*n*lambda.
One reference that may be worth adding is:
http://radiomobile.pe1mew.nl/?Calculations:Propagation_calculation:Fresnel_zones
150.227.15.253 (talk) 15:48, 4 February 2015 (UTC)
thread_title:  Confusions 

message: This article is wrong on many levels, including its reference to circular apertures and diffraction (concepts also associated with Fresnel, but having no relationship to Fresnel zones).  Specifically, the stated formula is trivially derived based on the difference between the direct path length and alternate path lengths being a multiple of a half-wavelength between two points (i.e., the transmitting and receiving antennas):
:F_n = \sqrt{\frac{n \lambda d_1 d_2}{d_1 + d_2}} 
where,
Fn = The nth Fresnel Zone radius in metres
d1 = The distance of P from one end in metres
d2 = The distance of P from the other end in metres
\lambda = The wavelength of the transmitted signal in metres
The various Fresnel zone ellipsoids do not "define volumes in the radiation pattern of a (usually) circular aperture":  they define the respective terminating surfaces for alternate path lengths that differ by the various multiples of a half-wavelength.
The Confusions section above alludes to this problem:  I am declaring BULLSHIT.  — Preceding unsigned comment added by 73.3.80.189 (talk) 05:02, 27 May 2015 (UTC)
replies: text: I would put it this way: the article tells you how to calculate an ellipsoid. What this has to do with wave transmission, reflection from obstacles, interference patterns, or anything like that, you'll have to learn elsewhere, because the author of this article is keeping pretty quiet on those points. But you can visualize the ellipsoid. 178.39.188.74 (talk) 15:50, 10 June 2015 (UTC)
thread_title:  One of the Worst Technical Articles on Fresnel Zones 

message: That was about the worst summary I've ever seen on an article.  I know what fresnel zones are, and even I was confused after reading it.  Bad phrasing, extraneous material, parenthetical extras, weasel words, ick.  I just rewrote it from scratch and then tried to see if there was anything I left out that was in the original text that could be salvaged.  I leave it up to someone else to make it more accurate, but I think now it is at least understandable. --ssd (talk) 04:50, 24 August 2015 (UTC)
replies: text: Oh, and the illustrations are probably wrong too. --ssd (talk) 04:54, 24 August 2015 (UTC)
thread_title:  resummarized 

message: Substituted at 15:36, 29 April 2016 (UTC)
thread_title: Assessment comment

message: The antennas are in the focal points of the Fresnel zone, not at its boundary, as stated in "The cross sectional radius of each Fresnel zone is ..., shrinking to a point at the antenna on each end." Petr Matas 14:37, 4 June 2016 (UTC)
thread_title:  Shape around the antenna 

message: The Fresnel zone is a difficult concept to visualize and understand because it involves quite a few different things such as phase, distance, deflection, cycles, multipath, polarity, etc., and it's even more difficult to explain.  It may be easier to understand by using an analogy with something that one can sense (e.g. sound waves) rather than something one can only imagine (e.g. radio waves).  I'll try an analogy for phase here.
Let's say Alfred is hammering a large nail into a hard piece of wood.  He is hammering a blow at exactly one strike per second.  Each strike of the hammer of the nail makes a very loud sound.  
Let's say Bill is standing next to Alfred.  We will call this Location X.  We can imagine Alfred's hammer is the transmitter and Bill's ears are the receiver.  Each blow of the hammer coincides with the sound Bill hears.  The sound is in-sync.  
Bill calls Alfred on his cell phone and Alfred answers it with his left hand as he hammers with his right hand.  Bill holds the phone to his left ear.  It's a perfect cellular network and the hammering Bill hears on the cell phone in his left ear is exactly in-sync with the actual sound Bill hears from his right ear.  He hears the hammer in both ears at the same time.
Now let's place Bill several hundred meters away from Alfred.  We will call this Location Y.
Alfred is still hammering, but now the sound doesn't coincide to the hammering.  When Alfred strikes the nail, Bill hears it with his left ear via the cell phone, but Bill hears nothing with his right ear.  But when Alfred lifts the hammer over his head, Bill finally hears the sound with his right ear.  This is obviously because the soundwave moves so much slower than a light wave or a radio wave.  As Alfred continues to hammer, the pattern remains the same.  Instead of hearing the hammer hit the nail once per second, he hears it hit the nail twice per second.  In fact, it sounds like the nail is being hit twice as fast.  We can say the sound is out of sync.  
Now let's move Bill several hundred meters away further still.  We will call this Location Z.  
Alfred is still hammering, but now everything seems fine just as it was in Location X.  Bill can see Alfred hammering, and he can hear Alfred hammering in both ears at the same time.  Each sound seems to coincide with the strike of the hammer.  The sound is in sync.  
The sounds are not the same sounds, they are not the same wave, but they are arriving at Bill's ears at the same time, and seem to be the same sounds.  
So this in-sync and out-of-sync might be one way to visualize the concept of in-phase and out-of-phase.  The sounds can be associated with the top of a cycle.  When the same transmission arrives at a receiver from two different directions due to deflection, it's important that they arrive in-phase.      — Preceding unsigned comment added by 96.231.98.253 (talk) 13:35, 8 January 2017 (UTC)
thread_title:  Phase Analogy 

message: The following regarding polarization reflects a less than complete understanding of electromagnetic theory, specifically, plane waves and polarization:
    Linear polarization — the sine wave moves on a plane
        Vertical polarization — the sine wave moves on a vertical plane
        Horizontal polarization — the sine wave moves on a horizontal plane
    Circular polarization — the sine wave moves in a tight three-dimensional helix as it leaves the transmitting antenna
        RHCP (right-hand circular polarization) — the sine wave moves clockwise as it leaves the transmitter
        LHCP (left-hand circular polarization) — the sine wave moves counter-clockwise
First, a plane wave's equiphase surface lies on a plane, a spherical wave's equiphase surface lies on a sphere, and a cylindrical wave's equiphase surface lies on a cylinder, regardless of whether the wave's electric-field vector changes its direction along a line (i.e., linear polarization) or rotates (i.e., elliptical or circular polarization).  In practice, all electromagnetic waves are essentially spherical waves when far enough away from the source, and over any sufficiently small observation region at that distance, are treated as locally planar.  The characterization of a circularly-polarized wave as moving "in a tight three-dimensional helix" is just wrong on many levels.  — Preceding unsigned comment added by 107.191.0.144 (talk) 19:24, 20 January 2018 (UTC)
replies: text: If you can provide a better description or explanation of polarization as it applies to a Fresnel Zone, please do so.  — Preceding unsigned comment added by Mark The Droner (talk • contribs) 00:08, 2 March 2018 (UTC)
replies: text: Removing blatantly incorrect statements is somewhat more important than replacing it with new information, IMO. This is an encyclopedia, and encyclopedias should be factually accurate. What was stated was not a description or explanation of polarization and was simply confusing and wrong. Footlessmouse (talk) 22:44, 21 August 2020 (UTC)
replies: text: There is nothing incorrect about the polarization description above.  It's very possible polarization does not apply to certain waves such as light waves or sound waves.  I don't know.  But I do know understanding polarization is very important regarding radio wave transmission and receiving.  And therefore it is immeasurably important when considering the Fresnel Zone since constructive and destructive interference within a certain zone would depend wholly on polarization.  Another point - if polarization didn't matter or didn't exist, the Soviets would not have developed circular polarization.  Specific polarizations have their advantages and disadvantages depending on the environment.  Mark The Droner (talk) 13:00, 20 April 2021 (UTC)
thread_title: Polarization

message: In the Clearance calculation section, the article states 40% is maximum allowable obstruction. What?! I guess we're talking about the ellipsoid? Are we talking about its volume or perhaps the "area" of the (curved?) lines between emitter and receiver? I really don't have any idea and the article just tosses it in as if it's obvious.98.21.213.85 (talk) 17:00, 8 February 2024 (UTC)
replies: text: Yes, it's difficult to understand.  The concept of the Fresnel Zone is very complex and difficult to explain.  You have to understand what polarity means and what "phase shift" means and why it happens and how it is related to the various Fresnel Zones.  You also have to have a clear understanding of what happens to polarity of a secondary signal when it is deflected within the first or second Fresnel zone.  The location of the object deflecting (above, below, right, left) matters greatly due to the signal's polarity.  If you understand polarity and phase shift, then everything will make sense.  The percentages are actually referring to phase shift of a late arriving or deflected secondary signal (other than the primary signal).  In other words, a little bit of phase shift is okay, but a lot is not okay.  If you understand that, then you'll understand why 20% is okay and 40% is worst case and might be tolerated but would be best to avoid if possible.  Anything more than 50% will cause major interference because the two signals (primary and secondary) will begin to cancel each other out.  The whole Fresnel Zone concept is difficult to wrap your head around.  It will generally take 2-3 reads of the page and some careful reflection to get it.  Once you get it, everything will make sense. Mark The Droner (talk) 23:34, 16 February 2024 (UTC)
thread_title:  40%