This is the
talk page for discussing improvements to the
Fresnel zone article. This is not a forum for general discussion of the article's subject. |
Article policies
|
Find sources: Google ( books · news · scholar · free images · WP refs) · FENS · JSTOR · TWL |
This article is rated Start-class on Wikipedia's
content assessment scale. It is of interest to the following WikiProjects: | ||||||||||||||||||||||||||||
|
needs more reliable sources....-- 24.144.100.184 02:10, 2 November 2006 (UTC)
what's the unit of r ? what about the fomula in metric ?
This article needs to be generalized to include the optics usage. (Yes, I know, sofixit and all that. I don't have time right now, so I'll just mention it here for now.)-- Srleffler 04:07, 21 March 2006 (UTC)
Any idea why the formula for the radius on this page gives a different multiplier (43.3 instead of 72.6)? EdDavies 12:36, 3 April 2006 (UTC)
The link to "this page" above appears to be no longer reachable, but I've seen 43.3 elsewhere and had the same question. See this page instead. The 43.3 multiplier (for the radius in feet and distance in miles) may be the "obstacle-free radius", which is often taken as 60% of the Fresnel zone radius. Can someone verify this? 70.89.158.189 21:50, 25 October 2006 (UTC) Gerald Reynolds
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)smoo222
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)Bill Shultz
This is written like a high-school science report. Please recast it in, at the very least, the passive voice.
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)
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)Smoo222
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)
It seems as if the pronounciation has been "fixed" in the Fresnel lens fashion while the correct (French!) pronounciation should probably be "/frɛnɛl/ fre-NELL"! 150.227.15.253 ( talk) 15:48, 4 February 2015 (UTC)
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)Douglas Liddell October 2, 2012
You are correct, this can be very confusing given that the information is erroneous. I saw a webpage that could be the source for this wrong information. I'll fix it to say 0-180 180-360 and so on and add that this is the path-length phase difference. Maxbezada ( talk) 20:47, 30 January 2013 (UTC)
Isn't this statement self contradictory: "If a reflective object is tangent to the 1st zone, the electromagnetic wave will be shifted 180o because of the increased path length, undergo an additional 180o phase shift due to the reflection, and reinforce the direct wave at the receiver. Consequently, there should be no reflective objects in the 1st Fresnel zone."
I read it to say that reflections tangent to the 1st zone reinforce the direct wave, so don't have any reflective objects there. Seems like they would enhance the signal at the receiver so you should have many. Jimbo ( talk) 01:57, 15 December 2016 (UTC)
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: [1] 150.227.15.253 ( talk) 15:48, 4 February 2015 (UTC)
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):
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
= 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)
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)
The comment(s) below were originally left at Talk:Fresnel zone/Comments, and are posted here for posterity. Following several discussions in past years, these subpages are now deprecated. The comments may be irrelevant or outdated; if so, please feel free to remove this section.
It is not necessary to explain this in such technical terms. |
Last edited at 23:53, 28 January 2007 (UTC). Substituted at 15:36, 29 April 2016 (UTC)
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)
Agreed. This should be corrected to state that radius of the first Fresnel Zone ellipsoid reduces to approximately one-fourth of a wavelength at locations near either of the antennae (focal points). Or for the n'th Fresnel zone: . The radius goes to zero at the points beyond each antenna. PhysBrain ( talk) 22:55, 10 October 2018 (UTC)
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) Mark The Droner ( talk) 22:55, 10 January 2017 (UTC)Mark The Droner
This page has had major issues for over ten years now. I've read it several times, and although parts of it are very well written, other parts, even within the well-written parts, are poorly or downright badly explained using incorrect words and numbers. So after studying it for some time, I took it upon myself to rewrite parts of the first two sections. I also added a Polarity section, as it is impossible to conclude what happens to a signal bouncing in the Fresnel zone without discussing its polarity. Finally, I really don't like either of the first two graphics. Although they do give the reader an idea of the different regions of the Fresnel zone, there are errors in each which confuse the reader. For example, the first graphic shows n=1, but "n" is nowhere to be found in the graphic. And nothing in the explanation below the graphic makes any sense to me. In the second graphic, the placement of the numbers 1, 2, and 3 are nothing short of horrible and will do nothing but completely confuse the reader. Also, it shows the antennas touching the second Fresnel zone which is impossible. I would redraw both of these graphics but I have no such tools to do so. Perhaps somebody with such illustrative graphic tools and a sound understanding of the Fresnel zone regions could step in and replace these graphics. — Preceding unsigned comment added by Mark The Droner ( talk • contribs) 12:03, 11 January 2017 (UTC)
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)
As a WISP operator of twenty years and counting, ten years supporting AT&T Long Lines microwave paths in the early '70s, and almost fifty years total of radio propagation experience ... I can objectively say the current page has many errors.
There are so many errors and misunderstandings, it's impossible to edit anything on the page. I may start a new paragraph called CORRECTIONS and let the erroneous information as is.
To start ... The equation assumes an isotropic starting and ending points which don't exist in the real world; it mentions obstructions in the path which is incorrect as the equation deals with reflections, not refractions or obstructions.
WISP operators treat the equation as the holy grail of propagation requirements, when in reality I have seen it in practice only once in my life. An operator asked my opinion of his path equations before spending $20,000 in a mid-span repeater link - I ran my version of the equation and showed his initial assessment was flawed ... I saved him $20,000.
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)
I'm adding a flashlight analogy to this talk section to help confused readers who wander in here understand Fresnel Zones. Do not confuse this analogy with the behavior of light waves, as this particular example is specific to radio waves transmitted in vertical polarity (know that most radio waves are transmitted in vertical polarity).
Stand in a rectangular room at ground level. We'll say this entire room is in the 1st Fresnel Zone. Your associate is standing in the far end of the room and is holding a flashlight. You stand in the other end of the room. The flashlight is the transmitter, your eyes are the receiver.
There is a large mirror laying on the floor.
Your associate moves the flashlight up and down and up and down in rhythm. This flashlight mimics the energy of any vertical dipole antenna which would transmit the energy up and down in the form of a sine wave. Your eyes see the light from the flashlight directly. This is the primary signal. The flashlight appears to be moving up and down and up and down in rhythm.
Now look in the mirror on the floor. The same flashlight is moving as seen in the mirror. But it does not appear to be moving up and down and up and down. Instead, the image is inverted. The flashlight appears to be moving down and up and down and up.
This is what happens to a vertical polarity wave when it deflects off the ground. The wave becomes inverted.
Both these "signals" are in the first Fresnel Zone which means, by definition, there is very little change in phase shift. That's because the distance between the lamp and the eyes is very close to the distance between the lamp and the mirror plus the mirror and the eyes. We could say, for example, the primary distance is 100 and the secondary distance is 100.1.
However, because of the deflection of the mirror, the secondary signal, which is inverted (or upside down) is completely out of phase with the primary signal, and effectively cancels out the sine wave of the primary signal. Hence, we have destructive interference due to the deflective surface of the mirror.
In the real world, an antenna installer facing such a problem might consider raising the length of the transmitting antenna, or receiving antenna, or both antennas, which would effectively move the deflective object out of the first Fresnel Zone and into the second Fresnel Zone.
Back to our experiment.
Now, remove the mirror. Run the same experiment. Ignoring the floor, walls, and ceiling, you have a wonderful line-of-sight signal with no interference. Perfect!
Next, place the mirror on the left wall. Run the same experiment. Now you can see the light moving in the mirror. Let's say this distance, again, is 100.1. This time, the flashlight appears to be moving up and down and up and down exactly in tandem with the "primary signal." Hence, the signal is even better than before, it's a stronger signal, and there is no phase shift problem. It's perfect. It's even better than perfect.
All the above has happened in the first Fresnel Zone and therefore is pretty straight forward.
Now let's look at the second Fresnel Zone. Here's where it gets tricky.
The primary signal is exactly the same as before. It's the same distance. Let's say it's 100.
But now the left wall of the house is gone and the mirror is hung on the left side stockade fence outside. The distance that the secondary signal travels is greater than before, it's now 100.5. And this is enough distance to cause a phase shift relative to the primary signal. The phase shift of the deflected secondary signal is one half wave length out of sync with the primary signal, and therefore, cancels out the primary wave and is destructive interference.
Now move the mirror onto the exterior wall of the neighbor's house. This represents the third Fresnel Zone. The distance of the secondary signal is now 101. This is enough distance to cause a phase shift of exactly one wave length - which means there is no effective phase shift. The arriving waves of the primary and secondary signals are exactly aligned. This is constructive interference.
One could create dozens of such scenarios, but the concept is the same. Mark The Droner ( talk)17 February 2024