Talk:Radio propagation

From Wikipedia, the free encyclopedia
Radio Portal
This article is supported by the Radio WikiProject.

This project provides a central approach to Radio-related subjects on Wikipedia.
Please participate by editing the article attached to this page and help us assess and improve articles to good and 1.0 standards. Visit the wikiproject page for more details.
B This article has been rated as B-Class on the Project's quality scale.
(If you rated the article please give a short summary at comments to explain the ratings and/or to identify the strengths and weaknesses.)
This article is within the scope of WikiProject Amateur radio, which collaborates on articles related to amateur radio technology, organizations, and activities. If you would like to participate, you can edit the article attached to this page, or visit the project page, where you can join the project and/or contribute to the discussion.
B This article has been rated as B-class on the quality scale.
Top This article has been rated as Top-importance on the importance scale.

plus the horizon distance corresponding to the height of the receiver antenna (with adjustments if the terrain is not flat).

Why removed? - Patrick 12:37 18 Jul 2003 (UTC)

Contents

[edit] Conduction?

Radio propagation is *influenced* by the conductive properties of the earth and ionosphere but does not rely on them. Radio waves propagate through a vacuum without movement of charge carriers.--Wtshymanski 20:37, 2 December 2005 (UTC)

Radio propagation is made by electrical conduction through a transmission medium. On the earth, electrical conduction in a media. In space, it is a vacuum. JDR 20:46, 2 December 2005 (UTC)

No, you are mistaken. Radio propagation does not occur by electrical conduction. In particular, when radio waves or light or any other EM radiation propagate in a vacuum there are no charge carriers moving, as Wtshymanski said. It's a little more complicated when a wave propagates through a medium, since then electrons in the medium do move in response to the wave, and this affects the propagation of the wave. The electrons do not, however, propagate along with the wave—they move perpendicularly to the direction of wave propagation. Fundamentally, propagation of EM waves is not an electrical conduction phenomenon. --Srleffler 22:49, 7 December 2005 (UTC)
Precisely. You can't see down a copper wire, for example. --Wtshymanski 03:44, 8 December 2005 (UTC)


Cleaned up the topics on troposcatter, meteor scatter, and diffraction, adding more generic descriptions of all three. Added separate category for meteor scatter. --Highsand 23:32, 2 September 2006 (UTC)


I think this article needs a bit of work - I'll just mention one sentence that jumped out at me.

"The far-field magnitudes of the electric and magnetic field components of electromagnetic radiation are equal,and their field strengths are inversely proportional to distance."

To me this is a nonsense sentence. How are A/m and V/m equal? —Preceding unsigned comment added by 71.63.206.237 (talk) 23:44, 10 November 2007 (UTC)

Agree, it makes no sense, nor is it relevant to the context. I have never really read this article until you brought that up. The equation adds nothing but 'intimidation'. The first paragraph is repetitious and self contradicting. This article is in need of attention, I wish I had nothing else to do right now. John 05:47, 12 November 2007 (UTC)

[edit] VSWR

I have to disagree that doubling the distance from the transmitter means the signal strength is reduced to (exactly) one quarter - the same cannot inherently be true in radio signal propagation - being that electromagnetic transmission is subject to the same velocity constraints as the speed of light (providing no distortion) - radio signal strength (VSWR) and distance (distortion) go hand in hand with power signal integrity - therefore transmission is reduced to "NEARLY" one quarter. --Lperez2029 23:06, 25 November 2006 (UTC)

It sounds like most of that whole paragraph was lifted from a textbook of some description, see my edit below. MonstaPro 03:08, 23 April 2007 (UTC)

[edit] Underwater propagation

It would be good to have some information about radio propagation through other media than Earth's atmosphere, in particular submarine radio propagation. 131.111.8.103 15:05, 11 March 2007 (UTC)

[edit] Inverse-square law, format of

Thought I'd rearrange it so the mathematical formula is easier to read. MonstaPro 03:08, 23 April 2007 (UTC)

Also added link to Google Maps as paradigm of increased stability in satellite links (yes, this occasionally conjecturous!). MonstaPro 03:15, 23 April 2007 (UTC)
What does Google Maps have to do with satellite link stability? --Drizzd 09:04, 14 June 2007 (UTC)

[edit] Signal strength(?) vs. field strength

(I first asked this over at Watt, but it was suggested that this might be a better venue P=)

Ignoring gains, losses and free-space loss for the moment, field strength is typically defined as \scriptstyle{V/m=\frac{V}{4\pi\,m}=\sqrt{\frac{4\pi 30W}{4\pi\,m^2}}=\frac{\sqrt{30W}}{m}\,\!} or \scriptstyle{W/m^2=\frac{W}{4\pi\,m^2}.\,\!} I realize field strength defines the surface area of a "power sphere" (inversely, \scriptstyle{W=\frac{W}{m^2}4\pi\,m^2\,\!}), but when attempting to define the strength of a signal at a receiver's input point (i.e., antenna connection point), isn't the situation like a ball resting on a flat surface?: The ball's contact to the surface is only at a single point (or very small surface area around the point), not the whole surface area of the ball, thus the more relevant measurement is the "strength/intensity" of the radius <touching the flat surface (i.e., the particular "flux line"). Wouldn't the strength of the flux line be \scriptstyle{\frac{W}{m}\,\!} and be considered the "signal strength" (signal strength seems to confirm that, but other articles and outside sources seem to use signal and field strength interchangeably)?

Let W_l=W/m=\frac{W}{m};\qquad\;W_a=W/m^2=\frac{W}{4\pi\,m^2}.\,\!

Let's say you have a 100kW station that is 1km away, giving the following results:

100,000W @ 1000m 100Wl .007958Wa

If you have ten other stations with different ERPs and distances, five equaling Wl and five Wa,

Line Area
10,000W @ 100m 100Wl .07958Wa
1000W @ 10m 100Wl .7958Wa
100W @ 1m 100Wl 7.958Wa
10W @ .1m 100Wl 79.58Wa
1W @ .01m 100Wl 795.8Wa
Line Area
1000W @ 100m 10Wl .007958Wa
10W @ 10m 1Wl .007958Wa
.1W @ 1m .1Wl .007958Wa
.001W @ .1m .01Wl .007958Wa
.00001W @ .01m .001Wl .007958Wa

and you measured the strength of the signal right at the antenna connection point (but not through any antenna, just "barefoot") of a receiver, which set of five stations would equal the same strength of the 100kW station, the 100Wl or .007958Wa set? If it is Wl, would it be the same for any antenna attached or, as antenna length/size increases, is that when area (i.e., Wa) comes into play? Now let's say there is an eleventh, distant station, say 500km away (well beyond its prescribed field strength), coming in via, say, either ducting or E-skip, wouldn't that be the same 100Wl or .007958Wa?
Finally, incorporating free-space loss (where λ is wavelength), W/m^2=\frac{W}{4\pi\,m^2}.\,\! becomes W/m^2=\frac{W}{4\pi\,m^2}\cdot\frac{\lambda^2}{4\pi}=W\left(\frac{\lambda}{4\pi\,m}\right)^2.\,\!.

Does that mean W/m\,\! becomes \frac{W\lambda}{\sqrt{4\pi}\,m}\,\! or just \frac{W\lambda}{m}\,\!?
 ~Kaimbridge~16:41, 24 April 2007 (UTC)

The strength of the "flux" or field strength, is actually measured in volts per meter, which as you indicate drops off inversely proportionally to the distance. So an antenna receving a signal will have a voltage produced that drops off proportionally to the distance from the transmitter. However the current available from the antenna is proportional to the voltage. The power available is proportional to the voltage multiplied by the current, so that the power from the receiving antennna still falls off inversely propoprtional to the square of the distance from the emiitter. GB

Right, but that's ignoring the actual question: Yes, field strength is usually measured in volts, but it is a constant that \scriptstyle{W=\frac{V^2}{\Omega}\,\!}. I guess the core question is why is it \scriptstyle{\frac{(V/m)^2}{\Omega}\,\!} and not the appearingly more obvious \scriptstyle{\frac{V^2}{\Omega\,m}\,\!}, as distance has nothing to do with the actual conversion of V to W?!? Relatedly, I find it misleading to define \scriptstyle{\operatorname{dB}V/m=20\log_{10}(V/m)\,\!}, as \scriptstyle{20\log_{10}(V/m)=10\log_{10}(V^2/m^2)\,\!}, thus \scriptstyle{\operatorname{dB}V/m\,\!} should actually be \scriptstyle{\operatorname{dB}(V/m)^2\,\!}!
But back to the original question, which way is more fundamentally "right" depends on which of the above two sets of five signals (100Wl or .007958Wa) is true——since only one of them can be! P=) ~Kaimbridge~19:29, 7 May 2007 (UTC)

[edit] Field Strength, Power Flux Density and Signal Strength

There are two inter-related parameters: Field Strength, which has the dimensions Volts per metre, and Power Flux Density, which has the dimensions Watts per square metre. They are related by the formula:

      E^2
PFD = --- 
      Zfs

Where:

  • PFD = Power Flux Density in Watts per square metre
  • E = Field Strength in Volts per metre
  • Zfs = Characteristic Impedance of Free Space, a constant with a value of 377 ohms.

The terms Field Strength and Power Flux Density have these dimensions and are precise engineering terms. The term Signal Strength is more vague. It could mean the same thing as field strength (i.e. relating to an electromagnetic wave ), or it could refer to a signal voltage in a circuit. --Harumphy 13:08, 15 June 2007 (UTC)


[edit] Proposal to Merge this with "Radio Waves"

Radio waves has an section on propagation that needs this material. There may be propagation of other than radio wave material that could be retained here? Ideas? John 19:15, 15 August 2007 (UTC)

  • Oppose Radio propogation is a seperate dicipline, and can go into far more detail on interaction of radio waves with the earth. The Radio waves article should be more general and higher level, including creation, detection, history, natural phenomena. There should be jst too much material to merge them together. Graeme Bartlett 21:54, 15 August 2007 (UTC)

[edit] Metor scatter

Removed the term 'base' and replaced with 'station' base sound's like 'CB slang' station is the proper term. Also removed reference to 'power input' satellite communications do not require allot of 'power' just allot of 'antenna' . --DP67 talk/contribs 14:32, 28 October 2007 (UTC)

fix header. --DP67 talk/contribs 14:46, 28 October 2007 (UTC)

[edit] "During solar cycle minimas, frequencies only up to about 15 MHz are usable"

This is bullcrap. We're in a solar minimum now. CQWW, the largest ham radio contest of the year, was the weekend just gone, and there was PLENTY of actvity on 21 and 28 MHz bands. True, it's mostly single hop, but to imply the bands are unusable is patently false. When the results and reviews of the contest are published this should be eminently confirmed.

Replace with "During solar minima, propagation of higher frequencies is generally worse." 128.232.250.254 12:24, 30 October 2007 (UTC)