Hi:I remember reading somewhere than ELF [Extremely Low Frequency] radiotransmission is inefficient because it requires to much power.If that is the case, wouldn't MW [Medium Wave] radio transmissionrequire even more power?MW and ELF are forms of electromagnetic radiation in the RF spectrum.An photon [or electromagnetic wave] of a higher-frequency has moreenergy than a photon of a lower-frequency.Let's say there are there are two radio transmitters, one emits 2 GHzwaves while the other emits 2 kHz waves. If the two radio transmittersuse the same modulation scheme [AM/FM, etc.] and emit the same amountof photons-per-second-per-square-meter, the 2 GHz transmitter will beusing more watts than the 2 kHz transmitter -- because a 2 GHz photonrequires more power to generate than a 2 kHZ photon. Right?So how would transmitting a lower-frequency radio wave require morepower than transmitting a higher-frequency radio wave?Thanks,Radium

I know this is the BASIC channel how ever, All I can say is WOW!-- "I'd rather have a bottle in front of me than a frontal lobotomy""Daily Thought: SOME PEOPLE ARE LIKE SLINKIES. NOT REALLY GOOD FOR ANYTHING BUT THEY BRING A SMILE TO YOUR FACE WHEN PUSHED DOWN THE STAIRS."

ELF radio needs more watts than MW radio?

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christofire 17 years ago

Well one sixteenth of a wavelength is pretty close to uniform in my book! All antennas exhibit some sort of phase centre, although for some it may appear diffuse when inspected at close range, but at distances further than that practical yardstick the deviation from a perfect spherical wave is small enough to be found _utterly insignificant_ in practice. Indeed, it follows that practical antennas do behave _sufficiently_ like point sources when inspected from a distance (>2D^2/lambda), albeit not-necessarily-isotropic ones. So link budgets for practical terrestrial and satellite microwave links have always been planned using the inverse square law and have always used radiation patterns measured beyond the yardstick distance (or measured closer and the results transformed to the 'far field'), and these patterns have always exhibited finite beamwidths and sidelobes. The proprietors of such links have found this method of planning to be accurate and repeatable, and the sizes of dishes (etc.) have correctly been chosen to achieve the required coverage areas, to yield the required signal-to-noise ratios, and so on.

I'm afraid my understanding is rooted in practical applications of the physics and you will never convince me that a practical microwave antenna can generate an ideal collimated beam, having a beamwidth of zero degrees and no sidelobes, or that a practical microwave link can be planned without using the inverse-square law, which you appeared to state and was the point about which I entered this thread. I suspect there are many others who read this Usenet group who share my understanding.

Please let this disagreement rest now. I have tried hard to retain objectivity but I fear if this branch to this thread is taken further along the same lines my objectivity may give way to facetiousness.

Chris

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jimp 17 years ago

Yes, it is pretty close, but it isn't EXACT, is it?

That's the point.

Yes, the difference falls below the measurement level under some set of conditions, but it isn't EXACT, is it?

My whole point is that the inverse square law is an EXACT theoretical relationship that is APPROACHED in the real and practical world, but is by no means universally applicable in the real and practical world under all conditions.

I'm just totally amazed so many people seem to have a problem distinguishing between the exact, theoretical, mathematical world and the real, practical, world where an approximation is sufficient to build something.

Jim Pennino Remove .spam.sux to reply.

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christofire 17 years ago

You're the one who wrote: 'it is practical to generate a beam that over the distances of interest is collimated well enough that the inverse square law does not strictly apply. Most real microwave links are that way' ... which is incorrect

... and: 'What I have never done is have occasion to use the inverse square law in an RF link calculation'

... both of _your_ statements seemingly relating to the 'real, practical world' and not 'the exact, theoretical, mathematical world' but applying nonsensical notions of perfectly collimated beams from antennas. So it seems the problem distinguishing between those two worlds is yours alone!

Over and out.

Chris

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Timo A. Nieminen 17 years ago

Worse than that, you only get exact inverse square behaviour for a monopole point source. Given that there are no electromagnetic wave monopole point sources ...

This is helped by us not having to be very far from the surface of the planet to not be on the planet. For practical beams, it's hard to get the distance to being really "astronomical", even at optical frequencies.

This thread is such a splendid example of a usenet tempest-in-a-teacup!

Timo

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Edward Green 17 years ago

Ok. So it was my chance to feel morally superior for a second for standing up for the underdog.

But really, to the extent there was a serious response, you might hope it would center on power requirements for transmitting a given measure of information, with some particular attention to the role of the radio frequency used (how's _that_ for a 19th century title?). Instead, it was a catalyst for a series of "I am more cleverer than you"'s about side-bands...

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Edward Green 17 years ago

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To my ignorant brain, there would seem to be two reasonable meanings to "inverse square law behavior".

One would be that you are sufficiently far from the source that the intensity of radiation is well approximated by an inverse square law through a full solid angle.

The other would be that you were at a distance from the (collimated) source such that the intensity of radiation was well approximated by the inverse square law _over a partial solid angle_ : for example, so that the radiation pattern were well approximated by a cone.

It's not clear to me if all sources have an inverse square law regime in the first sense for _any_ distance : can we get sufficiently far from a laser, for example, that it appears equally bright through a full solid angle, even behind the laser?

The answer to this question would seem to be "no", for the following reason: a perfect laser, or collimated monochromatic beam, would apparently have a distribution of photon momenta containing only one value of wave vector k. Obviously no real beam can have this property (for one thing, it would be a plane wave filling all of space, not a collimated beam with finite cross section anywhere). BUT, it is not necessary for photons to be emitted in this unphysical way in order for the answer to the question to be "no": it is only necessary for them to have some distribution which is _not_ uniform in solid angle! This non-uniformity in solid angle will persist to all distances, so the general source will never appear to be a point source, no matter how far we back up from it: at least not in the sense of radiating equally through a full solid angle. Instead, the far field intensity will have the limiting form:

f(phi,psi)/r^2

Am I correct, or am I all screwed up (with appropriate details, please :-).

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Edward Green 17 years ago

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Aha! My idle amateur speculations are corroborated by somebody who at least does a convincing simulation of an expert.

I have the feeling there was some confusion in this thread between a point source and an isotropic point source, and which one was implied when "inverse square law" was mentioned.

ng

That is admirable restraint.

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jmfbahciv 17 years ago

S*it happens. Use it as fertilizer and be productive.

/BAH

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ELF radio needs more watts than MW radio?

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