ELF radio needs more watts than MW radio?
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Green Xenon [Radium] - 25 Jul 2008 01:31 GMT Hi:
I remember reading somewhere than ELF [Extremely Low Frequency] radio transmission is inefficient because it requires to much power.
If that is the case, wouldn't MW [Medium Wave] radio transmission require 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 more energy than a photon of a lower-frequency.
Let's say there are there are two radio transmitters, one emits 2 GHz waves while the other emits 2 kHz waves. If the two radio transmitters use the same modulation scheme [AM/FM, etc.] and emit the same amount of photons-per-second-per-square-meter, the 2 GHz transmitter will be using more watts than the 2 kHz transmitter -- because a 2 GHz photon requires more power to generate than a 2 kHZ photon. Right?
So how would transmitting a lower-frequency radio wave require more power than transmitting a higher-frequency radio wave?
Thanks,
Radium
Jamie - 25 Jul 2008 02:55 GMT > Hi: > [quoted text clipped - 22 lines] > > Radium I know this is the BASIC channel how ever, All I can say is WOW!
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Sjouke Burry - 25 Jul 2008 03:19 GMT >> Hi: >> [quoted text clipped - 23 lines] >> Radium > I know this is the BASIC channel how ever, All I can say is WOW! What do you expect from radium trolling?
Eric Gisse - 25 Jul 2008 03:24 GMT On Jul 24, 4:31 pm, "Green Xenon [Radium]" <glucege...@gmail.com> wrote: [snip]
This kid suffers a terminal case of "doesn't know what he is talking about".
jimp@specsol.spam.sux.com - 25 Jul 2008 03:35 GMT In sci.physics Green Xenon [Radium] <glucegen1x@gmail.com> wrote:
> Hi:
> I remember reading somewhere than ELF [Extremely Low Frequency] radio > transmission is inefficient because it requires to much power. Then your memory isn't worth a crap or what you read was wrong.
<snip crap>
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Sam Wormley - 25 Jul 2008 05:14 GMT > I remember reading somewhere than ELF [Extremely Low Frequency] radio > transmission is inefficient because it requires to[o] much power. That's bullshit.
Spaceman - 25 Jul 2008 05:28 GMT > Hi: > > I remember reading somewhere than ELF [Extremely Low Frequency] radio > transmission is inefficient because it requires to much power. True and not true. Maybe this will help shed some ELF light on the subject. http://en.wikipedia.org/wiki/Extremely_low_frequency
 Signature James M Driscoll Jr Spaceman
Michael Black - 25 Jul 2008 05:38 GMT >> Hi: >> [quoted text clipped - 4 lines] > Maybe this will help shed some ELF light on the subject. > http://en.wikipedia.org/wiki/Extremely_low_frequency Of course, that's not an inefficiency of spectrum.
And the inefficiency of the long antennas means nothing since such low frequencies are used for specific purposes where such low frequencies are the only choices. Given that, the only choice is to use such low frequencies, or not communicate at all.
Michael
Edward Green - 03 Aug 2008 19:17 GMT On Jul 25, 12:28 am, "Spaceman" <space...@yourclockmalfunctioned.duh> wrote:
> > Hi: > [quoted text clipped - 3 lines] > True and not true. > Maybe this will help shed some ELF light on the subject.http://en.wikipedia.org/wiki/Extremely_low_frequency Congratulations. Out of messages 2-7 in this thread, yours was the first which at least tried to contribute some constructive information, instead of a random insult: that is, unless "bullshit", "snip crap" and "doesn't know what he is talking about" now constitute constructive information when written by approved posters.
Geez... I'm beginning to sound like Tom Potter -- but you guys should be ashamed of yourselves. The Wikipedia article does mention some efficiency issues, and you could have discussed whether these were intrinsic to any use of ELF, or just to the transmitting stations used by the USN.
Of course the issue of photon energy seems like a completely misguided idea... although it _might_ have been nice to discuss whether there were any basis for the implicit idea of "constant information per carrier photon", or not -- and why.
Since when does being on the right side of most discussions (let's say for the sake of argument) justify churlish, not to say loutish, behavior. You gentlemen can contribute much more than that. There are other readers besides the OP.
Benj - 03 Aug 2008 19:34 GMT > Since when does being on the right side of most discussions (let's say > for the sake of argument) justify churlish, not to say loutish, > behavior. You gentlemen can contribute much more than that. There > are other readers besides the OP. Come on! This was a "Radium" question! If you don't know that that means "open season", then search the archives a bit!
OK. We are all properly admonished. NOW!!! How about we start discussing the "Jewish Question" !!!! :-)
Edward Green - 03 Aug 2008 22:24 GMT > > Since when does being on the right side of most discussions (let's say > > for the sake of argument) justify churlish, not to say loutish, [quoted text clipped - 3 lines] > Come on! This was a "Radium" question! If you don't know that that > means "open season", then search the archives a bit! <snip provocative material>
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...
jmfbahciv - 04 Aug 2008 14:11 GMT >> Since when does being on the right side of most discussions (let's say >> for the sake of argument) justify churlish, not to say loutish, [quoted text clipped - 3 lines] > Come on! This was a "Radium" question! If you don't know that that > means "open season", then search the archives a bit! sh.t happens. Use it as fertilizer and be productive.
> OK. We are all properly admonished. NOW!!! How about we start > discussing the "Jewish Question" !!!! > :-) /BAH
Androcles - 25 Jul 2008 07:59 GMT | Hi: | [quoted text clipped - 22 lines] | | Radium You are making the classic mistake of confusing quality with quantity, energy with power. Which has more pressure, a 55,000 psi water jet or Niagara Falls? http://www.laserfab.co.uk/cnc_water_jet_cutting.htm
A cigar tip is red hot and will painfully burn your skin. Can you heat a cup of coffee with it?
How big is this ELF photon? http://www.crystalinks.com/solarnews.html Now that's extreme.
kronecker@yahoo.co.uk - 25 Jul 2008 09:40 GMT On Jul 25, 12:31 pm, "Green Xenon [Radium]" <glucege...@gmail.com> wrote:
> Hi: > [quoted text clipped - 22 lines] > > Radium There is the modulation method - AM or FM. FM is more efficient - at least narrowband FM. That's why we use FM for mobile transmitters. You could use supressed-carrier AM of course but that is a sod to demodulated.
The higher the frequency the shorter the distance it can travel for a given power. Therefore VLF can travel round the world and back again! trouble is you may need for ELF an aerial the size of a mountain range! For a 10 Gig frenquency you would need to pump out a heluva lot of power for it to go any distance. Inverse square law.
Would help of you went to study engineering at Uni - then most of your questions would be answered.
K.
jimp@specsol.spam.sux.com - 26 Jul 2008 00:05 GMT In sci.physics kronecker@yahoo.co.uk wrote:
> On Jul 25, 12:31 pm, "Green Xenon [Radium]" <glucege...@gmail.com> > wrote: [quoted text clipped - 24 lines] > > > > Radium
> There is the modulation method - AM or FM. FM is more efficient - at > least narrowband FM. That's why we use FM for mobile transmitters. You > could use supressed-carrier AM of course but that is a sod to > demodulated. Nonsense.
FM is common because it is intrinsically immune to impulse noise and cheap to implement.
Supressed carrier is trivial to demodulate these days but more expensive to do.
> The higher the frequency the shorter the distance it can travel for a > given power. Therefore VLF can travel round the world and back again! Nonsense.
Most long distance terrestrial communication is done on HF.
> trouble is you may need for ELF an aerial the size of a mountain > range! About the only thing you got right.
> For a 10 Gig frenquency you would need to pump out a heluva lot of > power for it to go any distance. Inverse square law. Nonsense.
At 10 Ghz, it is all line of sight and all in the antenna. It doesn't take much power to hear your own signal bounce off the moon at 10 Ghz.
All space communications take place in the Ghz.
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kronecker@yahoo.co.uk - 26 Jul 2008 09:45 GMT On Jul 26, 11:05 am, j...@specsol.spam.sux.com wrote:
> In sci.physics kronec...@yahoo.co.uk wrote: > > On Jul 25, 12:31 pm, "Green Xenon [Radium]" <glucege...@gmail.com> [quoted text clipped - 34 lines] > FM is common because it is intrinsically immune to impulse noise > and cheap to implement. It's cheap but not immune to noise and suseptable to multipath big- time.
> Supressed carrier is trivial to demodulate these days but more > expensive to do. Do tell how...It's not in any text book so maybe we can learn with your advanced knowledge.
> > The higher the frequency the shorter the distance it can travel for a > > given power. Therefore VLF can travel round the world and back again! > > Nonsense. > > Most long distance terrestrial communication is done on HF. You have never heard of the inverse square law obviously. High frequencies are line of site only and can go long distances because you pump out more power. You need to compare apples with apples.
> > trouble is you may need for ELF an aerial the size of a mountain > > range! > > About the only thing you got right. The only thing we agree on - you must be a physicist - no idea about engineering.
As for your moon thing - it's line of site again!! Try communicating from London to New York at 10GHz.
Idiot.
K.
jimp@specsol.spam.sux.com - 26 Jul 2008 17:05 GMT In sci.physics kronecker@yahoo.co.uk wrote:
> On Jul 26, 11:05 am, j...@specsol.spam.sux.com wrote: > > In sci.physics kronec...@yahoo.co.uk wrote: [quoted text clipped - 38 lines] > It's cheap but not immune to noise and suseptable to multipath big- > time. I never said supressed carrier wasn't immune to noise.
As for multipath, all modulation methods are susceptable to it. FM has a slight advantage there with discriminator capture.
> > Supressed carrier is trivial to demodulate these days but more > > expensive to do. > > > Do tell how...It's not in any text book so maybe we can learn with > your advanced knowledge. See any current amateur radio transceiver. There have been IC's to do it for decades.
> > > The higher the frequency the shorter the distance it can travel for a > > > given power. Therefore VLF can travel round the world and back again! [quoted text clipped - 7 lines] > because you pump out more power. You need to compare apples with > apples. The inverse square law applies to isotropic radiators. No real world RF antenna is an isotropic radiator.
Define "high frequencies".
Things don't become line of sight until about 50 Mhz. Most long distance terrestrial communications is done between about 5 Mhz and 30 Mhz, which is HF.
The typical amateur radio transceiver puts out 100 W max in the HF bands.
My log books, and the logs of 100s of thousands of amateur operators are full of contacts around the globe with far less power than 100 W in the 1.6 Mhz to 29 Mhz range.
> > > trouble is you may need for ELF an aerial the size of a mountain > > > range! [quoted text clipped - 3 lines] > The only thing we agree on - you must be a physicist - no idea about > engineering. No, I'm a BSEE and an amateur radio operator for 40 years.
Have you ever seen a HF transmitter much less operated one?
> As for your moon thing - it's line of site again!! Try communicating > from London to New York at 10GHz. That's exactly the point. You can't communicate anywhere that isn't line of sight much over about 100 Mhz no matter how much power you run unless you use some reflective technique such as tropo scatter.
You know nothing about RF communications.
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kronecker@yahoo.co.uk - 26 Jul 2008 22:16 GMT On Jul 27, 4:05 am, j...@specsol.spam.sux.com wrote:
> In sci.physics kronec...@yahoo.co.uk wrote: > > On Jul 26, 11:05 am, j...@specsol.spam.sux.com wrote: [quoted text clipped - 81 lines] > are full of contacts around the globe with far less power than 100 W > in the 1.6 Mhz to 29 Mhz range. No problem there.
> > > > trouble is you may need for ELF an aerial the size of a mountain > > > > range! [quoted text clipped - 21 lines] > > Remove .spam.sux to reply. Amateurs are the worst kind! There is not way to demodulate double side-band supressed carrier (esp at low SNRs). The only way (for analogue that is) is to recover the carrier and this cannot be done since the carrier aint there in the first place!
Try locking a PLL into supressed carrier. So you are talking complete rubbish.
Here is the basic euqation
m.cos(wmt)cos(wct) where wm and wc are the modulating and carrier frequencies and m is the amplitude. Actually what you are probably thinking of is where a BFO is multiplied into this. However this is not stable and not phase-locked to the carrier either.It needs constant adjustment though its a lot better than it was because of DDS and stable crystals that we didn't have a long time back. If you try and limit the signal to recover the carrier then it will work at high SNRs but not at low SNRs.
K.
Michael Black - 26 Jul 2008 23:04 GMT > On Jul 27, 4:05 am, j...@specsol.spam.sux.com wrote: >> In sci.physics kronec...@yahoo.co.uk wrote: [quoted text clipped - 115 lines] > The only way (for analogue that is) is to recover the carrier and this > cannot be done since the carrier aint there in the first place! Well there's a garbling, since it's far more common to see SSBsc, ie Single Sideband with suppressed carrier. Either someone started with a more complicated example for the sake of it, or it's suddenly been reinforced to support a false notion.
A single sideband, and before going on this tangent the talk was of ELF so if any voice modulation is going to go on down there it's going to be SSBsc, is really easy to demodulate. Beat a signal against it, and the sideband translates down to audio. No problem with mistuning, you simply live with an odd sounding signal, a little retuning will fix that.
Note that even if you started with a DSBsc signal, there are plenty of SSB receivers out there perfectly capable of stripping off the unwanted sideband and then the rest of the receiver treats it like it was an SSB signal. Indeed, the only difference is that you wasted the power used for the extra sideband. Sometimes that's fine, since it makes the transmitter simpler.
But even if the discussion truly was DSBsc, demodulation is easy, and has been well described for 50 years.
You don't look for the highly suppressed carrier, you get the information about where to place the locally placed carrier at the receiver by looking at the sidebands. Simple detectors of thirty years ago would take the IF signal in the receiver, and double it in frequency, giving a constant frequency, and divide it down by two to get the needed frequency and it's right there in the middle, derived from the sidebands.
More complicated methods use a dual channel arrangement, with the VCO locked to the outputs of the product detectors. Webb described a practical circuit in CQ magazine about 1957 or 58, and while it used a lot of tubes, it wasn't excessive. With solid state devices, it's far easier.
A lot of portable shortwave receivers made in the past thirty years use a synchronous detector, just what we are talking about, that work just like that 1957 circuit.
Michael
jimp@specsol.spam.sux.com - 26 Jul 2008 23:05 GMT In sci.physics kronecker@yahoo.co.uk wrote:
> Amateurs are the worst kind! There is not way to demodulate double > side-band supressed carrier (esp at low SNRs). Nonsense but irrelevant as virtually no one uses double side band supressed carrier and it has nothing whatsoever to do with the previous discussion.
Double side band was played with about 40 years ago and essentially abandoned as ssb is more efficient both in bandwidth and power.
Most all supressed carrier is done single side band.
Vestigial sideband is used extensively as in analog TV broadcast.
<snip babbling nonsense>
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Michael Black - 27 Jul 2008 00:50 GMT > In sci.physics kronecker@yahoo.co.uk wrote: > [quoted text clipped - 11 lines] > > Vestigial sideband is used extensively as in analog TV broadcast. Though oddly enough, the problem with SSB is that it's hard to tune. Not in terms of receiving something listenable to, but to tune it exactly. There is nothing to lock onto, so one always has to make do with "that's about right". It's fine for voice since mistuning only makes someone sound higher or lower pitched. But music is horrible since you do notice when it's mistuned.
The redundant sideband, as I posted about earlier, allows for perfect tuning of the reinserted carrier. Plus the redundant sideband, with the right detector, allows for a certain level of frequency diversity reception, and of course the redundancy means one sideband may arrive at your receiver without interference while you have to live with what you get if one sideband is sent.
The carrier is the main hog of power at the transmitter, eliminate it and you get a far bigger level of efficiency than going whole hog and getting rid of the extra sideband. Sending the extra sideband gives those advantages.
Michael
jimp@specsol.spam.sux.com - 27 Jul 2008 01:15 GMT In sci.physics Michael Black <et472@ncf.ca> wrote:
> > In sci.physics kronecker@yahoo.co.uk wrote: > > [quoted text clipped - 18 lines] > only makes someone sound higher or lower pitched. But music is > horrible since you do notice when it's mistuned. Why would anyone in their right mind transmit music with SSB?
Music is generally about fidelity which means bandwidth.
One of the primary reasons for using SSB is to reduce bandwidth.
> The redundant sideband, as I posted about earlier, allows for > perfect tuning of the reinserted carrier. Plus the redundant [quoted text clipped - 3 lines] > while you have to live with what you get if one sideband is > sent. I doubt you are going to see much benefit from a frequency diversity of 6 Khz at 10 Mhz.
> The carrier is the main hog of power at the transmitter, eliminate > it and you get a far bigger level of efficiency than going whole > hog and getting rid of the extra sideband. Sending the extra > sideband gives those advantages. Well, it all depends on what it is you are trying to achieve.
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Don Bowey - 27 Jul 2008 07:07 GMT On 7/26/08 5:15 PM, in article hoqsl5-mre.ln1@mail.specsol.com,
> In sci.physics Michael Black <et472@ncf.ca> wrote: > [quoted text clipped - 22 lines] > > Why would anyone in their right mind transmit music with SSB? Because they could, economically. Before the conversion to digital transmission, telecoms transmitted 20kHz audio via SSB coast-to-coast.
> Music is generally about fidelity which means bandwidth. Yes, but that doesn't preclude ssb when you know how to do it well.
> One of the primary reasons for using SSB is to reduce bandwidth. > [quoted text clipped - 15 lines] > > Well, it all depends on what it is you are trying to achieve. cliff wright - 27 Jul 2008 13:27 GMT > In sci.physics Michael Black <et472@ncf.ca> wrote: > [quoted text clipped - 44 lines] > > Well, it all depends on what it is you are trying to achieve. Stone me! What a mess we can get into sometimes here. Lets point out a few facts about VLF signals (below say 50KHz). 1. Yes indeed. You need a "dirty great" antenna. These frequencies have been used since almost the beginning of radio and antenna designs of enormous proportions have gone with them. In 1907/8 Marconi used 45KHz for transatlantic service from Ireland and used an antenna about 3Km long by 1Km wide with about 50 Kw of rotary gap spark power. Later the German staion at Nauen used directly generated 24 KHz and an antenna like a vast skeletal circus tent 200 m high at the centre and 75 m high at the edges covering many hectares of ground. This could be recieved by a crystal set in South America. Megawatts while common to overcome antenna losses are mainly used to make the service as absolutely reliable as possible. Also early means of generation like arc transmitters tended to be easy to build in high power forms. Like the US navy's 0.5MW staions built about the end of WW1, and the arc system was inherently limited to low frequencies. The signal is apparently ducted between the ground and lower ionospher which explains the world wide coverage. ALL these signals have to be low speed telegraphy of some kind. The bandwidth of the tuned antennas alone would preclude the use of modulated signals, and ther simply isn't the spectrum space for sidebands in their usual sense. The lowest frequency I have ever come across, and I have had a professional and amateur interest in it for many years is 9KHz although in practice around 12 KHz is getting near the practical limit. It is of course simply VERY much easier to radiate higher frequencies. As soon as Hams discovered the properties of short waves in the early 1920's most of this ELF disappeared except for specialist applications like submarine communication. BTW I've always wondered what sort of antenna the sub uses for reception? I bet that's still classified perhaps? Cliff Wright ZL1BDA ex G3NIA
christofire - 27 Jul 2008 16:51 GMT >> In sci.physics Michael Black <et472@ncf.ca> wrote: >> [quoted text clipped - 34 lines] > antenna the sub uses for reception? I bet that's still classified perhaps? > Cliff Wright ZL1BDA ex G3NIA Take a look at the RN Radar and Radio Museum pages at http://www.rnmuseumradarandcommunications2006.org.uk/PAGE%2032.htm. There's some information on a variety of different types of LF/VLF receiving antenna, the ALK for example, which apparently used loops attached to a buoy. Lots of reference to the 10 to 40 kHz frequency range here. Also it seems fairly well known that transmission of VLF signals to UK submarines was moved from BT's Rugby station to VTC's Anthorn station http://tx.mb21.co.uk/gallery/anthorn.php in 2003 or thereabouts. The frequency of these transmissions seems to be variously reported as 16 kHz, 19.6 kHz, etc. - perhaps it's FSK.
Incidentally, the remark above 'the arc system was inherently limited to low frequencies' is probably incorrect. The arc was just a means of making and breaking a circuit and history records that some of the earliest demonstrations of radio (Lodge and others) were carried out in lecture theatres using a sparking induction coil and a short dipole antenna (two plates) transmitting to a nearby loop antenna. The average wavelength may have been around a metre or less.
Chris
Michael Black - 27 Jul 2008 17:07 GMT "cliff wright" <c.c.wright@paradise.net.nz> wrote in message
> news:488c699f$1@clear.net.nz... >> [quoted text clipped - 47 lines] > frequency of these transmissions seems to be variously reported as 16 kHz, > 19.6 kHz, etc. - perhaps it's FSK. I seem to recall reading that submarines did generally come close to the surface to receive signals, so I think the buoy system makes sense. And of course, a lot of work has been done on making good receiving antennas at low frequencies, loops and even active elements. You lose too much trying to use a tiny antenna to transmit low frequencies, so you have to keep raising the power and at some point you get virtually no return for the increases. At the receiver, you use amplification to compensate, but it's easier to amplify small signals than power signals.
> Incidentally, the remark above 'the arc system was inherently limited to low > frequencies' is probably incorrect. The arc was just a means of making and [quoted text clipped - 3 lines] > plates) transmitting to a nearby loop antenna. The average wavelength may > have been around a metre or less. Spark seems to be what you are talking about, and yes it was inherently wideband.
The arc transmitter came a bit later, and was limited to relatively low freuqencies (though I'm not sure the limitation was noticed at the time; as previously posted for a while the higher frequencies were dismissed as "useless" so everyone hung around a relatively small slice of the spectrum). It was a real CW transmitter, the arc transmitter was a real oscillator.
I seem to recall there were also transmitters that used mechanical generators to cause a CW signal, and those obviously were limited to low frequencies.
Michael
John Larkin - 27 Jul 2008 19:26 GMT >"cliff wright" <c.c.wright@paradise.net.nz> wrote in message >> news:488c699f$1@clear.net.nz... [quoted text clipped - 81 lines] > > Michael Poulsen arcs typically worked in the 50-100 KHz range, up to a megawatt or so. The arc chamber dwarfed the operators and had massive air and water cooling bits.
Around 1927, Alexanderson alternators (rotating AC generators) were generating 400KW at 24 KHz and 1 KW at 200 KHz. Goldschmidt alternators hit numbers like 200KW at 50 KHz.
John
Edward Green - 03 Aug 2008 19:33 GMT > BTW I've always wondered what sort of > antenna the sub uses for reception? I bet that's still classified perhaps? http://publications.drdo.gov.in/gsdl/collect/defences/index/assoc/HASH01ec/da9eb 1dc.dir/doc.pdf
contains some interesting looking pictures.
Michael Black - 27 Jul 2008 16:54 GMT > In sci.physics Michael Black <et472@ncf.ca> wrote: > [quoted text clipped - 22 lines] > > Why would anyone in their right mind transmit music with SSB? Your efficiency, both in terms of power and in terms of spectrum utilization. You only need one sideband for content, so why send the extra sideband (redundancy aside). Almost forty years ago, shortwave broadcast stations did start talking about and/or playing with SSB, to make better use of their allocated spectrum. Shortwave broadcast stations transmit music as part of their programming.
> Music is generally about fidelity which means bandwidth. > > One of the primary reasons for using SSB is to reduce bandwidth. Which you get when you drop the other sideband, and instant halving of the bandwidth used.
There is nothing inherently narrow bandwidth about SSB. It does tend to be narrow because you are mostly transmitting only voice, and that doesn't take up much bandwidth. And it was certainly easier to restrict bandwdith with SSB, since in the early days those phasing methods were so limited that they couldn't deal with wide bandwidth, and it's easier to make a crystal filter that is narrow than wide.
So long as only voice was used for SSB, nobody gave this thought. But once shortwave broadcasters started to play with SSB, then of course they had to reveal that SSB could indeed be wide bandwidth (yet still narrower than an equivalent DSB signal, since the extra sideband is never sent).
>> The redundant sideband, as I posted about earlier, allows for >> perfect tuning of the reinserted carrier. Plus the redundant [quoted text clipped - 6 lines] > I doubt you are going to see much benefit from a frequency diversity > of 6 Khz at 10 Mhz. But that is precisely what causes a lot of problems with AM with full carrier over long distances. The sideband(s) arrive while the carrier fades, and then there's not enough carrier to properly demodulate it. You have to reinsert the carrier at the receiver, so all those synchronous detectors have determined where to place it by using the sidebands as the information.
This isn't a guess, DSB when demodulated properly is seen as a diversity method. Obviously not as good as transmitting on two very distinct frequencies, but it's there, and a lot simpler than having two transmitters.
Michael
jimp@specsol.spam.sux.com - 27 Jul 2008 17:35 GMT In sci.physics Michael Black <et472@ncf.ca> wrote:
> > In sci.physics Michael Black <et472@ncf.ca> wrote: > > [quoted text clipped - 29 lines] > with SSB, to make better use of their allocated spectrum. Shortwave > broadcast stations transmit music as part of their programming. Yeah, and it isn't what I would call listenable.
> > Music is generally about fidelity which means bandwidth. > > > > One of the primary reasons for using SSB is to reduce bandwidth. > > > Which you get when you drop the other sideband, and instant halving > of the bandwidth used.
> There is nothing inherently narrow bandwidth about SSB. It does > tend to be narrow because you are mostly transmitting only voice, [quoted text clipped - 3 lines] > with wide bandwidth, and it's easier to make a crystal filter > that is narrow than wide. Non sequitur.
First you say it is an instant half the bandwidth, then you say there is nothing inherently narrow about it.
You can't have it both ways.
The early days were a half century ago and crystal filters were mostly replaced with DSP decades ago.
> So long as only voice was used for SSB, nobody gave this thought. > But once shortwave broadcasters started to play with SSB, then > of course they had to reveal that SSB could indeed be wide bandwidth > (yet still narrower than an equivalent DSB signal, since the > extra sideband is never sent).
> >> The redundant sideband, as I posted about earlier, allows for > >> perfect tuning of the reinserted carrier. Plus the redundant [quoted text clipped - 13 lines] > those synchronous detectors have determined where to place it by > using the sidebands as the information. Non sequitur.
You've already said with the right detector you don't need the carrier at all.
> This isn't a guess, DSB when demodulated properly is seen as a diversity > method. Obviously not as good as transmitting on two very distinct > frequencies, but it's there, and a lot simpler than having two > transmitters.
> Michael
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Edward Green - 03 Aug 2008 19:41 GMT On Jul 27, 12:35 pm, j...@specsol.spam.sux.com wrote:
> In sci.physics Michael Black <et...@ncf.ca> wrote:
> > Which you get when you drop the other sideband, and instant halving > > of the bandwidth used. [quoted text clipped - 12 lines] > > You can't have it both ways. Makes sense to me. It's like saying that cutting a suit down the middle makes an instantly more narrow suit, but that there is nothing inherently narrow about a half-suit: it could be half a size 58, vs. a child's size.
Wim Lewis - 02 Aug 2008 09:24 GMT >In sci.physics kronecker@yahoo.co.uk wrote: >> You have never heard of the inverse square law obviously. High [quoted text clipped - 4 lines] >The inverse square law applies to isotropic radiators. No real world >RF antenna is an isotropic radiator. The inverse square law applies to anisotropic radiators, too.
 Signature Wim Lewis <wiml@hhhh.org>, Seattle, WA, USA. PGP keyID 27F772C1
Benj - 02 Aug 2008 10:08 GMT > In article <haurl5-jtg....@mail.specsol.com>, <j...@specsol.spam.sux.com> wrote: > >In sci.physics kronec...@yahoo.co.uk wrote: [quoted text clipped - 7 lines] > > The inverse square law applies to anisotropic radiators, too. I know this is a Radium thread so I hesitate....but you guys are REALLY in need of a radio engineer!
Yes, the inverse square law applies to anisotropic radiators as well IN THE FAR FIELD REGION!
In the "near field" or what some here have suggested is the "faster than light" region it doesn't apply.
VLF transmissions tend to bend around the earth which is one reason they are useful for long distances and found exclusive use in the early days of radio. Higher frequencies (so-called "short wave" ) were discovered to bounce off the ionosphere (more or less) and thus achieved popularity later for long distance transmission using that bounce. Higher frequencies are not reflected back so longer distances are harder to achieve. But this does now mean that over the horizon transmissions are not possible. The high frequency waves tend to be scattered by diffraction sending energy down below the horizon. This is the way that certain over the horizon radars (DEW line) work. But since most of the energy is NOT scattered, HUGE amounts of power are needed.
So.... What if we had just ONE photon at a frequency of ONE Hz, how much carrier power would be needed? Would it be less than the DEW radar? What if that one photon were single sideband?
John Larkin - 02 Aug 2008 17:52 GMT >> In article <haurl5-jtg....@mail.specsol.com>, <j...@specsol.spam.sux.com> wrote: >> >In sci.physics kronec...@yahoo.co.uk wrote: [quoted text clipped - 33 lines] >much carrier power would be needed? Would it be less than the DEW >radar? What if that one photon were single sideband? You are REALLY in need of a physicist!
1 photon per second at 1 Hz is a power level of 6e-34 watts. Somewhat less than the DEW line transmitters.
Single-photon SSB is meaningless.
John
Don Bowey - 02 Aug 2008 23:46 GMT On 8/2/08 2:08 AM, in article d475cb99-1c06-456f-8f29-23a84a7834d6@d77g2000hsb.googlegroups.com, "Benj" <bjacoby@iwaynet.net> wrote:
>> In article <haurl5-jtg....@mail.specsol.com>, <j...@specsol.spam.sux.com> >> wrote: [quoted text clipped - 30 lines] > since most of the energy is NOT scattered, HUGE amounts of power are > needed. I was on the Dew line Extension project in the Aleutians and worked with the tropo radios, etc. Five Watts (the exciter) into a 60 foot parabolic antenna gave us usable communications at 100+ miles, with occasional frequent deep fades. Twenty-five kW got rid of the fade effects.
> So.... What if we had just ONE photon at a frequency of ONE Hz, how > much carrier power would be needed? Would it be less than the DEW > radar? What if that one photon were single sideband? jimp@specsol.spam.sux.com - 02 Aug 2008 16:25 GMT In sci.physics Wim Lewis <wiml@hhhh.org> wrote:
> >In sci.physics kronecker@yahoo.co.uk wrote: > >> You have never heard of the inverse square law obviously. High [quoted text clipped - 4 lines] > >The inverse square law applies to isotropic radiators. No real world > >RF antenna is an isotropic radiator.
> The inverse square law applies to anisotropic radiators, too. So you are saying a perfectly collimated beam follows the inverse square law?
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Benj - 02 Aug 2008 17:20 GMT On Aug 2, 11:25 am, j...@specsol.spam.sux.com wrote:
> > >The inverse square law applies to isotropic radiators. No real world > > >RF antenna is an isotropic radiator. > > The inverse square law applies to anisotropic radiators, too. > > So you are saying a perfectly collimated beam follows the inverse > square law? Yes, he is, Jim! And the reason for that is because a "perfectly collimated" beam simply does not exist!
Even in the case of a single mode TEM 00 Gausian beam laser, the radiation spreads out in the far field. One can arrange things so that the narrow "waist" of the output beam occurs at a distance from the laser, and that beam SEEMS to not follow the inverse square relationship, but the fact is as I pointed out above, the seeming failure is due to being in what is essential the "near field" of the beam. At a great enough distance the beam expands.
jimp@specsol.spam.sux.com - 02 Aug 2008 19:55 GMT In sci.physics Benj <bjacoby@iwaynet.net> wrote:
> On Aug 2, 11:25?am, j...@specsol.spam.sux.com wrote:
> > > >The inverse square law applies to isotropic radiators. No real world > > > >RF antenna is an isotropic radiator. > > > The inverse square law applies to anisotropic radiators, too. > > > > So you are saying a perfectly collimated beam follows the inverse > > square law?
> Yes, he is, Jim! And the reason for that is because a "perfectly > collimated" beam simply does not exist! It certainly does mathematically.
> Even in the case of a single mode TEM 00 Gausian beam laser, the > radiation spreads out in the far field. One can arrange things so [quoted text clipped - 3 lines] > failure is due to being in what is essential the "near field" of the > beam. At a great enough distance the beam expands. If you want to talk practical, 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.
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John Larkin - 02 Aug 2008 21:01 GMT >In sci.physics Benj <bjacoby@iwaynet.net> wrote: >> On Aug 2, 11:25?am, j...@specsol.spam.sux.com wrote: [quoted text clipped - 10 lines] > >It certainly does mathematically. Not for a beam made of waves.
>> Even in the case of a single mode TEM 00 Gausian beam laser, the >> radiation spreads out in the far field. One can arrange things so [quoted text clipped - 9 lines] > >Most real microwave links are that way. Never heard of the Radar Equation? Or done a microwave link budget?
John
jimp@specsol.spam.sux.com - 02 Aug 2008 22:35 GMT In sci.physics John Larkin <jjlarkin@highnotlandthistechnologypart.com> wrote:
> >In sci.physics Benj <bjacoby@iwaynet.net> wrote: > >> On Aug 2, 11:25?am, j...@specsol.spam.sux.com wrote: [quoted text clipped - 10 lines] > > > >It certainly does mathematically.
> Not for a beam made of waves. What part of "mathematically" are you having trouble understanding?
> >> Even in the case of a single mode TEM 00 Gausian beam laser, the > >> radiation spreads out in the far field. One can arrange things so [quoted text clipped - 9 lines] > > > >Most real microwave links are that way.
> Never heard of the Radar Equation? Or done a microwave link budget? Yes, and many, many times.
What I have never done is have occasion to use the inverse square law in an RF link calculation.
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christofire - 02 Aug 2008 23:05 GMT > In sci.physics John Larkin <jjlarkin@highnotlandthistechnologypart.com> > wrote: [quoted text clipped - 41 lines] > > Jim Pennino Your results must have been very inaccurate.
The limited aperture dimensions of metal dishes, horns, splash-plates, helices, Yagis, etc., used for terrestrial microwave links, in terms of the wavelength, means beam widths are generally measured in degrees for frequencies below EHF. Furthermore, intensive frequency re-use has led to extensive use of absorber-lined tubs shrouding dishes to constrain sidelobe/backlobe levels - this technique can increase the edge taper, increasing the beamwidth for a given aperture size.
A beamwidth of one degree, or for that matter, any angle other than zero requires use of the inverse square law for PFD, or an inverse law for field strength.
Chris
John Larkin - 02 Aug 2008 23:14 GMT >In sci.physics John Larkin <jjlarkin@highnotlandthistechnologypart.com> wrote: > [quoted text clipped - 37 lines] >What I have never done is have occasion to use the inverse square law >in an RF link calculation. And you ignore distance?
(Microwave, not RF)
John
Sjouke Burry - 02 Aug 2008 23:46 GMT > In sci.physics John Larkin <jjlarkin@highnotlandthistechnologypart.com> wrote: > [quoted text clipped - 12 lines] > > What part of "mathematically" are you having trouble understanding? Math and nature sometimes fit nicely together, but are not the same. Any "REAL" beam does not have a 100% math description. To do that you have to know it all, and only trolls claim such. So your perfect beam does not exist. Of course you can think of one, you can almost make one, but in the end all math used in describing the real world is an approximation.
All beams have some divergence, and somewhat obey the inverse square law. Just try and aim any laserbeam on a target 3 miles away, and you will find out. Did that once, to make a "really" straight path at a 3 mile airfield. At 3 miles the laserbeam was about 3-6 yards diameter, and moving all over the place,because a mathematically rigid beam also does not exist.
Timo A. Nieminen - 02 Aug 2008 22:44 GMT > In sci.physics Benj <bjacoby@iwaynet.net> wrote: >> On Aug 2, 11:25?am, j...@specsol.spam.sux.com wrote: [quoted text clipped - 10 lines] > > It certainly does mathematically. Not even then, if the beam is required to have finite energy flux [1] and be a solution of the Helmholtz equation (or even the paraxial wave equation).
I think a better phrasing of your original point would have been "The inverse square law applies to the far field of a radiator." Don't like the inverse square law? Stay in the near field, where beams can be collimated rather than becoming spherical waves, or you can get faster than 1/r^2 fall-off from higher-order multipole sources.
But while many (including myself) are picking on the details, your original point that high frequencies can allow highly directive transmission is spot on. Also the point made by another that the Earth-ionosphere waveguide avoids the inverse square law too. Tough luck for the frequencies in the middle - all they're good for is broadcast transmission when you want more bandwidth than you're allowed at the lower frequencies.
[1] Infinite energy flux can deliver unto you infinite plane waves and Bessel beams, which are perfectly collimated, but not physically realisable.
 Signature Timo
jimp@specsol.spam.sux.com - 02 Aug 2008 23:35 GMT In sci.physics Timo A. Nieminen <timo@physics.uq.edu.au> wrote:
> > In sci.physics Benj <bjacoby@iwaynet.net> wrote: > >> On Aug 2, 11:25?am, j...@specsol.spam.sux.com wrote: [quoted text clipped - 10 lines] > > > > It certainly does mathematically.
> Not even then, if the beam is required to have finite energy flux [1] and > be a solution of the Helmholtz equation (or even the paraxial wave > equation).
> I think a better phrasing of your original point would have been "The > inverse square law applies to the far field of a radiator." Don't like the > inverse square law? Stay in the near field, where beams can be collimated > rather than becoming spherical waves, or you can get faster than 1/r^2 > fall-off from higher-order multipole sources.
> But while many (including myself) are picking on the details, your > original point that high frequencies can allow highly directive [quoted text clipped - 3 lines] > transmission when you want more bandwidth than you're allowed at the lower > frequencies.
> [1] Infinite energy flux can deliver unto you infinite plane waves and > Bessel beams, which are perfectly collimated, but not physically > realisable. Most reasonable response yet.
The geometry of the inverse square law is predicated upon radiation from a point source.
The field from a real radiator approaches following the inverse square law as the distance increases such that the divergence approaches that of a point source.
Given a good enough beam former, that distance may become an astronomical distance.
The collimating characteristics of many antennas are such that you can not be on this planet and be far enough away to have enough divergence to approximate inverse square law behavior.
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christofire - 02 Aug 2008 23:37 GMT > In sci.physics Timo A. Nieminen <timo@physics.uq.edu.au> wrote: > [quoted text clipped - 55 lines] > > Jim Pennino Sorry ... you have a screw loose.
Chris
jimp@specsol.spam.sux.com - 03 Aug 2008 00:25 GMT In sci.physics christofire <christofire@btinternet.com> wrote:
> > In sci.physics Timo A. Nieminen <timo@physics.uq.edu.au> wrote: > > [quoted text clipped - 55 lines] > > > > Jim Pennino
> Sorry ... you have a screw loose.
> Chris Sorry you can't understand junior high school geometry.
Let's try it this way:
The inverse square law is predicated upon radiation from a point source.
Any radiator with real area doesn't become a point until you are an infinite distance from it.
Therefor, for a real area at less than an infinite distance, the inverse square law is an approximation to what really happens.
The quality of the approximation depends on the real area and the distance from it.
Got it now?
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christofire - 03 Aug 2008 12:49 GMT > In sci.physics christofire <christofire@btinternet.com> wrote: - snip -
>> > The field from a real radiator approaches following the inverse >> > square law as the distance increases such that the divergence [quoted text clipped - 33 lines] > > Jim Pennino Radiation from an aperture antenna is a matter of diffraction. At any point of inspection, the field strength (and therefore the PFD) is the result of integrating contributions from the distribution of current in the aperture. If it's a one-dimensional aperture, like a thin wire dipole, then the integral is over that one dimension; if it's a two-dimensional aperture, like a dish, the integral is over both dimensions. This is a little more complicated than 'high school geometry' but the calculus involved can often be kept quite simple - it all depends on the form of the current distribution.
Away from the antenna, simple geometry indicates that the path lengths over which the different components act are different, so the result of the integral will depend on the distance between the point of inspection and the antenna. However, it has been found for many practical implementations of dish, etc., that when the difference between path lengths from the centre and edges of the (normal) aperture is one sixteenth of the wavelength then the result is essentially independent of the distance. The distance at which this condition is achieved is the well-known yardstick 2D^2/lambda, and, for example, this equates to 360 metres for a 3 metre diameter dish working at 6 GHz; perhaps the largest size used for terrestrial microwave links. Therefore, radiation pattern measurements need to be carried out using a measurement range at least this long for the patterns to be generally applicable.
Of course you have the prerogative to apply different interpretations of the physics to your own experiments and if your interpretation brings theory and practice together for the general case then it has some validity. What I've described above is a small part of the physics that's actually used in the design of communication links, broadcasting, radar, etc. There are well-known simplifications like 'ray optics' that neglect diffraction phenomena, but these are strictly applicable only to the case of infinitesimal wavelength, so they might be useable in optics but they cause approximation when applied to systems with apertures of only 60 wavelengths - the above example. When the radiation from a dish is predicted using ray optics, the ray paths from the feed to the reflector all end up parallel in the aperture so this gives the impression of a 'collimated' parallel-sided beam ... but this is a major approximation and is certainly not what is measured in practice.
Chris
jimp@specsol.spam.sux.com - 03 Aug 2008 18:05 GMT In sci.physics christofire <christofire@btinternet.com> wrote:
> > In sci.physics christofire <christofire@btinternet.com> wrote: > > > - snip -
> >> > The field from a real radiator approaches following the inverse > >> > square law as the distance increases such that the divergence [quoted text clipped - 33 lines] > > > > Jim Pennino
> Radiation from an aperture antenna is a matter of diffraction. At any point > of inspection, the field strength (and therefore the PFD) is the result of [quoted text clipped - 5 lines] > be kept quite simple - it all depends on the form of the current > distribution.
> Away from the antenna, simple geometry indicates that the path lengths over > which the different components act are different, so the result of the [quoted text clipped - 9 lines] > using a measurement range at least this long for the patterns to be > generally applicable.
> Of course you have the prerogative to apply different interpretations of the > physics to your own experiments and if your interpretation brings theory and [quoted text clipped - 10 lines] > 'collimated' parallel-sided beam ... but this is a major approximation and > is certainly not what is measured in practice.
> Chris The inverse square law is predicated upon a spherical wave front.
A true spherical wave front is obtained only from a point source.
Point sources don't exist in the real world.
At sufficient distance, the difference between the real wave front and a true spherical wave front approaches the limits of measurment, and therefor the inverse square law provides a value indistiguisable by MEASUREMENT from the "real" value at sufficient distance.
The use of the inverse square law is always an approximation in the real world, though granted sometimes a very good approximation.
The inverse square law is a THEORETICAL relationship and 2D^2/lambda is a PRACTICAL rule of thumb.
Got the point yet?
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christofire - 03 Aug 2008 20:17 GMT > In sci.physics christofire <christofire@btinternet.com> wrote: > [quoted text clipped - 78 lines] > > Jim Pennino 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
jimp@specsol.spam.sux.com - 03 Aug 2008 20:45 GMT In sci.physics christofire <christofire@btinternet.com> wrote:
> > In sci.physics christofire <christofire@btinternet.com> wrote: > > [quoted text clipped - 78 lines] > > > > Jim Pennino
> Well one sixteenth of a wavelength is pretty close to uniform in my book! Yes, it is pretty close, but it isn't EXACT, is it?
That's the point.
> 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 Yes, the difference falls below the measurement level under some set of conditions, but it isn't EXACT, is it?
> not-necessarily-isotropic ones. So link budgets for practical terrestrial > and satellite microwave links have always been planned using the inverse [quoted text clipped - 5 lines] > 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 [quoted text clipped - 3 lines] > about which I entered this thread. I suspect there are many others who read > this Usenet group who share my understanding. 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.
> 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 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.
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christofire - 03 Aug 2008 21:47 GMT > In sci.physics christofire <christofire@btinternet.com> wrote: >> > In sci.physics christofire <christofire@btinternet.com> wrote: [quoted text clipped - 8 lines] > practical, > world where an approximation is sufficient to build something. 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
Edward Green - 03 Aug 2008 22:59 GMT > 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 [quoted text clipped - 4 lines] > when inspected from a distance (>2D^2/lambda), albeit > not-necessarily-isotropic ones. 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.
<...>
> 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. That is admirable restraint.
Timo A. Nieminen - 03 Aug 2008 21:56 GMT > The geometry of the inverse square law is predicated upon radiation > from a point source. 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 ...
> The field from a real radiator approaches following the inverse > square law as the distance increases such that the divergence [quoted text clipped - 6 lines] > not be on this planet and be far enough away to have enough divergence > to approximate inverse square law behavior. 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!
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Edward Green - 03 Aug 2008 22:50 GMT On Aug 2, 6:35 pm, j...@specsol.spam.sux.com wrote:
> In sci.physics Timo A. Nieminen <t...@physics.uq.edu.au> wrote: > [quoted text clipped - 48 lines] > not be on this planet and be far enough away to have enough divergence > to approximate inverse square law behavior. 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 :-).
Jasen Betts - 03 Aug 2008 11:07 GMT > In sci.physics Benj <bjacoby@iwaynet.net> wrote: >> On Aug 2, 11:25?am, j...@specsol.spam.sux.com wrote: [quoted text clipped - 10 lines] > > It certainly does mathematically. only if it has infinite width.
consider the difraction that when your perfectly collimated beam passes through an aperature that exactly matches its size
contrast that with what happens when it doesn't.
Bye. Jasen
jimp@specsol.spam.sux.com - 03 Aug 2008 18:05 GMT In sci.physics Jasen Betts <jasen@xnet.co.nz> wrote:
> > In sci.physics Benj <bjacoby@iwaynet.net> wrote: > >> On Aug 2, 11:25?am, j...@specsol.spam.sux.com wrote: [quoted text clipped - 10 lines] > > > > It certainly does mathematically.
> only if it has infinite width. Which is quite possible mathematically.
> consider the difraction that when your perfectly collimated beam passes > through an aperature that exactly matches its size
> contrast that with what happens when it doesn't.
> Bye. > Jasen Concider what happens when the perfectly collimated beam impacts a half gallon of Chunky Monkey ice cream.
However, I'm not sure what either has to do with a wave front propogating in free space.
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Benj - 03 Aug 2008 19:45 GMT On Aug 3, 1:05 pm, j...@specsol.spam.sux.com wrote:
> In sci.physics Jasen Betts <ja...@xnet.co.nz> wrote: > [quoted text clipped - 27 lines] > However, I'm not sure what either has to do with a wave front propogating > in free space. Has everything to do with it. Mathematics is not "reality". In the real world, things are not "infinite" or "point sources" or other mathematical concepts. So for the most part this argument is stupid. But you were the one saying that the inverse square law did not apply to an anisotropic beam. We say it pretty much did at a distance. Then you say only at "infinity" and then we have to answer how close of an approximation to reality do you want your math to be? So you say how far if not infinity? Astronomical distances? And we say "could be", and we note the distance needed is that which takes you into the "far field" of the source. So how far away is that? Well how large is the source? If the source is "infinite" then a "perfectly collimated" beam is mathematically possible. But who cares since "infinite objects" are not possible. All you are doing is arguing how many angels can stand on the head of a pin!
The bottom line is all of this is that radiation sources creating propagating radiation can be replaced by an "equivalent aperture" at some point and as noted, an aperture source actually has a transform relationship between the aperture "source" and the propagating beam. At a sufficient distance the expansion of the beam is always dictated by those mathematics. Hence all arguments of "collimated beams" not following inverse square laws at some location is invalid. The questions of how closely are the laws followed or how far you have to be away are merely practical fine points. It is your "natural philosophy" that were are taking issue with here.
jimp@specsol.spam.sux.com - 03 Aug 2008 19:55 GMT In sci.physics Benj <bjacoby@iwaynet.net> wrote:
> On Aug 3, 1:05?pm, j...@specsol.spam.sux.com wrote: > > In sci.physics Jasen Betts <ja...@xnet.co.nz> wrote: [quoted text clipped - 28 lines] > > However, I'm not sure what either has to do with a wave front propogating > > in free space.
> Has everything to do with it. Mathematics is not "reality". Bingo, you finally got the point, almost.
Theoretical mathematics is not reality and the inverse square law is theoretical mathematics.
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Bob Masta - 25 Jul 2008 13:53 GMT >Hi: > >I remember reading somewhere than ELF [Extremely Low Frequency] radio >transmission is inefficient because it requires to much power. I suspect what you actually read about was the US military scheme to communicate with submerged submarines using high-power VLF that could penetrate seawater. At one point there were plans to use huge underground ore veins in Michigan's Upper Penninsula as the transmitter antenna. That might have been a tad inefficient!
There's also the issue of low carrier frequencies not supporting high symbol rates. I seem to recall that the submerged subs would get only code, at rates so slow even the rankest beginning amateur operator would have had no trouble keeping up... <g>
Best regards,
Bob Masta DAQARTA v4.00 Data AcQuisition And Real-Time Analysis www.daqarta.com Scope, Spectrum, Spectrogram, Sound Level Meter FREE Signal Generator Science with your sound card!
hhc314@yahoo.com - 25 Jul 2008 18:04 GMT > On Thu, 24 Jul 2008 17:31:55 -0700 (PDT), "Green Xenon [Radium]" > [quoted text clipped - 25 lines] > FREE Signal Generator > Science with your sound card! Shush Bob, you're deliving into 100% reliable long distance communications techniques first discovered and exploited in the 1950s and which have remained classified ever since. As you point out, the information transmission rate is incredibly low, but extremely reliable and nearly impossible to block or jam, and it reliably reaches every location on earth, surface, underwater, or underground, at extremely low baud rates.
Now what would anyone imagine what purpose such a limited system might be used for? This is precisely why VLF systems of this type remain highly classified. Yes, correct. They have only one practical application. Let's simply call it the Fed Ex principle, which applies to things that must be reliably delivered on schedule.
Harry C.
Igor - 26 Jul 2008 00:29 GMT On Jul 24, 8:31 pm, "Green Xenon [Radium]" <glucege...@gmail.com> wrote:
> Hi: > [quoted text clipped - 22 lines] > > Radium I'm not sure about output power, but I do know that the lower the frequency, the longer the wavelength, and hence the longer the transmitting antenna. ELF requires a humongous antenna.
Wim Lewis - 26 Jul 2008 10:38 GMT >I remember reading somewhere than ELF [Extremely Low Frequency] radio >transmission is inefficient because it requires to much power. > >If that is the case, wouldn't MW [Medium Wave] radio transmission >require even more power? ELF can be inefficient because the wavelength is so large that it's hard to build an efficient antenna for it. On the transmitting side, this means that only a small amount of the current sloshing around in the antenna gets coupled out into free-space radiation; as a result you need higher currents in the antenna to get the same radiated output, and that means more losses to things like resistive heating.
If you can build an antenna that's appropriately sized for the ELF wavelength (hundreds or thousands of kilometers) then you can avoid this. The US used to have a couple of giant ELF antennas in the Midwest; Wikipedia says they were disassembled earlier this decade. I don't know how submarines are signaled these days.
Anyway, MW has a shorter wavelength than ELF, so it's easier to build a good antenna for that band.
>An photon [or electromagnetic wave] of a higher-frequency has more >energy than a photon of a lower-frequency. [....] The important thing is usually not how many photons can be emitted, but how much energy can be picked up by the receiver compared to the amount of noise it's also picking up. The amount of energy per photon really isn't significant, at least not for radio. The energy of a single photon is insignificant compared to the power of the Force ... errr ... I mean, it's tiny compared to the amount of energy you need to be heard over the background noise.
 Signature Wim Lewis <wiml@hhhh.org>, Seattle, WA, USA. PGP keyID 27F772C1
John Larkin - 26 Jul 2008 17:23 GMT >Hi: > [quoted text clipped - 15 lines] >using more watts than the 2 kHz transmitter -- because a 2 GHz photon >requires more power to generate than a 2 kHZ photon. Right? Sure; for constant photon rates, one transmitter is outputting, say, 1 million watts, and the other is doing 1 watt. But broadcasters don't pay for photons, they pay for watts.
>So how would transmitting a lower-frequency radio wave require more >power than transmitting a higher-frequency radio wave? Any transmitter can run at any power level, and photons don't matter. The problem with ELF is that an efficient antenna is enormous, and elf waves don't bounce off the ionosphere like mw waves do, so most of the energy cruises right out into space.
Huge elf rigs were/are used for loran-C, WWVB, and communicating with atomic subs. They use inefficient ground-wave propagation, so need huge power levels. The loran-C station north of San Francisco is about 100 KHz at some megawatts peak power.
Really, you should look some of this stuff up.
http://en.wikipedia.org/wiki/LORAN
http://en.wikipedia.org/wiki/Longwave
Read "Tuxedo Park"...
http://en.wikipedia.org/wiki/Alfred_Lee_Loomis
John
Michael Black - 26 Jul 2008 19:12 GMT >> Hi: >> [quoted text clipped - 32 lines] > huge power levels. The loran-C station north of San Francisco is about > 100 KHz at some megawatts peak power. Not really. They use frequencies that will have good penetration, and which don't suffer much from radio conditions.
Because the frequencies are so low, that ground-wave will be considerable, while higher frequencies need to bounce off the ionosphere and such to get the same sort of distance. But that is unreliable, and of course often is dependent on the time of the day.
Those frequencies are terribly reliable, interference aside.
Once that choice is made, then they have to live with the inefficiencey. They have decided there is no other choice for their needs, and then compensate with the high power to overcome the inefficiency of the antennas.
As an example, there's a broadcast station in Ottawa on 580KHz that comes in fine during the day. But at night, they are required to cut back on their power, so the station goes away, not enough power for ground wave to Montreal, while bouncing off the ionosphere results in a bounce too far away. They have to cut back their power at night so the better propagation at night does not leave them with a booming signal bouncing off the ionosphere to interfere with all the other stations on that frequency. I'm sure there are plenty of locations much further away that can receive the station at night, that can't receive it at night (after all, I can hear plenty of AM broadcast stations at night from much further away that I could never hear during the daytime). I did receive the Ottawa station at night during one period a decade ago, when an emergency situation allowed them to run full power at night; ground wave reception was fine.
So they run WWVB and such at low frequencies so the reliable ground wave is used for really long distances (I can receive it fine here in Montreal, I've had an "atomic clock" for almost five years and rarely does it not sync up at night). One can argue that their high power is not just because of the inefficient antennas at 60KHz, but so it is receivable so far away, just like that Ottawa station where the ground wave signal disappears when they lower their power.
Michael
John Larkin - 26 Jul 2008 22:57 GMT >>> Hi: >>> [quoted text clipped - 70 lines] >but so it is receivable so far away, just like that Ottawa station >where the ground wave signal disappears when they lower their power. WWVB and Loran-C use low frequencies for phase stability. Ionosphere bounce has bad fading and erratic prop delay; ground wave is very lossy but is much more amplitude and phase stable. Both are being killed by GPS.
1000 watts is enough for SSB communications halfway around the world. You can't do that with a megawatt of ELF.
John
jimp@specsol.spam.sux.com - 26 Jul 2008 23:25 GMT In sci.physics John Larkin <jjlarkin@highnotlandthistechnologypart.com> wrote:
> WWVB and Loran-C use low frequencies for phase stability. Ionosphere > bounce has bad fading and erratic prop delay; ground wave is very > lossy but is much more amplitude and phase stable. Both are being > killed by GPS. Not quite; the decommisioning of the Loran system has been indefinetly delayed and the implementation of a new generation Loran system as a backup for GPS is under study.
> 1000 watts is enough for SSB communications halfway around the world. > You can't do that with a megawatt of ELF. Much less than 100 W is enough for SSB communications halfway around the world.
And depending on the state of the sun, 1 W is often more than enough.
 Signature Jim Pennino
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Edward Green - 03 Aug 2008 19:04 GMT On Jul 24, 8:31 pm, "Green Xenon [Radium]" <glucege...@gmail.com> wrote:
> Hi: > [quoted text clipped - 18 lines] > So how would transmitting a lower-frequency radio wave require more > power than transmitting a higher-frequency radio wave? The problem seems to be, assuming photons have something to do with it: in particu
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