VLF Transmission Using Soundcard

In summary, the conversation discusses the possibility of creating a VLF transmitter by running a software signal generator through a soundcard and into an antenna. It is also mentioned that playing two signals at once could result in a combination of the two waves, rather than two separate carrier waves. The use of bandpass filters at the receiver is suggested to separate these combined waves. The conversation then delves into the topic of PCM (Pulse Code Modulation) and its ability to create a carrier wave for a radio signal by pushing directly to an amplifier and antenna. It is clarified that a PCM signal is a digital representation of an analog signal and therefore, by definition, a square wave. The conversation also explores the potential use of OFDM (Orthogonal
  • #71
sru2 said:
I have a problem with Mr Shannon's work, or perhaps it is an interpretation of it. Read this:
How can R exceed C?

Let me provide an example. Lets assume I open a channel on 20Khz. I define 1 bit to be exactly 1/20000 per second, in other words, a single period. The maximum I can send information (i.e. R) is 1/20000 per second, but the capacity (i.e. C) is 20000 bits per second.

Now this may seem strange until I mention this part. What size is a photon?

A hertz is defined as cycles per second. So a 1Hz photon is the same size as a 20Khz photon, that is, they are both 1 light second long. The wavelength of 14,990m must be packed into that 1 light second. This means, at 20Khz, each period is:

299 792.458m / 14990m = 19.99m

The question then becomes, is a photon 1 light second long, or is it 19.99m long and 20000 represent a 20Khz signal?

Now if this seems odd, consider the following. If the photon was 1 light second long, then every signal, regardless of its frequency, would take the same amount of time to detect (i.e. at least 1 second). Thus, a photon must be a single period.

This means that Mr Shannon is wrong, or his work has been interpreted incorrectly.

Did he understand quantum theory?

Do you? I think this is a bit of a blind ally in your view of the system and that photons are not relevant to any of this classical / mathematical topic. You say that a photon is "a single period". Where is your justification for that? Single period of what? If you have a channel, operating at 10GHz and 1Hz wide, your photons are a million times the energy / frequency of photons in a similar channel that is operating at 1kHz. Your channel, in principle, need not involve EM in the link at all (apart from the EM forces involved between atoms); you could use modulated sound. Discussing the 'extent' of a photon is a very slippery slope. As far as I am aware, photons are regarded as point particles (of no extent), these days and that seems to fit the evidence.

What do you mean by that statement? Do you mean a bitstream in which the bits have a period of 1/20,000s? Do you mean that you are modulating the 20kHz carrier with a binary bitstream with a rate of 20kb/s? Shannon is not concerned with just a binary channel, in any case and the possible bits per channel Hz is not as simple as you imply. You can have a multilevel signal, operating in a given bandwidth and the limit arises, basically from the uncertainty in which discrete level to choose in your receiver in the presence of noise. (Hence the SNR factor). The limit is what you can achieve with infinite computing power / decoding time but the practical limits seem to be never better than than half the attainable limit (Not the Shannon limit).
 
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  • #72
Do you?

Yes, quite well actually.

You say that a photon is "a single period". Where is your justification for that? Single period of what?

I think you are focusing on the wrong thing. The point I was making is that frequencies are quanta of energy. Thus the minimum possible time it takes to detect a signal of a given frequency is the amount of time take to receive that quanta. Thus, the capacity of the channel is limited to how many bits can be represented by the number of quanta delivered in a given time. The maximum rate is always going to be 1 second divided by the frequency of the photon.

Thus, Shannon's work does not make sense as R can never exceed C, in fact, in an ideal scenario R = 1/C for any symbol based communication.


Discussing the 'extent' of a photon is a very slippery slope. As far as I am aware, photons are regarded as point particles (of no extent), these days and that seems to fit the evidence.

There is something very wrong with our notion of hertz, the notion of photon dimensions, speed and distance.

If you return to calculation that I showed in regards to the period of a wave, you will see that anything with a wavelength of less than 1m is deforming space to maintain a constant light second. It implies that speed and distance are variable depending upon energy level and time is a constant in any inertial frame of reference.

What do you mean by that statement? Do you mean a bitstream in which the bits have a period of 1/20,000s? Do you mean that you are modulating the 20kHz carrier with a binary bitstream with a rate of 20kb/s?

20Khz is how much arrives over one second. Each quanta, or packet, of energy carries a portion of that 20Khz in any given time. Let's assume that 20000 packets each carrying 1/20000 of the energy are delivered in 1 second. By dropping every other packing (i.e. not sending it), I have created a binary on-off pattern with a capacity of 20Kbits.

Shannon is not concerned with just a binary channel, in any case and the possible bits per channel Hz is not as simple as you imply. You can have a multilevel signal, operating in a given bandwidth and the limit arises, basically from the uncertainty in which discrete level to choose in your receiver in the presence of noise. (Hence the SNR factor). The limit is what you can achieve with infinite computing power / decoding time but the practical limits seem to be never better than than half the attainable limit (Not the Shannon limit).

It looks like it is based upon classical theory rather than modern quantum theory. As a result, it is no longer applicable.
 
  • #73
"Frequencies are quanta of energy"?? That's what photons are. 'Frequency' is a unit of 1/time.
Are you suggesting some sort of photon gun that can send or not send a photon, according to whether or not you want to send a 0 or a 1? As you know quantum theory "quite well", you will surely appreciate there is quite a lot a randomness / uncertainty involved in the emission of photons. How were you planning to sort out that problem?

I'm afraid that a lot what you write seems to be rather muddled. Do you have an answer for my question "single period of what?". You say I am focussing on the wrong thing but I at least deserve clarification of what you mean by "single period" and how it relates to a photon.
What do your symbols R and C stand for in your statements?

C = B log2 ( 1+S/N)

is how I remember the Shannon theorem,
where

C is the channel capacity in bits per second;
B is the bandwidth of the channel in hertz (passband bandwidth in case of a modulated signal);
S is the average received signal power over the bandwidth (in case of a modulated signal, often denoted C, i.e. modulated carrier), measured in watts (or volts squared);
N is the average noise or interference power over the bandwidth, measured in watts (or volts squared); and
S/N is the signal-to-noise ratio (SNR) or the carrier-to-noise ratio (CNR) of the communication signal to the Gaussian noise interference expressed as a linear power ratio (not as logarithmic decibels).

This says that C/B can be as big as you like if the S/N is big enough. It also implies that it is very dependent on S/N but independent of the symbols you choose to use.

Just because someone discovered quantum theory doesn't alter how your proposed, very classical sound card plus computer, is going to behave. Shannon seems to apply well enough to all the modern communications techniques that are used in 3G and 4G telephony so I reckon you can assume it's right for you too. You will need a lot more work to prove Shannon's work "doesn't make sense".

Also, I really would love to hear something substantial about your magnetic antenna noise reducer. You are strangely silent on that one.
 
  • #74
"Frequencies are quanta of energy"?? That's what photons are. 'Frequency' is a unit of 1/time.

In other words, a single period, as I described earlier. You said:

You say that a photon is "a single period". Where is your justification for that? Single period of what?

I think you are replying too fast and are confusing yourself.

Are you suggesting some sort of photon gun that can send or not send a photon, according to whether or not you want to send a 0 or a 1? As you know quantum theory "quite well", you will surely appreciate there is quite a lot a randomness / uncertainty involved in the emission of photons. How were you planning to sort out that problem?

Already done, at least in terms of light.

http://www.newscientist.com/article/dn7420-light-gun-fires-photons-one-by-one.html

In terms of radio, I'm sure its achievable, but its not important to this discussion at this time.

I'm afraid that a lot what you write seems to be rather muddled. Do you have an answer for my question "single period of what?". You say I am focussing on the wrong thing but I at least deserve clarification of what you mean by "single period" and how it relates to a photon.

I'm not muddled, you are. As I said, I think you were reading it too fast and got your wires crossed.

What do your symbols R and C stand for in your statements?

C = B log2 ( 1+S/N)

is how I remember the Shannon theorem,

That's Shannon-Hartley Theorem. It is derived from Channel capacity which you can read about here:

https://en.wikipedia.org/wiki/Channel_capacity

This leads to this:

https://en.wikipedia.org/wiki/Noisy-channel_coding_theorem

In this theorem, it states that R can exceed C, which is impossible from a quantum mechanics viewpoint.


Just because someone discovered quantum theory doesn't alter how your proposed, very classical sound card plus computer, is going to behave. Shannon seems to apply well enough to all the modern communications techniques that are used in 3G and 4G telephony so I reckon you can assume it's right for you too. You will need a lot more work to prove Shannon's work "doesn't make sense".

It would not apply to a sound card, but my software has applications beyond this project which make this very important.

There is a vast difference between scientific and practical engineering applications. Given this, the equations provided to engineers are simplifications of the scientific equations and this can impact your understanding a process.

Perhaps this description will give you a better understanding. Consider a signal between two antenna of one second in duration at 20Khz. What is actually happening? Well, if we examine it at a quantum level, we observe that there is not a single stream of photons. Because of the amplitude, we actually get a billions of streams of photons, all slightly out of phase, and each of these streams are photons of 20Khz. We need this to overcome losses in the transmission. In effect, each of these streams are independent channels.

In a lossless environment, we can use a single stream of photons and skip every other photon to create a square wave.

Thus the rate, which is one photon at a time, can never exceed the capacity which is the frequency.

How does this apply to a noisy channel?

Well, the capacity is billions of time larger than you think. The "capacity" referred to in the noisy channel coding theorem is scientifically wrong, its a classical notion.

By restricting yourself to this classical notion, you are in effect preventing yourself from improving upon radio sensitivity and design.

Also, I really would love to hear something substantial about your magnetic antenna noise reducer. You are strangely silent on that one.

It seems like you want to be confrontational for some reason. Why? As stated, the principle is correct, but it would take years of research to develop. In the end, it may prove more effective to cool the receiver.
 
  • #75
If the principle of this antenna is "correct" then there must be a peer reviewed reference to it. Without something like that, how can I accept your statement? If the antenna has low resistance then adding magnets or cooling it can have no effect on the noise that appears at the input. (The equivalent input noise resistor which, of course, may be helped along a bit by cooling it.)

What is a "single period"? I am not confused about the fact that you haven't defined it and that you use the words photon and frequency, apparently interchangeably. Where is a reference to the fact that it's "Already done, at least in terms of light."? I'd like to know where photons become binary digits. That confuses me and I'd need more than just your assurances that it's legit. Perhaps a reference might be appropriate here, too.

Re Shannon and noisy channels. All you (and that Wiki article) are saying is that it is, in principle, possible to approach arbitrarily close to the 'achievable limit' by appropriate coding. There is no evidence there that Shannon is "wrong" (unless you can show us some in yet another reference).

I thought this was a serious project until I discover that you are proposing to revolutionise antenna theory, information theory and a number of other things yet you base your experimental evidence on what you are getting with a proprietary sound card and software and a standard PC full of steaming circuitry, right next to your hopefully ultra low noise receiving system. And you say you can't even look at the raw data, apparently.
I hope you accept that the S in S/N is a pretty important factor in any comms system yet you seem to have no intention of investigating this until the last minute, assuming it will all be fine. Your last post seems to belittle Engineers as not capable of understanding basics. But they do tend to know how to get things to work.
 
  • #76
If the principle of this antenna is "correct" then there must be a peer reviewed reference to it. Without something like that, how can I accept your statement?

Don't, its a theory.

What is a "single period"? I am not confused about the fact that you haven't defined it and that you use the words photon and frequency, apparently interchangeably.

It was just an example to make the problem more manageable from a classical viewpoint. It doesn't mean anything.

I also got side-tracked with this one, its irrelevant to this discussion.

Where is a reference to the fact that it's "Already done, at least in terms of light."? I'd like to know where photons become binary digits. That confuses me and I'd need more than just your assurances that it's legit. Perhaps a reference might be appropriate here, too.

Its irrelevant to this discussion, but I will explain. A 20Khz photon cannot be a full 20Khz, as hertz is defined as the number of cycles per second. Thus, a photon is a fraction of 20Khz at a defined energy level. If we send these sequentially over the period of 1 second, we have delivered 20000 cycles. If we now skip every other photon in this one second period, we will have sent a square wave, or binary pulse. The capacity of such a photon stream, in terms of binary, is the number of photons.

Does that explain it better?

Re Shannon and noisy channels. All you (and that Wiki article) are saying is that it is, in principle, possible to approach arbitrarily close to the 'achievable limit' by appropriate coding. There is no evidence there that Shannon is "wrong" (unless you can show us some in yet another reference).

Given the above description, how can the rate (i.e. number of photons per second) exceed the capacity (i.e. total number of photons in one second)?

It can't, so the theorem is scientifically incorrect. From an engineering perspective, it may be functional but that's only because 'capacity' is incorrectly defined.

I thought this was a serious project until I discover that you are proposing to revolutionise antenna theory, information theory and a number of other things yet you base your experimental evidence on what you are getting with a proprietary sound card and software and a standard PC full of steaming circuitry, right next to your hopefully ultra low noise receiving system. And you say you can't even look at the raw data, apparently.

Nobody is revolutionizing anything, other than your understanding of science. You have developed the impression that your knowledge of engineering somehow means that you also understand science that underpins it.

Do you understand that engineering and the equations that an engineer learns are simplifications of the real scientific equations and processes?

In short, engineering is a dumbed down version of science to enable the functional use of equipment developed by scientists.

As for my sound card, that's an entirely different matter. You have a tendency to connect unrelated matters in your mind, which is a hallmark of mental illness.

I hope you accept that the S in S/N is a pretty important factor in any comms system yet you seem to have no intention of investigating this until the last minute, assuming it will all be fine. Your last post seems to belittle Engineers as not capable of understanding basics. But they do tend to know how to get things to work.

You have already been told the reason why. I will just amplify the signal above the noise. Its a non-issue in this project.
 
  • #77
So you are saying that most of the statements you have made are not relevant (reference-free). You seem to be totally confusing photons with information bits, with no referencing to justify it. I thought, from the title, that this thread was all supposed to be around a practical application of your sound card - which is why I have been referring to it. (Seems a reasonable thing to be doing). I may be nuts, but at least I am trying to keep to the point of this thread rather than calling on irrelevancies when challenged on a practical point.
I understand that amplifying signal plus noise doesn't help with the signal to noise ratio. I understand that cooling an antenna is less effective than cooling the front end of the receiver. I understand that signal level is important in assessing the probable signal to noise ratio. You have given me no reason to accept your more flamboyant ideas because none of them are supported by references. We're left with normal communications issues which, as a dumb engineer, I find can be quite challenging enough - and I think you will find that too.
good luck with it.
 
  • #78
So you are saying that most of the statements you have made are not relevant (reference-free). You seem to be totally confusing photons with information bits, with no referencing to justify it.

You do understand how digital information is represented?...you know, high/low, on/off, there/not there, photon/no photon...

This should be obvious, extremely obvious.

I thought, from the title, that this thread was all supposed to be around a practical application of your sound card - which is why I have been referring to it. (Seems a reasonable thing to be doing). I may be nuts, but at least I am trying to keep to the point of this thread rather than calling on irrelevancies when challenged on a practical point.

We got distracted when you brought up Mr Shannon. My software is designed for scientific applications as well as general radio DSP. As such, I had to point out that Shannon's work is not strictly accurate and more a rule of thumb. I probably delved a little to deep in quantum theory for you to keep up, but its quite accurate.


I understand that amplifying signal plus noise doesn't help with the signal to noise ratio.

Your understanding is wrong, at least in regards to this modulation scheme. The signal is not constant, its pulsed on and off. Thus, the only noise entering the signal, from the transmitter, will be during an on signal which only adds to the signal quality rather than interfering with it. The key is setting an appropriate noise floor at the receiver, hence the focus on reception.


I understand that cooling an antenna is less effective than cooling the front end of the receiver.

...again, your understanding is wrong. The whole system must be cooled until the signal gets into the digital domain. If you just cooled the front-end, then you would pick up thermal noise in the antenna and cables feeding into it. Your noise floor is only as good as the hottest part in the chain.

I understand that signal level is important in assessing the probable signal to noise ratio.

It depends on the system requirements. It becomes less of a concern with particular modulation schemes and the ability to amplify.

You have given me no reason to accept your more flamboyant ideas because none of them are supported by references.

Does your brain not work? Do you always defer to authority? Most of what I have said should be obvious, use a calculator, a pencil and a piece of paper. It shouldn't take long to confirm most of what I said.

We're left with normal communications issues which, as a dumb engineer, I find can be quite challenging enough - and I think you will find that too.

Its a walk in the park, set a noise floor on the receiver, squelch it and amplify the transmitter. I think you tend to over-complicate things.
 
  • #79
sru2 said:
You do understand how digital information is represented?...you know, high/low, on/off, there/not there, photon/no photon...
This should be obvious, extremely obvious.
We got distracted when you brought up Mr Shannon. My software is designed for scientific applications as well as general radio DSP. As such, I had to point out that Shannon's work is not strictly accurate and more a rule of thumb. I probably delved a little to deep in quantum theory for you to keep up, but its quite accurate.

Your understanding is wrong, at least in regards to this modulation scheme. The signal is not constant, its pulsed on and off. Thus, the only noise entering the signal, from the transmitter, will be during an on signal which only adds to the signal quality rather than interfering with it. The key is setting an appropriate noise floor at the receiver, hence the focus on reception.


...again, your understanding is wrong. The whole system must be cooled until the signal gets into the digital domain. If you just cooled the front-end, then you would pick up thermal noise in the antenna and cables feeding into it. Your noise floor is only as good as the hottest part in the chain.

It depends on the system requirements. It becomes less of a concern with particular modulation schemes and the ability to amplify.

Does your brain not work? Do you always defer to authority? Most of what I have said should be obvious, use a calculator, a pencil and a piece of paper. It shouldn't take long to confirm most of what I said.

Its a walk in the park, set a noise floor on the receiver, squelch it and amplify the transmitter. I think you tend to over-complicate things.

So you have a systen that can produce a single photon, to order? Amazing. (Reference?)

A reference for this could be helpful.. Interesting that Shannon's concepts can be ignored jusgt because someone uses the word "Quantum".

It's the noise entering the receiver that counts. Filtering your transmitted signal goes without saying and, whilst transmitted SNR needs to be high enough, that shouldn't be a problem for anyone with a knowledge of modulators. Noise at the input of the receiver is unpredictable so you can hardly 'mask it out'. As with all modulation systems, the signal is not constant - or it wouldn't contain any information.

So do they cool the whole of a giant radio telescope - or do they just cool the parametric converter at its focus? What planet did you get your radio engineering ideas from (too basic to be bothering with, no doubt)? I do know that 'they' are concerned with reducing the power that can reach the feed due to the 'hot earth' behind the dish because that has a low conductivity and appears as an additional noisy resistor in series with the first stage noise resistor. The metallic reflector has much less effect (as would your metal antenna) - this is pretty basic theory which always tends to apply.

Possibly not but I do try to maintain a level of politeness, even when dealing with questionable ideas. Given the choice between the majority, well established view which can be followed in detail, and with many references, compared with a totally unsubstantiated statement then, yes, I to tend to go with authority. It's only when someone has seriously proved themselves that they tend to be worth taking seriously. You have provided no evidence to support what you say so why should I go for it? Take a "piece of paper" and prove it for all of us.

Squelch is not a good idea for systems with a low SNR because they never turn on. In a high noise situation, the signal will not be identifiable until it has been 'processed' out of the noise so you can't tell when its there. No information whatsoever is available when squelch is operating. (More very basic stuff that can't be ignored).
Turning up the transmitter is fine as long as you don't care about the outgoing interference and can afford the equipment and the power. Have you a clue about the actual numbers involved? They could be another crucial factor - It's no good putting your fingers in your ears and going La La La about it. Sums and measurements are needed. A simple factor of 20dB in the link budget can make the difference between a 1W amplifier and a 100w amplifier. It may not be Rocket / Quantum Science but it's relevant to a practical system.
 
  • #80
So you have a systen that can produce a single photon, to order? Amazing. (Reference?)

That was not your original point, what was it you said:

You seem to be totally confusing photons with information bits

Let's just ignore that one shall we?

A reference for this could be helpful.. Interesting that Shannon's concepts can be ignored jusgt because someone uses the word "Quantum".

That's the difference between classical physics and quantum physics.

It's the noise entering the receiver that counts.

Thanks for repeating my reply. I did get it when I wrote it.

So do they cool the whole of a giant radio telescope - or do they just cool the parametric converter at its focus?

It would help, obviously some of the energy goes into heating the dish, rather than being reflected.

What planet did you get your radio engineering ideas from (too basic to be bothering with, no doubt)?

One where we obey the laws of physics, not make it up as we go along.

Possibly not but I do try to maintain a level of politeness, even when dealing with questionable ideas.

No you don't. You hide behind mockery and passive-aggressiveness to distract from comments which are clearly wrong.

Given the choice between the majority, well established view which can be followed in detail, and with many references, compared with a totally unsubstantiated statement then, yes, I to tend to go with authority. It's only when someone has seriously proved themselves that they tend to be worth taking seriously. You have provided no evidence to support what you say so why should I go for it? Take a "piece of paper" and prove it for all of us.

A prime example... :)

Squelch is not a good idea for systems with a low SNR because they never turn on. In a high noise situation, the signal will not be identifiable until it has been 'processed' out of the noise so you can't tell when its there. No information whatsoever is available when squelch is operating. (More very basic stuff that can't be ignored).

Whoever said there would be a low SNR?

Turning up the transmitter is fine as long as you don't care about the outgoing interference and can afford the equipment and the power. Have you a clue about the actual numbers involved? They could be another crucial factor - It's no good putting your fingers in your ears and going La La La about it. Sums and measurements are needed. A simple factor of 20dB in the link budget can make the difference between a 1W amplifier and a 100w amplifier. It may not be Rocket / Quantum Science but it's relevant to a practical system.

Given the distance and the background noise, it won't need that much of a boost.
 
  • #81
You must forgive my lack of credence for much of what you write. I didn't start off being so skeptical about your posts. Initially, I tried to make interested and helpful comments; it was an interesting idea. But you kept making more and more unjustified statements, apparently trawling the more fanciful regions in order to justify them. (Withdrawing them as easily as you made them in the first place). It is usually the mark of a viable idea that it tends to converge rather than what has happened with this one. We suddenly leap from a simple data processing and RF linking problem to quantum theory. And to what end? Just to avoid being wrong in some fundamental aspect of Comms engineering, I suspect.

The energy of a photon is hf (yes?). At low frequency the number of photons per Watt increases, yet is not the bandwidth of low frequency systems (i.e. the information capacity) less? You stated that a photon represents one bit of data. This is self contradictory because it implies that low frequency would carry more bits for the same power. Could you resolve this please?

"That's the difference between classical physics and quantum physics." What is the difference? Don't they both require some degree of proof and justification?
Do you really suggest that I / we all should just take your word for some of this without either some detailed calculations / experimental results or a decent reference? This is supposed to be a serious Science forum.

"obviously some of the energy goes into heating the dish, rather than being reflected"
Do you understand that a radio telescope is not heated up by the incident radiation it is being used to detect - it is at ambient temperature due to local heat sources (sun or surrounding air). But the temperature of the reflecting surface adds little noise because of the low resistance involved (reflection at a clean metallic surface). As I said before, this resistance component appears as a very low, hot resistance in series with the receiver input resistance. How much heating would you expect from the signal from a distant galaxy?

It is interesting and confusing that you were, at first, talking of very high data rates. Then you modified it to extremely low data rates. Now you say there will be no SNR problems. So why would you be needing such lengthy computation times? Surely it would only require a simple demodulator to extract a healthy data rate from a high level signal. Have you Any Idea what the insertion loss of your wireless connection is likely to be? If you insist that is is not a problem then is would be the first wireless communication link that didn't need to consider the problem of signal loss or noise and interference.
 
  • #82
It is usually the mark of a viable idea that it tends to converge rather than what has happened with this one. We suddenly leap from a simple data processing and RF linking problem to quantum theory. And to what end? Just to avoid being wrong in some fundamental aspect of Comms engineering, I suspect.

Not really. It has to do with DSP software I am writing and to clear up some misconceptions you have regarding Mr Shannon's work. With this new perspective, it should be easier to understand how we can develop high throughput at low frequencies.

The energy of a photon is hf (yes?). At low frequency the number of photons per Watt increases, yet is not the bandwidth of low frequency systems (i.e. the information capacity) less? You stated that a photon represents one bit of data. This is self contradictory because it implies that low frequency would carry more bits for the same power. Could you resolve this please?

What I said was that the presence (or lack) of a photon in a given time period can be used to represent binary signal.

Also, your claim that low frequency have more photons is completely wrong:

h = 6.63 x 10^(-34) 6.626068 × 10-34 m2 kg / s
c = 3.00 x 10^8 m/s 299 792 458 m / s

E = hc/λ

Joules per photon:
1Hz = 1.986445212595144e-25 Joules
2Hz = 9.93222606297572e-26 Joules
10Hz = 1.986445212595144e-26 Joules

In 1W (or J/s) number of photons per second:
1Hz = 5,034,118,200,992,686,002,853,829.9708804
2Hz = 10,068,236,401,985,372,005,707,659.941761
10Hz = 50,341,182,009,926,860,028,538,299.708804

"That's the difference between classical physics and quantum physics." What is the difference? Don't they both require some degree of proof and justification?

The difference is that classical equations are simplifications that do not take into account a wide range of factors. They are rough estimates that approximate a result suitable for a functional scenario. Quantum mechanics has different solutions which reveal different approaches that make the classical calculations redundant or simply incorrect.

Do you understand that a radio telescope is not heated up by the incident radiation it is being used to detect - it is at ambient temperature due to local heat sources (sun or surrounding air).

Its an interaction, an energy exchange, thus heat is produced. It is just lost at a rate equal to, or greater than, the incoming radiation which maintains its temperature.

How much heating would you expect from the signal from a distant galaxy?

Enough to knock a few dB of the detectable signal at least.

It is interesting and confusing that you were, at first, talking of very high data rates. Then you modified it to extremely low data rates. Now you say there will be no SNR problems.

The high data rates are provided by complex modulation schemes (i.e. multi-symbol). The lower value I provided is related to what I would call the base rate, which is the data rate provided by the simplest modulation scheme. SNR is not really an issue given the proximity.

So, just for you I did a few signal tests today using a DDS LCD1602 with a 2m bell wire cable, a 50 ohm resistor simulating a load at a distance of 4m. The average signal strength in the pictures below is -110dBm:

TkpIP.jpg


and then I moved up the frequency band:

DMNMC.jpg


So why would you be needing such lengthy computation times?

Its the frequency separation from the FFT...I did mention that signal detection is done in Frequency domain, rather than in the time domain?

Have you Any Idea what the insertion loss of your wireless connection is likely to be?

Considering that the system is coupling, then both the antenna and the receiver are a load. So, you need to get specific here. Anyway, look up the specs of that signal generator and you should get a good idea. The receiver is a tight coil of 500G of enameled magnetic wire (approx. 722m - 0.315mm / 30SWG / 28AWG).
 
  • #83
Both your arithmetic and the answer are nonsense, I'm afraid.
If the energy of a photon is hf then, if you double the frequency, you double the energy. It's Proportional! I have no idea what you did to obtain those numerical results but it has to throw all of your other assertions into doubt.

The energy of a photon at a frequency of 10Hz is 6.6E-33J. Of a photon of 10GHz, it is 6.6E-24J. Assuming that one could actually control exactly when a photon of 10GHz would be emitted, it would be interesting to know just how you would plan to detect it. Even if you were to get your numbers correct (multiplying h by f isn't too taxing, surely) then you'd have to admit that it may be just a tad difficult to make a practical detection of each of a stream of photons with those energies. The thermal energy in any detecting equipment is far higher at a temperature of 300K (or even 100K) than this. Which is why you can't make quantum measurements of such low frequency EM. Individual photons of optical frequencies can be detected but, even at IR, the problem gets very hard. You have a good few octaves to go to reach your VLF frequencies. Totally out of the question. QM is a total red herring in this context but you might at least get your statements correct when you invoke it.

You clearly haven't a clue about the noise contribution of a receiving antenna or you wouldn't write such nonsense about that, either. "Enough to knock a few dBs off"?? What's the point of an antenna with Gain if it just reduces the signal level? Go away and learn a bit about exactly what low noise design of receiving systems actually involves. If you gave it a second's thought, you would realize that the few fW arriving on an area of a reflector is not going to alter its temperature by a remotely detectable amount in the presence of a hot Earth nearby with which it is in thermal equilibrium.

Your signal processing is based on a temporal series of samples. This is time domain. How you choose to process those temporal later on doesn't affect that: it's just a way of looking at the process. The only way you could truly be operating in the frequency domain would be to use a large number of separate band-pass frequency domain filters and to observe the levels of signal from each. You use the term 'frequency domain' as if it's magic. If your processing takes a long time to process signals that are as high as you say they are then it is not very efficient and you should be looking at making it a bit less long winded. With a processor that will be operating at several GHz, you should have time to do plenty of processing of a signal of just a few kHz.


I asked you what the insertion loss was. You clearly don't know the meaning of the term but, instead of going and finding out, you give another nonsense reply.

Why do I need to look up the spec of a signal generator? If you tell me it has an output impedance of 50Ωand delivers +10dBmW (or whatever) into an impedance of 50 Ohms, that's all that need be said. Actual power delivered into the wire 'antenna' is pretty unfathmable, though. From your description, it is not possible to know what the setup is exactly. Can I assume that you have a length of wire connected to the sig gen output and a coil connected to the sound card input? There are no units on the graphs so we can't tell anything about what the axes represent. Are those supposed to be samples of CW signals and is that the ADC output? What is the frequency?

Btw, what is "magnetic wire"? Iron? I don't think you mean that. AWG 28 copper has a resistance of about 1Ω for every 5m so your 700m coil probably has quite a high resistance (140Ω +) and not ideal. I assume that, by "receiver" here, you mean 'receiving antenna'. The whole paragraph is rather garbled. All that we need to know is what the power into the receiver is for a given power from the signal generator. What power do you get at 1kHz, for instance - or, at least, what pk-pk Voltage are you getting (a more reliable quantity for your ADC to be measuring)? I suspect that it is the coil plus the whole of your PC, sound card and mains supply that is acting as a sort of antenna at some frequencies, actually, with a variety of coupling modes over the band..

You have a piece of equipment that operates in the real world. It has nothing to do with quantum communications and in no way looks at 'photons' on an individual basis. So why introduce that irrelevant topic? It just generates confusion.

Just looking at your very first post, I am further confused because it seems you are using the sound card as an output. Are you using one at each end of the link at this stage?
 
  • #84
Both your arithmetic and the answer are nonsense, I'm afraid.
If the energy of a photon is hf then, if you double the frequency, you double the energy. It's Proportional! I have no idea what you did to obtain those numerical results but it has to throw all of your other assertions into doubt.

Right, you are running off at the keyboard at this stage. Let's review the equations I provided you with and you can refer to this link for further examples:

http://electron6.phys.utk.edu/phys250/modules/module 1/photons.htm

h = 6.63 x 10^(-34) 6.626068 × 10-34 m2 kg / s
c = 3.00 x 10^8 m/s 299 792 458 m / s

E = hc/λ

Joules per photon:
1Hz = 1.986445212595144e-25 Joules (0.00000124 eV)
2Hz = 9.93222606297572e-26 Joules (0.00000062 eV)
10Hz = 1.986445212595144e-26 Joules (0.000000124 eV)

In 1W (or J/s) number of photons per second:
1Hz = 5,034,118,200,992,686,002,853,829.9708804
2Hz = 10,068,236,401,985,372,005,707,659.941761
10Hz = 50,341,182,009,926,860,028,538,299.708804

We calculate it as follows:

First we use the formula E = hc/λ which calculates the energy of a photon at a given wavelength:

E = Energy in Joules
h = Planck's constant
c = speed of light
λ = wavelength

E = ((6.626068 * 10e-34) * (299792458)) / wavelength
E = 1.986445212595144e-25 / wavelength

This provides us with the following figures for the energy of a photon at a given Hertz:

1Hz = 1.986445212595144e-25 Joules (0.00000124 eV)
2Hz = 9.93222606297572e-26 Joules (0.00000062 eV)
10Hz = 1.986445212595144e-26 Joules (0.000000124 eV)

1 Watt is the same as 1 Joules per second. So, to calculate how many photons are in 1 Watt at a given Hertz, emitted over 1 second, we use the following formula:

number of photons = 1 Watt (or 1 J/s) / energy of a photon (J)

As you can see from the above formula, the number of photons is dependent on frequency and the higher that frequency gets, the more photons it takes to produce 1 W of power.

Using this formula we get the following number of photons for a given frequency:

1Hz = 5,034,118,200,992,686,002,853,829.9708804
2Hz = 10,068,236,401,985,372,005,707,659.941761
10Hz = 50,341,182,009,926,860,028,538,299.708804

Now that you understand that the figures I presented are correct, we can move to calculating total theoretical capacity, per second, in binary in an out of phase 1W signal. This is simply the number of photons:

At 1Hz it is approx. 533 Exabytes
At 2Hz it is approx. 1.66 Zettabytes
At 10Hz...we don't even have a name for it.

Now let's really complicate matters. How many different photons exist between any given Hertz? The short answer is around about 10e29 before uncertainty comes into play.

So, between 1Hz and 2Hz, we have 10e29 channels whose capacity expands from 533 Exabytes to 1.66 Zettabytes as the frequency increases. As a minimum estimate (i.e. using the lowest channel capacity) the total theoretical capacity, for a binary signal, between 1Hz and 2Hz is:

5.33e+49 Exabytes.

I think that's enough quantum physics for the moment.
 
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  • #85
I think you have done very well to start off with a simple definition that energy is proportional to frequency and, going round the houses, managed to prove an inverse relationship. Other party tricks can show that 2=1.
That last post was just plain wrong.
Just look at E=hf. What does that tell you?
 
Last edited:
  • #86
Come on, I thought you would have spotted that one. I switched frequency for wavelength.

Couldn't resist. :)

E = hc/λ

E = Energy in Joules
h = Planck's constant
c = speed of light
λ = wavelength

λ 1Hz = 299,790,000m
λ 2Hz = 149,900,000m
λ 10Hz = 29,979,000m

E = ((6.626068 * 10e-34) * (299792458)) / wavelength
E = 1.986445212595144e-25 / wavelength

E 1Hz = 6.6261223276131425331065078888555e-34
E 2Hz = 1.3251802619046991327551701134089e-33
E 10Hz = 6.6261223276131425331065078888555e-33

number of photons per second = 1 Watt (or 1 J/s) / energy of a photon (J)

1Hz = 1.5091782954755973367955496869703e+33
2Hz = 7.54614318328803631827789112635e+32
10Hz = 1.5091782954755973367955496869703e+32

The total theoretical capacity for a binary 1W signal, over 1 second, is:

1Hz = 1.5091782954755973367955496869703e+33 bits
2Hz = 7.54614318328803631827789112635e+32 bits
10Hz = 1.5091782954755973367955496869703e+32 bits

With 10e29 different photons between Hertz, a 1Hz bandwidth signal of 1 W with a base frequency of 1Hz has the following minimum capacity:

7.54614318328803631827789112635e+32 * 10e29
= 7.54614318328803631827789112635e+61 bits

As we increase the frequency theoretical capacity drops. This is the point you wanted me to resolve earlier, there is nothing to resolve. The 'capacity' you are referring to is based upon modulation of an ever increasing frequency, which is a classical view point that really does not make sense at the quantum level.
 
  • #87
Why? I just can't believe you.
 
  • #88
Why? I just can't believe you.

Beer...it seemed funny at the time...but you have shown me the error of my ways. :)
 
  • #89
Yet again you make an error and then try to pass it off as a joke.
You really have no credibility for me any more.
 
  • #90
Yet again you make an error and then try to pass it off as a joke.
You really have no credibility for me any more.

So, how do explain this comment from you?

The energy of a photon is hf (yes?). At low frequency the number of photons per Watt increases, yet is not the bandwidth of low frequency systems (i.e. the information capacity) less? You stated that a photon represents one bit of data. This is self contradictory because it implies that low frequency would carry more bits for the same power. Could you resolve this please?

I just demonstrated that I was correct.
 
  • #91
Do you have trouble parsing sentences? My comment compared two contradictory notions. How can it be right or wrong?
You have, presumably, come to terms with the idea that low frequency photons are low energy so would not be a way of carrying data individually in the presence of thermal energy as they are undetectable. How would you propose to detect a photon with an energy level corresponding, not to an atomic transition, not to a molecular transition but the energy corresponding to a single free electron in a metal? This whole idea is nonsense and shows that you just know nothing of the real situation.
I ask, once more, do you have a single reference to support you?
 
  • #92
Do you have trouble parsing sentences? My comment compared two contradictory notions. How can it be right or wrong?

Let's review what you asked...

The energy of a photon is hf (yes?). At low frequency the number of photons per Watt increases, yet is not the bandwidth of low frequency systems (i.e. the information capacity) less?

The first portion of your question asserts a statement that is wrong. The bandwidth of low frequency systems is not less, it is more. The problem is photon creation and detection which leads to low bandwidths. Its an engineering issue, not a physics issue.

You stated that a photon represents one bit of data. This is self contradictory because it implies that low frequency would carry more bits for the same power. Could you resolve this please?

Where is the contradiction? This is accurate. There are more photons, thus more bits can be represented.


You have, presumably, come to terms with the idea that low frequency photons are low energy so would not be a way of carrying data individually in the presence of thermal energy as they are undetectable.

This is a different issue, an engineering issue. We have been able to detect single photons at visible or near-infrared wavelengths for over a decade now.

http://physicsworld.com/cws/article/news/2000/may/12/quantum-dots-detect-single-photons

It still does not change the theoretical capacity, just the practical detectable capacity. The latter changes with time, the former is an absolute.

How would you propose to detect a photon with an energy level corresponding, not to an atomic transition, not to a molecular transition but the energy corresponding to a single free electron in a metal?

Its an engineering issue that will be solved in time. This question is like asking someone from a hundred years ago to design a 4Ghz processor. Its not that it cannot be done, its just that there are a wide range of discoveries and inventions that need to occur before such a device become practical.

This whole idea is nonsense and shows that you just know nothing of the real situation.

There is nothing outrageous about setting an absolute limit on the total theoretical capacity of a channel at a given hertz and wattage.

I ask, once more, do you have a single reference to support you?

The math is in this thread, what more do you need?
 
  • #93
If the sample of your Maths is that you don't know the difference between proportion and inverse proportion then I don't think we can rely on it. I'm afraid I am going to have to invoke the rules of the Forum and say that we can't carry on unless you can furnish us with a reference or some reliable experimental evidence.
You are sure to throw your toys out of the pram about this but there is nothing else that I can do. Some people could read what you have written and run the risk of actually believing that it has some substance. We cannot have that.
No references means the thread is over as far as I'm concerned - and as far as any of the moderators are concerned too, I think.
 
  • #94
If the sample of your Maths is that you don't know the difference between proportion and inverse proportion then I don't think we can rely on it. I'm afraid I am going to have to invoke the rules of the Forum and say that we can't carry on unless you can furnish us with a reference or some reliable experimental evidence.

If you cannot comprehend the simple math, then a reference would be equally useless to you. Everything you need to confirm it is given in the post above.

https://www.physicsforums.com/showpost.php?p=3947771&postcount=86

You are sure to throw your toys out of the pram about this but there is nothing else that I can do. Some people could read what you have written and run the risk of actually believing that it has some substance. We cannot have that.

...even though it is accurate?

No references means the thread is over as far as I'm concerned - and as far as any of the moderators are concerned too, I think.

This thread is in relation to VLF comms, not photons. Just because it was shown that you do not really understand quantum mechanics, radio transmission, data transfer, data representation, etc., does not give you the right to sulk.

Lose the ego and just admit when you're wrong.
 
<h2>1. What is VLF transmission using soundcard?</h2><p>VLF transmission using soundcard is a method of sending very low frequency (VLF) radio signals through a computer's soundcard. This allows for the transmission of data or audio signals over long distances without the need for specialized equipment.</p><h2>2. How does VLF transmission using soundcard work?</h2><p>VLF transmission using soundcard works by converting the digital data or audio signals into analog signals that can be transmitted through the soundcard. The soundcard then amplifies and modulates the signals before sending them through an antenna.</p><h2>3. What are the advantages of VLF transmission using soundcard?</h2><p>Some advantages of VLF transmission using soundcard include its low cost, ease of use, and ability to transmit signals over long distances. It also does not require specialized equipment, making it accessible to a wider range of users.</p><h2>4. What are the limitations of VLF transmission using soundcard?</h2><p>One limitation of VLF transmission using soundcard is its low data transfer rate, which may not be suitable for transmitting large amounts of data. It also requires a clear line of sight between the transmitting and receiving antennas, making it susceptible to interference.</p><h2>5. What are the applications of VLF transmission using soundcard?</h2><p>VLF transmission using soundcard has various applications, including long-distance communication, data transfer, and remote sensing. It is also used in amateur radio, military communication, and scientific research, such as studying the Earth's ionosphere and magnetic field.</p>

1. What is VLF transmission using soundcard?

VLF transmission using soundcard is a method of sending very low frequency (VLF) radio signals through a computer's soundcard. This allows for the transmission of data or audio signals over long distances without the need for specialized equipment.

2. How does VLF transmission using soundcard work?

VLF transmission using soundcard works by converting the digital data or audio signals into analog signals that can be transmitted through the soundcard. The soundcard then amplifies and modulates the signals before sending them through an antenna.

3. What are the advantages of VLF transmission using soundcard?

Some advantages of VLF transmission using soundcard include its low cost, ease of use, and ability to transmit signals over long distances. It also does not require specialized equipment, making it accessible to a wider range of users.

4. What are the limitations of VLF transmission using soundcard?

One limitation of VLF transmission using soundcard is its low data transfer rate, which may not be suitable for transmitting large amounts of data. It also requires a clear line of sight between the transmitting and receiving antennas, making it susceptible to interference.

5. What are the applications of VLF transmission using soundcard?

VLF transmission using soundcard has various applications, including long-distance communication, data transfer, and remote sensing. It is also used in amateur radio, military communication, and scientific research, such as studying the Earth's ionosphere and magnetic field.

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