![]() |
Minimum photons-per-second [amplitude] required for 150 KHz?
Hi:
What is the minimum amount of photons-per-second needed for a 150 KHz AM radio carrier wave to transmit audio signals? Around 20,000-photons- per-second? Thanks, Radium |
Minimum photons-per-second [amplitude] required for 150 KHz?
"Radium" wrote in message ups.com... Hi: What is the minimum amount of photons-per-second needed for a 150 KHz AM radio carrier wave to transmit audio signals? Around 20,000-photons- per-second? Thanks, Radium The same as the minimum number of data bits required to represent this signal. The number of photons/data bits will vary depending on the complexity, frequency and depth of modulation of the carrier wave. Mike G0ULI |
Minimum photons-per-second [amplitude] required for 150 KHz?
"Radium" wrote in message ups.com... Hi: What is the minimum amount of photons-per-second needed for a 150 KHz AM radio carrier wave to transmit audio signals? Around 20,000-photons- per-second? Thanks, Radium you should have stayed with the alt.sci or sci.physics groups, you don't know what you are getting your self in for here! |
Minimum photons-per-second [amplitude] required for 150 KHz?
On Jun 17, 3:56 pm, "Mike Kaliski" wrote:
The same as the minimum number of data bits required to represent this signal. The number of photons/data bits will vary depending on the complexity, frequency and depth of modulation of the carrier wave. I am talking about an analog carrier wave. In the analog realm, there is no such thing as "bits". |
Minimum photons-per-second [amplitude] required for 150 KHz?
In rec.radio.amateur.antenna Radium wrote:
Hi: What is the minimum amount of photons-per-second needed for a 150 KHz AM radio carrier wave to transmit audio signals? Around 20,000-photons- per-second? The answer is not simple because any given photon only has one frequency and one energy. So at any given time, you need some number of photons at different frequencies to get the frequency components and some number of photons at each frequency component to the the amplitude components of the total signal. Did you get tired of everyone calling you a clueless moron on sci.physics and sci.physics.electromagnetics and think you would try here? -- Jim Pennino Remove .spam.sux to reply. |
Minimum photons-per-second [amplitude] required for 150 KHz?
On Jun 17, 3:56 pm, "Dave" wrote:
"Radium" wrote in message ups.com... Hi: What is the minimum amount of photons-per-second needed for a 150 KHz AM radio carrier wave to transmit audio signals? Around 20,000-photons- per-second? Thanks, Radium you should have stayed with the alt.sci or sci.physics groups, you don't know what you are getting your self in for here! Huh? |
Minimum photons-per-second [amplitude] required for 150 KHz?
On Jun 17, 4:05 pm, wrote:
So at any given time, you need some number of photons at different frequencies to get the frequency components and some number of photons at each frequency component to the the amplitude components of the total signal. Well, in FM the peak-to-peak amplitude remains constant but the energy [frequency] varies. In AM, the frequency remains constant but the peak to peak amplitude varies. |
Minimum photons-per-second [amplitude] required for 150 KHz?
"Radium" wrote in message oups.com... On Jun 17, 3:56 pm, "Mike Kaliski" wrote: The same as the minimum number of data bits required to represent this signal. The number of photons/data bits will vary depending on the complexity, frequency and depth of modulation of the carrier wave. I am talking about an analog carrier wave. In the analog realm, there is no such thing as "bits". Radium If a single photon can represent a single discrete energy level, then at some point you will have to translate your analogue signal into discrete photons. A process similar to digitising an audio signal to produce a CD. The number of photons will depend on the sampling rate chosen, the bandwidth and depth of modulation of the original signal and the fidelity with which you wish to reconstruct a representation of the original signal. 300,000 photons per second should do the trick, as that is the frequency of the original signal and each photon can represent the amplitude of each half of a single sine wave. It is standard practice to sample at least double the frequency of whatever you are trying to capture. Mike G0ULI |
Minimum photons-per-second [amplitude] required for 150 KHz?
In rec.radio.amateur.antenna Radium wrote:
On Jun 17, 3:56 pm, "Mike Kaliski" wrote: The same as the minimum number of data bits required to represent this signal. The number of photons/data bits will vary depending on the complexity, frequency and depth of modulation of the carrier wave. I am talking about an analog carrier wave. In the analog realm, there is no such thing as "bits". In the analog realm, there is no such thing as photons as they are discrete quanta. -- Jim Pennino Remove .spam.sux to reply. |
Minimum photons-per-second [amplitude] required for 150 KHz?
In rec.radio.amateur.antenna Radium wrote:
On Jun 17, 4:05 pm, wrote: So at any given time, you need some number of photons at different frequencies to get the frequency components and some number of photons at each frequency component to the the amplitude components of the total signal. Well, in FM the peak-to-peak amplitude remains constant but the energy [frequency] varies. In AM, the frequency remains constant but the peak to peak amplitude varies. You've never seen what an AM signal looks like on a spectrum analyzer, have you? Go look at: http://www.fas.org/man/dod-101/navy/docs/es310/AM.htm Hot flash for you, the AM modulation process creates other frequencies. If you only have one frequency, you don't have modulation of any kind. -- Jim Pennino Remove .spam.sux to reply. |
Minimum photons-per-second [amplitude] required for 150 KHz?
On Jun 17, 4:38 pm, "Mike Kaliski" wrote:
300,000 photons per second should do the trick, as that is the frequency of the original signal and each photon can represent the amplitude of each half of a single sine wave. So the amount of photons-per-second should be double the frequency of the carrier-wave? In my first post of the thread, I stated that the carrier frequency for the AM signal is 150 KHz. Each photon in that signal is 150 KHz. It's possible to have one 150 KHz photon, right? My question was relating the modulator wave. If I have using 150 KHz photons for my carrier-wave on AM radio, what is the minimum amount of photons-per-second I would require to transmit, a modulator-signal [through the 150 KHz carrier-signal] of 20 KHz? I am guessing 40,000 150-KHz-photons-per-seconds. Am I right? |
Minimum photons-per-second [amplitude] required for 150 KHz?
"Radium" wrote in message ups.com... On Jun 17, 4:38 pm, "Mike Kaliski" wrote: 300,000 photons per second should do the trick, as that is the frequency of the original signal and each photon can represent the amplitude of each half of a single sine wave. So the amount of photons-per-second should be double the frequency of the carrier-wave? In my first post of the thread, I stated that the carrier frequency for the AM signal is 150 KHz. Each photon in that signal is 150 KHz. It's possible to have one 150 KHz photon, right? My question was relating the modulator wave. If I have using 150 KHz photons for my carrier-wave on AM radio, what is the minimum amount of photons-per-second I would require to transmit, a modulator-signal [through the 150 KHz carrier-signal] of 20 KHz? I am guessing 40,000 150-KHz-photons-per-seconds. Am I right? Radium You can have a single photon oscillating at a frequency of 150,000 cycles per second. Measuring that photon will give you a sample of 1/150,000th of a second duration. If you want to do anything meaningful, you need to have a whole lot more photons. If you modulate a 150kHz carrier with a signal of 20kHz then the bandwidth of the signal will extend from 150 kHz -20 kHz to 150 kHz +20 kHz or from 130-170 kHz. Signals centred on 150 kHz represent just the carrier wave. Signals at 130 kHz and 170 kHz represent 100% modulation of the carrier wave. Now the modulation of the carrier wave is symmetrical about the center frequency, so you only need to measure one half. One way of recovering the signal is to measure the frequency of each photon between 130 and 150 kHz at a rate of 300,000 samples per second. The variation of each photon from the carrier frequency represents the modulation. A 20 kHz signal can be accurately represented using 40,000 samples, but this is different from detecting modulation on a higher frequency carrier wave. Mike G0ULI |
Minimum photons-per-second [amplitude] required for 150 KHz?
In rec.radio.amateur.antenna Radium wrote:
On Jun 17, 4:38 pm, "Mike Kaliski" wrote: 300,000 photons per second should do the trick, as that is the frequency of the original signal and each photon can represent the amplitude of each half of a single sine wave. So the amount of photons-per-second should be double the frequency of the carrier-wave? In my first post of the thread, I stated that the carrier frequency for the AM signal is 150 KHz. Each photon in that signal is 150 KHz. It's possible to have one 150 KHz photon, right? My question was relating the modulator wave. If I have using 150 KHz photons for my carrier-wave on AM radio, what is the minimum amount of photons-per-second I would require to transmit, a modulator-signal [through the 150 KHz carrier-signal] of 20 KHz? I am guessing 40,000 150-KHz-photons-per-seconds. Am I right? No. For AM with a 150 Khz carrier and a steady 20 Khz tone, you have to emit 1 130 Khz photon, 2 150 Khz photons, and 1 170 Khz photon approximately every 12 microseconds. -- Jim Pennino Remove .spam.sux to reply. |
Minimum photons-per-second [amplitude] required for 150 KHz?
wrote in message ... In rec.radio.amateur.antenna Radium wrote: On Jun 17, 4:38 pm, "Mike Kaliski" wrote: 300,000 photons per second should do the trick, as that is the frequency of the original signal and each photon can represent the amplitude of each half of a single sine wave. So the amount of photons-per-second should be double the frequency of the carrier-wave? In my first post of the thread, I stated that the carrier frequency for the AM signal is 150 KHz. Each photon in that signal is 150 KHz. It's possible to have one 150 KHz photon, right? My question was relating the modulator wave. If I have using 150 KHz photons for my carrier-wave on AM radio, what is the minimum amount of photons-per-second I would require to transmit, a modulator-signal [through the 150 KHz carrier-signal] of 20 KHz? I am guessing 40,000 150-KHz-photons-per-seconds. Am I right? No. For AM with a 150 Khz carrier and a steady 20 Khz tone, you have to emit 1 130 Khz photon, 2 150 Khz photons, and 1 170 Khz photon approximately every 12 microseconds. -- Jim Pennino Remove .spam.sux to reply. Nicely put Jim. Mike G0ULI |
Minimum photons-per-second [amplitude] required for 150 KHz?
In article . com,
Radium wrote: Snip Go away cross posting nut case. Plonk -- Telamon Ventura, California |
Minimum photons-per-second [amplitude] required for 150 KHz?
In article ,
"Mike Kaliski" wrote: Snip Plonk -- Telamon Ventura, California |
Minimum photons-per-second [amplitude] required for 150 KHz?
In article ,
wrote: Snip Plonk -- Telamon Ventura, California |
Minimum photons-per-second [amplitude] required for 150 KHz?
Telamon wrote:
In article , "Mike Kaliski" wrote: Ohh gawd, bizarre telemundo is back ... PLONKERS! JS |
Minimum photons-per-second [amplitude] required for 150 KHz?
On Jun 17, 3:56 pm, "Dave" wrote:
"Radium" wrote in message ups.com... Hi: What is the minimum amount of photons-per-second needed for a 150 KHz AM radio carrier wave to transmit audio signals? Around 20,000-photons- per-second? Thanks, Radium you should have stayed with the alt.sci or sci.physics groups, you don't know what you are getting your self in for here! as in 'go fish' ~ RHF |
Minimum photons-per-second [amplitude] required for 150 KHz?
On Jun 17, 4:45 pm, wrote:
In rec.radio.amateur.antenna Radium wrote: On Jun 17, 4:05 pm, wrote: So at any given time, you need some number of photons at different frequencies to get the frequency components and some number of photons at each frequency component to the the amplitude components of the total signal. Well, in FM the peak-to-peak amplitude remains constant but the energy [frequency] varies. In AM, the frequency remains constant but the peak to peak amplitude varies. You've never seen what an AM signal looks like on a spectrum analyzer, have you? Scroll down to "A More Realistic Spectrum" - Go look at:http://www.fas.org/man/dod-101/navy/docs/es310/AM.htm Check-out the two side-by-side Images Hot flash for you, the AM modulation process creates other frequencies. If you only have one frequency, you don't have modulation of any kind. -- Jim Pennino Remove .spam.sux to reply. |
Minimum photons-per-second [amplitude] required for 150 KHz?
On Sun, 17 Jun 2007 15:19:59 -0700, Radium
wrote: What is the minimum amount of photons-per-second needed for a 150 KHz AM radio carrier wave to transmit audio signals? Around 20,000-photons- per-second? That one is real simple: 2 photons (of appropriate amplitude, hence color) at most 3.333 (less would be better, but not too much less) microseconds apart. Feel free to desire more, but you asked for the minimum. If you want more audio (sideband) content, that will certainly drive up the count too. Now, how's your quantum efficiency these days? (Use it to boost the count higher.) 73's Richard Clark, KB7QHC |
Minimum photons-per-second [amplitude] required for 150 KHz?
In article . com,
Radium wrote: Hi: What is the minimum amount of photons-per-second needed for a 150 KHz AM radio carrier wave to transmit audio signals? Around 20,000-photons- per-second? 1. What is the energy of a photon at 150 kHz? 2. What is the minimum discernable signal in your receiving system? (How much power is needed at the receiver to overcome the internal noise of the receiver system and detect the signal?) 3. What signal to noise ratio makes for a tolerable listening condition? (How much more power than quetion #2 is needed at the receiver to decode the modulation and yield a usable signal?) Mark Zenier Googleproofaddress(account:mzenier provider:eskimo domain:com) |
Minimum photons-per-second [amplitude] required for 150 KHz?
In article .com,
Denny wrote: On Jun 17, 10:37 pm, Richard Clark wrote: On Sun, 17 Jun 2007 15:19:59 -0700, Radium wrote: Snip I need a beer... Nope. You just need to be added to the kill file. Cheers. Plonk -- Telamon Ventura, California |
Minimum photons-per-second [amplitude] required for 150 KHz?
On Jun 18, 11:05 am, (Mark Zenier) wrote:
1. What is the energy of a photon at 150 kHz? 6.2 X 10^-10 eV |
Minimum photons-per-second [amplitude] required for 150 KHz?
John Smith I wrote:
Telamon wrote: In article , "Mike Kaliski" wrote: Ohh gawd, bizarre telemundo is back ... PLONKERS! JS The irony is he's proudly plonking those who are actually more on topic with this 'photon' thread than he often is in other threads, where the subject is really OT. |
Minimum photons-per-second [amplitude] required for 150 KHz?
In article .com,
Denny wrote: Snip plonk -- Telamon Ventura, California |
Minimum photons-per-second [amplitude] required for 150 KHz?
In article ,
Plonk Dodger wrote: Snip Plonk -- Telamon Ventura, California |
Minimum photons-per-second [amplitude] required for 150 KHz?
Telamon wrote:
... plonk My gawd man! Is your quest to plonk the world ... ;-) JS |
Minimum photons-per-second [amplitude] required for 150 KHz?
In rec.radio.amateur.antenna Denny wrote:
On Jun 18, 10:53 pm, Telamon wrote: In article .com, Denny wrote: On Jun 17, 10:37 pm, Richard Clark wrote: On Sun, 17 Jun 2007 15:19:59 -0700, Radium wrote: Snip I need a beer... Nope. You just need to be added to the kill file. Cheers. Plonk -- Telamon Ventura, California Oh gawd, now my life is over.... denny / k8do Maybe not. Let's see how many more times you are added to his killfile. -- Jim Pennino Remove .spam.sux to reply. |
Minimum photons-per-second [amplitude] required for 150 KHz?
In article . com,
Radium wrote: On Jun 18, 11:05 am, (Mark Zenier) wrote: 1. What is the energy of a photon at 150 kHz? 6.2 X 10^-10 eV You didn't answer the other questions. How much power does a good radio need to get a listenable signal? Here's some numbers input impedance 50 ohms Noise Figure 5 db Bandwidth 6 kHz signal to noise ratio 40 dB (This would be what some of the newsgroups listeners here would want if they had one of their pretty damn good radios listening to a broadcast station. Really picky ones would probably want a 8-15 kHz bandwidth with a 60 dB s/N ratio). There are equations out there that will give you how much power you need for this signal... Mark Zenier Googleproofaddress(account:mzenier provider:eskimo domain:com) |
Minimum photons-per-second [amplitude] required for 150 KHz?
Telamon wrote:
In article .com, Denny wrote: Snip plonk Anyone who would do that deserves a Bigger SNIP and a Monster PLONK tom K0TAR |
Minimum photons-per-second [amplitude] required for 150 KHz?
Telamon wrote:
In article .com, Denny wrote: Snip plonk Do you only post here to plonk people? I guess that could be fun.... ;^) - 73 de Mike KB3EIa - |
Minimum photons-per-second [amplitude] required for 150 KHz?
|
Minimum photons-per-second [amplitude] required for 150 KHz?
On Jun 21, 1:08 pm, Michael Coslo wrote:
Telamon wrote: In article .com, Denny wrote: Snip plonk Do you only post here to plonk people? I guess that could be fun.... ;^) - 73 de Mike KB3EIa - Thats about it. Don't mind him. He was abused as a child. He's plonked me twice in the last 5 years or so, and I'm sure this post will bring me up to three.. Which begs the answer... If I was plonked before, how can I be replonked, unless I was unplonked.. I noticed he had to plonk Denny twice.. I guess his plonking device is about as effective as viagra on a 98 year old.. I call him Telaprick... And yes, he has worked hard in the past to deserve the new name I have bestowed upon him. I quit posting to the shortwave group about a year ago, cuz I got tired of dealing with a couple of brain dead morons that think they own the whole show over there. And if you post anything that is even slightly technical, they will whine and complain that it's too much for the average shortwave listener to handle, or comprehend. So I decided, $%&* em if they can't take a joke. The only thing that goes "plonk" around here are the turds that fall into the toilet each morning after I've had my first cup of go juice. Only a whiny bitchette would go around constantly "plonking" people. It adds more clutter than the posts he is plonking.. MK |
Minimum photons-per-second [amplitude] required for 150 KHz?
In article ,
Michael Coslo wrote: Telamon wrote: In article .com, Denny wrote: Snip plonk Do you only post here to plonk people? I guess that could be fun.... ;^) Nope. I just figure that if a person will entertain a Troll then they are not worth reading. Plonk -- Telamon Ventura, California |
Minimum photons-per-second [amplitude] required for 150 KHz?
In article ,
Tom Ring wrote: Telamon wrote: In article .com, Denny wrote: Snip plonk Anyone who would do that deserves a Bigger SNIP and a Monster PLONK Good for you. Plonk -- Telamon Ventura, California |
Minimum photons-per-second [amplitude] required for 150 KHz?
In article .com,
Radium wrote: On Jun 20, 1:01 pm, (Mark Zenier) wrote: In article . com, Radium wrote: On Jun 18, 11:05 am, (Mark Zenier) wrote: 1. What is the energy of a photon at 150 kHz? 6.2 X 10^-10 eV You didn't answer the other questions. How much power does a good radio need to get a listenable signal? Here's some numbers input impedance 50 ohms Noise Figure 5 db Bandwidth 6 kHz signal to noise ratio 40 dB (This would be what some of the newsgroups listeners here would want if they had one of their pretty damn good radios listening to a broadcast station. Really picky ones would probably want a 8-15 kHz bandwidth with a 60 dB s/N ratio). There are equations out there that will give you how much power you need for this signal... I guess the dynamic range needs to be at least 140 dB [to match human ear's loudness perception] and bandwidth at least 40 KHz [to match human's pitch perception]. I am not sure about the input impedance or noise figure. We're not talking about human perceptions, we're talking about a radio. Anyway, how much power does a radio signal to need to be picked up? To start with, there is electronics noise. This comes from temperature of the circuit, and how wide the frequency response. The equation is kTB, where k is Boltzmann's Constant, T it the absolute temperature in Kelvin, and B is the bandwidth in Hertz. This is called the thermal noise power. Next, there's a fudge factor, called either the Noise Factor or the same thing stated in decibels, the Noise Figure. It's determined by the quality of the transistors (or other active amplifying element) and the topology of the circuit. It specifies how much more noise is added in the circuit. The best way to get it is to measure it. (5db, about a factor of 3, would be for a pretty good shortwave radio). Multiply thermal noise power by the noise factor, and you find the "noise floor", or "minimum detectable signal" for a circuit. But that doesn't do you much good, because all you can do with that is, maybe, tell if there is a signal there, or perhaps not. So you need more power in the signal to overcome the noise. For reasonable audio, a factor of 10,000 for power is in the ball park. (This is usually expressed as decibels, ie. 40 dB). So here's a cut and paste from a unix program that does calulations with stated dimensions, along with some comments. [mzenier@localhost mzenier]$ units --verbose 1948 units, 71 prefixes, 28 functions You have: boltzmann * 290 kelvin * 6 kilohertz * 3 * 10000 Boltzmann's constant * room temperature * bandwidth * Noise Factor * 40 dB You want: watts boltzmann * 290 kelvin * 6 kilohertz * 3 * 10000 = 7.2069944e-13 watts This is how much power is needed to receive the signal with a signal to noise ratio of 40 dB in a shortwave radio. You have: 150 kilohertz h frequency * Planck's Constant You want: joules 150 kilohertz h = 9.9391031e-29 joules Energy per photon at 150 kHz You have: (boltzmann * 290 kelvin * 6 kilohertz * 3 * 10000)/(150 kilohertz h) You want: hertz (boltzmann * 290 kelvin * 6 kilohertz * 3 * 10000)/(150 kilohertz h) = 7.2511517e+15 hertz Divide the power needed by the energy in each photon, and you get the number of Photons per second at 150 kHz to give a 40 dB signal to noise ratio in a pretty good quality receiver with a 6 kilohertz bandwidth. 7.2511517e+15 is a lot more than 20,000. (And a good demonstration why they don't use the math for quantum physics for radio frequency calculations.) Mark Zenier Googleproofaddress(account:mzenier provider:eskimo domain:com) |
All times are GMT +1. The time now is 09:55 AM. |
Powered by vBulletin® Copyright ©2000 - 2025, Jelsoft Enterprises Ltd.
RadioBanter.com