"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