Thankyou Chris, and Simon,
It was kind of you to take time out of your marking schedule to calculate
If I was transmitting a signal I wouldn't consciously decide how often to
transmit photons, but I assume from what you say that emitting one photon
per cycle that is the lowest continuous wave that one could transmit, and
that a less frequent transmission would become interrupted, or amplitude
modulated, although the amplitude modulation can't be in the conventional
sense because that would broaden the spectrum and I would have thought that
a single photon must be a single frequency - i.e. no bandwidth.
If I had a 1 Watt transmitter and fed it into a 240dB attenuator the output
would be less than 12.3x10^-24W. If this were fed into an antenna then some
cycles would occur before a photon could be emitted, now I'm wondering what
happens to the power while it's building up, because a photon ought to be
phase-closed - i.e. the cycle should be complete. This quantisation must
play havoc with the calculation of things like radiation resistance, because
at time no radiation occurs at times. Me thinks some very strange things
must happen at low power levels.
Hmmm, I also should be working ... !
73 and thanks once again.
----- Original Message -----
From: "Chris Trayner" <[email protected]>
To: <[email protected]>
Sent: Tuesday, June 01, 2010 4:05 PM
Subject: Re: LF: QRO / QRP
Thanks for the question - it's an excuse not to settle down to the marking
there is a lower
level of QRP which cannot be received by ANY receiver, no matter how well
endowed with antennas and low noise amplifiers - simply because it is less
than 1 photon.
I wonder how small it is ?
The energy of a photon is E = h f
where h is Planck's constant, 6.626 . 10^-34 Js
and f is frequency, Hz.
(Physicists call frequency nu, but that is harder to typeset than f.)
Thus the energy of a photon at 136kHz is about 9 . 10^-29 Joules.
To get a power in watts you have to decide how often you are going to emit
those photons - you could emit one every few days and get an arbitrarily low
To get some figure, suppose we emit one every cycle, i.e. 136k of them per
Then the power is P = E / T (T the period of the cycle), i.e. P = E f
For 136kHz this gives a power of 1.23 . 10^-23 W, i.e. 12.3 . 10^-24, i.e
If your transmitter was only one million-million-millionth of a percent
efficient, you could power it off the dynamic microphone you spoke into.
Ho, hum, back to the real world of marking.
73, Chris G4OKW
Dr Chris Trayner
School of Electronic & Electrical Engineering,
The University of Leeds,
Leeds LS2 9JT, United Kingdom
Tel: +44 113 34 32053
Fax: +44 113 34 32032