Return-Path: Received: from mtain-ma03.r1000.mx.aol.com (mtain-ma03.r1000.mx.aol.com [172.29.96.11]) by air-md07.mail.aol.com (v129.4) with ESMTP id MAILINMD071-8b8c4c05347039b; Tue, 01 Jun 2010 12:25:20 -0400 Received: from post.thorcom.com (post.thorcom.com [193.82.116.20]) by mtain-ma03.r1000.mx.aol.com (Internet Inbound) with ESMTP id 60CAD38000180; Tue, 1 Jun 2010 12:25:19 -0400 (EDT) Received: from majordom by post.thorcom.com with local (Exim 4.14) id 1OJUGO-0006Hw-RP for rs_out_1@blacksheep.org; Tue, 01 Jun 2010 17:24:24 +0100 Received: from [193.82.116.32] (helo=relay1.thorcom.net) by post.thorcom.com with esmtp (Exim 4.14) id 1OJUGO-0006Hn-EL for rsgb_lf_group@blacksheep.org; Tue, 01 Jun 2010 17:24:24 +0100 Received: from v-smtp-auth-relay-5.gradwell.net ([79.135.125.99]) by relay1.thorcom.net with esmtp (Exim 4.63) (envelope-from ) id 1OJUGM-0004vc-8P for rsgb_lf_group@blacksheep.org; Tue, 01 Jun 2010 17:24:24 +0100 Received: from host81-149-112-44.in-addr.btopenworld.com ([81.149.112.44] helo=hugh country=GB ident=ashford) by v-smtp-auth-relay-5.gradwell.net with esmtpa (Gradwell gwh-smtpd 1.290) id 4c05341e.283b.4 for rsgb_lf_group@blacksheep.org; Tue, 1 Jun 2010 17:23:58 +0100 (envelope-sender ) Message-ID: From: "Hugh_m0wye" To: References: <4C0505F3.23023.3198CFC@mike.dennison.ntlworld.com>,<1OJR3q-0nK4tk0@fwd09.t-online.de> <81893174-4350-4FE1-8D70-4FF4EBB44BCE@leeds.ac.uk> <13C79C2C1C7C41FB801BC4911C404C52@Hugh> <57EF78DF-0EB3-4719-A5D4-57966E86C1BE@leeds.ac.uk> Date: Tue, 1 Jun 2010 17:23:32 +0100 MIME-Version: 1.0 X-Priority: 3 X-MSMail-Priority: Normal X-Mailer: Microsoft Outlook Express 6.00.2900.5931 X-MimeOLE: Produced By Microsoft MimeOLE V6.00.2900.5931 X-Spam-Score: 3.4 (+++) X-Spam-Report: autolearn=disabled,FORGED_MUA_OUTLOOK=3.36 Subject: Re: LF: QRO / QRP Content-Type: text/plain; format=flowed; charset="iso-8859-1"; reply-type=original Content-Transfer-Encoding: 7bit X-Spam-Checker-Version: SpamAssassin 2.63 (2004-01-11) on post.thorcom.com X-Spam-Level: * X-Spam-Status: No, hits=1.6 required=5.0 tests=FORGED_MUA_OUTLOOK autolearn=no version=2.63 X-SA-Exim-Scanned: Yes Sender: owner-rsgb_lf_group@blacksheep.org Precedence: bulk Reply-To: rsgb_lf_group@blacksheep.org X-Listname: rsgb_lf_group X-SA-Exim-Rcpt-To: rs_out_1@blacksheep.org X-SA-Exim-Scanned: No; SAEximRunCond expanded to false x-aol-global-disposition: G x-aol-sid: 3039ac1d600b4c05346f77a3 X-AOL-IP: 193.82.116.20 Thankyou Chris, and Simon, It was kind of you to take time out of your marking schedule to calculate that. 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. Hugh M0WYE ----- Original Message ----- From: "Chris Trayner" To: Sent: Tuesday, June 01, 2010 4:05 PM Subject: Re: LF: QRO / QRP Hi Hugh, Thanks for the question - it's an excuse not to settle down to the marking again ;-) > 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 power. To get some figure, suppose we emit one every cycle, i.e. 136k of them per second. 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 12.3 pico-pico-watts. 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