Return-Path: Received: (qmail 10210 invoked from network); 2 Jan 2002 22:59:03 -0000 Received: from unknown (HELO murphys-inbound.services.quay.plus.net) (212.159.14.225) by excalibur-qfe1-smtp-plusnet.harl.plus.net with SMTP; 2 Jan 2002 22:59:03 -0000 X-Priority: 3 X-MSMail-Priority: Normal Received: (qmail 9972 invoked from network); 2 Jan 2002 22:59:03 -0000 X-MimeOLE: Produced By Microsoft MimeOLE V6.00.2800.1106 Received: from unknown (HELO post.thorcom.com) (212.172.148.70) by murphys.services.quay.plus.net with SMTP; 2 Jan 2002 22:59:03 -0000 Received: from majordom by post.thorcom.com with local (Exim 3.33 #2) id 16LuIj-0007mu-00 for rsgb_lf_group-outgoing@blacksheep.org; Wed, 02 Jan 2002 22:56:01 +0000 Received: from imo-r01.mx.aol.com ([152.163.225.97]) by post.thorcom.com with esmtp (Exim 3.33 #2) id 16LuIi-0007mp-00 for rsgb_lf_group@blacksheep.org; Wed, 02 Jan 2002 22:56:00 +0000 Received: from DL4YHF@aol.com by imo-r01.mx.aol.com (mail_out_v31_r1.9.) id l.13.4590407 (3934) for ; Wed, 2 Jan 2002 17:55:10 -0500 (EST) From: DL4YHF@aol.com Message-ID: <13.4590407.2964e9ce@aol.com> Date: Wed, 2 Jan 2002 17:55:10 EST Subject: Re: LF: Very slow DFCW To: rsgb_lf_group@blacksheep.org MIME-Version: 1.0 Content-Type: text/html; charset=windows-1252 X-Mailer: AOL 6.0 for Windows DE sub 10510 Precedence: bulk Reply-To: rsgb_lf_group@blacksheep.org X-Listname: rsgb_lf_group Sender: Content-transfer-encoding: 8bit Hi Steve,

>> If you start off with a 650Hz tone sampled at 11025Hz with a record size of 16384 you have a record length of 1.49 seconds (BW = 0.67Hz). Averaging or skipping samples could be done down to maybe a 2205Hz sample rate, but that would only get a record length of about 7.5Hz (BW = 0.134Hz). The only way I can see that 5.3mHz can be achieved is if you mix down the input signal to close to zero frequency (say 20Hz) and low-pass filter to eliminate aliasing. This would make the 16384 record length long enough for mHz range resolutions. Is this how it is done ?
<<<

Basically, yes.
In fact the audio signal sampled at 11025 Hz is first multiplied with a two-phase "Local Oscillator" signal, with the frequency set for the center of the display (with complex numbers, you may call the output I / Q samples).
Then, the sample rate is reduced ("decimated") in a chain of "decimation stages" (don't know if thats the correct term). Each decimation stage is not only a "sample rate divider" but real a low pass filter (for anti-aliasing, not just an average filter).
Assume each stage divides the sample rate by two:
The first decimation stages has 11025 samples/second "in"
and produces 5512.5 samples/second.
The second 5512.5 samples/second "in" and puts 2756 samples/second out,
etc etc etc.
After a 7 of such stages there are 11025/(2^7) = 11025/128 = 86.13 samples per second (still I/Q samples). This 86.13 Hz sample rate goes into a 16k FFT (for example), giving a frequency resolution of 5.3mHz.
You could go much higher in FFT resolution but it takes an utterly long time to fill the FFT buffer.
Example: To put 16384 samples from the decimated signal (86.13 Hz) in a buffer, it takes 190 seconds.
I think ARGO and SPECTRAN work similar, with the difference that you don't have to think about these "details" there.. but Spectrum Lab was never intended to be user friendly HI.

Hope this answers your question a bit.

Good luck and happy experimenting !
73 Wolf DL4YHF.

P.S: Phase Meter..

Alan - in fact I have been playing with a "phase sensitive waterfall display" which Markus DF6NM suggested over half a year ago. Indeed, the result of the FFT calculation carries the amplitude and phase, but it's not easy to interpret the phase values and putting subsequent FFT results together (all I got up to now is a very colorfull display which looks like "moiré effect" on a TV screen. But with a bit of inspiration, there will be more soon... :-)