Message-ID: <001501c3bf39$aedd0fb0$0401a8c0@vaiosn>
From: "Stewart Nelson" <sn@scgroup.com>
To: rsgb_lf_group@blacksheep.org
References: <01C3BF2E.2774C230.g4jnt@thersgb.net>
Date: Wed, 10 Dec 2003 08:21:34 -0800
MIME-Version: 1.0
Subject: Re: LF: Related Technical Query - Soundcard calibrator
Content-Type: text/plain; charset=iso-8859-1; format=flowed
Content-Transfer-Encoding: 8bit
Sender: <owner-rsgb_lf_group@blacksheep.org>
Precedence: bulk
Reply-To: rsgb_lf_group@blacksheep.org

Hi Andy,

IMO, you are being too much of a perfectionist to worry about
sin(x)/x.  I believe that x must be pi * fsig / fsample,
because the spectrum goes to zero at multiples of fsample.
So at 20 kHz the spectral density is 20*log(sin(pi*20/5000)/(pi*20/5000)).
This is about 0.00023 dB down from the level near DC.

The remaining question is what's the scale factor.
I believe that you can ignore the high frequency behavior,
and pretend that the power is spread over half the sample
frequency.  If your shift register puts out +/- 1V square
wave into 1 ohm (total RMS is one watt), then the low
frequency power density is 10*log(1/2.5E6) = -64 dBW/Hz.

My math is really weak, so I verified this by simulating a
random square wave, and looking at the spectrum with Cool Edit.

73,

Stewart KK7KA

----- Original Message ----- 
From: "Andy talbot" <g4jnt@thersgb.net>
To: <rsgb_lf_group@blacksheep.org>
Sent: Wednesday, December 10, 2003 6:59 AM
Subject: RE: LF: Related Technical Query - Soundcard calibrator


> Yes, the power spectrum of  a SIN(X) / X  pattern.
> 
> To Brian 'GVB, the limits will be (theoretically at least) - infinity to 
> infinity, but in practice a few lobes will be sufficient, say about +/- 5
> 
> Andy