16-20 bits per integer digit sounds reasonable for double precision FP ,as opposed to the 5 bits which I suggested (from dividing the max exponent by FFT size).
Is any further clarification possible, without going into too much detail? THX David ---------------------------------------- > Date: Sun, 29 Oct 2006 12:00:22 -0500 > From: [EMAIL PROTECTED] > To: [email protected] > Subject: Re: [Prime] FFT size > > Quoting david eddy <[EMAIL PROTECTED]>: > > > > > I apologize if I am being naive here, but is it fair > > to say that for an FFT size of 1024K (2^20), a number > > (MOD the exponent) is split into the > > 2^20 digits (base 32) for input? > > > > The output is then 2^20 floating point numbers, > > each of which is rounded to the nearest integer? > > It's usually 16-20 bits per 'digit', but otherwise yes. > The number of bits per digit can be a constant, but in the > case of the discrete weighted transform it varies from > digit to digit. > > jasonp > > ------------------------------------------------------ > This message was sent using BOO.net's Webmail. > http://www.boo.net/ > _______________________________________________ > Prime mailing list > [email protected] > http://hogranch.com/mailman/listinfo/prime _________________________________________________________________ Be one of the first to try Windows Live Mail. http://ideas.live.com/programpage.aspx?versionId=5d21c51a-b161-4314-9b0e-4911fb2b2e6d _______________________________________________ Prime mailing list [email protected] http://hogranch.com/mailman/listinfo/prime
