
Rockbox mail archiveSubject: Re: Re: berzerk peak meter? Alpha testers wanted.Re: Re: berzerk peak meter? Alpha testers wanted.
From: Paul Suade <paul.suade_at_laposte.net>
Date: Tue, 24 Sep 2002 20:04:51 +0200 Should be the last one. I did a mistake and strange enough, when I tried to used to compute j like gf in the example in http://www.dattalo.com/technical/theory/logs.html I found less accurate results than the first one despite of my mistake. Really a surprise... In fact, using "log2_table[j];" for the fractional part of log2(x) seems more accurate than "(log2_table[j] + ((x&4095)*(log2_table[j+1]log2_table[j]))>>12);". I'm really puzzled because the second method should be more accurate than the first one by linearly smothing the scale effect of the first method !? it should be another problem of overflowing... Nonetheless, if precision is enough, using only the first method has the real benefit to make fflog2 even faster to compute. By the way, I give here two ways to compute log10, you'de better to use the one which gets rid of fflog2 and uses log10_table instead since it has the same result than (fflog2(x)*LOG2_10)>>12 but faster (no multiply at all).  Original Message  From: "Paul Suade" <paul.suade_at_laposte.net> To: <rockbox_at_cool.haxx.se> Sent: Tuesday, September 24, 2002 3:44 PM Subject: Re: Re: berzerk peak meter? Alpha testers wanted. > Overflow bug fixed. > > Added fflog10 based on fflog2. > > It uses now a fixed point 4.12 bits integer instead of a fixed point 16.16 > bits integer. > > Example of results : > > fflog10(32767) = 4.50146 > fflog10(20479) = 4.28882 > fflog10(17408) = 4.24048 > fflog10(1279) = 3.08472 > fflog10(264) = 2.4082 > fflog10(72) = 1.85718 > fflog10(18) = 1.25513 > fflog10(15) = 1.17603 > Press ENTER to continue...
Page was last modified "Mon Nov 16 10:57:21 2020" The Rockbox Crew  Privacy Policy 