Re: Reordering question

On Apr 21, 2004, at 1:09 AM, Grigory A. wrote:
> Now I am looking on
> void III_reorder(mad_fixed_t xr[576], struct channel const *channel, 
> unsigned char const sfbwidth[39])
>
> C implementation takes ~30 000 cycles - it is too much. So could you
> tell me can I use for reordering this one simple scheme
>
> for sb=0 to 31 do //For all 32 subbands
>    for window=0 to 2 do //For each of the 3 windows
>        for ws=0 to 5 do //For each sample within a window
>             Xro[18*sb + ss] = Xdis[18*sb + window + 3*ws] //Reorder
>        end for
>    end for
> end for
>
> If answer is "no" - so could you explain the role of
> unsigned char const sfbwidth[39]
> in original MAD reordering function
> very shortly of course.

A couple of notes:

- Only short block scalefactor bands should be reordered. So, if it is 
a mixed block, only subbands 2-31 should be reordered. (This is the 
purpose of the 18*sb expression, because long block scalefactor bands 
0-35 correspond to subbands 0-1.)

- The sfbwidth array provides the width of each scalefactor band (per 
window), and varies according to sampling frequency and whether the 
block is a normal short block or a mixed block.

- The purpose of reordering is to put the samples into subband order, 
with three windows per subband, and six samples per window (3*6 = 18, 
the width of one subband). Thus scalefactor bands with more than six 
samples per window will be spread across several subbands (placed into 
the correct window), and scalefactor bands with fewer than six samples 
per window will be coalesced with neighboring band windows.

-- 
Rob Leslie
[email protected]