I''m talking about writing a bitrate converter, not using some wave editor, which I am perfectly capable of doing myself, thank you very much . Anyway, you''d think that 8 bit to 16 bit conversion should be lossless and work as simply as: 16bitvalue = 8bitvalue << 8, but it''s not - for some reason I''m getting heavy noise instead, which only very remotely resembles the original signal. In other words, the following doesn''t work:

char signal8bit[] = { ... };
short* signal16bit = new short[numsamples];

for(int i = 0; i < numsamples; i++)
   signal16bit[i] = signal8bit[i] << 8;

Similarly, downsampling shouldn''t be much other than 8bitvalue = 16bitvalue >> 8, should it? Maybe I''m an idiot, but what''s the catch here?

Maybe you need to use unsigned data types?

Nope - signedness doesn''t have anything to do with this. After a little bit of reverse engineering, this seems to be the solution:

16bitvalue = (8bitvalue + 128) << 8;

Or, in the generic form:

//SBD - source bit depth//DBD - destination bit depth//val - sample to convertint RedefineBitDepth(int val, int SBD, int DBD){	//upgrade	if(DBD > SBD) return ((val + (1 << SBD - 1)) << (DBD - SBD));	//downgrade	else return ((val - (1 << SBD - 1)) >> (SBD - DBD));}

I think so at least... If anyone spots an error, let me know.

You''re programming could be absolutely correct, I really don''t know. But anything in 8 bit is gonna give you noise in the background, it''s one of those reasons why we have 16, and soon 32 bit :D

There is some audio theory (that I don''t know) that you have to take into consideration when changing bit rates in a file. The waveform is greatly affected by any changes made to it.

Doesn''t help, but maybe, i unno.

cheers!

the noise you get has to do with the SQNR (signal-to-quantization-noise-ratio /or something) the more bitrate you use the more range of values to quantize the signal you will have.

8 bit -> 256 values of amplitude
16 bit -> 65536 values

this amount of values manage the same signal you are using to record, so in 8 bit you have less positions to quantize.

i hope my english explain it well, if not, search for "sqnr".