Hi Up for a difficult question about bit and how it works? We know that sample rate sets the bandwith of the system by mesuring differences in the voltage level over time. In order for the desired frequency bandwith to be faithfully encoded in the digital domain, the selected sample rate must be twice as high as the higeste frequency to be recorded. We also know that higher bit the more detail mesurement. Since 16 bit represent 6 dB x bit + 1,8dB we can say that the dynamic range of a 16 bit signal is 97.8 and for 24 bit it is 145.8. We also know that the differences in the samples mesured would be in the LSB ( least significant bit). Because of the binural code a code can only be 0 or 1, therefore there will be problems when a samples is between what the converts can't handle. So far so good. What we have a problem understanding here is: Is the amplitude scale for the mesuring constant, meaning 16 bit and 24 bit has the same roof, but more posibilities for 24 between this scale. Ergo a more precise mesurament. Why doesn't 24 bit signal distort at the same volume as 16 bit when thay have the same roof? Or does the scale expands upward to have more headroom for 24 bit? But then there will also be few changes when we talk about possible sample values. Another way to see it this is not a correct way mathwise but ok example) 16 bit is 1 cm with 10 possibilities inside 24 bit is 1 cm with 20 possibilities inside = more possibilites, same roof. or 16 bit is 1 cm with 10 possibilities inside 24 bit is 2 cm with 20 possibilities inside = higher roof, same possibilities. Am I confusing you? So am I!!!! All help are greatly appreaciated. Best regards Bluemusic

Since max level relates directly to voltage(say 10v is max level) the 24 bit simply gives finer discrimination between 0v and 10v. Right? Did I say the out loud ?

Keeping with your example, I think if the range of human hearing is 1cm then 16bit would include the top 70% of that scale. and 24 bit would include the top 90% At least that's the way I've always understood it.

The point is that no matter what sound you are recording you adjust the roof so that it fits into your 16 or 24 bits. The loudest sound gets recorded as all 1s and the quietest nonzero sound gets recorded as 000...0001. The absolute volume doesn't matter. Just the ratio of loud to quiet. So your first example/analogy is the right one. To over simplify a bit (reading Terry Pratchett causes me to make bad puns) the (maximal) dynamic range is the ratio between the loudest possible sound and the quietest possible nonzero sound. Since you have finer gradations with 24 bits you have a smaller nonzero quietest sound compared to the loudest sound. Since you have 8 more bits, you have a 6X8=48dB bigger ratio between he quietest and loudest sound.

The max level (in dBu) has nothing to do with the number of bits the levels are encoded to - it is determined purely by the design of the equipment. Each piece of digital equipment has full-scale level specifications associated with its inputs and outputs. The absolute values of these are commonly quoted in dBu (for professional gear). This maximum value is the full-scale voltage that can be input or output without clipping, and is known as 0dBFS for that input or output. Equipment that has the same maximum levels will have the same 0dBFS, but could have different bit depths. A 24-bit ADC with a +20dBu full scale has 48dB more dynamic range than a 16-bit ADC with the same +20dBu full scale, since the extra 8 bits at 6dB/bit account for the additional 48dB. By the way, you posted much the same question in two different forums on this board. You get more sympathetic and coherent replies if you ask a question in just one forum that is relevant to that topic.

Ahh sorry. Wasn't sure wich forum to attend.. I think I am closer now. Consider this thread dead and reply to the other one..