Did you know that you probably just are listening to 19 or 18 bit sound?

>boray, i thought that there was no loss of sound quality
>while doing digital bouncing.

In theory you will never notice if you don't make an extreme test, like having a track recorded, set the fader at 1% of unity gain (probably the lowest you can set it to while still sounding), bounce to another track. Then use every gain you have to get it up to level again (you probably have to make several bounces to get it to the same volume again). Compare to the old track. If you do that test with a 24 bit mixer, 24 bit (non floating point) recording format, a 24 bit sound recorded on the track to start with, then what you will have on the new track will have the same dynamics as for a 18 bit sound. Will you be able to tell that it sounds worse? If you have great monitors and great ears, then maybe! In practise you will never do something like that anyway.

>If the mixer90%of the things you are talking about i do not
>have the knowledge to challenge(although they sound
>pretty wild) but if you are correct than a few of the basic
>concepts i(and many others) have come to know are totally
>wrong.
>
>at this point either you could be insane...or i could be
>really dumb.

Or you have just spent too much time with people like BlueBear who think they know everything, but who has no deep knowledge. He has read a book about digital recording and now he thinks he knows everything. But hey, sometimes you need to think too. Sometimes it's not enough to just repeat what is stated in a book.


You need to learn and understand this to get what I'm saying:

How binary values works.
How waveforms are presented in binary values.
How an A/D converter works.
How a waveform's amplitude is changed by altering the sound's volume (or the other way around).
How a D/A converter works.

You don't need to know how a digital mixer works (how the binary values are handled), because in the end, it will end up as binary waveform data, either when you store it (if the format not is floating point) or when it reaches the D/A. That will allways happen regardless of how the mixing is performed.

/Anders
 
Son of Mixerman said:
Can someone how to change the amplitude without changing the volume?

SoMm

With a compressor you can change the percieved volume without making the overall waveform amplitude higher, but a fader's job is just to change the volume, not the dynamics. A sound's volume and it's waveform amplitude (and it's fader setting) are lineary connected.
 
Blue Bear Sound said:
What's really funny is that you've managed to get some people to believe your crap..... which is both sad and humourous at the same time....

:rolleyes:
...and exciting!!! :eek:
 
Boray said:
With a compressor you can change the percieved volume without making the overall waveform amplitude higher, ......

I take that back!!! The "overall sum" of the waveform will be higher of course....
 
Oh My Freaking God

Boray, you are confused. Very very confused.

If your house has 16 light fixtures in it, of which 10 are turned on right now, do the other six exist? Same thing applies digitally.

Let's take the following example. I'll demonstrate 16 bits, big endian for simplicity:

1010111101110000

The above word (2 bytes) is a 16bit binary representation of a number. It is not 12 bit. It remains 16bit even though the LSBs are 0.

The next example is 12 bit

101011110111

Even though I cut and paste the first 12 bits from that 16bit number above, and although they are mathamatically equal, they are not the same! The 16 bit value will always have the extra 'zeros' padded there, even if they are not used.

By the way, what digital audio software did you write in the late 80s? Can I have a copy? The source code? What was it called, and what company distrubuted it?
 
Re: Oh My Freaking God

DavidChristophe said:
Boray, you are confused. Very very confused.

If your house has 16 light fixtures in it, of which 10 are turned on right now, do the other six exist? Same thing applies digitally.

Let's take the following example. I'll demonstrate 16 bits, big endian for simplicity:

1010111101110000

The above word (2 bytes) is a 16bit binary representation of a number. It is not 12 bit. It remains 16bit even though the LSBs are 0.

The next example is 12 bit

101011110111

Even though I cut and paste the first 12 bits from that 16bit number above, and although they are mathamatically equal, they are not the same! The 16 bit value will always have the extra 'zeros' padded there, even if they are not used.

By the way, what digital audio software did you write in the late 80s? Can I have a copy? The source code? What was it called, and what company distrubuted it?

The title of this thread is "Did you know that you probably just are listening to 19 or 18 bit sound?". In your analogy, when 6 of the 16 light bulbes are turned off, do you still see any light from them? Of course not! You could just as well just had 10 light bulbes to reproduce the same light. The same applies to a D/A. If you not are using all of it's resolution, then the sound from it could just as well had been represented with a D/A that has a lower resolution.

The program i referred to was called "Tape Sampler" and was a 1 bit sampler for the Commodore64 to record audio from the data tape recorder. It worked but it didn't sound very good. Do you still like to have it? I have made programs for A/Ds and D/As as well later on... But they were technical measurement applications not involved in audio.

/Anders
 
Re: Re: Oh My Freaking God

Boray said:
The program i referred to was called "Tape Sampler" and was a 1 bit sampler for the Commodore64 to record audio from the data tape recorder. It worked but it didn't sound very good.
No surprise there..........!

:rolleyes:
 
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