Executive Summary on How Digital Audio works

[davedvdy]

Re: Executive Summary on How Digital Audio works

Question: If I record audio on an Mbox at 24 bit instead of 16, will it still sound better when put on a CD when I bounce to 16 bit/ 44.1k? How would the data transfer down still sound better?

 

[Robert Delahunt]

Re: Executive Summary on How Digital Audio works

I started writing this in the "modellers topped out" thread to address some misunderstandings about how digital audio works, and decided that it deserves it's own thread. Here you go:

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Here is the executive summary of how digital audio works:

Think of a sine wave on a chart (if you don't know what a sine wave is, think of an S turned on its side). This is a sound wave in the air, which is what our ears hear. Now imagine if under the curve of the wave, you drew straight vertical lines that went from the bottom of the page to the bottom of the curve. Finally, draw a straight line between the tops of each of the lines. You will notice that the "connect the dots" line is pretty close to, but not exactly the same as, the original line.

This is how sound is represented digitally. The computer has a list of numbers, each representing one of these lines, and when they are played through the D/A converter (which essentially "connects the dots") the result is an audio signal. The lines are referred to as samples. Software that processes audio is essentially just changing the value of those samples.

Digital audio quality is measured using two values: sample rate, and sample size.

The sample rate is the number of samples per second, measured in Hertz. The more samples there are, the closer the "connect the dots" line will be to the real audio wave. This is why higher sample rates are preferable. CD audio has a sample rate of 44.1 kHz, or 44,100 samples per second.

Sample size is exactly that: the size of each sample. As I said, a sample is simply a number; the sample size determines how big that number is allowed to be. The bigger the number, the more precise the sample will be. Imagine you had to spell your name on the top of a desk using building blocks. If you used a small number of really large blocks, it would be difficult to make your name readable. With lots of small blocks, it would look much better. Now apply this analogy to the lines under the sine wave: if the lines were made of only a few big blocks, they would not accurately represent the wave. This would be a small sample size.

One thing that confuses a lot of people is that sample size is measured in bits. For example, CD audio has a 16 bit sample size. This does NOT mean there are only 16 levels of sound possible in a sample. 16 bits refers to the size of the number in binary, which happens to be 65,536. This means that the sample can have over sixty-five thousand possible values.

Combining sample rate and sample size, we can get an idea of just how much data is used to process the audio. Using the CD example again (16 bit 44.1 kHz), we know that there are 44,100 samples, each with up to 65,536 possible values, per second of audio. That's a lot of precision! Yes, it is still not a perfect curve, but it is so close our ears can't tell.

I hope that is a good explanation.

Uh, no, should be 16 bit = 16 resistors in the analog-to-digital input section, 16kbit would be 16,000 of them. It's mainly for reading the amplitude of the sound. I could be wrong, but the USAF taught us (in circuit diagram form) analog-to-digital converters, and I could swear that the main thing was that the # resistors determined the # possible values, which got converted into digital "bits" of 16 bit = a 16 character binary value for which resistors were "on" or "off" due to the input power.

It's not really that big a deal though, just trying to get clarification. Am I right?

 

[Robert Delahunt]

Re: Executive Summary on How Digital Audio works

Question: If I record audio on an Mbox at 24 bit instead of 16, will it still sound better when put on a CD when I bounce to 16 bit/ 44.1k? How would the data transfer down still sound better?

If you record in something higher than what the resulting media is, yes, it will get down-converted. However, you want to start off with as pure / clean / complex signal as you can, because you can always lose quality, but you can never gain it. Try this: take mp3, encode to 16bit or something smaller than it is, listen to it and hear the quality loss, then encode at higher bit rate that same mp3 and listen again: there should be no difference because quality was already lost and cannot be gotten back.

Hope that makes sense...

 

[GuitarStv]

Re: Executive Summary on How Digital Audio works

Uh, no, should be 16 bit = 16 resistors in the analog-to-digital input section, 16kbit would be 16,000 of them. It's mainly for reading the amplitude of the sound. I could be wrong, but the USAF taught us (in circuit diagram form) analog-to-digital converters, and I could swear that the main thing was that the # resistors determined the # possible values, which got converted into digital "bits" of 16 bit = a 16 character binary value for which resistors were "on" or "off" due to the input power.

It's not really that big a deal though, just trying to get clarification. Am I right?

Sorry, not quite correct. 16 bits refers to the fact that each digital grouping sent representing the volume of the wave being converted has 16 binary digits. For example:

1 bit grouping (can hold a number equal to 2^1):
0 = 0
1 = 1

2^1 = 2, there are two possible values

4 bit grouping (can hold a number equal to 2^4):
0000 = 0
0001 = 1
0010 = 2
0011 = 3
0100 = 4
0101 = 5
0110 = 6
0111 = 7
1000 = 8
1001 = 9
1010 = 10
1011 = 11
1100 = 12
1101 = 13
1110 = 14
1111 = 15

2^4 = 16, there are 16 possible values

16 bit grouping (can hold a number equal to 2^16):
0000000000000000 = 0
0000000000000001 = 1
0000000000000010 = 2
0000000000000011 = 3
. . .
1111111111111111 = 65535

2^16 = 65536, there are 65536 possible values

Therefore in a 16 000 bit digital grouping you'ld have 2^16 000 digits (which ends up being a ridiculously large number). I can't remember how inputs are usually stored, but I would assume that there's some sort of combination of capacitors and transistors - but you may be right about the resistor thing.

 

[Robert Delahunt]

Re: Executive Summary on How Digital Audio works

Sorry, not quite correct. 16 bits refers to the fact that each digital grouping sent representing the volume of the wave being converted has 16 binary digits. For example:

1 bit grouping (can hold a number equal to 2^1):
0 = 0
1 = 1

2^1 = 2, there are two possible values

4 bit grouping (can hold a number equal to 2^4):
0000 = 0
0001 = 1
0010 = 2
0011 = 3
0100 = 4
0101 = 5
0110 = 6
0111 = 7
1000 = 8
1001 = 9
1010 = 10
1011 = 11
1100 = 12
1101 = 13
1110 = 14
1111 = 15

2^4 = 16, there are 16 possible values

16 bit grouping (can hold a number equal to 2^16):
0000000000000000 = 0
0000000000000001 = 1
0000000000000010 = 2
0000000000000011 = 3
. . .
1111111111111111 = 65535

2^16 = 65536, there are 65536 possible values

Therefore in a 16 000 bit digital grouping you'ld have 2^16 000 digits (which ends up being a ridiculously large number). I can't remember how inputs are usually stored, but I would assume that there's some sort of combination of capacitors and transistors - but you may be right about the resistor thing.

Digital to Analog Wiki

I'm very confident that each bit = one resistor (i.e. each bit ends up being one digit in binary). So there are 65,536 possible values. I guess I was just not getting that part: I knew it, just didn't see it yet

But back to the thing about downsampling, it's like any information converted to digital, whether video or audio. You want to start off with the most information, whether that's a camera set to max resolution, or the most pure audio (and/or highest bit rate possible). However, with audio CDs, yes, if you have more info than you started with, you downsample to the CD audio format, which is 16bit 44.1kHz etc.

So this is why resizing a JPG to a higher resolution, resampling to a higher bit format, etc, don't result in an increase in quality or information. My advice: start with the highest quality (equipment, guitar, amp, etc) you can afford.

 

[davedvdy]

Re: Executive Summary on How Digital Audio works

If you record in something higher than what the resulting media is, yes, it will get down-converted. However, you want to start off with as pure / clean / complex signal as you can, because you can always lose quality, but you can never gain it. Try this: take mp3, encode to 16bit or something smaller than it is, listen to it and hear the quality loss, then encode at higher bit rate that same mp3 and listen again: there should be no difference because quality was already lost and cannot be gotten back.

Hope that makes sense...

That does make a lot of sense, thank you. I can see how the result will be better starting at a higher resolution of sound.