96khz?

[cashlz]

In somewhat simple terms could someone explain to me exactly what the sampling rate is and how it works. I hope I am even asking this question correctly. Thanks once again

[woodman]

In somewhat simple terms could someone explain to me exactly what the sampling rate is and how it works. I hope I am even asking this question correctly. Thanks once again

I'd be better able to explain this to you if I only knew how to draw diagrams directly into the computer which I don't. My computer literacy is not what I wish it were. Having said that, let me try to make this concept understandable to you.

An analog audio waveform is comprised of a multitude of up and down "squiggles" (for want of a better term). These up and down excursions are constantly varying in both speed (frequency) and amplitude (volume).

In order to convert this analog waveform into the digital format, the waveform is "sampled" by an A to D converter at TWICE the frequency of the highest pitched sound that a given format is capable of. In the conventional "Redbook" CD format, this "sampling rate" is 44.1 thousand times per second, which means that the format can store and reproduce any frquency up to 22KHz (22 thousand "cycles" per second which is higher than any human that I know can hear). This sampling yields a voltage value of the waveform at that instant in time. It is translated into a 16 bit "word" of 1s and 0s which can represent any of a possible 65,536 different values.

In order to convert this digital signal back into analog form (so that it can be heard as sound), a D to A converter is used that takes each digital "word" sample and converts it to a voltage value that is theoretically the exact same value as the original analog signal voltage that the sample represents. Therefore, it's possible to reconstruct the original audio waveform to its original state - recreating the original sound in (almost) perfect fashion.

I hope this made the whole concept a bit clearer for you

[cashlz]

Ok, I gotcha that makes alot of sense and I suppose thats why its called "true" 96. Thanks alot!

[maxg]

I'd be better able to explain this to you if I only knew how to draw diagrams directly into the computer which I don't. My computer literacy is not what I wish it were. Having said that, let me try to make this concept understandable to you.

An analog audio waveform is comprised of a multitude of up and down "squiggles" (for want of a better term). These up and down excursions are constantly varying in both speed (frequency) and amplitude (volume).

In order to convert this analog waveform into the digital format, the waveform is "sampled" by an A to D converter at TWICE the frequency of the highest pitched sound that a given format is capable of. In the conventional "Redbook" CD format, this "sampling rate" is 44.1 thousand times per second, which means that the format can store and reproduce any frquency up to 22KHz (22 thousand "cycles" per second which is higher than any human that I know can hear). This sampling yields a voltage value of the waveform at that instant in time. It is translated into a 16 bit "word" of 1s and 0s which can represent any of a possible 65,536 different values.

In order to convert this digital signal back into analog form (so that it can be heard as sound), a D to A converter is used that takes each digital "word" sample and converts it to a voltage value that is theoretically the exact same value as the original analog signal voltage that the sample represents. Therefore, it's possible to reconstruct the original audio waveform to its original state - recreating the original sound in (almost) perfect fashion.

I hope this made the whole concept a bit clearer for you

If I have understood what you are saying there are 65,535 different possible values a given point on a waveform can have within a range of 0(?) - 22,000 Hz on a CD. This implies to me that the accuracy of the representation of that waveform is limited.

This would mean that for 96KHz with a 16 bit word the accuracy of the representation of the waveform would be worse as the range is now 96,000/2 or 48 KHz to be represented with the same 65,535 possible values.

However if we use a 24 bit word (with 16,777,215 possible values) the accuracy of the representation of the waveform is higher.

Correct? Or am I a mile off base again?

[E-Stat]

In somewhat simple terms could someone explain to me exactly what the sampling rate is and how it works. I hope I am even asking this question correctly. Thanks once again

I'll offer perhaps an ever simpler explanation: sampling is analagous to a connect-the-dots picture. For every small sample of time, numbers or "dots" are assigned to the waveform. In context to your reference to a higher than Redbook standard rate, the problem with RB is that under certain conditions, there are not enough dots to look like a smooth line. During very quiet passages, the number of bits firing (again the dots) is greatly reduced, causing the RB standard to go deaf. Similarly, while the RB standard can handle simple waveforms at very high frequencies, complex harmonic overtones produced by multiple instruments overwhelms the fixed budget of dots available. Hence the reason behind the high resolution DVD-A and SACD standards.

rw

[mtrycraft]

In context to your reference to a higher than Redbook standard rate, the problem with RB is that under certain conditions, there are not enough dots to look like a smooth line.

Well, your analogy is a non starter. There are no dots to connect at the CD player output, it is a continuous wave. THe Red book is sufficient.

During very quiet passages, the number of bits firing (again the dots) is greatly reduced, causing the RB standard to go deaf.

Nonsense.

Similarly, while the RB standard can handle simple waveforms at very high frequencies, complex harmonic overtones produced by multiple instruments overwhelms the fixed budget of dots available.

More nonsense.



Hence the reason behind the high resolution DVD-A and SACD standards.

rw


No, that is not the reason at all. Maybe a good marketing ploy, yest, to appease the audiophile community. Mastering and archiving and multi channel playback is the reasons for DVD-A.

[E-Stat]

There are no dots to connect at the CD player output, it is a continuous wave. THe Red book is sufficient.

Ah yes, the ever perspicacious comments from our resident ditch digger. I have no doubt that RB is sufficient for you.

rw

[WmAx]

Ah yes, the ever perspicacious comments from our resident ditch digger. I have no doubt that RB is sufficient for you.

rw

If this is true, I am his apprentice-in-training. He digs, I clean his shovel. :-)

Seriously, though, RB standard is adequate for me, except for that it only allows for two channels. For some music, multi-channel is very important, IMO.(New age, pop, etc. that has begun to implement such surreal sound effects in it intended to be played over multichannel as released in it's SCAD and/or DVD-A version(s) .)

-Chris

[mtrycraft]

He digs, I clean his shovel. :-)
-Chris

Great. As you can see, it meeds a lot of cleaning

[mtrycraft]

Ah yes, the ever perspicacious comments from our resident ditch digger. I have no doubt that RB is sufficient for you.

rw

Why didn't you address the issues you messed up?

[E-Stat]

Why didn't you address the issues you messed up?

I'll let Sir Terrence enlighten you instead. Good luck.

Will SACD make vinyl obsolete?

rw

[WmAx]

, the problem with RB is that under certain conditions, there are not enough dots to look like a smooth line.
rw

This would, in a crude way, be true if the output stage did not include an anti-alias filter. Of course, any properly designed cd player has such a low pass filter. Therfor the result is a smoothed sinewave, as it should be. Their is a trend of some 'exotic' cd players to not include an antialias filter(they claim this, though, still a mild low pass filter(1st order), otherwise the aliased signal could possibly damage electronics and/or speakers--this insufficient slope will not provide adequate filtering of aliased data to prevent audible side-effects), the aliased bi-products can result in audible inter-modular artifacts when combined with the main audible passband. These would most probably be identified in a blind test easily! :-)
-Chris

[mtrycraft]

These would most probably be identified in a blind test easily! :-)
-Chris

But he would have no idea about that. He abhors blind tests and runs from them as if they were the plague.

[FLZapped]

In somewhat simple terms could someone explain to me exactly what the sampling rate is and how it works. I hope I am even asking this question correctly. Thanks once again

It's simple. You have the title "96kHz" - which is 96,000 cycles per second....in sampling terms, that means you're going to slice something up into 96,000 pieces every second.

Then, each one of those slices is assigned a number to represent the sample's position(yes, it's referenced to something). So now you have a numeric representation of you're original waveform and you can store it on a digital storage medium, like a CD. (BTW - those "slices" are not sequential on a CD, so if you get a scratch, it doesn't wipe out a large chunk of continuous data, but that's another topic of discussion.)

You do the opposite to reassemble it. Just take those numbers, put the sample back into position according to it's value and glue all of them back together, that's the basics.

-Bruce