24/192 music downloads are useless??

[cpp]

You only need it greater than 2x, hence the 44.1 although it could have been 44 or 48, not sure why 44.1

Interesting how 44.1 was established

http://www.cs.columbia.edu/~hgs/audio/44.1.html

http://www.indiana.edu/~emusic/etext...er5_rate.shtml

[mtrycrafts]

Interesting how 44.1 was established

http://www.cs.columbia.edu/~hgs/audio/44.1.html

Chapter Five: Principles of Digital Audio

How interesting, thanks.

[audiofox]

No, that is absolutely not true. You only need it greater than 2x, hence the 44.1 although it could have been 44 or 48, not sure why 44.1.

The 2X factor is the referred to as the Nyquist rate, a key design rule for digital signal processing system:

Nyquist rate - Wikipedia, the free encyclopedia

Note that this is typically used as the minimum required rate, but many digital communications systems exceed that rate if other factors affect the signal quality or particular digital modulation formats are used, so it is one of several variables that are traded against each other when optimizing a digital comm or signal processing system.

[avnetguy]

No, that is absolutely not true. You only need it greater than 2x, hence the 44.1 although it could have been 44 or 48, not sure why 44.1.

Remember I'm only talking about digital reconstruction, not about what can actually be heard by human ears. When sampling an input signal with the frequency at half the sampling rate the best achieved would be a triangle wave (samples taken at peaks) and the worst would be no waveform (samples taken at zero crossing), this depends on how your sample clock matched up with the incoming signal. Now as the input signal frequency decreases you are guaranteed to have one of the two points being non-zero and the frequency component is therefore retained. Now all input frequencies from the sampling rate / 2 (22050) to sampling rate / 3 (14700) will be represented by only two points placed "somewhere" on the slopes of the sine wave. So my question here is will those two digital points provide an accurate representation of the amplitude of that input sine wave?

Steve

[avnetguy]

Exactly, but my point in all of this is how will we know that lower bitrates and word lengths are sufficient if hardware, in general, is the limiting technology.

DJ

I don't think you'll have to worry about it as the higher standards are already established, both sampling rate (192 or 96 kHz) and resolution (24 bit) are in place for all AVRs that to can use DTS-HD Master Audio.

Of course you'll have to wait at least a few years before the 10TB ipod's and such come out, then the music industry will magically recognize the need for higher rates and of course charge higher prices.

Steve

[cpp]

Of course you'll have to wait at least a few years before the 10TB ipod's and such come out, then the music industry will magically recognize the need for higher rates and of course charge higher prices

Of course by then you will be older, your hearing will fade ( trust me your hearing fades with time) and the price of these cool tricks will be sky high and by then you will not hear the difference between 16/44.1 and 24/96

[PENG]

So my question here is will those two digital points provide an accurate representation of the amplitude of that input sine wave?

Steve

Theoretically yes if you know the frequency, but I think you know that already, so what is the hidden question?

[avnetguy]

Theoretically yes if you know the frequency, but I think you know that already, so what is the hidden question?

Now my answer to that would be no, I really don't see how you could consistantly capture the actual amplitude with only two points per cycle. In my mind you'd see the frequency with a varied amplitude as the two points move along the sine wave on each following cycle.

Going back to my previous statement, if I now have 4 samples per cycle on the sine wave I'd end up with a pretty good approximation of the amplitude regardless of their position on the sine wave.

So no hidden questions, it just looks to me like the representation of input frequencies from the sampling rate / 2 down to sampling rate / 4 will be far from a good approximation. So is my thinking process off here?

Steve

P.S. Gotta take the dog to the park now but maybe there is an easy way to prove/disprove this? How about testing with a lower sampling rate (4 kHz) and an input sine wave signal with a frequency that falls more into our normal hearing range (1000-2000Hz).

[avnetguy]

So here is a quick visual example that shows the difference in the actual amplitude captured.

Both sides show sine waves (top to bottom) at 3900, 3500, 3000, 2500 Hz.
The left side is sampled at 16000 and the right side 8000.



So visually it is pretty obvious at the sample rate of 8000 the amplitude of the signal is not represented very well as your input signal approaches 4000.

Steve

[PENG]

So visually it is pretty obvious at the sample rate of 8000 the amplitude of the signal is not represented very well as your input signal approaches 4000.

Steve

Can you link us to your source? Just looking at them I don't know what those graphs represent. They don't look like sine waves, but they can be resolved into sine waves of fundamental frequencies and harmonics. All I can say is that as long as you don't sample at zero crossing as you mentiioned before, two points in a cycle should allow you to reconstruct the original sine wave in theory. In practice there are many factors that influence how difficult it is to reconstruct the original signal accurately, but it is theoretically possible and we know it is happening otherwise we won't have CDs.

Take a look of the formula here and you should be able to see that it can done given that you have two values within one cylcle and the frequency is known.

Sine wave - Wikipedia, the free encyclopedia