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Digital Audio Signals and How They Really Behave In the Real World - Page 2

post #16 of 27

The background noise seems to be pretty static. There's a blip every couple of seconds though which seems to be interference?

I guess the soundcard has just better shielding, voltage regulators etc. Also the D1 doesn't use ESS chips, does it?


Edited by xnor - 3/1/13 at 11:07am
post #17 of 27

That was very interesting!  And very well presented!

post #18 of 27
Quote:
Originally Posted by xnor View Post

The background noise seems to be pretty static. There's a blip every couple of seconds though which seems to be interference?

 

There is some clicking sound more than 10 times per second, at somewhat irregular intervals. However, it is mono (the same clicking is on both channels always at the same time), which does indeed suggest that it resulted from badly implemented grounding or power supply filtering on the motherboard, rather than some pseudo-random DAC artifact in the ALC887.

 

ESS probably used the worst competing product (which they forgot to name) they could find for their presentation, so given that the artifacts are still always under -100 dBFS, it is not likely to be much of an issue in practice with any decent DAC.


Edited by stv014 - 3/1/13 at 12:09pm
post #19 of 27

Awesome video, thanks for sharing

post #20 of 27

I couldn't help but notice that his output o'scope was set on AC coupling.  

 

I'm not suggestion that would have made a big difference, or even any difference at all, but I am curious to know what it would have looked like if set to DC coupling.

post #21 of 27

IIRC, the lowest frequency shown is 1 kHz, so I don't think you'd see any difference at all.

post #22 of 27

You're probably right.

post #23 of 27

Sorry to rez such an old thread.

 

At about 6:50 into the video, Monty points out that there is only one possible band limited signal that passes through all the sample points and that therefore the band limited sample is the only possible output.  I agree with Monty, this is not at all obvious.  Can anyone explain how this works?  How does the playback signal "know" how to find its way exactly through the sample points to form the original signal?      
 

post #24 of 27

the precondition is that the the input analog signal has been low pass filtered to have no content at or higher than the Nyquist frequency before digitization

 

then when the discrete time digital signal is converted back to analog the goal is to find the "best fit" continuous analog signal thru the points that has no frequency content above Nyquist

 

it turns out that a linear low pass filter with sharp cutoff just below Nyquist does the job

 

technologically we can upsample in the DAC to do a large part of this sharp cutoff filter in the in digital domain but at much higer sample rate

"interpolating" the best value for the intermediate points with a weighted average of some number of delayed digital samples giving this curve fit to the original lower rate digital points

the more points used in the curve fit the better you can interpolate the "best" continuous curve

 

we still need a analog low pass "reconstruction/image reject" filter but it can be much simpler when fed with the higher upsampled and digitally interpolated signal

post #25 of 27

Thanks for the reply!  I have to admit it's mostly going over my head. 

 

I knew before that the Nyquist frequency had to be twice the frequency of the signal we want to sample.  I guess what I didn't know is that you must first band limit the analog data for the digital sampling to work.  I'm still not clear how the output waveform knows to conform itself to the input waveform.  Is the output wave form simply the only energy state possible?  That is, given the discrete samples that have been recorded, will any path through them that is not the original waveform require a higher energy state than is available to the system because we applied the low pass filter to the analog signal before sampling it?  
 

post #26 of 27
Quote:
Originally Posted by BlindInOneEar View Post

I'm still not clear how the output waveform knows to conform itself to the input waveform.  Is the output wave form simply the only energy state possible?  That is, given the discrete samples that have been recorded, will any path through them that is not the original waveform require a higher energy state than is available to the system because we applied the low pass filter to the analog signal before sampling it?  

 

 

The reconstructed path is the one that has the smallest bandwidth. Any other path will generate frequencies that are higher than half the sampling frequency. So, it is more of a "frequency" state than "energy" state.

 

Actually, the output waveform has not to "know" anything: just removing all frequencies above half the sampling frequency is the same thing as generating that optimal reconstructed path.

post #27 of 27

To make things simpler think of a single sine wave. A low frequency sine wave will consist of hundreds of samples so even if you connected them with straight lines it wouldn't look too bad. This is the mistake that everyone who doesn't understand digital audio makes.

The samples are not connected by straight lines or in a staircase like fashion, but there has to be a continuous sine wave with some frequency X that has to match the sample values.

 

As you increase frequency, connecting samples with straight lines will break down even visually. But you can still fit only one sine wave in there since we have to have more than 2 samples per cycle.

Exactly two samples doesn't work because think what would happen if the sample value was taken exactly at the zero-crossing of the sine wave - all sample values would be zero which is silence and not a tone.

If you have less than 2 samples per cycle you end up with aliasing. That is frequencies above half the sampling rate will be "mirrored back" to below the sampling rate.

 

 

That is why the low-pass filtering is absolutely necessary. In the image above a low pass filter would attenuate the blue sine wave because its frequency is greater than the sampling rate.


Edited by xnor - 6/10/13 at 2:39am
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