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Author Topic: Is PCM a Tremolo Machine ?  (Read 26605 times)

Zoesch

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Re: Is PCM a Tremolo Machine ?
« Reply #15 on: June 15, 2004, 06:14:54 PM »

Nika Aldrich wrote on Wed, 16 June 2004 05:09

Do tell.  Find the phrase in Shannon that says that it has to be a stationary waveform and we'll see who, exactly, is distorting the words.



That was Fourier, not Shannon
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Chuck

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Re: Is PCM a Tremolo Machine ?
« Reply #16 on: June 15, 2004, 07:18:41 PM »

Hi Nika, and ... Zoesch,,

I used the PCM1704 behind the DF1704. But stay cool, any of your 'el cheapo' sigma delta stuff will put out the same results. Please... just take a look on a scope.

Shannon's proof is for signals that have a Fourier transform.

A signal that is not stationary does not have a fourier transform, or can you please tell me the Fourier transform of "Kid Charlemagne'" ???

The ringing of any interpolation filter has only one intent. It helps to reduce the amplitude modulation of a stationary signal and this works best, if the signal's samples are analyzed from minus infinity to plus infinity.

I am not breaking Shannon. I am just telling that Shannon's prove cannot be used for speech or music, as those a not stationary signals.

Zoesch, as I told earlier, the diagrams are samples from a mathematically pure sine wave. I don't understand your argument, as there is no argument. Please think about it.

Charles Smile

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Nika Aldrich

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Re: Is PCM a Tremolo Machine ?
« Reply #17 on: June 15, 2004, 08:07:02 PM »

Chuck wrote on Wed, 16 June 2004 00:18

Shannon's proof is for signals that have a Fourier transform.


Fourier's theorem states that any waveform can be broken into discreet frequency components and phase components.  The transform is the mathematical tool that does this.  That means that ANY waveform can be broken into frequency and phase components and that ANY waveform has a Fourier transform and if you don't understand this I would quickly stop posting, pull that ludicrous paper off your website, back up, and do some very, very fundamental research.  ALL waveforms have Fourier Transforms.  Can you find me one that doesn't?  Can you email me a waveform's data or formula that you don't believe has a Fourier transform?  

Quote:

A signal that is not stationary does not have a fourier transform,


This is so deplorably false that I don't know where to go with this.  

Quote:

 or can you please tell me the Fourier transform of "Kid Charlemagne'" ???


If it is a waveform then, per Fourier's theorem, it can be broken into frequency and phase components.  If it can be broken into frequency and phase components then it has a Fourier transform.  It is that simple.  Sure, it's transform will be more complex than that of a sine wave, but it does, most definitely have a Fourier transform.

Quote:

The ringing of any interpolation filter has only one intent. It helps to reduce the amplitude modulation of a stationary signal and this works best, if the signal's samples are analyzed from minus infinity to plus infinity.


No no no no no.  A sampled waveform has no amplitude modulation.  It is a mere representation of a waveform that passes through its sample points, but it is not, unto itself, a waveform, as you are trying to make it.  It is only a representation.  It HAS to be reconstructed.  I'm afraid that this line of thought, however, is far too advanced if you don't understand Fourier above, so let's hammer that out before we continue on with Shannon.  You can't understand Shannon until you understand Fourier.  You clearly have such a misguided view of Fourier that it will be completely counterintuitive to attempt to understand Fourier applied, as in Shannon.

Quote:

I am not breaking Shannon. I am just telling that Shannon's prove cannot be used for speech or music, as those a not stationary signals.


I asked you before and I ask you again - show me ANYWHERE that Shannon (or Fourier for that matter) specifies that their theorems only work with stationary waveforms.  Show me.  I'm waiting.  Go.  

Nika
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sfdennis

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Re: Is PCM a Tremolo Machine ?
« Reply #18 on: June 15, 2004, 09:03:50 PM »

Charles,
I stepped into this little quagmire of yours in another thread and decided to back out 'cuz I didn't have time to do the research then. I withdrew my post in that other thread since I hadn’t really done the proper background work. But, since you've brought this up again, here goes.

No. PCM is not a termolo machine.

Stefan and Nika have pretty much nailed it. You are not using a reconstruction filter, and the signal you are examining with your scope is the output of a zero-order-hold DAC with no reconstruction whatsoever. So I’m not at all surprised that you came to those conclusions. A couple of things should have clued you in.
Chuck wrote on Tue, 15 June 2004 11:21

I have used one of the best reconstruction filters available, the DF1704. You think you can up with something better.


The DF1704 is NOT a reconstruction filter. It is a digital—read, discrete timeinterpolation filter. Its primary application appears to be for oversampling before input to the PCM1704 which does not do on-board sample rate conversion. That’s a feature that was added to later delta sigma chips.

The point here is that reconstruction filters are continuous time beasts. If you look again at the interpolation formula on your web page, you’ll note that n is a discrete variable, but t is continuous. No digital filter will ever give you continuous t—not ever.

Sure, you could stay in the discrete time domain all you want, upsample to light in a vain attempt to simulate a continuous time signal, but that’s silly. Just stick a reasonable analog filter at the output of your DAC and many of the effects you’re observing will go away. I thought that the MFB filter described in the eval board of the 1704 was pretty reasonable.

Second clue. If you look at where your DF1704 goes in the TI diagrams, you’ll note that it feeds the DAC—it can’t possibly be a reconstruction filter. Reconstruction filters go after the DAC.

After writing that, I also have to say that it is a good thing to explore the limitations in our applications and implementations of the sampling theorem. So in general, I am sympathetic with what you’re trying to do. It is true that all of our filters contain imperfections. It is true that no real implementation of the reconstruction equation can yield exact results (we’d need infinite precision machines). In spite of that, these errors can be rendered insignificantly small with proper design. And the key to good design is how to make those errors virtually vanish.

I believe that you’re wrong this time, but you shouldn’t be discouraged. There aren’t many posters that actually pick up a scope and try to understand what they’re seeing. That’s a rare and valuable trait and should be encouraged.

-Dennis
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chrisj

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Re: Is PCM a Tremolo Machine ?
« Reply #19 on: June 15, 2004, 09:13:53 PM »

What people are saying here is that in order to put a signal through THAT sequence of sampled points, within a certain bandwidth (which is part of the sampling/reconstructing system), the resulting wave HAS to be a continuous, non-tremelo volume even if the apparent sampled points are not.
In other words there's only one wave that can do that (up that near to the Nyquist frequency) and it doesn't tremelo. If it did the points would be slightly different.
What YOU are saying is that since real-world systems aren't really perfect, the reconstructed wave isn't either. You're right there, and it's a fair concern, although since you're not reconstructing anything what you're showing on the scope is pretty moot. You're orders of magnitude off from what the real distortion products are. If you run a reconstruction filter your output is still not going to be perfect, but it'll be a lot more relevant. How about trying that?

debuys

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Re: Is PCM a Tremolo Machine ?
« Reply #20 on: June 16, 2004, 12:21:41 AM »

These discussions never cease to amaze me. First, I end up learning a great deal about theory and a bit about human nature. Second, I eventually become anoyed.

I see (with a few exceptions) absolutely valid data and arguments, but I miss the point. There always seems to be an agenda by the author. YES, there are limitations to digital reproduction of a signal. YES, there are inherent inacuracies. What I never see in these discussions is a comparison.

What's the alternative? What happens when you do get an accurate reproduction of a 22khz sine wave? Other than lab equipment what makes anything close to a 22khz sine wave? I asked my dog, but I'm out of treats and since she can't even tell me I'm not sure I should give her one.

What is up with the "pictures" of sounds we cant hear? The only time I use a scope and a signal generator is when something sounds wrong to me.

I see too much irony in these debates. So, if I did have an all analog system what would the medium be? Subs and vinyl dont get allong and I cant hear past 20k even if tape hiss didn't cover it up.

I can't pretend to have even part of the knowledge of many of the folks that participate in these debates but I would like to see what's important adressed. I find it moderately amuseing people spend this amount of time attempting to prove to me that my typical music gear stinks useing math and graphs.

I'd love to see these claims substantiated by a trained EAR. Here are my minimum specs for these said ears: nearly flat hearing from 20 20khz with a margin adjusted for the 4k range and abillity to transcribe 4 part choral harmonies and percussion at reasonable tempos.

If I even saw one truly objective paper on this I'd take it to heart rather than just learn some pure science.

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Zoesch

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Re: Is PCM a Tremolo Machine ?
« Reply #21 on: June 16, 2004, 12:32:47 AM »

Nika Aldrich wrote on Wed, 16 June 2004 10:07

Fourier's theorem states that any waveform can be broken into discreet frequency components and phase components.  The transform is the mathematical tool that does this.  That means that ANY waveform can be broken into frequency and phase components and that ANY waveform has a Fourier transform and if you don't understand this I would quickly stop posting, pull that ludicrous paper off your website, back up, and do some very, very fundamental research.  ALL waveforms have Fourier Transforms.  Can you find me one that doesn't?  Can you email me a waveform's data or formula that you don't believe has a Fourier transform?  



Wrong...

For a waveform to have a fourier series it has satisfy Dirchlet conditions (Let f be a piecewise regular real-valued function defined on some interval [a,b], such that f has only a finite number of discontinuities and extrema in [a,b]. Then the Fourier series of this function converges to f when f is continuous and to the arithmetic mean of the left-handed and right-handed limit of f at a point where it is discontinuous) or in simplese... a finite number of discontinuities which means that signal is Lebesgue integrable (Or Riemann integrable, as Lebesgue integrals equal Riemann integrals, not the other way around) and its integral will not diverge.

Something as simple as f(x)=x when x goes from -infinite to +infinite has no Fourier transform (It does integrate, but the integral does not converge), another one is cos(1/x) as x approaches zero the function becomes infinitely discontinuous... both examples CAN'T be found in nature (Both violate the second law of thermodynamics if they were to be a system's transfer function), there's more of course...

So it's not any signal, it's any signal that satisfies Dirichlet conditions, so the real answer is MOST functions can be proven to have a Fourier series, and ALMOST ALL real world functions will satisfy Dirichlet conditions and thus have a Fourier seriers.

Not that this changes Chuck's pseudo-research BS, but we have to be clear and objective.
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Chuck

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Re: Is PCM a Tremolo Machine ?
« Reply #22 on: June 16, 2004, 01:19:24 AM »

Hi Zoesch,,

my old math book tells me, that f(x) has to fulfill Dirichlet conditions in every finite interval (-L,L).

And one of the Dirichlet conditions is that f(x) must be periodic with 2L outside of (-L,L).

That sounds to me, as if f(x) has to repeat infinitely, as it must be periodic for fullfilling the Dirichlet condition.

This I would call a stationary signal.

I think that speech or music signals do not fulfill the Dirichlet condition.

Can you clear that up ?

Charles Smile
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Chuck

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Re: Is PCM a Tremolo Machine ?
« Reply #23 on: June 16, 2004, 01:32:58 AM »

On the practical side,,

yes, the DF1704 is an oversampling filter. Yes it feeds the PCM1704 DAC. Yes I have gentle analog filters behind the DAC.

The idea of an oversampling filter actually IS that you don't need an analog brickwall filter behind your DAC anymore.

Before more criticise my measurements and/or equipment, could someone please make a simple measurement with any favorite DAC on a scope. Take f.e. a 21kHz sine wave, and show me that it is not amplitude modulated !!!

If you do not have it on a test CD, you can generate the sine with almost any wave-program, burn a CD and put it in your CD-player.

I am not measuring with poor equipment. I have tried several DACs of my own make and other products. I have tried with old and modern R2Rs and modern sigma-delta DACs (f.e. DSD1791).

The outcome is always the same, with the exception that it can be much worse than the scope-shots that I show on my site.

Have fun with reality.

Charles Smile
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Chuck

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Another way of looking at it...
« Reply #24 on: June 16, 2004, 01:58:45 AM »

Hi Stefan, Nika...

I found the following interesting explanation on Wikipedia.org.

It is a different viewpoint, but I think it has to do with the argument:

"A well-known consequence of the sampling theorem is that a signal cannot be both bandlimited and time-limited. To see why, assume that such a signal exists, and sample it faster than the Nyquist frequency. These finitely many time-domain coefficients should define the entire signal. Equivalently, the entire spectrum of the bandlimited signal should be expressible in terms of the finitely many time-domain coefficients obtained from sampling the signal. Mathematically this is equivalent to requiring that a (trigonometric) polynomial can have infinitely many zeros since the bandlimited signal must be zero on an interval beyond a critical frequency which has infinitely many points. However, it is well-known that polynomials do not have more zeros than their orders due to the fundamental theorem of algebra. This contradiction shows that our original assumption that a time-limited and bandlimited signal exists is incorrect." -end of quote-

This sounds a little funny, in that it says that a signal must be infinite (not time-limited) in order to be bandlimited.

Well, for being used with Shannon's theorem of being completely describable by samples of at least (for Nika) twice the speed, it must be bandlimited.

But if it is bandlimited, it cannot be time-limited anymore. Call it stationary, if you desire so.

Does that mean, that it is not possible to perfectly bandlimit a piece of music (that I call truly time-limited) ?

And does it mean that if I manage to perfectly bandlimit that piece, that it will ring on forever ?

So, as I cannot truly bandlimit it, if I do not want it to go on forever, I cannot meet the sampling requirements.

And if I am able to bandlimit it 100% the music is gone, as I have built an oscillator.

So how can we make use of Shannons theorem with music?  

Wow. I was not that wrong. I think I am really in the sweet spot of it...

Charles Smile
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Zoesch

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Re: Is PCM a Tremolo Machine ?
« Reply #25 on: June 16, 2004, 02:04:26 AM »

Chuck wrote on Wed, 16 June 2004 15:19

Hi Zoesch,,

my old math book tells me, that f(x) has to fulfill Dirichlet conditions in every finite interval (-L,L).



And most audio signals do as they exhibit periodicity within intervals and they are mostly continuous.

Quote:


And one of the Dirichlet conditions is that f(x) must be periodic with 2L outside of (-L,L).



Yes, but the period doesn't have to be 2L, as long as it's periodic in that interval, it can even have a single period over time.

Quote:


That sounds to me, as if f(x) has to repeat infinitely, as it must be periodic for fullfilling the Dirichlet condition.

This I would call a stationary signal.



No, not infinitely, but it needs to show periodicity within the interval, and yes, that would be a stationary signal as its statistical parameters will be constant over time

Quote:


I think that speech or music signals do not fulfill the Dirichlet condition.



They do, they are stationary, you can easily figure out that the interval in which they fulfill dirchlet conditions. Turbulence flows for example are not stationary (As their statistical components vary randomly over time) but their functions do fulfill the Dirichlet conditions.

BTW, apologies if I sounded harsh in the past, I don't think that your approach is correct (Just because you are missing the whole reconstruction part of the chain and that's why you are getting aliasing modulation) but I don't think you are the chump others are making you to be... and yes you are on to something here, you just need to focus the research appropriately.
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chrisj

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Re: Is PCM a Tremolo Machine ?
« Reply #26 on: June 16, 2004, 04:14:24 AM »

I'm reminded of when I tried the Audacity sinc interpolation SRC (also known as 'Secret Rabbit Code') on tone sweeps...
I could make a tone sweep that would go from 10K to 24K at 48K sampling rate. It would have enough artifacts that you could hear the tone go up, and see it on the meters.
Convert it to 44.1K using the proper SRC, and you could still see it but could no longer hear it at the top of its sweep because the SRC was cleaning up the problems with the original tonesweep's synthesis- very unexpected.
I wonder if that means you can clean up this sort of problem with arbitrarily small amounts of downsampling? 44.05K anyone?
For what it's worth, doing the process properly involves a huge amount of processing overhead, suggesting that the desirable result is only achieved by using a LOT of adjacent samples in the interpolation. I really suspect that it doesn't take anything remotely like infinite samples to get a fantastic result- the trouble is that reconstruction filters aren't using anything remotely like infinite samples, if the CPU load of that process is any indication.

sfdennis

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Re: Is PCM a Tremolo Machine ?
« Reply #27 on: June 16, 2004, 08:35:24 AM »

Chuck wrote on Tue, 15 June 2004 22:32

 Before more criticise my measurements and/or equipment, could someone please make a simple measurement with any favorite DAC on a scope.


I have, and without a proper reconstruction filter I see exactly what you described, and it is totally meaningless for many reasons. For this thread the most important one is that a DAC w/o a reconstruction filter is a zero-order-hold device. With a reconstruction filter, all I see are filter artifacts (phase distortion, mostly) that diminish as the sampling rate goes up or the signal frequency goes down. No amplitude modulation whatsoever.

Chuck wrote on Tue, 15 June 2004 22:32


The idea of an oversampling filter actually IS that you don't need an analog brickwall filter behind your DAC anymore.



Perhaps not a 'brickwall' filter (zero-width transition band), but you still need an analog (continuous time) reconstruction filter. When you interpolate to a higher rate, then the analog filter can have a wider transition band—provided you actually take the output at the higher rate. It is the increased width of the transition band that allows gentler filters. And you only get that if you actually operate the DAC at a higher rate.

Chuck wrote on Tue, 15 June 2004 22:32

 Yes I have gentle analog filters behind the DAC.


Ok, so three questions for you. What was the output rate of the DAC when you took the measurements? Second, what was the corner frequency, filter order, and stopband attenuation of your gentle analog filter?

I seem to recall from the other thread it was 44.1KS/s. If my memory is correct, then you really didn’t get rid of the rather tight analog reconstruction filter design problem at 44.1KS/s. At that CD rates, the reconstruction filter design problem is pretty unforgiving. The transition band from 20kHz to Nyquist is only about 4.6%. And you have to get the stopband attenuation down to about -90dB for 16 bit quantization to avoid aliasing. That's a very constrained design space if you also care about ripple and phase distortion. You don't have many choices.

I did ask those questions in earnest,and hope that you can see your way into posting answers.

-Dennis
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Chuck

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Re: Is PCM a Tremolo Machine ?
« Reply #28 on: June 16, 2004, 09:25:24 AM »

sfdennis wrote on Wed, 16 June 2004 14:35


Ok, so three questions for you. What was the output rate of the DAC when you took the measurements? Second, what was the corner frequency, filter order, and stopband attenuation of your gentle analog filter?

I seem to recall from the other thread it was 44.1KS/s. If my memory is correct, then you really didn't get rid of the rather tight analog reconstruction filter design problem at 44.1KS/s. At that CD rates, the reconstruction filter design problem is pretty unforgiving. The transition band from 20kHz to Nyquist is only about 4.6%. And you have to get the stopband attenuation down to about -90dB for 16 bit quantization to avoid aliasing. That's a very constrained design space if you also care about ripple and phase distortion. You don't have many choices.

I did ask those questions in earnest,and hope that you can see your way into posting answers.

-Dennis



Hi Dennis,,

as I am using the DF1704 8x oversampling filter,  my DAC's input is 8x44.1 = 352,80

As I recall I am using a small cap in the I/V feedback path, plus a simple RC filter between I/V conversion and output buffers. So this is not much more than 6dB.

I cannot use any steeper filter, as the product supports sample-rates up to 768kHz ( 96kHz x 8 ).

You will not get any better results with any DAC on the market.

And if you get a DAC doing something like a perfect 21kHz sine at 44.1 sample-rate, it will sound awful, as it will ring very very long.

As I posted earlier, a result of Shannon's theory is, that the more perfect you are able to limit the bandwidth, the less time-limited (ringing) your signal becomes.

Any kind of oversampling is an attempt to reduce one kind of distortion (beating) at the expense of another kind of distortion (ringing). So you can choose what kind of distortion you would like more.

You can eliminate both kinds of distortion if you record with a high enough sample-rate (4..5 x 44.1kHz) and don't use any bandlimiting during recording and also no oversampling during playback. The "reconstruction" filter will be of low order, as you only have to filter the odd order square products. In this case you would not even call it "reconstruction filter" as reconstruction is not necessary because you have enough samples per wave. This way your beat products (amplitude modulation) will be much smaller than with 44.1 plus oversampling AND you will have extremely short ringing. This is what is called a time-correct reproduction.

Of course, you cannot get to that high level of sound quality with sigma-delta converters, as they all use oversampling.

At least some have switchable filter characteristics. If you are able to give that a try, you will realize that the slow filter always sounds better than the sharp filter, although the sine-representation looks bad on the scope.

Charles Smile
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Nika Aldrich

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Re: Is PCM a Tremolo Machine ?
« Reply #29 on: June 16, 2004, 10:21:36 AM »

Zoesch wrote on Wed, 16 June 2004 05:32


Wrong...


Zoesch,

Good call.  I should have clarified - real, continuous waveforms, or something like that.  Of course cos(1/x) is not continuous at 0 (divide by 0), and other exceptions apply.  We should clarify for those watching that even a waveform that is on the zero crossing is "continuous."  A non-continuous waveform would be one where, for a certain value(s) of X, NO Y value exists.

Nika.

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