Idea for anti jitter filter

For example in my non os DAC the jitter critical signal is LE, which is 44,1 kHz.

If the signal would not be a square but a sine, there would be sidebands and / or noise visible with a good spectrum analyzer, caused by the jitter.

If that sine runs through a narrow bandpass filter that is tuned to the fundamental, ideally a crystal filter, the sidebands and the noise will be strongly attenuated.

Same will apply to a square wave that will additionally be converted to a sine, but could be shaped back to a square by a comparator.

All IMHO.
 
all this assumes that the discs are pressed without jitter - I know the normal pressing technique for most commercial discs has a lot of variance in the pit size. This is one of the problems they try to overcome with XRCD and other similar methods - to eliminate the jitter at the mastering and manufacturing stage. No point trying to eliminate jitter in the playback if it is already there in the production.

my opinion, or course.
 
Bernhard said:


For example in my non os DAC the jitter critical signal is LE, which is 44,1 kHz.

If the signal would not be a square but a sine, there would be sidebands and / or noise visible with a good spectrum analyzer, caused by the jitter.

If that sine runs through a narrow bandpass filter that is tuned to the fundamental, ideally a crystal filter, the sidebands and the noise will be strongly attenuated.

Same will apply to a square wave that will additionally be converted to a sine, but could be shaped back to a square by a comparator.

All IMHO.

Hi,

I don't understand why you need to go to a sine first and back to a square if you want to do the filtering?

Jan Didden
 
Bernhard said:


For example in my non os DAC the jitter critical signal is LE, which is 44,1 kHz.

If the signal would not be a square but a sine, there would be sidebands and / or noise visible with a good spectrum analyzer, caused by the jitter.

If that sine runs through a narrow bandpass filter that is tuned to the fundamental, ideally a crystal filter, the sidebands and the noise will be strongly attenuated.

Same will apply to a square wave that will additionally be converted to a sine, but could be shaped back to a square by a comparator.

All IMHO.
You don't understand what jitter is. A pure sine wave has no side bands.
If you have a low jitter master clock in your player the jitter of LE will be low enough in NON-OS. Jitter is just timing errors but these are far less critical in NON-OS compared to 8x over-sampling.
 
Pulse-R said:
all this assumes that the discs are pressed without jitter - I know the normal pressing technique for most commercial discs has a lot of variance in the pit size. This is one of the problems they try to overcome with XRCD and other similar methods - to eliminate the jitter at the mastering and manufacturing stage. No point trying to eliminate jitter in the playback if it is already there in the production.

my opinion, or course.


All bits from a CD are stored in a DRAM and read out by the masterclock of the player.
At worst the pit-jitter is a secondary effect.
The kind of DAC, IV-converter and analog stage and filtering also play an important role in sound quality.
 
Re: Re: Idea for anti jitter filter

QSerraTico_Tico said:

You don't understand what jitter is. A pure sine wave has no side bands.
If you have a low jitter master clock in your player the jitter of LE will be low enough in NON-OS. Jitter is just timing errors but these are far less critical in NON-OS compared to 8x over-sampling.

A tip: never say that people don't understand things because you don't understand their post ;)

I know a pure sine has no jitter. But you go from square to sine, that sine still has the jitter of the square. That is what you say in your post. Then you go filter the sine to get rid of the jitter sidebands. Then you go back to a square, and that will again introduce jitter. That's working backwards.

If you filter the sidebands from the square, you don't need the final conversion from sine back to square. Less jitter.

Jan Didden
 
Every digital input has a comparator range and some input noise. One has to make sure that the transition between the 0 logical input voltage and the 1 logical level input voltage should be very fast. This can be provided with a fast rise-time digital signal (LE). In this case the jitter caused by the transition will be low. The LE signal should also have low jitter. If you have a low-jitter master clock, the LE coming from the previous stage (error corrector or digital filter) can be synchronized to it. I use the following method:
The LE goes on the input of a 74HCT74. The low-jitter master clock (4.2366 MHz in my case, but can be any frequency that is integer multiple of 44.1 kHz) goes to the Clock input. The Q output goes to the DAC. Something like this:

http://tube.fw.hu/no_oversampling.pdf
 
janneman said:


jitter sidebands are not harmonics of the square wave frequency.

I know, but if you try to remove jitter sidebands * , also the harmonics of the square wave will be removed --> sine --> comparator --> square


* could it work ?

http://electronicdesign.com/Articles/Index.cfm?AD=1&AD=1&ArticleID=6297

QUOTE from the link:

Furthermore, driving the crystal filter with a square-wave test signal only increases the 2nd-harmonic content by a fraction of a decibel.

/QUOTE

So if the driving square wave has timing errors, will the resulting sine wave still have those errors or not ?
 
Re: Re: Re: Idea for anti jitter filter

janneman said:


A tip: never say that people don't understand things because you don't understand their post ;)

I know a pure sine has no jitter. But you go from square to sine, that sine still has the jitter of the square. That is what you say in your post. Then you go filter the sine to get rid of the jitter sidebands. Then you go back to a square, and that will again introduce jitter. That's working backwards.

If you filter the sidebands from the square, you don't need the final conversion from sine back to square. Less jitter.

Jan Didden

Ok, I stay out of this!

:smash:
 
I guess I'm missing something here. Jitter reduction methods are well understood and reasonably simple.

If you re-clock a digital data stream, the end result jitter is solely the jitter inherent in the re-clocking source. What Levinson does in their 36, 36S, 360 and 360S DACS is use a relatively fast phase locked loop to extract the original data. They then use this "pliable" clock to write data to a buffer. Data is read out of the buffer with a very low jitter clock. Now the only jitter left in the data stream is whatever level of jitter the low jitter clock had. All other jitter errors are not reduced, they are gone.

Unlike an analog signal, it is actually possible to improve a digital copy, remove jitter present in the original digital source (assuming only that the peak to peak jitter is less than a clock cycle), remove any jitter in the original source, actually remove jitter caused by cables, slicers etc..

Many tests have been performed that show a recovered CD signal usually has an error rate of zero (per song). So there is little danger of adding errors with a re-clocking scheme.

Many people have noticed that in many cases a digital copy of a CD sounded better than the original. This is possible because the information is not carried in the shape of the waveform as it is for an analog signal. The information is carried in such a way that if your receiver can clearly distinguish between ones and zeros the original source can be recovered even from a badly degraded digital waveform.
 
Finally someone who knows what he's talking about!

hermanv said:
If you re-clock a digital data stream, the end result jitter is solely the jitter inherent in the re-clocking source. What Levinson does in their 36, 36S, 360 and 360S DACS is use a relatively fast phase locked loop to extract the original data. They then use this "pliable" clock to write data to a buffer. Data is read out of the buffer with a very low jitter clock. Now the only jitter left in the data stream is whatever level of jitter the low jitter clock had. All other jitter errors are not reduced, they are gone.

That's how I understood it too.

hermanv said:
Unlike an analog signal, it is actually possible to improve a digital copy, remove jitter present in the original digital source (assuming only that the peak to peak jitter is less than a clock cycle), remove any jitter in the original source, actually remove jitter caused by cables, slicers etc..

Which means that any jitter caused upstream by the transport is also gone. In my book, that means that any money you spend to rigidize the transport, put it on springs, use expensive cables to the DAC etc are literally a waste of money.

hermanv said:
Many people have noticed that in many cases a digital copy of a CD sounded better than the original. This is possible because the information is not carried in the shape of the waveform as it is for an analog signal. The information is carried in such a way that if your receiver can clearly distinguish between ones and zeros the original source can be recovered even from a badly degraded digital waveform.

Amen!

Jan Didden
 
Benhard,

I think I know what you mean. Get a (phase-)noisy signal and pass it through a very narrow band filter, in order to remove the noise and get only the fundamental frequency. The problem is that the phase noise produces sidebands only if you do a spectrum analysis that implies time average. There is always a single discrete frequency present in any moment, not a continuous spectrum of frequencies. That is, the zero crossing of the signal leads or lags in reference to the ideal (jitter-free) signal, that is causing the instantenous frequency different from the ideal. In the time-domain that is jitter, in the frequency domain that is noise sidebands.
Still it is possible to use a high-Q filter to remove the sidebands, according to your idea. This is because the filter stores energy, and there is oscillation in the filter even if the input signal is removed (or its instantenous frequency changed by the phase noise a very little bit). This is a method for producing a low-jitter master clock, but how do you make a square wave from it and keep its low jitter, that is another question.
Here is my idea: make a crystal oscillator with tube. Put the high-Q filter in the anode. Amplify the signal into clipping with a second tube (maybe use a second high-Q filter in the anode, too). Capacitor-couple the signal to a zener that clamps it to TTL level. You get a square wave with very fast rise and sharp edges. Put it physically very close to the stage where it is used. A lot of solutions from the HF communication technology can be used here, especially from the pre-solid-state era. Jitter used to be a problem in the SSB (single sideband) amateur communication.
 
oshifis said:
Benhard,

I think I know what you mean. Get a (phase-)noisy signal and pass it through a very narrow band filter, in order to remove the noise and get only the fundamental frequency. The problem is that the phase noise produces sidebands only if you do a spectrum analysis that implies time average. There is always a single discrete frequency present in any moment, not a continuous spectrum of frequencies. That is, the zero crossing of the signal leads or lags in reference to the ideal (jitter-free) signal, that is causing the instantenous frequency different from the ideal. In the time-domain that is jitter, in the frequency domain that is noise sidebands.
Still it is possible to use a high-Q filter to remove the sidebands, according to your idea. This is because the filter stores energy, and there is oscillation in the filter even if the input signal is removed (or its instantenous frequency changed by the phase noise a very little bit). This is a method for producing a low-jitter master clock, but how do you make a square wave from it and keep its low jitter, that is another question.
Here is my idea: make a crystal oscillator with tube. Put the high-Q filter in the anode. Amplify the signal into clipping with a second tube (maybe use a second high-Q filter in the anode, too). Capacitor-couple the signal to a zener that clamps it to TTL level. You get a square wave with very fast rise and sharp edges. Put it physically very close to the stage where it is used. A lot of solutions from the HF communication technology can be used here, especially from the pre-solid-state era. Jitter used to be a problem in the SSB (single sideband) amateur communication.


Your filter, if it is narrow enough, which it probably isn't, also has to be rock-stable for its center frequency. It will not be perfect, the center frequency will 'jitter'. Yes, jitter. You carefully trade one problem for another. It isn't THAT simple, or I would have done it already :D .

Jan Didden
 
oshifis said:
The problem is that the phase noise produces sidebands only if you do a spectrum analysis that implies time average. There is always a single discrete frequency present in any moment, not a continuous spectrum of frequencies.

I almost always do averaging to reduce noise floor.

And a clean jitter free sine will not produce sidebands, so nothing wrong in my posting.
Sorry, I just can not write down everything starting from Adam & Eva.

oshifis said:


Still it is possible to use a high-Q filter to remove the sidebands, according to your idea. This is because the filter stores energy, and there is oscillation in the filter even if the input signal is removed (or its instantenous frequency changed by the phase noise a very little bit). This is a method for producing a low-jitter master clock, but how do you make a square wave from it and keep its low jitter, that is another question.


The crystal filter that I proposed in my first posting is a high-Q filter,

quote from the posted link:

the −6-dB bandwidth is 144 Hz, equating to a crystal with a Q in excess of 45,000.

That is for a 5 MHz crystal. Very narrow.

The whole thing could be compared to a turntable. The direct drive motor is performing horrible but has low torque, the heavy table does it all.

To shape the sine back to a square is no problem, use a precision comparator as done for example in a Kwak-clock.
 
janneman said:



Your filter, if it is narrow enough, which it probably isn't, also has to be rock-stable for its center frequency. It will not be perfect, the center frequency will 'jitter'. Yes, jitter. You carefully trade one problem for another. It isn't THAT simple, or I would have done it already :D .

The crystal filter is rock steady. No jitter. :D

And filters with multiple crystals are even narrower :D

You just didn't have the idea :D

I would have done it already but I don't have the damn crystal. :whazzat: