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View Full Version : Will I add jitter by using an external USB hard drive



Kellen
2009-09-17, 09:35
Sorry in advance if this shows my stupidity on the subject matter but .....

I am running out of storage space and am considering buying an external USB hard drive to use in place of the internal hard drive that I now use.

Before doing so I was curious if doing so would add jitter to my system as I'd be using a USB cable to transfer the music files from the external hard drive to the computer.

Does this process add jitter and if not why?

Thanks

pfarrell
2009-09-17, 09:44
Kellen wrote:
> Does this process add jitter and if not why?

No.

Jitter is a theoretical term used to separate audiofools from their
money. It can theoretically only exist between the creation of an
digital audio signal and the DAC. Nothing in the TCP/IP or hard disk
world can even have a theoretical impact.

Don't worry.


--
Pat Farrell
http://www.pfarrell.com/

Phil Leigh
2009-09-17, 11:12
Jitter is real (it can be measured). The real issue is "to what extent does it matter?".

The SB jitter is quite low, the touch is lower still and the TP is very low.

Anyway, as Pat said, "No". The beauty of the SB approach is that there is no way to add jitter into the replay chain until you get inside the SB itself and use either the internal DAC or the digital outputs. Nothing on your PC, ethernet, wi-fi etc can have ANY impact on jitter. None.

Modern DAC's deal with any residual jitter quite effectively in most cases anyway.

There are ways to virtually eliminate replay jitter using an External DAC and an i2s or wordclock connection.

pfarrell
2009-09-17, 11:27
Phil Leigh wrote:
> Jitter is real (it can be measured). The real issue is "to what extent
> does it matter?".

And IMHO, it rarely matters, and never matters as much as audiofools
think it does.

> Modern DAC's deal with any residual jitter quite effectively in most
> cases anyway.

So for nearly all audio things costing more than $50, its not an issue.


--
Pat Farrell
http://www.pfarrell.com/

Nonreality
2009-09-19, 03:17
I run my whole library, both flac and mp3,from an external drive and never see a problem with sound quality. I think pfarrell is correct but then again my hearing might not be capable of detecting problems.

funkstar
2009-09-19, 04:50
I run my whole library, both flac and mp3,from an external drive and never see a problem with sound quality. I think pfarrell is correct but then again my hearing might not be capable of detecting problems.
This is correct. The only place jitter can occur is between the player and the stereo, nothing before, (from the hard drive, to the PC, to the network to the player) can introduce or have an effect on jitter .

egd
2009-09-19, 06:57
The only place jitter can occur is between the player and the stereoaah, but for the vacuum between an audiofool's ears :D

dsdreamer
2009-09-19, 13:09
This is correct. The only place jitter can occur is between the player and the stereo, nothing before, (from the hard drive, to the PC, to the network to the player) can introduce or have an effect on jitter .

It is true that nothing can add jitter before the player, but once a signal is in the analog domain, jitter (i.e., rapid variation of group delay in the audio band) is hard to introduce in reasonbly designed, linear circuits operating with well-regulated power supply rails.

The place where jitter is most likely to be introduced is at the point where digital samples are converted to analog i.e., in the DAC. Any modulation of the effective DAC clock signal edge transitions will have the effect of causing phase/frequency modulation of the every sinewave that comprises the overall music signal.


V(k) = A0*sin(2*pi*f0*k*(T+dTJ(k)))

V(k) is the kth sample of a sinewave of nominal frequency f0 to be reproduced, from T-spaced PCM samples, where the time between samples (typically 1/44100 seconds) is modulated by some time-variant jitter represented by dTJ.

The equivalece to an FM waveform can be seen by rewriting the equation as follows:


V(k) = A0*sin(2*pi*k*T*(f0+f0*dTJ(k)/T)))


dTJ(k) can be deterministic in nature (e.g. a 60Hz sinewave due to imperfect power supply regulation), data related (different bit patterns cause the clock edge to be early or late to different degrees) or random (residual phase noise on the clock signal) or a combination of all three.

I am not going to get into the religious/tribal invective about how much jitter matters.