"The light is refracted within the cable, spdif uses a multi-mode cable so the light is refrated randomly and there are many different paths the light can take in the pipe , but if the cable moves the light path changes, sometimes slighty altering the length of the paths. Hence if there is vibration on the optical fibre (subwoofer induced...etc) the time base of the digital signal at the reciver is not steady because the transmision distance is constantly changing.

This error is Jitter and is usually very slight. Jitter can be eliminated by most DAC circuits by using a buffer and a crystal, to correct the time base. The buffer stores the bits and a crystal clock is used to read the data off the buffer at a regular rate."

"sometimes slighty altering the length of the paths"

source of figures used in my extrapolations = this web page

I must be missing something here. Speed of light = 186,000+ miles per second, IIRC. Upper limit of human hearing = ~ 20,000 cps. Average length of optical cable in home system =~ 10 ft. 5,280 feet = 1 mile.

5280 ft x 186000 miles = 982,080,000 feet.

A one cycle per second wave (analog) would travel that many feet in a second.

Divide that by 20,000 cycles and you get 49104 feet as the length of a 20,000 cycle per second wave as light (the shortest audible analog wave).

There are 1200cent in one octave. Average users can tolerate deviances of up to 50cent from a base note, while professional musicians pick up differences between 10 - 20 cent. A well-trained musician with good discriminating abilities can pick up differences of as little as 5cent.

5 / 1200 = 1/240th of an octave as the smallest interval discernable within an octave, where the highest "normal" octave (a"") has "a" at 3,950 cps and notes double in frequency 7,900 cps are generally only heard as harmonics.

Assuming the best ear, listening for variations in the highest octave, can discern 1/240th of the octave as a discrete variation. That is a variation of no less than 16 cps.

So Joe Pro, musician, listening to a note at 7,900 cps detects a 16 cps variation. How much of a different distance must the light travel to reach that variation?
124313.92 feet per wave of 7,900 cps
124062.65 feet per wave of 7,916 cps

difference = 251.27 feet

To add this distance, the light in the 10' cable would have to bounce within the cable more than 12 times.

So, for appreciable "jitter", on part of the wave the light would have perfect throughput, and then quickly degrade to a reflection requiring the light to bounce in the cable more than 12 times before completing its journey.

I'm not saying that this isn't possible, or that I might be off the mark in my thinking, but the math doesn't seem to support the concept of "jitter" for analog sound in optical cables. If you can keep from wincing at the pun, could someone enlighten me? Is there encoding in optical signals that is more sensitive?

Coaxial is cheaper anyway. TOSlink provides absolutely no advantage whatsoever. Therefore coaxial=much better.

For uncompressed audio on an SPDIF cable you transmit 32 bits per sample (even if only 16 bits are used for audio), 64 bits per stereo pair, and at 44.1khz x 64 = 2822400 bits per second. So the signal travelling down the fibre is opproximatly 3MHz and could be suseptible to Jitter. As far as I know DTS and AC-3 signals have similar bandwidths to PCM. In fact as data rates increase for 24-bit/96khz, multichannel, Meridian loss-less packing, SACD, etc.. jitter could be more of a problem on optical connectors.

My math ablity is not good enough to work out the quantity of virabtion on the fibre needed to cause errors on a 3Mhz signal, but it is possible.

Whether or not jitter is audible is another argument altogether. But you do come across people who if there is a measurable error in audio they clame to be able to hear it. This could open up the 44.1khz sampling rate vs 96Khz sampling rate aurgument.

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