(originally posted here∞)

Originally posted by aeberbach:
Billybones, you might want to get a proper 75 ohm digital cable. It doesn't have to be an expensive one but S/PDIF over coax does have slightly different needs than a simple shielded conductor like a phono-phono. You can make one yourself fairly easily too.

Peole who really know this stuff have posted long explanations here in the past and they made sense. I wish search worked better.

It's pretty simple. When the wavelength of a signal gets close to the same order of magnitude as the length of the cable, some side effects apply that are usually modeled as a "transmission line" (Google that term). We can no longer assume that the voltage and current are the same across a piece of resistanceless wire -- they vary with time, and therefore position. Think of the function of displacement in a string you're shaking: when the string is short and the shaking is fast, you start to appreciate that the mechanical wave has finite speed, i.e. you can notice the intermediate steps between moving one end of the string and seeing the other end move.

So anyway, the most notable thing for digital circuits is that there is only one frequency for a given type of string depending on its "flexibility" -- for electronics substitute the capacitance+inductance generated by the materials & geometry, where the current and voltage will be in phase (and the impedance is not imaginary, but no complex math allowed ;) ). This is called the characteristic impedance. When you hook up different kinds of strings (including their connectors) you create reflections and standing waves and other bad juju.

Imagine you want to send a very simple message to a person on the other end of your hybrid line: "I'm shaking my hand at a constant 5 cycles/sec." They have to guess this without watching you. Will it be easier with 3 pieces of fishing line tied together, or with a fishing line tied to a braided rope tied to a steel cable? In the latter case, you'll get a lot of your energy absorbed and reflected at the junction points, and the output waveform probably won't be recognizable.


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