What makes an interconnect sound like it does?

"What makes speakers sound different?"

Glad you asked. Not because I think this is an appropriate place to discuss this question (that's for another thread), but becuase your question facilitates a good analogy.

I can make some generalizations about the sound of a speaker that is based upon an electro-static design vs. a speaker based upon dynamic electro-mechanical design. A sealed "acousic suspension" design vs. a ported "bass reflex" design. 1st-order crossovers vs. 4th-order corssovers. Etc...

Now, are these generalizations accurate all of the time? Nope - that's why they are called "generalizations." However, it does give you things to consider. I think it would be interesting to discuss generalizations about cable design.

Loontoon,

As counter-intuituive as is may sound, I don't believe you *really* want to impedance match your components. Modern solid-state devices transfer voltage between products, not power. Optimum power transfer requires impedance matching. Optimum voltage transfer does not.

Today’s products have high input impedances and low output impedances. These are compatible with each other. Low impedance output stages drive high impedance input stages. This way, there is no loading, or signal loss, between stages. No longer concerned about the transfer of power, today’s low output/high input impedances allow the almost lossless transfer of signal voltages.

For example, the positive and negative outputs of the driving unit have an output impedance labeled ROUT. Each input has an impedance labeled RIN. Typically these are around 100 Ω for ROUT and 20k Ω for RIN. Georg Ohm taught us that 100 Ω driving 20k Ω (looking only at one side for simplicity) creates a voltage divider, but a very small one (-0.04 dB). This illustrates the above point about achieving almost perfect voltage transfer, if impedance matching is not done.

If it is done, you lose half your signal. Here’s how: impedance matching these units involves adding 100 Ω resistors (equal to ROUT) to each input (paralleling RIN). The new input impedance now becomes essentially the same as the output impedance (100 Ω in parallel with 20 kΩ equals 99.5 Ω), therefore matching. Applying Ohm’s law to this new circuit tells us that 100 Ω driving 100 Ω creates a voltage divider of ½. That is, ½ of our signal drops across ROUT and ½ drops across RIN, for a voltage loss of 6 dB. We lose half our signal in heat across ROUT. Not a terribly desirable thing to do.

Of course, I could be wrong. What the hell do I know?