Quote:

Originally Posted by

**Bill Fitzmaurice**
The only correlation between cable impedance and its effect on the response of the speaker is that impedance is the sum of resistance, capacitive reactance and inductive reactance. You seem to be heading generally, in the correct direction, but I would suggest that you're askew. DC resistance can in no way, be directly summed to AC capacitance or inductance scores (at least not for any useful purpose), and if they could, they would tell you nothing about the AC impedance qualities of a given length of cable. It you're suggesting that DC resistance, and DC capacitance, and DC inductance can some how be directly summed to provide a depiction of AC impedance qualities, then I fear that you're even more askew. I'm not quite sure what it is your claiming here, but you do seem to me on the right path, but far from it's end.

Since each of the individual components of impedance affects the result in a different manner one can't consider the impedance alone. For sinusoidal linearity and frequency analysis (Non-FFT), impedance is the sole determinate and this test metric is just about the oldest I'm aware of. If your now covertly speaking to phase errors, you'd be correct.

You could have a thousand cables all with the same impedance, each with different resistance, capacitance and inductance, so each would have a different transfer function. You seem to be keeping DC resistance in your calculations (I might add that you have also failed to cite a single frequency or, speak to their importance.), when they don't have a direct relevance to AC waveforms. Also, if all of the cables had the same impedance scores across the entire audio band, that would suggest the same sinusoidal plotting of the frequency response for each cable. They may in fact have very different phase and group delay qualities, but they would appear on a bench to have identical frequency responses, when impedance has been used to plot such. These timing errors effect the tonal qualities by reducing the actual power (P) score (true Wattage behind each frequency). When reactive loads are present like capacitance or inductance, they cause reactive power (Q), not true power (P) to become present at some points. This type of power actually counters the effects of the true power (P) applied to a circuit. The significance of this is when only reactive power (Q) is present, no work can be performed. The difference between the two (S) can be divided by Q to estimate a devices efficiency. The more true power present, the better. To further educate: when purely inductive or capacitive loads are present, they will effect a 90-degree phase shift. In pure unmitigated occurrences as these, an amplifier will not output Wattage, because in such a scenario, the circuit is only seeing reactive power (Q). Outside of a pure state, their properties can be used and intermingle with other filters, to produce may different types of hybrid filters, with various orders and types of filtration, which will permit work/ P, but they still produce phase shifts, which result in various slopes of power decay (read between the lines - I'm providing a little more insight into why inductance and capacitance, reduce over power and bring about frequency attenuation) . However, phase shifts/errors are still afoot due to the shift between the Voltage and the Current waveforms, caused by the presence of these reactants. So what this adds up to, is that cables can effect the Wattage output of all frequencies, or just various frequencies, or a mix of both, if they add significant inductance, capacitance or impedance into the final stage of electrical signal delivery. Frankly, they can directly mess with passive crossovers designs (I think some of you, might have already gathered why). This is why it's requisite that only those with the appropriate training and equipment draw **'professional'** conclusions about transfer functions.

Here's an example that might bring this closer to understand, for more readers:

Energy being emitted out of an output, is mixed with Q & P: Q is properly recorded as VAr and P as Watts; both have coulombs in them, but the Q Voltage wave is out of phase with the Q current wave, for a given frequency(s), creating reactive power, thus mitigating its ability to perform work. P. both waves are in sufficient alignment, to produce work.

So, if a newbie was to apply their limited knowledge of Watts Laws, they would likely seek to determine the useable Wattage by measuring the output Voltage, hopefully using at least an o-scope, making their measurement just before the onset of clipping. Take that Voltage, square it and divide it by the drivers nominal impedance rating, and believe that they now have a good idea of the Wattage output, at whatever the frequency that they injected. A real keener, would likely take several measurements, at different frequencies, and plot them, thinking even better things about the accuracy of his calculations. One may even use a current clamp, to gain a more accurate current score and simply multiply the measured Voltage by the measured current... However, most if not all will have failed to realize the phase errors and their significance. Q must be subtracted from P, to determine S, the total usable power before clipping. So how does one determine Q, then take it to RMS Power? Good question, one that I will answer in due coarse (by the way, this also takes into understanding the pros and cons to current leading voltage and vice versa, and provides insight as to how it's possible for seemly properly functioning amplifiers to sound different - it's all about the phase) .

So now, you should be realizing a host of new possibilities.

All this aside, this kind of stuff, is at the edge of true, electrical engineering, therefore, it does become taxing to learn, and does require a real commitment of ones time. So please, just trust me when I say, as I have said, many times now, cables under 3 meters are not likely to produce audible differences from each other, regardless of the fact that they maybe electrically different (it is possible but of low probability, but not zero). It should be a given by now that the AWG must be appropriate - 14 Gauge can handle the Wattage output of any AVR, @ 3 meters or less in length.

If you just want to learn this stuff for the sake of learning, that's a whole different ball of wax, and I say welcome aboard. But for most hobbyists, I think doing so would be a waste of time, IMO.

When calculating the transfer function of a cable the individual contributions of resistance, capacitance and inductance are separately calculated and summed to give the final result. It should now be obvious to you that this is, at the very minimum, is incomplete and therefore does not support of your position, as being factual.

These calculations are quite elementary. Perhaps what you have been referring to is, but accurate calculations are not!

Resistance, capacitance and inductance specs of many wire varieties are available from reputable manufacturers, like Belden, so in less time than it took me to write this reply I could have plugged values into a calc and determined the exact effect on the frequency response by a given cable of a given length on a given speaker. Exactness is impossible through any means of measurement. The means that you have stated, would only produce crude estimates, ones that I currently suspect would produce unusable results, for use in objectively comparing short runs of audio cable.