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67.164.165.165
In Reply to: RE: Teach me! posted by Don T on March 20, 2008 at 14:42:22
My read on his comment is that he measures at each end of the cable and considers the difference as the cable contribution.
Yes, understood. But the key here is that this measurement will reflect the (alleged) distortion of the cable alone if and only if the distortion of the amplifier is zero.
Let me try to clarify this.
1) Hooking up a distorting amplifier to a distortionless cable and distortionless load will give rise to a distorted current through the loop, due only to amplifier distortion.
2) If the impedance of the leg of the cable one is measuring across is called Z, the frequency component of the voltage across that leg will be I_loop*Z, where I_loop stands for the frequency component of the distorted current, and Z is the impedance of the cable at that frequency . If we want the fundamental and each harmonic of this voltage, we need to consider that Z varies with frequency and use the appropriate Z value for the frequency of the harmonic being considered.
3) The impedance Z may not be constant with frequency. It's likely to be inductive, which means it has a minimum at low frequency and increases as the frequency increases. One result of this is that the percent distortion of the voltage across this impedance could be higher than the percent distortion of the amplifier itself, even though the cable is (by assumption) distortionless.
Let's take a specific example. Let's say the amplifier has 0.1 percent distortion (1 part in 1000) and that the distortion consists entirely of third harmonic. Let's say the amp is putting out a current of 1 Ampere for the fundamental component. This means that the third harmonic is 1/1000 of that, or 1 mA. Let's say we're testing at 10 kHz, and the series impedance of the leg of the cable we're measuring is 0.05 Ohms at 10 kHz, and 0.1 Ohms at 30 kHz (the third harmonic). Thus, the 10 kHz component (the fundamental) of the voltage across the cable will be 1 Amp times 0.05 Ohms = 0.05 Volts. At 30 kHz, the voltage will be 1 mA times 0.1 Ohms or 0.1 mV. The percent distortion across the cable will be 0.1 mV / 0.05 Volts, which is 2 parts in 1000 - twice that of the amp! And yet, if the current were undistorted (which would require a distortionless amplifier), then the distortion of the voltage across the cable would be zero.
This is the trap that the original poster has fallen into. It's not immediately obvious at all, so it's understandable that it could happen.
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