<font color='#000000'>HI Bruno
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Guest : Hi John,
We are only looking at the voltage to be dropped over the shield of one cable, not of an entire loop and its constituent cables.
The loop formed by the amplifier power cord will broadcast a field...the loop formed by the shield, the source ground, and the amp ground will intercept the amp haversines and any audio signal injected back into the line, the coupling coeff. based on the non integral number of twists in the power cord feed to the amp. The signal wire will do the same..diff is, the signal wire will produce the loop voltage across the amp input, whereas the ground will divide..
That is what I was referring to, as you mentioned power conditioning at the end of that paragraph.
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Quote </td></tr><tr><td id="QUOTE">The cable can be seen as a transformer, where the primary is the shield and the secondary is the signal wire.</td></tr></table>
No, it cannot. There is no internal magnetic field within a cylindrical conductor..so the braid to internal conductor cannot act as a transformer.
The only time that can occur, is if the center conductor and shield are not concentric, like at a tight bend or squashed cable.
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Quote </td></tr><tr><td id="QUOTE">If you inject current through the shield of a coax cable, it will develop a voltage across the resistive portion and one across the inductive portion. Like with a transformer, the inductive portion couples, the resistive portion doesn't.</td></tr></table>
Check Orstead's law...the integral within the shield shows no field is present..wish I could post pics here, I have some really neat analysis runs.
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Quote </td></tr><tr><td id="QUOTE">You can do a simple test. Short the input side of the cable. Connect a current source between the ends of the shield. Measure (differentially!) the output signal of the cable. Also measure the voltage from end to end on the shield. You will find that the voltage on the output side will correspond to the voltage dropped across the resistance of the shield only (ie is constant with respect to frequency) while the end-to-end voltage of the shield increases with frequency (is inductive).
I've done this test to verify theory, and found practice was fully in compliance.</td></tr></table>
I love your test...very cool...wish I could draw pictures here..
When you measure the output end of the cable, you are actually using the signal wire to get at the OTHER end of the coaxial braid without intercepting any of the field being generated outside of the braid..You have, in effect, created a coaxial current viewing resistor, with the shield as the resistive element. I have designed all my cvr's using that technique, as well as used that in my work environment.
When you measure across the shield from the outside, you create a big intercept loop..
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Quote </td></tr><tr><td id="QUOTE">I presume that once the semantics are cleared out, you will agree with me. If not, I have managed to surprise you with new information!</td></tr></table>
It would have been better if I could have posted pics, but nobody has provided me the info to do so here..
Bruno...If you're game, we can do some very interesting work..I have some rather incredible knowledge resources here to tap, you have some really good equipment.
I am allowed to co-author, I just can't do work for financial gain without gov't approval..I make the assumption that you're published work is that of you're employer?
Cheers,
John</font>
