twisted pair question with transmission

[coppercone2]

SO kind of related to PCB layout maybe, where its said that higher frequencies tend to follow under the conductor.

Which lead me to think of this idea...

If you have a twisted pair, lets say a 100 ohm impedance. Of signal + gnd.

Then you have a separate ground wire, some distance away from this twisted pair, say 1 inch.

If you generalize it to twin lead (I think the math is similar), you get for 16 awg wire, two wires that are basically right up to each other being 100 ohms.

Now you have the parallel ground wire 1 inch away, meaning the impedance of this twin lead (red spiral wire segment and the free negative wire) is ~450 ohms.

is it correct to think if you give this wire a signal, its a voltage divider? so the majority of the current will flow through the twisted pair black wire and a small portion on the strait wire?

[T3sl4co1l]

N conductors in constant cross-section gives a symmetrical NxN matrix(?) of transmission lines, yes.

For the N=3 case, and for asymmetrical distances (close pair plus one distant), the coefficients end up such that the pair-ness is easily recognizable, and we might express that close-together pair as an equivalent CM+DM representation (assuming for a moment, we have absolute voltages per conductor), where the CM is basically saying the two lines act together as one, which then becomes DM with respect to the remaining (more distant) conductor.  We of course have two degrees of freedom (two differential modes) for three conductors.  Note that we can convert any three-conductor scenario in this way, it doesn't matter what its relative impedances are; we aren't losing any degrees of freedom by thinking about two different kinds of differences, as long as we have a linearly independent basis using them.  And we can have different impedances and velocities for each of these, since in general we could wrap some conductors with dielectric and others not (different, within reason, of course).

Because the close-in pair CM acts as a thicker conductor against the remaining one, Z will be a bit lower than the twin-lead case, and also E-field balance at a distance won't be perfect; but yeah, it's actually three wires in a cable, of course it's asymmetrical, and presumably the signals would be balanced in amplitude to reflect that (or if nothing else, by running the cable through a choke core to improve balance).

Tim

[coppercone2]

how well does the bootleg impedance calculation stand up though? very error?

i am gonna make this with zip ties and measure it on a segment I think

how many feet long should it be to see something interesting? I have bootleg PVC wire I HATE