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Differential mode transmission lines with no groundplane

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R

Roy McCammon

Jan 1, 1970
0
Muriel said:
Roy,

Thanks for your explanation of "the basic that is not so basic as we
use to think".. For the first time, I'm beggining to understand the
meaning of "ground plane", specially with the concept of R.O.C. (very
good indeed!)..

Now, the mathematical part: So, I have two vectors of input and two
vectors of output. With them, I can extract the "transfer function" of
the pair of conductors? Is that right?

minor didactic point: you have two voltages at the input
which are components of a 2D vector, but other wise that is
right.
And, depending on this transfer function, it generates more or less
common or differential mode, right?

"generates" can mean several things, so I prefer to say
that the linear system (the transmission system) yields
Eigen modes that in many useful cases are exactly or
approximately the common mode and the differential mode.
Please, if you can, send me a numerical example of our hypothetical
situation, to make things clearer. If it's difficult, you can e-mail
it right to me...

OK, this is a sort of analogous dc circuits. I don't
recall if you said you are familiar with dc circuits, but
here goes.

First a nice system

V1in ---- 45 ohms -----+----- 45 ohms ---- V1out
|
10 ohms
|
V2in ---- 45 ohms -----+----- 45 ohms ---- V2out


ground ----------------------------------- ground.

so, if you do the equations you get

(V1out - V2out) = 0.1(V1in - V2in) and

(V1out + V2 out) = 1.0(V1out + V1out)

so in this system, the Eigen vectors are (1,-) and (1,1)
and the associated Eigen values are 0.1 and 1.0
or as an electrical engineer would say, the differential
mode gain is 0.1 and the common mode gain is unity.

If you put in V1in = -V2in = 1, the common mode
of the input vector is zero and the differential
mode is 2. So we expect the output common mode
to also be 0 and the output differential mode will
be +0.2


Here is a system with a slight impairment


V1in ---- 45 ohms -----+----- 45 ohms ---- V1out
|
10 ohms
|
V2in ---- 45 ohms -----+-+--- 45 ohms ---- V2out
|
1000 ohms
|
ground ------------------+---------------- ground.

In this system, a pure differential mode input will
produce and output that is mostly differential mode,
but there will be a small common mode component.
Like wise, an input that is purely common mode
V1in = V2in will produce and output that is mostly
common mode, but it will have a little bit of
differential mode. The system does have eigen
vectors and they are almost but not quite the common
mode and the differential mode. At this point, it
becomes a matter of judgment as to whether you
can ignore the slight deviations or not.




--
Achilles: I wish my wish would not be granted.
< an undescribable event occurs >
Achilles: What happened? Where's my Genie?
Tortoise: Our context got restored incorrectly.
Achilles: What does that cryptic comment mean?
Tortoise: The system crashed.


To email me send to :

rb <my last name> AT ieee DOT org
 
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