Differential mode transmission lines with no groundplane

[Roy McCammon]

Muriel wrote:

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.

I think others have set up the physical situation.
Just to repeat, there are three "conductors". Two
are the intentional transmission line. The third
is called ground or return or neutral.

V1in -------------------------------- V1out
V2in -------------------------------- V2out
ground -------------------------------- ground

"Ground" might be an actual intentional conductor
that runs along with the other two and it might be
a large conducting plane of copper and it might
just be the Earth itself and in some cases we
take ground to be a conducting sphere with infinite
radius of which we are inside, or as I like to say it
the "Rest of Creation" (the ROC). For convenience,
lets say that our intentional pair has an input side
(the left side) and an output side (the right side).
In general, V1out depends on both V1in and V2in.
Likewise, V2out depends on both V1in and V2in.

So here is the vector space stuff. Take the pair
of voltages at the input to be a vector in 2D
space (V1in,V2in) and the pair of voltages at the
output to also be a vector in 2D space (V1out,V2out).
This may be obvious and it may not be obvious, but
the output vector is just a linear transformation
of the input vector. This is profound. If true,
then all the truths that you know about vector
spaces and linear algebra must apply to these voltages.

So, if we have a linear vector space transformation,
then it probably has Eigen vectors. In particular,
if the the two lines of the pair are identical and
symmetrically placed with respect to the ground
conductor, then the Eigen vectors are the common mode
(V1in + V2in) and the differential mode (V1in - V2in).

Now, if the two lines are slightly different from
each other, the Eigen vectors (or Eigen modes) are
not exactly the common mode and the differential mode,
but often those are close enough that we just go ahead
and pretend that they are.

So if we put in a V1in, V2in pair of voltages, we
can decompose those two voltages into the common
mode and the differential mode, multiply by the
respective Eigen values (common mode gain and
differential mode gain) and we know what comes out.

--
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