Differential mode transmission lines with no groundplane

[The Technical Manager]

Is a two conductor transmission line such as the 300 ohm parallel flat
type cable used for VHF antennas being used to convey a differential
mode signal that is at a large (approximated as infinite) distance from
a groundplane capable of conveying a common mode signal as well ? A
common mode signal can only be conveyed if one or both of the current
carrying conductors is in proximity to a grounded structure. A two
conductor transmission line conveying a differential mode signal has a
virtual groundplane between its two conductors but does a common mode
signal view this virtual groundplane in the same way as it views a real
groundplane and travels along the transmission line or is it suppressed
from the lack of a real groundplane ?

 

[Roy McCammon]

The Technical Manager wrote:

Is a two conductor transmission line such as the 300 ohm parallel flat
type cable used for VHF antennas being used to convey a differential
mode signal that is at a large (approximated as infinite) distance from
a groundplane capable of conveying a common mode signal as well ? A
common mode signal can only be conveyed if one or both of the current
carrying conductors is in proximity to a grounded structure. A two
conductor transmission line conveying a differential mode signal has a
virtual groundplane between its two conductors but does a common mode
signal view this virtual groundplane in the same way as it views a real
groundplane and travels along the transmission line or is it suppressed
from the lack of a real groundplane ?

you may want to post this one on one of the ham radio ng's

I believe that yes it can carry a common mode signal
and that that is usually undesirable and that it
is usually suppressed with a balun.

--
local optimization seldom leads to global optimization

my e-mail address is: rb <my last name> AT ieee DOT org

 

[Nosko S.]

My additions to Kevin's comments shown as ***

Kevin G. Rhoads <Kevin.Rhoads@Dartmouth.edu> wrote in message
news:3F0B01B7.8CF340F8@Dartmouth.edu...

Is a two conductor transmission line such as the 300 ohm parallel flat
type cable used for VHF antennas being used to convey a differential
mode signal that is at a large (approximated as infinite) distance from
a groundplane capable of conveying a common mode signal as well ?
Yes.

*** Yup.

A
common mode signal can only be conveyed if one or both of the current
carrying conductors is in proximity to a grounded structure.
No it doesn't need a grounded structure, just a continuous return
path for the return current. A grounded structure can serve
that purpose, but so can other things. Continuity of the return
current pathway is the key. (For AC signals, continuity can
be achieved through capacitive coupling for part or all of the
return leg.)

*** The point of contention here is the word "proximity".

A) _IF_ you are talking about traveling waves along this "common mode"
structure (And I assume you are) , then it will need a "near" return or
"ground structure" to support the traveling wave as you surmise. A single
wire over a ground plane will have a charactersitic impedance. The two
conductors (with the same signal) can be thought of as a wider single
conductor over a ground plane. This is a common transmission line structure
and it is referred to as a "micro strip" and has been modeled and a
characteristic impedance can be predicted (a flat strip, not two
conductors).
So, you can send another RF signal along as a "common mode" signal if
you provide a return conductor a controlled distance away -- such as a
shield around the twin-lead. This would make it roughly equivalent to coax
or perhaps a "stripline" also called "strip transmission line". The latter
being a "center" conductor (usually flat) between two "close" ground planes.
You'd have to match impedances as usual.


B) If, on the ther hand, you are talking about a low frequency/aka
non-traveling wave, then any conducting return path will suffice. Like the
old telegraph lines which had a single wite over ling distances between
stations and used the earth as the return. So you can send power to a
preamp that way or audio to a whatever also.

I'm not exactaly sure where Kevin is going with the 'capacitive coupling'
idea.

A two
conductor transmission line conveying a differential mode signal has a
virtual groundplane between its two conductors
Yes.

*** Yup!

but does a common mode
signal view this virtual groundplane in the same way as it views a real
groundplane
No. Differential mode virtual groundplanes will not serve as current
returns for common mode signals. A "virtual" anything is an imaginary
construct used to simplify the physics or the math (or both) for a
specific
set of conditions (boundary conditions, signal forms and symmetries &c).
The symmetry plane of the differential signals on the balanced wiring
has no such mathematically necessary relation to common mode signals,
and cannot be used as a simplifying model element when dealing
with common mode or any mode other than balanced differential.

*** The differential mode "virtual ground plane" you refer to appears only
for / to the differential signal. It is there because the differential
signal is always "equal and opposite" and therefore there is a place
equidistant from both conductors where there is no electrical field
potential due to this signal.
NOTE the last four words of the last sentence. It is non-existent for the
common mode signal. Maybe this is what Kevin said.

(I'd have to think about the magnetic field, but gut feel says it is also
opposite & therefore has a virtual "magnetic plane")

Please don't start off on a tangent about G-line.


--
[please clip quoted postings down to the relevant parts, thank you]


Steve Nosko...Pardon my paranoia, but my email has no U's in it:
steuve.Nouskouwicz@moutoroula.com

 

[Alan Boswell]

Tech
The virtual ground exists only for the differential mode, so the common
mode needs a ground return path. Using multiple modes on transmission
lines has been common practice on telephone circuits since about 1930,
and the normal way to do it is to have two separate transmission lines,
one circuit per line, and then to have another circuit which works
between the common modes on each line. With good hybrid transformers,
there is almost no crosstalk. This idea can be built up, so that with
eight wires you can get seven circuits along them, and so on.
Alan



The Technical Manager wrote:

Is a two conductor transmission line such as the 300 ohm parallel flat
type cable used for VHF antennas being used to convey a differential
mode signal that is at a large (approximated as infinite) distance from
a groundplane capable of conveying a common mode signal as well ? A
common mode signal can only be conveyed if one or both of the current
carrying conductors is in proximity to a grounded structure. A two
conductor transmission line conveying a differential mode signal has a
virtual groundplane between its two conductors but does a common mode
signal view this virtual groundplane in the same way as it views a real
groundplane and travels along the transmission line or is it suppressed
from the lack of a real groundplane ?

 

[Kevin G. Rhoads]

*** The differential mode "virtual ground plane" you refer to appears only
for / to the differential signal. It is there because the differential
signal is always "equal and opposite" and therefore there is a place
equidistant from both conductors where there is no electrical field
potential due to this signal.
NOTE the last four words of the last sentence. It is non-existent for the
common mode signal. Maybe this is what Kevin said.

That was what I was tyring to say, in different words. Maybe between
your explanation and mine the OP can get it. I think your version
is probably clearer on the EE stuff, I was trained as a theoretical
mathematician first and then went into EE and I was replying more
in math terms, which is probably less useful for most people.

Kevin

 

[John Larkin]

A
common mode signal can only be conveyed if one or both of the current
carrying conductors is in proximity to a grounded structure.
No it doesn't need a grounded structure, just a continuous return
path for the return current. A grounded structure can serve
that purpose, but so can other things. Continuity of the return
current pathway is the key. (For AC signals, continuity can
be achieved through capacitive coupling for part or all of the
return leg.)

*** The point of contention here is the word "proximity".

A) _IF_ you are talking about traveling waves along this "common mode"
structure (And I assume you are) , then it will need a "near" return or
"ground structure" to support the traveling wave as you surmise. A single
wire over a ground plane will have a charactersitic impedance. The two
conductors (with the same signal) can be thought of as a wider single
conductor over a ground plane. This is a common transmission line structure
and it is referred to as a "micro strip" and has been modeled and a
characteristic impedance can be predicted (a flat strip, not two
conductors).
So, you can send another RF signal along as a "common mode" signal if
you provide a return conductor a controlled distance away -- such as a
shield around the twin-lead. This would make it roughly equivalent to coax
or perhaps a "stripline" also called "strip transmission line". The latter
being a "center" conductor (usually flat) between two "close" ground planes.
You'd have to match impedances as usual.

The limiting case is when the return conductor is the universe. This
is called a "G-line". Imagine a coax that gradually widens, like a
funnel, until the center-shield separation is so large that it's
essentially infinite, and the signal travels along a single wire in
free space. A funnel on each end makes a smooth transition from coax
into/out of a single-wire G-line.

Here's Ralph's:

http://home.att.net/~ralph.hartwell/hamradio/tower.htm


John

 

[The Technical Manager]

"Nosko S." wrote:

My additions to Kevin's comments shown as ***

Kevin G. Rhoads <Kevin.Rhoads@Dartmouth.edu> wrote in message
news:3F0B01B7.8CF340F8@Dartmouth.edu...
Is a two conductor transmission line such as the 300 ohm parallel flat
type cable used for VHF antennas being used to convey a differential
mode signal that is at a large (approximated as infinite) distance from
a groundplane capable of conveying a common mode signal as well ?
Yes.

*** Yup.


A
common mode signal can only be conveyed if one or both of the current
carrying conductors is in proximity to a grounded structure.
No it doesn't need a grounded structure, just a continuous return
path for the return current. A grounded structure can serve
that purpose, but so can other things. Continuity of the return
current pathway is the key. (For AC signals, continuity can
be achieved through capacitive coupling for part or all of the
return leg.)

This is RF and signals see transmission line structure and not wires. Unless
the return path for the common mode signal actually is the virtual groundplane
then it would have to use a third conductor and a third conductor at any finite
distance from the two wire transmission line will create a transmission line
type structure between the third conductor and the two wires of the
transmission line.

*** The point of contention here is the word "proximity".

A) _IF_ you are talking about traveling waves along this "common mode"
structure (And I assume you are) , then it will need a "near" return or
"ground structure" to support the traveling wave as you surmise. A single
wire over a ground plane will have a charactersitic impedance. The two
conductors (with the same signal) can be thought of as a wider single
conductor over a ground plane. This is a common transmission line structure
and it is referred to as a "micro strip" and has been modeled and a
characteristic impedance can be predicted (a flat strip, not two
conductors).
So, you can send another RF signal along as a "common mode" signal if
you provide a return conductor a controlled distance away -- such as a
shield around the twin-lead. This would make it roughly equivalent to coax
or perhaps a "stripline" also called "strip transmission line". The latter
being a "center" conductor (usually flat) between two "close" ground planes.
You'd have to match impedances as usual.

B) If, on the ther hand, you are talking about a low frequency/aka
non-traveling wave, then any conducting return path will suffice. Like the
old telegraph lines which had a single wite over ling distances between
stations and used the earth as the return. So you can send power to a
preamp that way or audio to a whatever also.

No the transmission line is being used at RF and seen by both common and
differential mode signals as a transmission line and not a piece of wire.

I'm not exactaly sure where Kevin is going with the 'capacitive coupling'
idea.


A two
conductor transmission line conveying a differential mode signal has a
virtual groundplane between its two conductors
Yes.

*** Yup!

but does a common mode
signal view this virtual groundplane in the same way as it views a real
groundplane
No. Differential mode virtual groundplanes will not serve as current
returns for common mode signals. A "virtual" anything is an imaginary
construct used to simplify the physics or the math (or both) for a
specific
set of conditions (boundary conditions, signal forms and symmetries &c).
The symmetry plane of the differential signals on the balanced wiring
has no such mathematically necessary relation to common mode signals,
and cannot be used as a simplifying model element when dealing
with common mode or any mode other than balanced differential.

*** The differential mode "virtual ground plane" you refer to appears only
for / to the differential signal. It is there because the differential
signal is always "equal and opposite" and therefore there is a place
equidistant from both conductors where there is no electrical field
potential due to this signal.
NOTE the last four words of the last sentence. It is non-existent for the
common mode signal. Maybe this is what Kevin said.

Thats what I was thinking. In theory the virtual groundplane can be at a
completely different potential to that of a real groundplane.

(I'd have to think about the magnetic field, but gut feel says it is also
opposite & therefore has a virtual "magnetic plane")

Assume there are no magnetic materials in close proximity to the transmission
line so that the problem can be looked at in terms of its electric fields.

 

[Nosko S.]

The Technical Manager <techman@yahoo.co.uk> wrote in message
news:3F0DC488.72191869@yahoo.co.uk...

"Nosko S." wrote:


This is RF and signals see transmission line structure and not wires.

I'm not getting what your point is here. I think you are saying first,
that the common mode signal you want to send is RF. Therefore, we are indeed
discussing my first situation -- namely the possibility of symultaneous
differential mode and common mode transmission lines in a single structure.

RE: "not wires"
The transmision line we are discussing in this case *is* composed of
wires, so that *is* what there is to "see". Placement of "wires" or if you
prefer "conductors" in a regular configuration produces a structure capable
of supporting RF waves and exhibiting wave-like behavior and we call these
things "transmission lines".

Unless the return path for the common mode signal actually is the
virtual groundplane then it would have to use a third conductor and
a third conductor at any finite distance from the two wire transmission
line will create a transmission line type structure between the third
conductor and the two wires of the transmission line.

Here we start to get a bit sticky, as you will see, as I develop the
concept of the common mode transmission line, but you are saying what I
previously said.
However, the "virtual groundplane" only exists for the differental
signal. It is non existent for the common mode signal so the following
makes no sense:
"...the return path for the common mode signal actually
is the virtual groundplane "
Ignore the virtual groundplane in discussions of the common mode of
transmission.
The rest of this paragraph is what I had said "...a third conductor at
any finite ...will create a transmission line..."

No the transmission line is being used at RF and seen by both common and
differential mode signals as a transmission line and not a piece of wire.
I'll take this as further confirmation that the common mode signal is

indeed RF and you are trying to determine if a common mode transmissin line
is possible here. I say yes -- by adding another "side" to the structure
which will act as the "ground" part for the common mode transmission line.
It only needs to be another, excuse my choice wire, but it is easy to
visualize a third wire along with the twin-lead. This would make it
"Tri-Lead". In fact the third wire could be the "hot" side and the two
wires of the twin-lead could act as the "ground" side. In reality, this
would amount to another quasi-balanced transmission. Pictures might help,
but I'm not going to attempt it.
If on the other hand, we were to put a shield all around the twin-lead's
two conductors (Which, BTW does exist) The new un-balanced transmission
line would more resemble coax than a balanced line -- except that the common
mode "center conductor" would consist of two parallel conductors.

Does that help any???

--
[please clip quoted postings down to the relevant parts, thank you]

--
'guards
..Steve K
.. ga 9
.. rb D
.. ag C
.. e I

I'm not exactaly sure where Kevin is going with the 'capacitive
coupling'
idea.


A two
conductor transmission line conveying a differential mode signal has
a
virtual groundplane between its two conductors
Yes.

*** Yup!

but does a common mode
signal view this virtual groundplane in the same way as it views a
real
groundplane
No. Differential mode virtual groundplanes will not serve as current
returns for common mode signals. A "virtual" anything is an imaginary
construct used to simplify the physics or the math (or both) for a
specific
set of conditions (boundary conditions, signal forms and symmetries
&c).
The symmetry plane of the differential signals on the balanced wiring
has no such mathematically necessary relation to common mode signals,
and cannot be used as a simplifying model element when dealing
with common mode or any mode other than balanced differential.

*** The differential mode "virtual ground plane" you refer to appears
only
for / to the differential signal. It is there because the differential
signal is always "equal and opposite" and therefore there is a place
equidistant from both conductors where there is no electrical field
potential due to this signal.
NOTE the last four words of the last sentence. It is non-existent for
the
common mode signal. Maybe this is what Kevin said.


Thats what I was thinking. In theory the virtual groundplane can be at a
completely different potential to that of a real groundplane.


(I'd have to think about the magnetic field, but gut feel says it is
also
opposite & therefore has a virtual "magnetic plane")

Assume there are no magnetic materials in close proximity to the
transmission
line so that the problem can be looked at in terms of its electric fields.

 

[Muriel]

Hi,

I was reading this thread and I found it opportune to ask some "basic"
questions about some concepts, that I think that are not to "basic"...

Thinking in electrical circuit theory terms, what is a _ground
plane_?? Is it just model or has a physical effect indeed?

The concepts: differential and common mode. These signals are in all
"real" circuits, or are just model used to represent non-intuitive
concepts? How they are generated/created?

Just to begin the discussion...

All the very best!!

Muriel


The Technical Manager <techman@yahoo.co.uk> wrote in message news:<3F099637.78E179D6@yahoo.co.uk>...

Is a two conductor transmission line such as the 300 ohm parallel flat
type cable used for VHF antennas being used to convey a differential
mode signal that is at a large (approximated as infinite) distance from
a groundplane capable of conveying a common mode signal as well ? A
common mode signal can only be conveyed if one or both of the current
carrying conductors is in proximity to a grounded structure. A two
conductor transmission line conveying a differential mode signal has a
virtual groundplane between its two conductors but does a common mode
signal view this virtual groundplane in the same way as it views a real
groundplane and travels along the transmission line or is it suppressed
from the lack of a real groundplane ?

 

[Joseph.D.Warner]

Muriel wrote:

Hi,

I was reading this thread and I found it opportune to ask some "basic"
questions about some concepts, that I think that are not to "basic"...

Thinking in electrical circuit theory terms, what is a _ground
plane_?? Is it just model or has a physical effect indeed?

Ground planes are real. Best examples are in microwave circuits where
the ground plane is usually very close to the transmission line in a
stripline circuit. Look up references.

Usually at low frequencies the ground plane is considered to be at
infinity for most circuits, except for transmission lines and coaxially
to multi-axial cables.

 

[Fred Abse]

On Wed, 23 Jul 2003 20:50:05 +0100, Muriel wrote:

My
specific interest is in switched mode power supplies, its EMI
generations mechanism and the resulting interference.

There must be dozens of books on the subject. Try Googling for "EMC"

--
Then there's duct tape ...
(Garrison Keillor)
nofr@sbhevre.pbzchyvax.pb.hx

 

[Muriel]

Steve,

Thanks for your explanation... But I think it really made me figure
out the real question I want to know:

- I know the definition of ground plane you gave below. My question
is: When I don't have a "large conducting plane or flat surface", what
(or where) is my reference???

- I understood the explanation of differential and common mode signals
too... But, when you study electromagnetic compatibility (EMC), you
learn that both kinds of signals are present in your feeding cable
(that come from mains, let's us suppose...). Why the common mode
signal is there, if it should not be there (theoretically)??? What is
the "cause" (thinking in terms of cause and effect) of this kind of
sinal propagating in the lines?????

Now, I think we're getting closer to the answers...

Best Regards,

Muriel


"Nosko S." <sorry.nospam@here.com> wrote in message news:<bfpijd$4br$1@localhost.localdomain>...

Muriel <mblfloripa@yahoo.com.br> wrote in message
news:7717a6d1.0307221327.62ba8904@posting.google.com...
...Thinking in electrical circuit theory terms, what is a _ground
plane_?? Is it just model or has a physical effect indeed?

A ground plane is a (relatively) large conducting plane or flat surface.
It is real. Think of a plate of copper. If we assign this to our reference
potential, we can think of it as the ground plane. That is, it is at zero
potential and all of our circuit voltages are referenced to it.


The concepts: differential and common mode. These signals are in all
"real" circuits, or are just model used to represent non-intuitive
concepts? How they are generated/created?

These are also real. Given two conductors, if you apply a +V to one
and -V to the other this is a differential signal(assuming |-V| = |+V|).
They are used all over the place. Television 300 ohm "twin-lead"
transmission line. The power lines running all over the country - and into
the house (at least in the US).

Given two conductors, if you apply +2V to BOTH conductors (as if they
are one) and use such a ground plane as a reference, this is a common mode
signal.


--
'guards
.Steve K
. ga 9
. rb D
. ag C
. e I

 

[The Technical Manager]

Muriel wrote:

Steve,

Thanks for your explanation... But I think it really made me figure
out the real question I want to know:

- I know the definition of ground plane you gave below. My question
is: When I don't have a "large conducting plane or flat surface", what
(or where) is my reference???

- I understood the explanation of differential and common mode signals
too... But, when you study electromagnetic compatibility (EMC), you
learn that both kinds of signals are present in your feeding cable
(that come from mains, let's us suppose...). Why the common mode
signal is there, if it should not be there (theoretically)??? What is
the "cause" (thinking in terms of cause and effect) of this kind of
sinal propagating in the lines?????

Now, I think we're getting closer to the answers...

Even worse is that in a device which actually includes cables, connectors and transmission lines is
that some of the differential signal gets transformed into a common mode signal and some of the
common mode signal gets transferred into a differential mode signal. Differential mode circuits are
officially defined in terms of their 4 transfer gains (Add, Acc, Acd and Adc) as described by
Middlebrook in his book on differential amplifiers. Judging from the lack of literature which
mentions the common to differential and differential to common mode transfers then I suspect that
it is not a well known phenomena by much of the electronics community.

 

[Muriel]

Techman,

Thanks for your contribution... As I thought, the transformation of
common mode to differential mode is a mistery (at least to us!).

Can someone help us with these doubts??

Regards,

Muriel

The Technical Manager <techman@yahoo.co.uk> wrote in message news:<3F254F79.F61C2E48@yahoo.co.uk>...

Even worse is that in a device which actually includes cables, connectors and transmission lines is
that some of the differential signal gets transformed into a common mode signal and some of the
common mode signal gets transferred into a differential mode signal. Differential mode circuits are
officially defined in terms of their 4 transfer gains (Add, Acc, Acd and Adc) as described by
Middlebrook in his book on differential amplifiers. Judging from the lack of literature which
mentions the common to differential and differential to common mode transfers then I suspect that
it is not a well known phenomena by much of the electronics community.

 

[Roy McCammon]

Muriel wrote:

Techman,

Thanks for your contribution... As I thought, the transformation of
common mode to differential mode is a mistery (at least to us!).

here is an equivalent circuit



.---------+------ 100 ohms------100 ohms ------
| |
Vcommon | Vdiff
| |
| '------ 100 ohms---+--100 ohms ------
| |
| 99900 ohms
| |
'----------------------------+--- round

1 volt of common mode voltage at the left
becomes 1 mV of differential voltage plus .9995 volts
of common mode voltage at the left.

Are you familiar with linear algebra and eigen vectors?







--
local optimization seldom leads to global optimization

my e-mail address is: rb <my last name> AT ieee DOT org

 

[Muriel]

Roy,

Please, don't come with "ready answers"... This way, this thread looks
like a i]undergraduated class where the teacher wants to show its very
deep knowledge to his humble pupil... That a thing for past
centuries...

Until now, everybody answered sincerely, wanting to know a little bit
more (not everything) about this subject and its premisses.

I'm asking questions about physical explanations to this phenomena:
common mode currents and differential mode currents. Don't come with a
bad designed draft of a circuit trying to tell me that I missed some
classes of algebra... Man, this is ridiculous.

Sincerely,

Muriel


Roy McCammon <rbmccammon@mmm.com> wrote in message news:<3F268801.5070403@mmm.com>...

Muriel wrote:
Techman,

Thanks for your contribution... As I thought, the transformation of
common mode to differential mode is a mistery (at least to us!).

here is an equivalent circuit



.---------+------ 100 ohms------100 ohms ------
| |
Vcommon | Vdiff
| |
| '------ 100 ohms---+--100 ohms ------
| |
| 99900 ohms
| |
'----------------------------+--- round

1 volt of common mode voltage at the left
becomes 1 mV of differential voltage plus .9995 volts
of common mode voltage at the left.

Are you familiar with linear algebra and eigen vectors?

 

[The Technical Manager]

Muriel wrote:

Techman,

Thanks for your contribution... As I thought, the transformation of
common mode to differential mode is a mistery (at least to us!).

Can someone help us with these doubts??

Quite often they are caused by imbalances in the two "paths" that convey the common and differential
mode signals through the various circuits of the system. In other words the effects on a single ended
signal conveyed using one path is different to that when the single ended signal is conveyed using the
second path. Imbalances include things like transmission lines being of different characteristic
impedances for each path in the same section of the circuit, amplifiers having a different gain for a
signal applied to one input than the other, transmission lines where one either leaks of picks up
interference signals more than the other in the same section of circuit or baluns that have slight
amplitude or phase imbalances on the balanced side.

You can make a start by analysing a differential amplifier where the voltage gain on input 1 with input
2 grounded is slightly different to the voltage gain on input 2 with input 1 grounded. Now apply both
differential mode and common mode signals to this amplifier and find out exactly what the output is.

 

[Muriel]

Roy,

First, Sorry if I seemed aggresive... trying to communicate with
written words sometimes is complicated... I'm a just a student, always
trying to learn more, and when people come with answers like yours I
always have the feeling that this someone is getting out of the focus
of the question... But it's Ok! I'm learning...

Well, I actually did not understand your schematic circuit... Could
you give an example to make things clearer?? Well, I had my classes of
algebra yet, and eingenvectors are not monsters to me (at least, I
think!!

I'm waiting for your replies...

Regards,

Muriel


Roy McCammon <rbmccammon@ieee.org> wrote in message news:<3F27588D.8030300@ieee.org>...

Patience, eager student.

If my response sounded arrogant or condescending,
I apologize profusely, for no such intent existed.

Think of me as a dottering old fool who sometimes
asks painfully obvious questions.

Now, please tell this dottering old fool if
you understood the circuit and if you are
comfortable with Eigen vectors.


Muriel wrote:
Roy,

Please, don't come with "ready answers"... This way, this thread looks
like a i]undergraduated class where the teacher wants to show its very
deep knowledge to his humble pupil... That a thing for past
centuries...

Until now, everybody answered sincerely, wanting to know a little bit
more (not everything) about this subject and its premisses.

I'm asking questions about physical explanations to this phenomena:
common mode currents and differential mode currents. Don't come with a
bad designed draft of a circuit trying to tell me that I missed some
classes of algebra... Man, this is ridiculous.

Sincerely,

Muriel


Roy McCammon <rbmccammon@mmm.com> wrote in message news:<3F268801.5070403@mmm.com>...

Muriel wrote:

Techman,

Thanks for your contribution... As I thought, the transformation of
common mode to differential mode is a mistery (at least to us!).

here is an equivalent circuit



.---------+------ 100 ohms------100 ohms ------
| |
Vcommon | Vdiff
| |
| '------ 100 ohms---+--100 ohms ------
| |
| 99900 ohms
| |
'----------------------------+--- round

1 volt of common mode voltage at the left
becomes 1 mV of differential voltage plus .9995 volts
of common mode voltage at the left.

Are you familiar with linear algebra and eigen vectors?




--
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 :

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[Roy McCammon]

Muriel wrote:

Well, I actually did not understand your schematic circuit... Could
you give an example to make things clearer?? Well, I had my classes of
algebra yet, and eingenvectors are not monsters to me (at least, I
think!!

That gives us two ways to proceed. Lets do the vector
space stuff first. By the way, I am a terrible two
fingered typist, so if something doesn't make sense,
there is a good possibility that I miswrote. Sometimes
I come across like I have rabies or some dementia.

.---------+------ 100 ohms------100 ohms ------
| |

Vcommon | Vdiff

| |
| '------ 100 ohms---+--100 ohms ------
| |
| 99900 ohms
| |
'----------------------------+--- round

1 volt of common mode voltage at the left
becomes 1 mV of differential voltage plus .9995 volts
of common mode voltage at the left.

Well, let me get into the Eigen vector first.

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.

I'll stop there. Let me know if this make sense.







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

 

[Muriel]

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?

And, depending on this transfer function, it generates more or less
common or differential mode, right?

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...

Thanks in advance and Regards!

Muriel


Roy McCammon <rbmccammon@ieee.org> wrote in message news:<3F289838.9090000@ieee.org>...

Muriel wrote:

Well, I actually did not understand your schematic circuit... Could
you give an example to make things clearer?? Well, I had my classes of
algebra yet, and eingenvectors are not monsters to me (at least, I
think!!

That gives us two ways to proceed. Lets do the vector
space stuff first. By the way, I am a terrible two
fingered typist, so if something doesn't make sense,
there is a good possibility that I miswrote. Sometimes
I come across like I have rabies or some dementia.


.---------+------ 100 ohms------100 ohms ------
| |

Vcommon | Vdiff

| |
| '------ 100 ohms---+--100 ohms ------
| |
| 99900 ohms
| |
'----------------------------+--- round

1 volt of common mode voltage at the left
becomes 1 mV of differential voltage plus .9995 volts
of common mode voltage at the left.

Well, let me get into the Eigen vector first.

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.

I'll stop there. Let me know if this make sense.







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