Yesterday, while repairing my antenna, something came to my mind I had never
focused on before.

Let us consider a bipole, that is a "black box" having TWO terminals and
including plain passive elements only (like capacitors, inductors, ... , no
diodes or other special devices), arranged the way you prefer, it does not
matter.

In my mind it was quite clear that, when fitting such a bipole into a circuit,
the sense makes no difference, i.e. one can reverse the two terminals with no
consequence. As a matter of fact, the bipole has an equivalent impedance that
remains the same independently of the way it is put in the circuit.

Yesterday a case occurred to me in which this is not actually true.

Instead of directly telling which it is, just for fun I wonder whether anyone
can figure out a case in which a bipole may not be reversed without
consequences. Not difficult, but it anyway requires some thinking.

Although probably unnecessary, let me recall that a filter is typically a
THREE-terminal device (IN, OUT, GROUND), not a TWO-terminal one.

73

Tony I0JX
Rome, Italy

On 31 ene, 21:20, "Antonio Vernucci" wrote:
Yesterday, while repairing my antenna, something came to my mind I had never
focused on before.

Let us consider a bipole, that is a "black box" having TWO terminals and
including plain passive elements only (like capacitors, inductors, ... , no
diodes or other special devices), arranged the way you prefer, it does not
matter.

In my mind it was quite clear that, when fitting such a bipole into a circuit,
the sense makes no difference, i.e. one can reverse the two terminals with no
consequence. As a matter of fact, the bipole has an equivalent impedance that
remains the same independently of the way it is put in the circuit.

Yesterday a case occurred to me in which this is not actually true.

Instead of directly telling which it is, just for fun I wonder whether anyone
can figure out a case in which a bipole may not be reversed without
consequences. Not difficult, but it anyway requires some thinking.

Although probably unnecessary, let me recall that a filter is typically a
THREE-terminal device (IN, OUT, GROUND), not a TWO-terminal one.

73

Tony I0JX
Rome, Italy

Hello Tony,

The answer may be in your own text (the ground issue). There may be a
third path via ground (capacitive coupling).

You can add a very good common mode choke at the input terminal of you
bipole. In that case the path via ground is blocked.

Try a very simple bipole: metallic case connected to terminal 1,
terminal 2 connected to nothing. When the center conductor of your
source is connected to terminal 1, you have the ground path.

The effect of reverse connection will be less when you use a very
small battery powered source that is completely floating. Measuring
data you can get out of it via optical link...

Best regards,

Wim
PA3DJS
www.tetech.nl
without abc, PM will reach me.

The answer may be in your own text (the ground issue). There may be a
third path via ground (capacitive coupling).

Hi Wim,

I acknowledege that your example leads to an asymmetric bipole, that one cannot
reverse it without consequences.

However, in the case which occurred to me, there is no need to assume the
existence of a third path via ground for justifying the asymmetry. Actually the
asimmetry remains even if the bipole would be in free space!

Forgive me if I do not immediately reveal my case, but I would like to see if
there are some more guesses before closing the issue.

73

Tony I0JX

On Jan 31, 9:38*pm, "Antonio Vernucci" wrote:
The answer may be in your own text (the ground issue). There may be a
third path via ground (capacitive coupling).

Hi Wim,

I acknowledege that your example leads to an asymmetric bipole, that one cannot
reverse it without consequences.

However, in the case which occurred to me, there is no need to assume the
existence of a third path via ground for justifying the asymmetry. Actually the
asimmetry remains even if the bipole would be in free space!

Forgive me if I do not immediately reveal my case, but I would like to see if
there are some more guesses before closing the issue.

73

Tony I0JX

as long as it is ONLY 2 ports, has ONLY passive linear components, AND
is small enough to be considered a lumped element, then you can always
reverse the terminals and not know the difference. The most
frequently violated condition is the last one, put in a piece of coax
with that is long enough to be measured at the highest frequency you
will use and all bets are off.

K1TTT wrote:


as long as it is ONLY 2 ports, has ONLY passive linear components, AND
is small enough to be considered a lumped element, then you can always
reverse the terminals and not know the difference. The most
frequently violated condition is the last one, put in a piece of coax
with that is long enough to be measured at the highest frequency you
will use and all bets are off.

An electrolytic capacitor acts differently if reverse biased. The results
are usually bad. It does qualify as two leaded, passive and linear when used
properly.

On Mon, 31 Jan 2011 17:23:33 -0600, joe wrote:

An electrolytic capacitor acts differently if reverse biased. The results
are usually bad. It does qualify as two leaded, passive and linear when used
properly.

Hi Joe,

The same can be said of a diode. Hence the electrolytic capacitor is
non-linear, which violates the premise.

73's
Richard Clark, KB7QHC

On Jan 31, 11:23*pm, joe wrote:
K1TTT wrote:

as long as it is ONLY 2 ports, has ONLY passive linear components, AND
is small enough to be considered a lumped element, then you can always
reverse the terminals and not know the difference. *The most
frequently violated condition is the last one, put in a piece of coax
with that is long enough to be measured at the highest frequency you
will use and all bets are off.

An electrolytic capacitor acts differently if reverse biased. The results
are usually bad. It does qualify as two leaded, passive and linear when used
properly.

an electrolytic that acts that way is not a linear component.

As long as it is ONLY 2 ports, has ONLY passive linear components, AND
is small enough to be considered a lumped element, then you can always
reverse the terminals and not know the difference. The most
frequently violated condition is the last one, put in a piece of coax
with that is long enough to be measured at the highest frequency you
will use and all bets are off.

Could you please explain me the meaning of your sentence:

"put in a piece of coax with that is long enough to be measured at the highest
frequency you will use and all bets are off"

as I have difficulties to interpret it due to my non-mother tongue english.

73

Tony I0JX
Rome, Italy

On Feb 1, 6:24*pm, "Antonio Vernucci" wrote:
As long as it is ONLY 2 ports, has ONLY passive linear components, AND
is small enough to be considered a lumped element, then you can always
reverse the terminals and not know the difference. *The most
frequently violated condition is the last one, put in a piece of coax
with that is long enough to be measured at the highest frequency you
will use and all bets are off.

Could you please explain me the meaning of your sentence:

"put in a piece of coax with that is long enough to be measured at the highest
frequency you will use and all bets are off"

as I have difficulties to interpret it due to my non-mother tongue english.

73

Tony I0JX
Rome, Italy

the general case is, if any part of the circuit is more than a small
fraction of a wavelength in size you may be able to detect the
difference between the ports. one common way to do that is to try to
measure a circuit that has a long piece of coax in it, the results may
be very different when you reverse the terminals.

"Antonio Vernucci" ha scritto nel messaggio
...
Forgive me if I do not immediately reveal my case, but I would like to see
if there are some more guesses before closing the issue.

From my very poor knowledge of trasmission lines...
... reversing inner and outer conductor of a coaxial cable ??


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