A small riddle, just for fun
 

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

A small riddle, just for fun
 

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.

A small riddle, just for fun
 

On Feb 1, 6:18*pm, K7ITM wrote:
On Jan 31, 12:20*pm, "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

I have a circuit I've been working on lately which has a simple series
LC in it, no other connection to the node between the inductor and
capacitor. *It turns out that the order of the inductor and capacitor
makes a big difference in the circuit performance. *I anticipated that
it would, and put them in the intuitively obvious order, only to find
out that it was the wrong order! *A proper model cleared things up
quite nicely. *However, in no way would I call that particular part of
the circuit a "two terminal" network. *The effect is the same as Wim
mentioned.

Cheers,
Tom

then there is either something wrong with your components or you have
too much coupling the the surrounding environment... so you really had
a 3 terminal network using the stray capacitance or inductance.

A small riddle, just for fun
 

On Feb 1, 3:24*pm, K1TTT wrote:
On Feb 1, 6:18*pm, K7ITM wrote:



On Jan 31, 12:20*pm, "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

I have a circuit I've been working on lately which has a simple series
LC in it, no other connection to the node between the inductor and
capacitor. *It turns out that the order of the inductor and capacitor
makes a big difference in the circuit performance. *I anticipated that
it would, and put them in the intuitively obvious order, only to find
out that it was the wrong order! *A proper model cleared things up
quite nicely. *However, in no way would I call that particular part of
the circuit a "two terminal" network. *The effect is the same as Wim
mentioned.

Cheers,
Tom

then there is either something wrong with your components or you have
too much coupling the the surrounding environment...

If by "too much" you mean "more than I wanted," you're absolutely
right. But of course with real components, there is no such thing as
zero coupling to the surrounding environment. As with all real
circuits, there are tradeoffs: high coil Q results in a large coil,
which in turn results in more capacitance to the shield than a smaller
low-Q coil would have. The shield is required for other reasons...

... so you really had
a 3 terminal network using the stray capacitance or inductance.

Exactly. Just what I said: the effect is the same as Wim mentioned.

Cheers,
Tom

A small riddle, just for fun
 

On Tue, 1 Feb 2011 15:24:41 -0800 (PST), K1TTT wrote:

then there is either something wrong with your components or you have
too much coupling the the surrounding environment... so you really had
a 3 terminal network using the stray capacitance or inductance.

This can be neutralized employing a "Wagner Earth." For inductors,
there is the Modified Owen method.

Basically, problems of this sort have been handled with driven shields
(a variant of the Wagner Earth/Ground). This exactly century-old
topic ( "Wagner Earth") is rarely discussed outside of precision
measurement or very high voltage handling circuits, however.

73's
Richard Clark, KB7QHC

A small riddle, just for fun
 

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.


You have pinpointed the correct issue: lumped elements.

The story began when, a few days ago, I was going to replace a trap of my HF
yagi. Not to make mistakes, I consulted the antenna assembly manual where I
found a big banner: do not invert traps otherwise the antenna will not work.

So, I thought, this is a case in which a bipole cannot be inverted.

This is clearly due to the fact that the external body of the trap (an aluminuim
can about 2 feet long), which contains two coils resonated at different
frequencies by means of built-in capacitors, is effectively part of the antenna
radiating element. So, the trap is a bipole not only comprising lumped elements,
and that is the reason why it cannot be inverted.

73

Tony I0JX
Rome, Italy

A small riddle, just for fun
 

Wim,

the story began when, a few days ago, I was going to replace a trap of my HF
yagi. Not to make mistakes, I consulted the antenna assembly manual where I
found a big banner: do not invert traps otherwise the antenna will not work.

So, I thought, this is a case in which a bipole cannot be inverted.

This is clearly due to the fact that the external body of the trap (an aluminuim
can about 2 feet long), which contains two coils resonated at different
frequencies by means of built-in capacitors, is effectively part of the antenna
radiating element. So, the trap is a bipole not only comprising lumped elements,
and that is the reason, I believe, why it cannot be inverted.

73

Tony I0JX
Rome, Italy

A small riddle, just for fun
 

Tom,

the story began when, a few days ago, I was going to replace a trap of my HF
yagi. Not to make mistakes, I consulted the antenna assembly manual where I
found a big banner: do not invert traps otherwise the antenna will not work.

So, I thought, this is a case in which a bipole cannot be inverted.

This is clearly due to the fact that the external body of the trap (an aluminuim
can about 2 feet long), which contains two coils resonated at different
frequencies by means of built-in capacitors, is effectively part of the antenna
radiating element. So, the trap is a bipole not only comprising lumped elements,
and that is the reason why it cannot be inverted.

So, as K1TTT has pointed out, a bipole can be inverted without consequences only
if it has only 2 ports, has only passive linear components, and is small enough
to be considered a lumped element.

73

Tony I0JX
Rome, Italy

A small riddle, just for fun
 

On 2 feb, 09:37, "Antonio Vernucci" wrote:
Wim,

the story began when, a few days ago, I was going to replace a trap of my HF
yagi. Not to make mistakes, I consulted the antenna assembly manual where I
found a big banner: do not invert traps otherwise the antenna will not work.

So, I thought, this is a case in which a bipole cannot be inverted.

This is clearly due to the fact that the external body of the trap (an aluminuim
can about 2 feet long), which contains two coils resonated at different
frequencies by means of built-in capacitors, is effectively part of the antenna
radiating element. So, the trap is a bipole not only comprising lumped elements,
and that is the reason, I believe, why it cannot be inverted.

73

Tony I0JX
Rome, Italy

Hello Tony,

So it isn't a bipole.

The tube containing the trap is a (short) common mode transmission
line that has effect on its environment (like a metal case with its
capacitive effect on ground). The trap inside the metal tube is in
series with the end of the short transmission line.

The ground in this case is the boundary where the near field is no
longer dominant w.r.t. to the radiating field. Think of a circular
ground at 0.16lambda from the structure. So one end of your bipole
has more ground to this virtual ground and is therefore a 3-pole
device.

For the limitation on lumped circuit approach, it is not important
whether or not it contributes to the overall radiation of a
structure. When you make a floating ground out of 4 quarter-wave
radials (monopole?), this structure has minor influence on the far
field pattern of the quarter wave (or better half wave) radiator above
it.

Best regards,


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

A small riddle, just for fun
 

"Wimpie" napisal w wiadomosci
...
On 2 feb, 09:37, "Antonio Vernucci" wrote:

For the limitation on lumped circuit approach, it is not important
whether or not it contributes to the overall radiation of a
structure. When you make a floating ground out of 4 quarter-wave
radials (monopole?), this structure has minor influence on the far
field pattern of the quarter wave (or better half wave) radiator above
it.


In http://www.padrak.com/ine/FARADAY1.html Faraday wrote:

"The view which I am so bold to put forth considers, therefore, radiation as
a kind of species of vibration in the lines of force which are known to
connect particles and also masses of matter together. It endeavors to
dismiss the aether, but not the vibration. The kind of vibration which, I
believe, can alone account for the wonderful, varied, and beautiful
phaenomena of polarization, is not the same as that which occurs on the
surface of disturbed water, or the waves of sound in gases or liquids, for
the vibrations in these cases are direct, or to and from the centre of
action, whereas the former are lateral. It seems to me, that the resultant
of two or more lines of force is in an apt condition for that action which
may be considered as equivalent to a lateral vibration; whereas a uniform
medium, like the aether, does not appear apt, or more apt than air or
water."

Is your 4 quarter-wave radials (monopole?), the same like Faraday's lines?
Two or more monopoles produces "polarised waves".

Two or more monopoles "is in an apt condition for that action which may be
considered as equivalent to a lateral vibration;".

Some textbooks say that Faraday suggested the transverse (angular)
wibrations.
Did he?
S*