"Roy Lewallen" bravely wrote to "All" (07 Apr 05 15:08:40)
--- on the heady topic of " VF, low-loss line, high-impedence line -
relationship"

RL From: Roy Lewallen
RL Xref: aeinews rec.radio.amateur.antenna:28088

RL Asimov wrote:

Since a portion of the EM field in open wire line is free to travel
outside the conductor into the environment then we may safely assume
there is an exchange between the environment and the conductor.

RL If the conductors are perfectly conducting, no part of the field at
RL all exists within the conductor. With good conductors like copper and
RL at HF and above, there's very little penetration of the conductor by
RL the fields, either electric or magnetic.

Is there any electron current in the conductor or not?


RL First of all, a mismatch doesn't cause loss.

An impedance mismatch in any medium causes a scattering of the energy.
Of course, it isn't a net loss as far as the universe is concerned but
some of the energy doesn't arrive where it was intended.


RL Secondly, as I explained in my last posting, the characteristic
RL impedance of a transmission line isn't the same thing as the
RL characteristic impedance of free space.

May I suggest you make up your mind whether the electric energy is
travelling in a conductive medium or not?



RL It has to do with the reflective
coefficient where the energy is returned.

RL Well, no. There isn't a bundle of energy trying to escape the line and
RL bouncing off the air, or bouncing off the air as it travels along the
RL line, or bouncing off the conductors into the air. So reflection
RL coefficient isn't applicable here.

What makes you so sure?


RL I'm afraid that the conclusions you've reached about loss and
RL characteristic impedance are based on a poor understanding of
RL fundamental transmission line operation. The result is some
RL conclusions that are, and are well known to be, untrue.

I think you are only concerned with modeling of transmission lines as
lumped constants but models can only go so far in explaining how
something works. Models are like analogies and we all know no analogy
is perfect even this one.

A*s*i*m*o*v

.... Anyone not wearing 2,000,000 sunblock is gonna have a REAL_ BAD_ DAY_7

On Fri, 08 Apr 2005 06:42:05 -0700, Wes Stewart
wrote:

On Fri, 08 Apr 2005 06:05:49 -0700, Wes Stewart
wrote:


I'll try to scan it to pdf and post is somewhere.

http://www.qsl.net/n7ws/Sterba_Openwire.pdf

My first scan didn't have the appendix.

It's there now.

On Fri, 08 Apr 2005 10:19:31 -0500, Cecil Moore
wrote:

Wes Stewart wrote:
http://www.qsl.net/n7ws/Sterba_Openwire.pdf

There seems to be a dotted line for feedline
radiation going to zero as feedline length
goes to zero. :-)

So it would seem. QRX, Reg will explain. :-)

Reg Edwards wrote:
In summary, the system as a whole BEHAVES as if there is NO radiation
from the line itself ...

How about BPL? (The Devil made me do it.)
--
73, Cecil http://www.qsl.net/w5dxp




----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==----
http://www.newsfeeds.com The #1 Newsgroup Service in the World! 100,000 Newsgroups
---= East/West-Coast Server Farms - Total Privacy via Encryption =---

Ian, G3SEK wrote:
"Now tell us about radiation from the line."

Reg`s problem has a line as long as it is wide. That would make a very
small square loop if I read right. The same current flows in all 4
sides.

According to Arnold King:
Beta is a factor = 2 pi/Lambda
Say Lambda=1, then Beta=2pi

Radiation resistance= 20 times Beta to the 4th times A squared

Say the sides of the tiny square loop are 0.01Lambda, then A, the loop`s
enclosed area is
0.0001

Radiation resistance =
20 (1555) (10 tp -8th)

=31100 (10 to -8th)
=0.00031 ohm

Radiated power = Isquared (radiation resistance)
Radiated power = negligible

Best regards, Richard Harrison, KB5WZI

On Fri, 08 Apr 2005 08:54:11 -0700, Wes Stewart
wrote:

On Fri, 08 Apr 2005 06:42:05 -0700, Wes Stewart
wrote:

On Fri, 08 Apr 2005 06:05:49 -0700, Wes Stewart
wrote:


I'll try to scan it to pdf and post is somewhere.

http://www.qsl.net/n7ws/Sterba_Openwire.pdf

My first scan didn't have the appendix.

It's there now.

Hi Wes,

Could I trouble you for a copy of section IV that continues on beyond
page 1180? I am interested in the discussion relating to Figure 12 so
you can restrict your efforts to that if this section runs on too
long.

Thanks,
Richard, KB7QHC

Reg Edwards wrote:

You've told us about radiation from the connections to the generator
and
the termination.

Now tell us about radiation from the line.

=================================

Ian, you are falling into the same sort of trap as old wives who
imagine most radiation comes from the middle 1/3rd of a dipole because
that's where most of the current is.

It is self-misleading to consider the various parts of a radiating
system to be separate components which are capable of radiating
independently of each other. They can't.

Actually they can, because that isn't the same as saying...

A system's behaviour must
be treated as a whole.

That is true, of course. Every component of an antenna (or in this case,
a parallel-wire transmission line) interacts with every other component.
The totality of those interactions is what determines how the RF voltage
and current will distribute themselves along the wires.

But once you know the magnitude and phase of the current in each small
segment of the antenna (which need not depend on theory or modeling - in
principle you could go around and measure it) then you have taken
complete account of the interactions. The radiated field from the whole
antenna is then the sum of the fields from the individual components
radiating independently.

However, we weren't originally talking about that...


We have already discussed that the power radiated from a generator +
twin-line + load is a constant and is independent of line length.

No, you have only asserted that.

Total power radiated is equal to that radiated from a wire having a
length equal to line spacing with a radiation resistance appropriate
to that length. The location of the radiator, insofar as the
far-field is concerned, can be considered to be at the load. The
current which flows in the radiator is the same as that flowing in a
matched load. And the load current is independent of line length.

Only if there are no radiative losses from the line itself - and you
have only asserted that, not proved it.

Mathematically, the only way for the total power radiated to remain
constant and independent of line length is for zero radiation from the
line.

Well obviously - but that is a circular argument, based entirely on your
assertion that the power delivered to the load is independent of the
line length.


In summary, the system as a whole BEHAVES as if there is NO radiation
from the line itself - only from fictitious very short monopoles (or
dipoles?) at its ends.

Sorry, but the "behaves as if" argument doesn't wash, because those
short monopoles are real. Since the line spacing is non-zero, those
short transverse sections must always exist, both in practice and in
your circuit model. Each section carries RF current, so it radiates - no
question about that, but it is entirely an end effect. It has nothing
whatever to do with radiation from the main line.

Looking edge-on at the line, we have two conductors carrying equal and
opposite currents, but one is slightly farther away than the other so
their transverse radiated fields do not quite cancel out.

The only question is mathematical: how does the small loss of energy
through radiation translate into a dB/m or dB/wavelength loss along the
transmission line?



--
73 from Ian G3SEK 'In Practice' columnist for RadCom (RSGB)
http://www.ifwtech.co.uk/g3sek

Ian, G3SEK wrote:
"---how does the small loss of energy through radiation translate into
dBm or dB/wavelength loss along a transmission line?"

It is the reverse of a beverage antenna, which is a sort of single-wire
transmission line above the earth in its simple configuration. The
Beverage is a horizontal wire sensitive to vertically polarized waves.
It is working with the wave throughout its travel along its length.

The Beverage is vertically polarized because that is the direction of
the electric field between its conductors, the wire and the earth
beneath the wire.

The direction of the electric field in a parallel-wire transmission line
is from wire to wire. The effective radiator length of this polarization
is the line spacing. This is short compared to the length of the
transmission line in nearly all cases. The radiation is not emerging
from the end of the transmission line. It radiates slightly all along
the line as the wave navigates the line, much as the Beverage gathers
energy slightly as the wave sweeps along its length.

Best regards, Richard Harrison, KB5WZI

"Ian White G3SEK" wrote in message
...
Reg Edwards wrote:

You've told us about radiation from the connections to the
generator
and
the termination.

Now tell us about radiation from the line.

=================================

Ian, you are falling into the same sort of trap as old wives who
imagine most radiation comes from the middle 1/3rd of a dipole
because
that's where most of the current is.

It is self-misleading to consider the various parts of a radiating
system to be separate components which are capable of radiating
independently of each other. They can't.

Actually they can, because that isn't the same as saying...

A system's behaviour must
be treated as a whole.

That is true, of course. Every component of an antenna (or in this
case,
a parallel-wire transmission line) interacts with every other
component.
The totality of those interactions is what determines how the RF
voltage
and current will distribute themselves along the wires.

But once you know the magnitude and phase of the current in each
small
segment of the antenna (which need not depend on theory or
modeling - in
principle you could go around and measure it) then you have taken
complete account of the interactions. The radiated field from the
whole
antenna is then the sum of the fields from the individual components
radiating independently.

However, we weren't originally talking about that...


We have already discussed that the power radiated from a generator
+
twin-line + load is a constant and is independent of line length.

No, you have only asserted that.

Total power radiated is equal to that radiated from a wire having a
length equal to line spacing with a radiation resistance
appropriate
to that length. The location of the radiator, insofar as the
far-field is concerned, can be considered to be at the load. The
current which flows in the radiator is the same as that flowing in
a
matched load. And the load current is independent of line length.

Only if there are no radiative losses from the line itself - and you
have only asserted that, not proved it.

Mathematically, the only way for the total power radiated to remain
constant and independent of line length is for zero radiation from
the
line.

Well obviously - but that is a circular argument, based entirely on
your
assertion that the power delivered to the load is independent of the
line length.


In summary, the system as a whole BEHAVES as if there is NO
radiation
from the line itself - only from fictitious very short monopoles
(or
dipoles?) at its ends.

Sorry, but the "behaves as if" argument doesn't wash, because those
short monopoles are real. Since the line spacing is non-zero, those
short transverse sections must always exist, both in practice and in
your circuit model. Each section carries RF current, so it
radiates - no
question about that, but it is entirely an end effect. It has
nothing
whatever to do with radiation from the main line.

Looking edge-on at the line, we have two conductors carrying equal
and
opposite currents, but one is slightly farther away than the other
so
their transverse radiated fields do not quite cancel out.

The only question is mathematical: how does the small loss of energy
through radiation translate into a dB/m or dB/wavelength loss along
the
transmission line?



--
73 from Ian G3SEK 'In Practice' columnist for RadCom (RSGB)
http://www.ifwtech.co.uk/g3sek

Asimov wrote:
"Roy Lewallen" bravely wrote to "All" (07 Apr 05 15:08:40)
--- on the heady topic of " VF, low-loss line, high-impedence line -
relationship"

RL From: Roy Lewallen
RL Xref: aeinews rec.radio.amateur.antenna:28088

RL Asimov wrote:

Since a portion of the EM field in open wire line is free to travel
outside the conductor into the environment then we may safely assume
there is an exchange between the environment and the conductor.

RL If the conductors are perfectly conducting, no part of the field at
RL all exists within the conductor. With good conductors like copper and
RL at HF and above, there's very little penetration of the conductor by
RL the fields, either electric or magnetic.

Is there any electron current in the conductor or not?

In a perfect conductor, no. In a real but good conductor like copper,
it's confined to a very thin layer at the surface. Look up "skin effect"
in any electromagnetics or basic text on RF, or google it.



RL First of all, a mismatch doesn't cause loss.

An impedance mismatch in any medium causes a scattering of the energy.
Of course, it isn't a net loss as far as the universe is concerned but
some of the energy doesn't arrive where it was intended.

If I connect a 50 ohm source to a one wavelength, 300 ohm transmission
line and connect the other end of that line to a 50 ohm resistor,
there's a 6:1 mismatch at both ends of the line. The power supplied by
the source and the power delivered to the load are exactly the same as
if I had used a 50 ohm line instead. This is, of course, overlooking
resitive loss in the line. If you consider the resistive loss, it can be
greater in one line than the other (the 300 ohm line might be less
lossy), depending on the physical construction of the line.

No loss is caused by the mismatch. No "scattering of energy occurs". All
of the energy from the source arrives at the load, where it was intended.



RL Secondly, as I explained in my last posting, the characteristic
RL impedance of a transmission line isn't the same thing as the
RL characteristic impedance of free space.

May I suggest you make up your mind whether the electric energy is
travelling in a conductive medium or not?

I'm sorry, I don't understand what you're asking. No RF energy exists in
or travels in a perfect conductor.


RL It has to do with the reflective
coefficient where the energy is returned.

RL Well, no. There isn't a bundle of energy trying to escape the line and
RL bouncing off the air, or bouncing off the air as it travels along the
RL line, or bouncing off the conductors into the air. So reflection
RL coefficient isn't applicable here.

What makes you so sure?

A basic understanding of electromagnetics derived from an electrical
engineering education, extensive additional reading and study, and about
30 years of engineering design experience including design of microwave
and very high speed time domain circuitry.


RL I'm afraid that the conclusions you've reached about loss and
RL characteristic impedance are based on a poor understanding of
RL fundamental transmission line operation. The result is some
RL conclusions that are, and are well known to be, untrue.

I think you are only concerned with modeling of transmission lines as
lumped constants but models can only go so far in explaining how
something works. Models are like analogies and we all know no analogy
is perfect even this one.

I have no idea what makes you think that my modeling or understanding of
transmission lines is limited to lumped constant models -- it's
certainly not true. Indeed, no analogy is perfect, but some are
certainly better than others, and some are demonstrably false. I'm
afraid that some of the ones you've put forth are in the latter category.

Roy Lewallen, W7EL