On Fri, 26 Nov 2004 22:05:13 +0000 (UTC), "Reg Edwards"
wrote:

Bob, your program can probably calculate the input impedance, Rin + j*Xin,
of a line having Zo = Ro + j*Xo, with given attenuation Alpha dB, and given
phase-shift Beta radians, with a terminating impedance Rt + j*Xt.

Which is a commonly needed quantity on the way to calculating the ultimate,
all-important, single number, transmission efficiency.

But can you do it with nothing at hand except a Smith Chart?

And the answer is...







an unequivocal NO.

However, just to satisfy my curiousity, exactly which of your
beautiful, zipped up Pascal programs will do that for me?
Bob, W9DMK, Dahlgren, VA
http://www.qsl.net/w9dmk

Bob, your program can probably calculate the input impedance, Rin +
j*Xin,
of a line having Zo = Ro + j*Xo, with given attenuation Alpha dB, and
given
phase-shift Beta radians, with a terminating impedance Rt + j*Xt.

Which is a commonly needed quantity on the way to calculating the
ultimate,
all-important, single number, transmission efficiency.

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

Bob asked -
However, just to satisfy my curiousity, exactly which of your
beautiful, zipped up Pascal programs will do that for me?

===============================================
Bob,

I thought nobody would ever ask. But they've been available from my website
for years.

Look at Programs -

RJELINE2
RJELINE3
RJELINE4
COAXPAIR
COAXRATE

and give yourself a few practical examples.

There is a one-line description after each program's name on the download
page in my website.

The above programs are dedicated to transmission lines. Input data includes
one or two physical dimensions which avoids restriction to particular
type-of-line numbers. Users are given a free hand to design cables to their
own specifications. Nevertheless, they are practical in nature and simple to
use. There are other programs which incorporate the same calculations but
which are not explicitly apparent to the user.

They use exact classical transmission line formulae and so are as accurate
as the input data over the stated frequency ranges. Usually from power
frequencies up to UHF. They take skin effect and the increase in inductance
at low frequencies and other subtle factors such as conductor proximity
effect in twin-lines in their stride.

They are good enough for the highest precision engineering applications. I
have not disclosed the source code to prevent it falling into the hands of
argumentative vandals, so-called guru's, and technically ignorant old-wives
who would ruin the programs' reputation, not forgetting mine, for
RELIABILITY. Reliability is Quality versus Time.

For references I quote my only tutors - Ohm, Ampere and Volta.

When considering transmission lines you can check your's and other programs
against mine for accuracy with confidence. You will discover the effects of
both your known and other, unsuspected approximations.

Readers should bear in mind I'm not getting paid for this.

Bob, I'm on MontGras, Chilean, Reserve Merlot, tonight. Nuff said.
----
.................................................. ..........
Regards from Reg, G4FGQ
For Free Radio Design Software go to
http://www.btinternet.com/~g4fgq.regp
.................................................. ..........

On Sun, 28 Nov 2004 02:03:57 +0000 (UTC), "Reg Edwards"
wrote:

Bob, I'm on MontGras, Chilean, Reserve Merlot, tonight. Nuff said.

There's nothing that can top vintage Reg.
Tnx, I'll visit the site and pick up my free samples.

Tschuss!
Bob, W9DMK, Dahlgren, VA
http://www.qsl.net/w9dmk

Who the heck are Ramo and Whinnery. Never heard of them! Presumably,
because you refer to them, they are or were people who make or made a living
out of re-iterating old wive's tales in book-form.

It was obvious I introduced G = C * R / L simply to show that a line's Zo
can be purely resistive even when it is NOT lossless. It can have any loss
you like.

Apparently you have not yet grasped the idea.

And, despite what R and W or YOU may have to say on the subject, it is an
exact expression at all frequencies from DC to almost infinity.

My only references are Ohm, Ampere and Volta who I'm sure you have heard of.

But no hard feelings. ;o)

Tonight I'm on Chilean, dry, 2004, MontGras, Reserve Chardonnay. I didn't
choose it myself. My loving daughter does my shopping. But it's quite a
pleasant, refreshing plonk.
----
Regards, Reg.

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

"Cecil Moore" wrote
Reg Edwards wrote:
The condition for which Zo of a transmission line is always purely
resistive
(Zo = Ro) is extremely simple. It is -

G = C * R / L

Wonder why Ramo and Whinnery say that's an approximation for low-loss
lines? If the R+jwL angle is equal to the G+jwC angle, doesn't that
make Z0 purely resistive?
--
73, Cecil

"Reg Edwards" wrote in message
...

Who the heck are Ramo and Whinnery. Never heard of them! Presumably,
because you refer to them, they are or were people who make or made a
living
out of re-iterating old wive's tales in book-form.

It was obvious I introduced G = C * R / L simply to show that a line's Zo
can be purely resistive even when it is NOT lossless. It can have any
loss
you like.

Apparently you have not yet grasped the idea.

And, despite what R and W or YOU may have to say on the subject, it is an
exact expression at all frequencies from DC to almost infinity.

My only references are Ohm, Ampere and Volta who I'm sure you have heard
of.

But no hard feelings. ;o)

Tonight I'm on Chilean, dry, 2004, MontGras, Reserve Chardonnay. I didn't
choose it myself. My loving daughter does my shopping. But it's quite a
pleasant, refreshing plonk.
----
Regards, Reg.

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

"Cecil Moore" wrote
Reg Edwards wrote:
The condition for which Zo of a transmission line is always purely
resistive
(Zo = Ro) is extremely simple. It is -

G = C * R / L

Wonder why Ramo and Whinnery say that's an approximation for low-loss
lines? If the R+jwL angle is equal to the G+jwC angle, doesn't that
make Z0 purely resistive?
--
73, Cecil



Fields & Waves in Communication Electronics, by S. Ramo, J.R. Whinnery, and
T. Van Duzer, Wiley, 3rd edition, 1994

107 proof Baker's for me

73, H.

Reg Edwards wrote:

Who the heck are Ramo and Whinnery. Never heard of them! Presumably,
because you refer to them, they are or were people who make or made a living
out of re-iterating old wive's tales in book-form.

It was obvious I introduced G = C * R / L simply to show that a line's Zo
can be purely resistive even when it is NOT lossless. It can have any loss
you like.

Apparently you have not yet grasped the idea.

Ramo and Whinnery are the authors of my 50's college textbook on fields
and waves. Of course it could be a misprint, but they say your above
formula is an approximation that is good for low-loss lines.

Apparently, something additional happens for high-loss lines. Chipman
seems to agree with Ramo and Whinnery when he introduces some additional
interference terms (discussed some time ago on this newsgroup). At the
time, I didn't realize the additional terms were interference terms but
the impedance of the load apparently somehow interacts with the
characteristic impedance of the high-loss transmission line to upset
the ideal relationships in your equation above.
--
73, Cecil http://www.qsl.net/w5dxp

Cecil Moore wrote:
Reg Edwards wrote:

It was obvious I introduced G = C * R / L simply to show that a line's Zo
can be purely resistive even when it is NOT lossless. It can have any
loss
you like.


Ramo and Whinnery are the authors of my 50's college textbook on fields
and waves. Of course it could be a misprint, but they say your above
formula is an approximation that is good for low-loss lines.

Certainly good at HF and UHF when the skin depth is likely to be a small
fraction of the conductor radius.

Apparently, something additional happens for high-loss lines.

Not so much high loss, as low frequency. Both L and R are frequency
dependent assuming normal (non superconducting) metallic conductors. G
and C may have a frequency dependency depending on the dielectric
characteristics.

Once the frequency is high enough so that the current can be considered
to flow only on the skin of the conductor, the effective AC resistance
is proportional to the square root of the frequency and the inductance
is constant. At frequencies below the above defined 'critical
frequency', the internal inductance must be considered as well as the
complicated frequency dependence of resistance.

Chipman
seems to agree with Ramo and Whinnery when he introduces some additional
interference terms (discussed some time ago on this newsgroup).

Yep.

At the
time, I didn't realize the additional terms were interference terms but
the impedance of the load apparently somehow interacts with the
characteristic impedance of the high-loss transmission line to upset
the ideal relationships in your equation above.

The relationship is correct for all frequencies and standing wave ratios
as long as the correct frequency dependent values of transmission line
parameters are used. The wave equation still describes the relationship
between current and voltage. The additional 'interference' terms appear
when calculating the energy distribution and loss characteristics.

bart
wb6hqk

Bart Rowlett wrote:
The relationship is correct for all frequencies and standing wave ratios
as long as the correct frequency dependent values of transmission line
parameters are used.

Here's the equation that Ramo and Whinnery says is an approximation
for low-loss lines.

Z0 = SQRT(L/C)[1 + j(G/2wC - R/2wL)]

If G = C * R / L then Z0 = SQRT(L/C)

So why did Ramo and Whinnery say it is an approximation for low-loss
lines?
--
73, Cecil http://www.qsl.net/w5dxp

On Sun, 28 Nov 2004 23:01:04 -0600, Cecil Moore
wrote:
Here's the equation that Ramo and Whinnery says is an approximation
for low-loss lines.
So you keep saying. Is it that difficult to find their exact solution
for any lines?

Richard Clark wrote:

Cecil Moore wrote:
Here's the equation that Ramo and Whinnery says is an approximation
for low-loss lines.

So you keep saying.

This is the first time I have posted the equation.

Is it that difficult to find their exact solution
for any lines?

Maybe the math is easier for the approximation?
--
73, Cecil http://www.qsl.net/w5dxp