Reg Edwards wrote:
Or even more simple, for Zo to be purely resistive, G = C*R/L

In "Transmission Lines" by Chipman, he gives an example where
R = 0.1 ohm/m and G = 0.9 micromhos/m. For Z0 to be a purely
resistive 50 ohms, G would have to be 40 micromhos/m making
the transmission line considerably more lossy just to achieve
a purely resistive Z0. Real world transmission lines rarely
have a purely resistive characteristic impedance.

The formula for the attenuation factor is R/2*Z0 + G*Z0/2
That's 0.001 + 0.0000225, so you can see that G has negligible
effect on losses, i.e. virtually all losses in the above
example are series I^2*R losses.

The attenuation factor is 0.0010225 for both the voltage and
current so it's obvious that the current attenuation is caused
by the series I^2*R losses, the same thing that causes the
voltage attenuation.
--
73, Cecil, W5DXP

David Ryeburn wrote:

Cecil Moore wrote:
I am assuming lossy lines, where R G, as is typical of transmission
lines used at HF frequencies. So exactly where does my argument fall flat?

If R is measured in ohms and G is measured in siemans (or mhos) how can
you say one of them is much larger than the other?

Sorry, should have said, "... where R/Z0 G*Z0, as is typical of
transmission lines used at HF frequencies."
--
73, Cecil, W5DXP

If R is measured in ohms and G is measured in siemans (or mhos) how can
you say one of them is much larger than the other?

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

Its easy for Cecil. He doesn't have the slightest trouble.

On Fri, 19 Nov 2004 23:48:22 -0600, Cecil Moore
wrote:
I am assuming lossy lines, where R G, as is typical of transmission
how can you say one of them is much larger than the other?
Sorry, should have said, "... where R/Z0 G*Z0
just a matter of Z0² ... no big deal
perhaps it shoulda been R· Z0² G ;-)

Reg Edwards wrote:
If R is measured in ohms and G is measured in siemans (or mhos) how can
you say one of them is much larger than the other?

Its easy for Cecil. He doesn't have the slightest trouble.

Just forgot to render them both dimensionless with the Z0
term. I suspect you knew what I meant anyway. :-)
--
73, Cecil, W5DXP

Gene Fuller wrote:
As for the E- and H-fields, this just gets more amusing by the minute.
The second paragraph is even more curious. Do you have a reference
for this migration of energy from the H-field to supply the suffering
E-field?

Here's the reference: From "Fields and Waves" by Ramo and Whinnery.
Take a close look at the exponential transmission line equations
for flat lines (no reflections):

V = Vmax(e^-az)(e^wt-bz) (1)

I = Vmax(e^-az)(e^wt-bz)/Z0 (2)

'a' (alpha) is the attenuation factor. The two equations are identical
except for the Z0 term. If you divide equation (1) by equation (2),
you get Z0. In a flat transmission line (no reflections) the current
is ALWAYS equal to the voltage divided by the characteristic impedance
of the transmission line. The voltage and current are attenuated by
EXACTLY the same factor. If the voltage drops because of I^2*R losses,
the current must decrease by exactly the same percentage. (I have avoided
calling it a current drop so it wouldn't upset you.)

Since the attenuation factor is R/2*Z0 + G*Z0/2 and since, for most
transmission lines used on HF, R/2*Z0 G*Z0/2, the current attenuation
is caused by the series I^2*R drop in the voltage and the V/I=Z0 ratio
that must be maintained - pretty simple logic.

I must have missed that day in class.

Yep, you must have. But it's not too late to learn what you missed
that day. :-)
--
73, Cecil, W5DXP

Cecil,

Nice try.

I cannot find the word "migrate" in your reference. However, I do find
several examples of your own interpretations mixed in.

When in doubt, change the subject?

More rattlesnake physics?

73,
Gene
W4SZ

Cecil Moore wrote:
Gene Fuller wrote:

As for the E- and H-fields, this just gets more amusing by the minute.

The second paragraph is even more curious. Do you have a reference
for this migration of energy from the H-field to supply the
suffering E-field?


Here's the reference: From "Fields and Waves" by Ramo and Whinnery.
Take a close look at the exponential transmission line equations
for flat lines (no reflections):

V = Vmax(e^-az)(e^wt-bz) (1)

I = Vmax(e^-az)(e^wt-bz)/Z0 (2)

'a' (alpha) is the attenuation factor. The two equations are identical
except for the Z0 term. If you divide equation (1) by equation (2),
you get Z0. In a flat transmission line (no reflections) the current
is ALWAYS equal to the voltage divided by the characteristic impedance
of the transmission line. The voltage and current are attenuated by
EXACTLY the same factor. If the voltage drops because of I^2*R losses,
the current must decrease by exactly the same percentage. (I have avoided
calling it a current drop so it wouldn't upset you.)

Since the attenuation factor is R/2*Z0 + G*Z0/2 and since, for most
transmission lines used on HF, R/2*Z0 G*Z0/2, the current attenuation
is caused by the series I^2*R drop in the voltage and the V/I=Z0 ratio
that must be maintained - pretty simple logic.

I must have missed that day in class.


Yep, you must have. But it's not too late to learn what you missed
that day. :-)
--
73, Cecil, W5DXP

Gene Fuller wrote:
I cannot find the word "migrate" in your reference. However, I do find
several examples of your own interpretations mixed in.

You expect me to remember the exact wording after 40 years?
Perhaps your prof, like mine, told you to skip sections 1.22-
1.28 in Ramo and Whinnery. I didn't take his advice - I read
them.

So Gene, please point out the error in my logic. If the
current attenuation is primarily from R, the series
resistance, what other explanation can there be than
energy is being supplied by the H-field to the sagging
E-field? Can you think of any other rational, logical,
non-emotional, technical explanation, given these
exponential transmission line equations?

V = Vmax(e^-az)*e^j(wt-bz)

I = Vmax(e^-az)*e^j(wt-bz)/Z0

V/I = Z0

(I think you are just trying to punish me for saying my
dog has a soul. :-)
--
73, Cecil, W5DXP

Cec, I was well aware of what Chipman was about to write years before he
wrote his most excellent, most reliable book on the subject of transmission
lines. There are are very few errors. All of those which I have found can be
attributed to the printer. But you don't find these unless you actually use
the book, fully understand the book, and do some practical sums.

There are far too many people who use books as bibles because they have
found them verbally convincing but who have never actually used them by
inserting practical engineering numbers. No-one unfamiliar with numbers can
call himself an engineer. More likely he is a plagiarising Old Wife.

I do not worship Chipman. I think he is still plodding around in his 90's. I
don't worship anybody.

Whilst on the subject of reliability, I have very recently had a serious
accident.

My corkscrew broke.

It can only have been due to metal fatigue and all the use its had.
Probably made in Taiwan or Korea.

For 5 days, for one reason or another, I have been confined to the house
without the opportunity to replace corkscrews. Yet there have been 4
unopened bottles of perfectly good wine in the cooler, with clean glasses.
Calamity! Absolute misery!

But my long mechanical engineering experience came to the rescue. I
discovered a 1/4" Philips screwdriver and a hammer. With lots of hammering I
eventually drove the cork (of a bottle of Premieres Cotes de Bordeaux,
sweet-white), inside the bottle.

The cork floats on the top of the wine and there's few problems with
pouring. The cork remains quite intact. No contaminating bits to spit out.

So everything is now back to normal. I shortly expect to obtain a new
corkscrew - this time with a spare.

Hic!
----
Reg.



"Cecil Moore" wrote in message
...
Reg Edwards wrote:
Write it down. It will then be obvious, to make the angle of Zo equal
to
zero it is necessary only that the angle of R+j*Omega*L be made equal to
the
angle of G+j*Omega*C.

Reg, there is no contradiction between what I have said and what you have
said.

Continuing the discussion - For average transmission lines used on HF
frequencies, "... the value of G ... is likely to be too small to affect
the
attenuation factor ..." Quote from "Transmission Lines" by Chipman, page
94.

Some of Chipman's calculations indicate that, for a typical 10 MHz
example,
R is about 0.1 ohm/meter while G is about 0.9 micromhos/meter. That's
about
a 100,000:1 ratio making G negligible as far as attenuation factor goes.

The attenuation factor depends almost entirely on R, the series resistance
parameter. G, the parallel conductance parameter, has a negligible effect
on the attenuation factor at HF.

Since, at HF, the attenuation factor consists almost entirely of series
resistance, and since the attenuation factor is identical for voltage and
current, it logically follows that the series resistance is primarily
responsible for the attenuation of the current.

Or even more simple, for Zo to be purely resistive, G = C*R/L

Actually, that is only an approximation for low-loss lines.
--
73, Cecil, W5DXP

Reg Edwards wrote:
There are far too many people who use books as bibles ...

Ain't that the truth? I post "1+1=2" and somebody wants a reference.
I just bought "Mathematics From the Birth of Numbers" by Jan Gullberg.
In addition to information on virtually all branches of mathematics,
it gives the history of the branches. It's really interesting.

With lots of hammering I
eventually drove the cork (of a bottle of Premieres Cotes de Bordeaux,
sweet-white), inside the bottle.

At the beer/wine busts at Texas A&M during the 50's, nobody
could ever remember to bring a corkscrew so that's the way
we did it. Sometimes we forgot a bottle opener and used the
bumper of my old '49 Chevvy to open the beer.
--
73, Cecil, http://www.qsl.net/w5dxp