(Robert Lay W9DMK) wrote in message ...


The condition for which Zo of a transmission line is always purely resistive
(Zo = Ro) is extremely simple. It is -

G = C * R / L

where G is shunt conductance, C is shunt capacitance, R is series
resistance, L is series inductance, all per unit length of line.


Well, that is pretty entertaining and
interesting, i will admit. However, the
result isn't very practical...

Most texts just assume lossless conditions,
so that R and G equal zero, so that Zo is
again, always purely real, and only dependant
on the distributed inductance and capacitance
of the transmission line.



Which applies to any line length, at any frequency from DC to UHF.

It is a shortcoming of the Smith Chart, with Zo always equal to Ro, that it
does not make you aware of this and can lead you up the garden path if you
are not careful. As has recently occurred on this newsgroup.

Don't get me wrong. I'm not against Smith Charts. They are graphically
educational within their limitations.
----
Reg, G4FGQ

Dear Reg,

You say that it is a shortcoming of the Smith Chart that Zo equals Ro.
However, I think that is either a misunderstanding or just misleading.
The Smith Chart only constrains the normalizing quantity to be purely
resistive - not the characteristic impedance of a particular
transmission line being shown on that chart. My program, SmartSmith,
for example, allows the user to specify both an Ro and an Xo term for
all transmission line sections.

When it's all said and done, the Smith Chart only implements the
transmission line equation (as shown on pages 24-10 and 27-29 in the
17th Edition of The ARRL Antenna Book).



You know, this seems to be quite a theory
problem regarding the Smith Chart. This goes back to
the heated argument on the Gamma reflection
coefficient using the conjugate of Zo in the
numerator.
Since i trust Les Besser more than
other people, i'm gonna assume that
you indeed CAN normalize a Smith chart
to a complex Zo. That is, a Zo that
is not a purely real impedance.



Slick

On 23 Nov 2004 07:36:11 -0800, (Dr. Slick) wrote:

[snip]
| Since i trust Les Besser more than
|other people,

Heh heh. Reminds me... I took his video taped course, "RF/Microwave
Transistor Amplifier Design".

I was watching one tape at home while the XYL was making dinner. She
could hear the audio and finally asked me to turn it off since it was
making *her* fall asleep. [g].

Really smart guy, but not a very dynamic speaker.

Jim Kelley wrote:
I guess one could infer that that if G / C R / L, and R + jwL G +
jwC, then perhaps there are losses. I would only add that there are
probably also small currents in shunt distributed along the line.

If R 0 and G 0, then there are losses. The only time a line is
lossless is when R = G = 0 which, according to Reg, is only in my wet
dreams about circles on Smith Charts. :-) For real world transmission
lines at HF, (usually) R/Z0 G*Z0. When I was a member of the high
speed cable group at Intel, I remember test leads designed for R/Z0=G*Z0
but they were expensive special order devices.

We apparently are more successful at designing very good dielectrics
than in finding an economically feasible conductor with a couple of
magnitudes less resistance than copper. Thus our ordinary transmission
lines have a lot more series resistance than shunt conductance, especially
open-wire transmission lines in free space. :-)
--
73, Cecil http://www.qsl.net/w5dxp

Gene Fuller wrote:
Do you s'pose that if the equality is perfect for zero-loss lines then
maybe it is an useful approximation for low-loss lines?

Do you really think R&W were proposing that this simple relationship is
more appropriate for low loss lines than for zero loss lines?

Nope, exactly the opposite. Apparently, they were proposing that this
simple relationship doesn't hold for highly lossy lines. Chipman also
has something to say about highly lossy lines.
--
73, Cecil http://www.qsl.net/w5dxp

Dr. Slick wrote:
Well, that is pretty entertaining and
interesting, i will admit. However, the
result isn't very practical...

On the contrary, the result is extremely practical and
isn't very ideal, just like real-world physics. :-)
--
73, Cecil http://www.qsl.net/w5dxp

Wes Stewart wrote in message . ..
On 23 Nov 2004 07:36:11 -0800, (Dr. Slick) wrote:

[snip]
| Since i trust Les Besser more than
|other people,

Heh heh. Reminds me... I took his video taped course, "RF/Microwave
Transistor Amplifier Design".

I was watching one tape at home while the XYL was making dinner. She
could hear the audio and finally asked me to turn it off since it was
making *her* fall asleep. [g].

Really smart guy, but not a very dynamic speaker.


Hehe! Yeah, also not exactly the
sexiest guy on the planet, is he!

I have his full RF Fund. course One
and Two on tape, like 12 tapes in all.

We should have a Les Besser Get
together sometime. Wooohaaa!

Fun!

Slick

Cecil Moore wrote in message ...
Dr. Slick wrote:
Well, that is pretty entertaining and
interesting, i will admit. However, the
result isn't very practical...

On the contrary, the result is extremely practical and
isn't very ideal, just like real-world physics. :-)


When was the last time you used

"G = C * R / L" for anything?


Slick

Dr. Slick wrote:
When was the last time you used
"G = C * R / L" for anything?

Yesterday.

Cecil Moore wrote:
Dr. Slick wrote:

When was the last time you used
"G = C * R / L" for anything?


Yesterday.

Just curious, Cecil. What were you using a distortionless line
for?
Tom Donaly, KA6RUH

Tom Donaly wrote:
Cecil Moore wrote:

Dr. Slick wrote:
When was the last time you used
"G = C * R / L" for anything?

Yesterday.

Just curious, Cecil. What were you using a distortionless line
for?

Methinks you jumped to conclusions. Slick didn't ask when was
the last time I used a distortionless line. He asked when was
the last time I used the equation "G=C*R/L" for anything. I'm
using it right now. I used it yesterday to refresh my memory
about distortionless lines which I did use quite often before
I retired from Intel in 1998. If I presently held a job in the
cable modem group, I suppose I would still be using them.
--
73, Cecil http://www.qsl.net/w5dxp