Not having much else to do at present I thought I would make a comment on Zo
and Ro of transmission lines. For entertainment and educational value, of
course, if you like that sort of thing.

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.

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

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 http://www.qsl.net/w5dxp

Cecil,

Try it.

I believe you will find that your equality requirement on angles reduces
to precisely the simple equation offer by Reg.

73,
Gene
W4SZ

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 http://www.qsl.net/w5dxp

Gene Fuller wrote:
Try it.

I believe you will find that your equality requirement on angles reduces
to precisely the simple equation offer by Reg.

Exactly! That's why I wonder why Ramo and Whinnery said it's an approximation.

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 http://www.qsl.net/w5dxp

Cecil,

They were undoubtedly confused by their models, and they could not deal
with reality.

73,
Gene
W4SZ

Cecil Moore wrote:

Gene Fuller wrote:

Try it.

I believe you will find that your equality requirement on angles
reduces to precisely the simple equation offer by Reg.


Exactly! That's why I wonder why Ramo and Whinnery said it's an
approximation.

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 http://www.qsl.net/w5dxp

Gene Fuller wrote:
They were undoubtedly confused by their models, and they could not deal
with reality.

Thanks Gene, I really appreciate it when you contribute something
techincal.
--
73, Cecil http://www.qsl.net/w5dxp

Gene Fuller wrote:

Cecil,

They were undoubtedly confused by their models, and they could not deal
with reality.

73,
Gene
W4SZ

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.

73, ac6xg

Cecil Moore wrote:

Gene Fuller wrote:

Try it.

I believe you will find that your equality requirement on angles
reduces to precisely the simple equation offer by Reg.



Exactly! That's why I wonder why Ramo and Whinnery said it's an
approximation.

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 http://www.qsl.net/w5dxp

Cecil,

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?

73,
Gene
W4SZ

Cecil Moore wrote:
Gene Fuller wrote:

Try it.

I believe you will find that your equality requirement on angles
reduces to precisely the simple equation offer by Reg.


Exactly! That's why I wonder why Ramo and Whinnery said it's an
approximation.

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 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

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