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

On Mon, 22 Nov 2004 22:08:29 +0000 (UTC), "Reg Edwards"
wrote:


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

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).

With my respects and best wishes,


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

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

For those with a mathematical bent, alternatively stated, for the angle of
Zo to be zero, or for Zo to be purely resistive -

L * G = C * R

or

L / C = R / G

which relationship is derived from -

Zo = Squareroot( Z / Y )

Where line series impedance Z = R + j * Omega * L

and line shunt admittance Y = G + j * Omega * C

provided the angle of Z is equal to the angle of Y.

Which makes Zo = Ro + j * ZERO

QED
------
Reg, G4FGQ

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