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

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

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