Tom Bruhns wrote:
(Note that our resolution of measured voltage and current into the two
modes generally assumes that we know the line's Zo, and the degree to
which we don't know that will introduce an error in our determination
of rho. But that's a whole 'nuther topic...)

To my way of thinking, rho is entirely dependent upon the impedances,
and the voltages (reflected voltages in particular) are dependent upon
rho. Not the other way around.

73, Jim AC6XG

Jim Kelley wrote:
To my way of thinking, rho is entirely dependent upon the impedances,
and the voltages (reflected voltages in particular) are dependent upon
rho. Not the other way around.

Even when the impedances are only V/I ratios? Seems like circular
logic to me. The V/I ratio causes rho which causes the voltage???
--
73, Cecil http://www.qsl.net/w5dxp



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Cecil Moore wrote:

Jim Kelley wrote:
To my way of thinking, rho is entirely dependent upon the impedances,
and the voltages (reflected voltages in particular) are dependent upon
rho. Not the other way around.

The V/I ratio causes rho which causes the voltage???

Nope. Rho is not dependent upon V/I ratios other than those at real
physical impedance discontinuities. I think you know that. V/I ratios
can vary with position along the line and are not constrained to
equalling Z0 or Zl.

I'm curious why you would ask.

73, Jim AC6XG

Jim Kelley wrote:

Cecil Moore wrote:
The V/I ratio causes rho which causes the voltage???

Nope. Rho is not dependent upon V/I ratios other than those at real
physical impedance discontinuities. I think you know that. V/I ratios
can vary with position along the line and are not constrained to
equalling Z0 or Zl.

Consider the following:

Source---50 ohm feedline---+---1/2WL 150 ohm---50 ohm load

Isn't the 50 ohms that causes rho=0 on the 50 ohm feedline
simply the V/I ratio at point '+'?
--
73, Cecil http://www.qsl.net/w5dxp



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Cecil Moore wrote:
Consider the following:

Source---50 ohm feedline---+---1/2WL 150 ohm---50 ohm load

Isn't the 50 ohms that causes rho=0 on the 50 ohm feedline
simply the V/I ratio at point '+'?

The nature of things a point '+' are undefined, so I can't address
that. But according to the way you defined the problem, the
characteristic impedance of the 50 ohm feedline is 50 ohms. That sets
the V/I ratio. The impedance is determined by the distributed
capacitances and inductances of the transmission line - not by the
voltage you put across it. Is there some other way I'm supposed to look
at it?

73, Jim AC6XG

Jim Kelley wrote:

Cecil Moore wrote:
Consider the following:

Source---50 ohm feedline---+---1/2WL 150 ohm---50 ohm load

Isn't the 50 ohms that causes rho=0 on the 50 ohm feedline
simply the V/I ratio at point '+'?

The nature of things a point '+' are undefined,

Nope, they are not. The V/I ratio at '+' equals 50 ohms.

so I can't address
that. But according to the way you defined the problem, the
characteristic impedance of the 50 ohm feedline is 50 ohms. That sets
the V/I ratio. The impedance is determined by the distributed
capacitances and inductances of the transmission line - not by the
voltage you put across it. Is there some other way I'm supposed to look
at it?

There are no reflections on the 50 ohm feedline because it "sees" 50
ohms at point '+'. The 50 ohms seen at point '+' is a V/I ratio equal
to 50 ohms. So V affects rho. And rho causes that same V? See
the circular logic?
--
73, Cecil http://www.qsl.net/w5dxp



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Cecil Moore wrote:

Jim Kelley wrote:
The nature of things a point '+' are undefined,

Nope, they are not.

There can exist no real point where the characteristic line impedance is
both 50 ohms and 150 ohms.

There are no reflections on the 50 ohm feedline because it "sees" 50
ohms at point '+'.

Oh my gawd, somebody has brainwashed our Cecil! Maybe the pod people
have invaded Texas! :-)

What about cancelled reflected waves, destructive interference and all
that?

73, Jim AC6XG

Cecil Moore wrote:
Isn't the 50 ohms that causes rho=0 on the 50 ohm feedline
simply the V/I ratio at point '+'?

I see now. Your interested in something else here, I think. The rho
for the whole network which includes both impedance discontinuities is
indeed zero. We've talked about that before. But the rho for the
single discontinuity at '+' is not equal to zero. The reflected
impedance (the load impedance, repeated a half wavelength away) is not
considered in the evaluation of rho at '+'. It is the characteristic
impedance of the line that is considered. You would agree, no?

73, Jim AC6XG

Jim Kelley wrote:

Cecil Moore wrote:
Isn't the 50 ohms that causes rho=0 on the 50 ohm feedline
simply the V/I ratio at point '+'?

I see now. Your interested in something else here, I think. The rho
for the whole network which includes both impedance discontinuities is
indeed zero. We've talked about that before. But the rho for the
single discontinuity at '+' is not equal to zero. The reflected
impedance (the load impedance, repeated a half wavelength away) is not
considered in the evaluation of rho at '+'. It is the characteristic
impedance of the line that is considered. You would agree, no?

I would agree if you were talking about s11. But rho on the coax is zero.
The impedance at '+' is 50 ohms. rho = (50-50)/(50+50) = sqrt(0/Pfwd) = 0
at point '+'. And that 50 ohms is a V/I ratio which, I assume, you would
agree cannot cause a voltage.
--
73, Cecil http://www.qsl.net/w5dxp



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Cecil Moore wrote:
But rho on the coax is zero.

Coax doesn't have a 'rho' - unless it's broken or damaged coax.

The impedance at '+' is 50 ohms.

By virtue of reflection from the 50 ohm load.

rho = (50-50)/(50+50) = sqrt(0/Pfwd) = 0
at point '+'.

The same is true at points to the left of point '+' as well. So what?

And that 50 ohms is a V/I ratio which, I assume, you would
agree cannot cause a voltage.

As far as I know, V/I ratios don't "cause" anything.

73, Jim AC6XG