Jim Kelley wrote in message ...
Ur right, thanks Dave. I meant to say voltage rather than power. Let
me ask the question properly.

Tom,
Since rho represents the fraction of forward voltage that is reflected,
what does a negative value for rho indicate?

Jim,

I suppose your question has been answered sufficiently (Thanks, Roy,
Cecil and Tam), but I'd like to offer a bit different viewpoint than
is implied by your "fraction that is reflected."

I prefer to think of it not as "a fraction that's reflected" but
rather as a resolution of a particular voltage and current into two
modes. There are two modes of propagation supported by TEM line, one
in each direction along the line. If you excite a line to
steady-state at one frequency, there will be some sinusoidal current
at each point along the line, and some sinusoidal voltage across the
line at each point along its length. (You can have a load at one end
and a source at the other, or two sources each with its own internal
impedance, one at each end, so long as they are on the same
frequency.) That set of voltages and currents can be resolved
mathematically into two components, one corresponding to the mode of
propagation in one direction and one corresponding to the mode in the
other direction. Rho is simply a number representing that resolution
at the point on the line where that rho is measured (or calculated).
It's a complex number because it represents both phase and amplitude.

(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...)

Cheers,
Tom

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:
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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