Cecil Moore wrote:

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

They sometimes cause 'rho' which then becomes an end
result and not the cause of anything.

I disagree. Reflections are caused by real impedances not reflected
ones.
Have you changed your mind about this?

At a two-port
network with reflections, rho usually cannot be calculated
from the physical impedances involved.

It can certainly be done using the optical formulas for a pair of
boundaries.

For the two boundaries as a network, and we call rho at the first
boundary r12 and rho at the second boundary r23 then
rho(network) = (r12 + r23)/(1 + r12*r23) = 0. Note that if we use your
value for r12, the network generates a reflection. I note the utility
of negative rho in this example.

But, with a transmission line at odd multiples of lambda/4, rho for the
network would be at a maximum and the network equation would be
(r12 - r23)/(1 - r12*r23). In such a case you'd want to use a load
impedance that would provide a r23 of +.5. (x - 150)/(x + 150) = .5,
so x = 450.

The moral is be
careful about saying that rho causes anything. Rho may
be only the end result of everything.

No question that rho is the end result of a ratio of impedances. It's
been my view that, like the V/I ratios we were speaking about, rho is
not a cause but a result.

73, Jim AC6XG

Jim Kelley wrote:
Cecil Moore wrote:

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

They sometimes cause 'rho' which then becomes an end
result and not the cause of anything.

I disagree. Reflections are caused by real impedances not reflected
ones. Have you changed your mind about this?

Where did I say rho causes reflections? I didn't! Your statement
does NOT disagree with my statement.

At a two-port
network with reflections, rho usually cannot be calculated
from the physical impedances involved.

It can certainly be done using the optical formulas for a pair of
boundaries.

Unfortunately, only the index of refraction of one of the boundaries
is known in my statement above. The index of refraction of the second
boundary is unknown, i.e. only the impedances at the two-port network
are known - the load is unknown.

No question that rho is the end result of a ratio of impedances. It's
been my view that, like the V/I ratios we were speaking about, rho is
not a cause but a result.

But earlier, I thought you said rho caused a result.
--
73, Cecil http://www.qsl.net/w5dxp



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Cecil Moore wrote:

Unfortunately, only the index of refraction of one of the boundaries
is known in my statement above. The index of refraction of the second
boundary is unknown, i.e. only the impedances at the two-port network
are known - the load is unknown.

Only if you've forgotten what you said it was. :-) If it's unknown,
how could you have known what it was a half wavelength away? We are
speaking about the problem you posed yesterday, right?

No question that rho is the end result of a ratio of impedances. It's
been my view that, like the V/I ratios we were speaking about, rho is
not a cause but a result.

But earlier, I thought you said rho caused a result.

I'll repost what I said. If you find something in error, please
advise. Thanks.

"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:
If it's unknown,
how could you have known what it was a half wavelength away? We are
speaking about the problem you posed yesterday, right?

No, we are speaking about a statement I made unrelated to the problem
I posted. For that statement, the length of the feedline is unknown
and the load is unknown. What is known is the forward power and reflected
power on each side of the impedance discontinuity.

No question that rho is the end result of a ratio of impedances. It's
been my view that, like the V/I ratios we were speaking about, rho is
not a cause but a result.

But earlier, I thought you said rho caused a result.

"To my way of thinking, rho is entirely dependent upon the impedances,
and the voltages (reflected voltages in particular) are dependent upon
rho.

You said "rho is not a cause but a result" but then implied that
voltages are caused by (dependent upon) rho. Seems to me, rho
cannot both cause a voltage and be caused by a (voltage divided
by a current) which is an impedance upon which rho is dependent.
--
73, Cecil, W5DXP

Cecil Moore wrote:

For that statement, the length of the feedline is unknown
and the load is unknown.

I was commenting on this, the subject of our conversation:

"Consider the following:

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

As I was saying, for the two boundaries as a network, and we call rho at
the first boundary r12 and rho at the second boundary r23 then
rho(network) = (r12 + r23)/(1 + r12*r23) = 0. Note that if we use your
value for r12, the network generates a reflection. I note the utility
of negative rho in this example.

Seems to me, rho
cannot both cause a voltage and be caused by a (voltage divided
by a current) which is an impedance upon which rho is dependent.

Voltages on a transmission line do not determine reflection
coefficients.
Reflection coefficients are determined by characteristic impedances, not
virtual ones.

73, Jim AC6XG

Jim Kelley wrote:
. . .
Voltages on a transmission line do not determine reflection
coefficients.
Reflection coefficients are determined by characteristic impedances, not
virtual ones.

73, Jim AC6XG

I disagree with this. When applied to transmission lines, the (voltage)
reflection coefficient is, as far as I can tell, universally defined as
the ratio of reflected to forward voltage to reverse voltage at a point.
So a reflection coefficient can be, and often is, calculated for every
point along a line, not just at discontinuities or points of actual
reflection. This can be done with nothing more than the knowledge of the
values of forward and reflected voltages at the point of calculation.

As it turns out, the value of the reflection coefficient at any point
will be equal to (Z - Z0) / (Z + Z0), where Z is the impedance seen
looking down the line toward the load at the point of calculation. I'm
very leery of the use of "virtual" anything, since it often adds an
unnecessary level of confusion. But if I were to calculate a reflection
coefficient at some point along a continuous line, I could replace the
remainder of the line and the load with a lumped load of impedance Z,
and maintain exactly the same reflection coefficient and forward and
reverse traveling waves in the remaining line. I wouldn't object, then,
if someone would say that there was a "virtual impedance" of Z at that
point when the line was intact, since all properties prior to that point
are unchanged if a lumped Z of that value is substituted for the
remainder. (I personally wouldn't call it "virtual" -- I'd just call it
the Z at that point, since it's the ratio of V to I there.) The point is
that the reflection coefficient was the same before and after the
substitution of the remaining line with a real lumped Z. Before the
substitution, reflection was occurring at the load. After, the
reflection is occurring at the new, substituted Z load. Yet the
reflection coefficient and traveling waves remain the same on the
remaining line.

A reflection coefficient isn't the cause of anything. It's simply a
calculated quantity used for computational and conceptual convenience.
Only an impedance discontinuity causes reflections, but we can calculate
a reflection coefficient at any point we choose, with its value being
well defined and unambiguous.

Roy Lewallen, W7EL

Roy Lewallen wrote:

Jim Kelley wrote:
. . .
Voltages on a transmission line do not determine reflection
coefficients.
Reflection coefficients are determined by characteristic impedances, not
virtual ones.

73, Jim AC6XG

I disagree with this. When applied to transmission lines, the (voltage)
reflection coefficient is, as far as I can tell, universally defined as
the ratio of reflected to forward voltage to reverse voltage at a point.

That rho is equivalent to that ratio of voltages is not in dispute. I
might dispute that it's 'defined' by that ratio. We agree the
reflection is caused by an impedance discontinuity. It is the
relationship of those impedances that determines how much of an incident
voltage will be reflected. From my perspective, one builds a network of
impedances in order to achieve the desired voltage relationships. But
one cannot build voltage relationships in order to obtain a network of
impedances.

Maybe it's another chicken and egg argument.

Only an impedance discontinuity causes reflections, but we can calculate
a reflection coefficient at any point we choose, with its value being
well defined and unambiguous.

Wouldn't the most well defined and unabiguous be at a point of
reflection? ;-)

73, Jim AC6XG

I disagree with this. When applied to transmission lines, the (voltage)
reflection coefficient is, as far as I can tell, universally defined as
the ratio of reflected to forward voltage to reverse voltage at a point.
So a reflection coefficient can be, and often is, calculated for every
point along a line, not just at discontinuities or points of actual
reflection.

This can be done with nothing more than the knowledge of the
values of forward and reflected voltages at the point of calculation.

=============================

Sorry! Just to continue and further confuse the haggling, the forward
voltages are unknown because one does not know, in the case of amateur
systems, what is the internal voltage and internal impedance of the
transmitter.

It is this unknown voltage and internal impedance which the so-called SWR
(Rho) meter merely ASSUMES.

Jim Kelley wrote:


Cecil Moore wrote:
For that statement, the length of the feedline is unknown
and the load is unknown.

I was commenting on this, the subject of our conversation:

My statement that you objected to didn't have anything to
do with the following.

"Consider the following:

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

Reflection coefficients are determined by characteristic impedances, not
virtual ones.

On the contrary, the reflection coefficient, rho, at '+' in
the example above, is NOT determined by the characteristic
impedances. Above, rho (looking at '+' from the left) is determined
by taking the square root of (Pref/Pfwd) = 0. (150-50)/(150+50) is
NOT rho. I was mistaken to call that quantity "rho" in my article.
That quantity that I called "rho" is actually 's11' and I need to
update my article.
--
73, Cecil, W5DXP

Cecil Moore wrote:

Jim Kelley wrote:
Reflection coefficients are determined by characteristic impedances, not
virtual ones.

On the contrary, the reflection coefficient, rho, at '+' in
the example above, is NOT determined by the characteristic
impedances.

I just showed you how characteristic impedances are used to calculate
the reflection coefficient at '+'. But you can wish it into the
cornfield if you like, Anthony. :-)

(150-50)/(150+50) is
NOT rho.

Is it the reflection coefficient for a 50 ohm to 150 ohm impedance
discontinuity?

I was mistaken to call that quantity "rho" in my article.
That quantity that I called "rho" is actually 's11' and I need to
update my article.

Since S-parameters were never even mentioned in your article, updating
it seems an understatement.

73, Jim AC6XG