For Vfwd1 = 1 volt or the value of your choice at the input end of the
line, fill in the following blanks:

At the input end of the lossy line:

Vref1 = volts
Vfwd2 = volts
Vref2 = volts
Ifwd1 = amps
Iref1 = amps
Ifwd2 = amps
Iref2 = amps
V = volts
I = amps
P = watts

At the output end of the lossy line:

Vfwd1 = volts
Vref1 = volts
Vfwd2 = volts
Vref2 = volts
Ifwd1 = amps
Iref1 = amps
Ifwd2 = amps
Iref2 = amps
V = volts
I = amps
P = watts

Feel free to toss in values for as many Pf's and Pr's as you think there
are, at both ends of the line.

That's what you're going to have to do to get me to pay attention to
you. Volts, amps, watts. If it's beneath your dignity, too complicated,
too simple, too much work, too boring, or any other excuse, that's
perfectly ok. I'll just continue ignoring your hand-waving. (Which is,
come to think of it, probably a great relief anyway.)

Roy Lewallen, W7EL

Cecil Moore wrote:
Roy Lewallen wrote:

Frankly, I haven't paid any attention to your ducking, dodging, and
hand-waving. You haven't been able to produce an analysis showing the
voltages, currents, and powers in the same simple circuit I analyzed.
As far as I'm concerned, nothing you've posted constitutes a proof of
anything.


If one leads a horse to water and it refuses to drink, digging another
lake is just a waste of time. Analyze the following and see if you
get the same results. I'm betting you will catch your error.

Z0=68-j39
---lossy feedline---+---1WL 50 ohm lossless line---10+j50 load
Vfwd1-- Vfwd2--
--Vref1 --Vref2

Vfwd1 = Vfwd1*rho1 + Vfwd1*tau1 one forward, one reflected component

Vref2 = Vref2*rho2 + Vref2*tau2 one forward, one reflected component

Vref1 = Vfwd1*rho1 + Vref2*tau2 both reflected components added together

Vfwd2 = Vfwd1*tau1 + Vref2*rho2 both forward components added together

Note that every voltage has two components. You chose only one component
for your Vfwd. You ignored the other component of Vfwd. Hint: you cannot
have voltages left over from calculating the total forward voltage and
the total reflected voltage.

On Fri, 05 Sep 2003 00:28:43 -0500, Cecil Moore
wrote:

If one leads a horse to water and it refuses to drink, digging another
lake is just a waste of time.

Hi you two,

The Navy had a way of dealing with these academic issues that put both
of you to shame.

I once offered that stale observation of "leading a horse to water..."
to a Chief Bos'un's Mate. He, understandably, was not amused by the
lack of imagination, insight, invention, intelligence, or what-ever
was suppose to be the hallmark of my specialty and simply replied with
a ho-hum:
"You hold its head under water and suck on its ass."

Seems no one is interested in actually getting anything accomplished
when trading complaints is obviously more fun. ;-)

73's
Richard Clark, KB7QHC

Roy Lewallen wrote:
That's what you're going to have to do to get me to pay attention to
you. Volts, amps, watts.

Roy, you are the one who made the mistake. I have told you exactly
what your mistake is. The rest is your problem, not mine.

Once again there are four component voltages and four component currents.
Forward voltage and current contain TWO terms, not one term as you assert.
Reflected voltage and current contain TWO terms, not one term as you assert.

I'm at work and don't remember all the values in the experiment but
here is something similar which I posted yesterday.

----lossy feedline---+---1WL 50 ohm lossless feedline---10+j60 load
Pfwd1-- Pfwd2--
--Pref1 --Pref2

The ratio of Pref2 to Pfwd2 is 0.7225 on the 50 ohm line.

(Pfwd1 - Pref1) = (Pfwd2 - Pref2)

I don't remember the Z0 of the lossy line but the reflection coefficient
can be calculated to solve the problem. The main thing to realize from
the above is Pfwd2 Pref2. Therefore, Pfwd1 Pref1, i.e. total reflected
power is ALWAYS less than total forward power when dealing with passive
loads.

You have terms left over. There should be no terms left over. There
should be one forward voltage, one forward current, one reflected
voltage, and one reflected current. Until you collect all the terms,
your analysis will contain an error. Why are you so dead set against
collecting like terms?
--
73, Cecil, W5DXP

No, I don't have a problem, nor did I make a mistake. I presented an
example with voltages, currents, and powers that are all
self-consistent, obey Ohm's and Kirchoff's laws, and don't violate any
physical laws. It simply uses those laws, equations that can be found in
nearly any transmission line text, and arithmetic. The analysis is
correct as written.

You've demonstrated that you're unable to produce a similar analysis
which can be fit into your conception of how things work. If your
problem is dyslexia or unfamiliarity with complex arithmetic, all you
have to do is say so. Neither is anything to be ashamed of.

I now return the readers to the standard flurry of hand-waving,
objecting, dodging, and excuses. But don't expect an analysis,
derivation, or proof. And I'll be back outta here.

Roy Lewallen, W7EL

W5DXP wrote:

Roy, you are the one who made the mistake. I have told you exactly
what your mistake is. The rest is your problem, not mine.
. . .

Roy Lewallen wrote:
No, I don't have a problem, nor did I make a mistake. I presented an
example with voltages, currents, and powers that are all
self-consistent, obey Ohm's and Kirchoff's laws, and don't violate any
physical laws. It simply uses those laws, equations that can be found in
nearly any transmission line text, and arithmetic. The analysis is
correct as written.

Nope, it isn't and after following all those laws, you violated one
you should have learned in the 4th grade, i.e. to collect like terms.

You've demonstrated that you're unable to produce a similar analysis
which can be fit into your conception of how things work.

No I haven't. I have produced a simple logical analysis that proved
yours to be wrong. Do you really believe yourself incapable of
making a conceptual error? In the following:

---lossy line---x---1WL 50 ohm lossless line---10+j50 load
Pfwd1-- Pfwd2-- = 30.53W rho=0.82 at 88.9 deg
--Pref1 --Pref2 = 20.53W

I measure ten volts across the ten ohm resistor. I measure 30.53W
forward on the 50 ohm line and 20.53W reflected on the 50 ohm line.
Since the 50 ohm line is lossless, the forward voltage is in phase
with the forward current and the reflected voltage is in phase
with the reflected current. The values of voltages and currents on
the 50 ohm line are easy to calculate. The load reflection coefficient
is easy to calculate. Analysis from 'x' to the load is a no-brainer.

The forward power on the 50 ohm line is 10W less than the reflected
power on the 50 ohm line, i.e. Pfwd2-Pref12 = 10W just as it should.

At point 'x', on the source side of 'x', Pfwd1-Pref1 MUST equal
that same 10W. It doesn't matter what happens between 'x' and the
source. At point 'x', Pfwd1 simply cannot be less than Pref1.

Conditions are the same whether the 1WL of lossless 50 ohm feedline
is in the circuit or out of the circuit. You have not taken all the
forward and reflected terms into account. You have violated something
you should have learned in the 4th grade, i.e. to collect like terms.
Instead, you threw away half of the forward terms and half of the
reflected terms. No wonder you got it wrong.
--
73, Cecil http://www.qsl.net/w5dxp



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Cecil Moore wrote:

Roy Lewallen wrote:
No, I don't have a problem, nor did I make a mistake. I presented an
example with voltages, currents, and powers that are all
self-consistent, obey Ohm's and Kirchoff's laws, and don't violate any
physical laws. It simply uses those laws, equations that can be found in
nearly any transmission line text, and arithmetic. The analysis is
correct as written.

Nope, it isn't and after following all those laws, you violated one
you should have learned in the 4th grade, i.e. to collect like terms.

You've demonstrated that you're unable to produce a similar analysis
which can be fit into your conception of how things work.

No I haven't. I have produced a simple logical analysis that proved
yours to be wrong. Do you really believe yourself incapable of
making a conceptual error? In the following:

---lossy line---x---1WL 50 ohm lossless line---10+j50 load
Pfwd1-- Pfwd2-- = 30.53W rho=0.82 at 88.9 deg
--Pref1 --Pref2 = 20.53W

I measure ten volts across the ten ohm resistor. I measure 30.53W
forward on the 50 ohm line and 20.53W reflected on the 50 ohm line.
Since the 50 ohm line is lossless, the forward voltage is in phase
with the forward current and the reflected voltage is in phase
with the reflected current. The values of voltages and currents on
the 50 ohm line are easy to calculate. The load reflection coefficient
is easy to calculate. Analysis from 'x' to the load is a no-brainer.

The forward power on the 50 ohm line is 10W less than the reflected
power on the 50 ohm line, i.e. Pfwd2-Pref12 = 10W just as it should.

At point 'x', on the source side of 'x', Pfwd1-Pref1 MUST equal
that same 10W. It doesn't matter what happens between 'x' and the
source. At point 'x', Pfwd1 simply cannot be less than Pref1.

I observe that you make the assertion, but fail to do the arithmetic
to demonstrate the assertion, while Roy does the arithmetic which
seems to contradict your assertion. There is a strong suspicion that
the reason you don't do the arithmetic (since you do seem fond of
numerical examples) in this case is that you have been unable to make
arithmetic produce the desired result.

Note that a requirement for Vr to be greater than Vf is that the
reactive component of the load has a different sign than the
reactive component of the line. So the problem specification is
incomplete.

Since the reactances on each side of 'x' have different signs,
the circuit looks like it might be somewhat resonant and it
should not be a surprise when resonant circuits produce higher
voltages.

....Keith

wrote:
I observe that you make the assertion, but fail to do the arithmetic
to demonstrate the assertion, while Roy does the arithmetic which
seems to contradict your assertion. There is a strong suspicion that
the reason you don't do the arithmetic (since you do seem fond of
numerical examples) in this case is that you have been unable to make
arithmetic produce the desired result.

I don't need to waste my time on arithmetic to prove that the
principle of conservation of energy is valid. Anyone who believes
that the reflected power in a passive circuit can ever be greater
than the forward power is simply NUTS and should resign from the
human race!

Roy's assertion that reflected power is greater than forward power
is simply the result of omniscientism, defined by a refusal to
collect like terms in the manner that his 4th grade teacher taught
him to do. :-)
--
73, Cecil http://www.qsl.net/w5dxp



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wrote:
Perhaps completing the arithmetic would demonstrate whether the
definitions you think represent Pfwd and Prev actually do.

It ain't rocket science. Everything flowing toward the load is Pfwd.
Everything flowing toward the source is Pref. There's no third term
because there's no third direction. The directional possibilities of
power in a transmission line = binary. There's no magic third term
pointing toward Alpha Centauri.
--
73, Cecil http://www.qsl.net/w5dxp



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I'm trying to stay out of this, but can't avoid mentioning that although
Cecil insists on only one Pref and one Pfwd, he has different criteria
for voltage waves, and requires multiple Vfwd and Vref. So there are
four directions for voltage, but only two for power?

Roy Lewallen, W7EL

Cecil Moore wrote:
wrote:

Perhaps completing the arithmetic would demonstrate whether the
definitions you think represent Pfwd and Prev actually do.


It ain't rocket science. Everything flowing toward the load is Pfwd.
Everything flowing toward the source is Pref. There's no third term
because there's no third direction. The directional possibilities of
power in a transmission line = binary. There's no magic third term
pointing toward Alpha Centauri.

Cecil Moore wrote:

wrote:
Perhaps completing the arithmetic would demonstrate whether the
definitions you think represent Pfwd and Prev actually do.

It ain't rocket science. Everything flowing toward the load is Pfwd.
Everything flowing toward the source is Pref.

Well, I suppose. If that is the extent of the concept of Pfwd and Prev
its pretty simple.

But without some equations defining Pfwd and Prev, the concept is
not particularly useful since there are no predictions which you
can make from it.

....Keith