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



-----= Posted via Newsfeeds.Com, Uncensored Usenet News =-----
http://www.newsfeeds.com - The #1 Newsgroup Service in the World!
-----== Over 100,000 Newsgroups - 19 Different Servers! =-----

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



-----= Posted via Newsfeeds.Com, Uncensored Usenet News =-----
http://www.newsfeeds.com - The #1 Newsgroup Service in the World!
-----== Over 100,000 Newsgroups - 19 Different Servers! =-----