The first example was much too easy. How about this one?

---50 ohm feedline---+---300 ohm feedline---
Pfwd1=100w-- Pfwd2 not given--
--Pref1=0w --Pref2 not given

Given a Z0-match at point '+':
Solve for Vfwd1, Ifwd1, Vref1, Iref1, Pfwd2, Vfwd2, Ifwd2,
Pref2, Vref2, Iref2, including magnitudes and phase angles
for all voltages and currents. Source is unknown. Load is
unknown. Lengths of feedlines are unknown.

Who thinks this one is impossible to solve?
--
73, Cecil http://www.qsl.net/w5dxp


----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==----
http://www.newsfeeds.com The #1 Newsgroup Service in the World! 100,000 Newsgroups
---= East/West-Coast Server Farms - Total Privacy via Encryption =---

Just hook it up and see if it works.
If not, change one the "thingys" and try again.
Repeat until it works.
end loop


"Cecil Moore" wrote in message
...
The first example was much too easy. How about this one?

---50 ohm feedline---+---300 ohm feedline---
Pfwd1=100w-- Pfwd2 not given--
--Pref1=0w --Pref2 not given

Given a Z0-match at point '+':
Solve for Vfwd1, Ifwd1, Vref1, Iref1, Pfwd2, Vfwd2, Ifwd2,
Pref2, Vref2, Iref2, including magnitudes and phase angles
for all voltages and currents. Source is unknown. Load is
unknown. Lengths of feedlines are unknown.

Who thinks this one is impossible to solve?
--
73, Cecil http://www.qsl.net/w5dxp


----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet
News==----
http://www.newsfeeds.com The #1 Newsgroup Service in the World! 100,000
Newsgroups
---= East/West-Coast Server Farms - Total Privacy via Encryption =---

Hal:

Could you explain that in less technical terms?

John

"Hal Rosser" wrote in message
...
Just hook it up and see if it works.
If not, change one the "thingys" and try again.
Repeat until it works.
end loop


"Cecil Moore" wrote in message
...
The first example was much too easy. How about this one?

---50 ohm feedline---+---300 ohm feedline---
Pfwd1=100w-- Pfwd2 not given--
--Pref1=0w --Pref2 not given

Given a Z0-match at point '+':
Solve for Vfwd1, Ifwd1, Vref1, Iref1, Pfwd2, Vfwd2, Ifwd2,
Pref2, Vref2, Iref2, including magnitudes and phase angles
for all voltages and currents. Source is unknown. Load is
unknown. Lengths of feedlines are unknown.

Who thinks this one is impossible to solve?
--
73, Cecil http://www.qsl.net/w5dxp


----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure
Usenet
News==----
http://www.newsfeeds.com The #1 Newsgroup Service in the World!
100,000
Newsgroups
---= East/West-Coast Server Farms - Total Privacy via Encryption
=---

Cecil Moore wrote:

The first example was much too easy. How about this one?

---50 ohm feedline---+---300 ohm feedline---
Pfwd1=100w-- Pfwd2 not given--
--Pref1=0w --Pref2 not given

Given a Z0-match at point '+':
Solve for Vfwd1, Ifwd1, Vref1, Iref1, Pfwd2, Vfwd2, Ifwd2,
Pref2, Vref2, Iref2, including magnitudes and phase angles
for all voltages and currents. Source is unknown. Load is
unknown. Lengths of feedlines are unknown.

Who thinks this one is impossible to solve?

I just received an email from someone who solved it. Now
who says an energy analysis is impossible or meaningless?
--
73, Cecil http://www.qsl.net/w5dxp


----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==----
http://www.newsfeeds.com The #1 Newsgroup Service in the World! 100,000 Newsgroups
---= East/West-Coast Server Farms - Total Privacy via Encryption =---

On Sat, 02 Jul 2005 08:21:07 -0500, Cecil Moore
wrote:

The first example was much too easy. How about this one?

---50 ohm feedline---+---300 ohm feedline---
Pfwd1=100w-- Pfwd2 not given--
--Pref1=0w --Pref2 not given

Given a Z0-match at point '+':
Solve for Vfwd1, Ifwd1, Vref1, Iref1, Pfwd2, Vfwd2, Ifwd2,
Pref2, Vref2, Iref2, including magnitudes and phase angles
for all voltages and currents. Source is unknown. Load is
unknown. Lengths of feedlines are unknown.

Without disclosing the answers or the exact procedure for solving the
"brain teaser", I would like to draw attention to some of the implicit
relationships that "ought" to help.
1) It is assumed that both feelines have purely resistive
characteristic impedances (imaginary component, Xo, is zero).
2) Regardless of the length of the 300 ohm line and its termination
impedance, the standing wave pattern and the voltages and currents,
both incident and reflected as a function of distance x along that
line are determined completely by the requirement/condition that there
is a Z0 match at point "+".
3) There are an infinite number of lengths of the 300 ohm line and a
corresponding infinite number of termination impedances for that line
that will produce a Z0 match at point "+". However, because of (2),
above, some of those combinations are well known combinations with
well understood results (e.g., odd multiple of quarter wavelength or
an integer number of half wavelengths).
4) Due to conditions (1) and (2) above, the phase relations between
all of the voltages and currents immediately adjacent to either side
of point "+" are trivial (i.e., any two quantities chosen will be
either exactly in phase or exactly 180 degrees out of phase with one
another).

Due to (3) and (4) above, it would seem that an arbitrary choice of
either a quarter wave line with an 1800 ohm termination or a half wave
line with a 50 ohm termination would provide a convenient example with
which to begin an analysis. However, that is not necessary and only
provides a crutch to get off dead center.

If all of the above elements are kept in mind, then it becomes a
matter of solving a simple algebraic relationship involving 4
equations with 4 unknowns (the incident and reflected voltages and
currents at the right hand side of point "+").

The actual numerical answer to such a problem is irrelevant. The
points to be learned from all this are really the implicit
relationships (2), (3) and (4) above. Without an understanding of
those points, it is virtually impossible to even know where to start.
I think that is the real point that Cecil is trying to make.

Bob, W9DMK, Dahlgren, VA
Replace "nobody" with my callsign for e-mail
http://www.qsl.net/w9dmk
http://zaffora/f2o.org/W9DMK/W9dmk.html

W9DMK (Robert Lay) wrote:
The actual numerical answer to such a problem is irrelevant. The
points to be learned from all this are really the implicit
relationships (2), (3) and (4) above. Without an understanding of
those points, it is virtually impossible to even know where to start.
I think that is the real point that Cecil is trying to make.

Actually, it is one mm broader than that. In the above analysis,
an energy analysis works just as well as any other, contrary to
the Sacred Cow Lamentations I and II of some experts on this
newsgroup. There is so much redundancy built into the voltage,
current, and power relationships in a transmission line that there
are a number of valid ways to skin the cat. An energy analysis is
one of those valid ways. Two people have sent me emails with
correct solutions. Here's how to approach the solution from an
energy standpoint.

---50 ohm feedline---+---300 ohm feedline---
Pfwd1=100w-- Pfwd2 not given--
--Pref1=0w --Pref2 not given

rho=250/350=0.7143, rho^2 = 0.51, (1-rho^2) = 0.49
rho^2 is the power reflection coefficient.
(1-rho^2) is the power transmission coefficient.

Pfwd1*rho^2 = 100*0.51 = 51w reflected back toward the source
at the match point. My article labels that quantity 'P3'

Pfwd1*(1-rho^2) = 100*0.49 = 49 watts transmitted through the
match point toward the load. My article labels that quantity
'P1' (as does Dr. Best's QEX article).

For a match to exist Pref2(1-rho^2) must equal 51w, the part
of Pref2 transmitted back through the match point, i.e. not
re-reflected. My article labels that quantity 'P4'

That makes Pref2 = 51w/0.49 = 104.1w, and makes
Pref2(rho^2) = 53.1w, the part initially re-reflected. My article
labels that quantity 'P2' as does Dr. Best's QEX article.

So Pfwd2 = P1 + P2 + P3 + P4 = 49w + 53.1w + 51w + 51w = 204.1w

Who said powers can never be added? Pfwd2 is indeed 204.1w.

Now that we know all the powers (without knowing a single voltage)
we can calculate the voltages and currents whose phase angles
are all either zero degrees or 180 degrees. As Bob sez, phase
angles are trivial at a Z0-match point.

Is there anybody out there who still believes that an energy
analysis is impossible and/or "gobbledegook"?

Incidentally, the two 51w component powers represent the amount
of destructive interference energy involved in wave cancellation
and the amount of constructive interference energy re-reflected
toward the load as a result of that wave cancellation. This is
something that Dr. Best completely missed in his QEX article.
He correctly identified P1 and P2 but completely ignored P3 and P4.
Thus he came up with the equation: Ptot = 75w + 8.33w = 133.33w.
Remember that argument on this newsgroup from spring of 2001?
--
73, Cecil http://www.qsl.net/w5dxp


----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==----
http://www.newsfeeds.com The #1 Newsgroup Service in the World! 100,000 Newsgroups
---= East/West-Coast Server Farms - Total Privacy via Encryption =---

I would have used the term "Whatchacallit", but "thingy" seems to work.

"John Smith" wrote in message
...
Hal:

Could you explain that in less technical terms?

John

"Hal Rosser" wrote in message
...
Just hook it up and see if it works.
If not, change one the "thingys" and try again.
Repeat until it works.
end loop


"Cecil Moore" wrote in message
...
The first example was much too easy. How about this one?

---50 ohm feedline---+---300 ohm feedline---
Pfwd1=100w-- Pfwd2 not given--
--Pref1=0w --Pref2 not given

Given a Z0-match at point '+':
Solve for Vfwd1, Ifwd1, Vref1, Iref1, Pfwd2, Vfwd2, Ifwd2,
Pref2, Vref2, Iref2, including magnitudes and phase angles
for all voltages and currents. Source is unknown. Load is
unknown. Lengths of feedlines are unknown.

Who thinks this one is impossible to solve?
--
73, Cecil http://www.qsl.net/w5dxp


----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure
Usenet
News==----
http://www.newsfeeds.com The #1 Newsgroup Service in the World!
100,000
Newsgroups
---= East/West-Coast Server Farms - Total Privacy via Encryption
=---

W9DMK (Robert Lay) wrote:
On Sat, 02 Jul 2005 08:21:07 -0500, Cecil Moore
wrote:


The first example was much too easy. How about this one?

---50 ohm feedline---+---300 ohm feedline---
Pfwd1=100w-- Pfwd2 not given--
--Pref1=0w --Pref2 not given

Given a Z0-match at point '+':
Solve for Vfwd1, Ifwd1, Vref1, Iref1, Pfwd2, Vfwd2, Ifwd2,
Pref2, Vref2, Iref2, including magnitudes and phase angles
for all voltages and currents. Source is unknown. Load is
unknown. Lengths of feedlines are unknown.


Without disclosing the answers or the exact procedure for solving the
"brain teaser", I would like to draw attention to some of the implicit
relationships that "ought" to help.
1) It is assumed that both feelines have purely resistive
characteristic impedances (imaginary component, Xo, is zero).
2) Regardless of the length of the 300 ohm line and its termination
impedance, the standing wave pattern and the voltages and currents,
both incident and reflected as a function of distance x along that
line are determined completely by the requirement/condition that there
is a Z0 match at point "+".
3) There are an infinite number of lengths of the 300 ohm line and a
corresponding infinite number of termination impedances for that line
that will produce a Z0 match at point "+". However, because of (2),
above, some of those combinations are well known combinations with
well understood results (e.g., odd multiple of quarter wavelength or
an integer number of half wavelengths).
4) Due to conditions (1) and (2) above, the phase relations between
all of the voltages and currents immediately adjacent to either side
of point "+" are trivial (i.e., any two quantities chosen will be
either exactly in phase or exactly 180 degrees out of phase with one
another).

Due to (3) and (4) above, it would seem that an arbitrary choice of
either a quarter wave line with an 1800 ohm termination or a half wave
line with a 50 ohm termination would provide a convenient example with
which to begin an analysis. However, that is not necessary and only
provides a crutch to get off dead center.

If all of the above elements are kept in mind, then it becomes a
matter of solving a simple algebraic relationship involving 4
equations with 4 unknowns (the incident and reflected voltages and
currents at the right hand side of point "+").

The actual numerical answer to such a problem is irrelevant. The
points to be learned from all this are really the implicit
relationships (2), (3) and (4) above. Without an understanding of
those points, it is virtually impossible to even know where to start.
I think that is the real point that Cecil is trying to make.

Bob, W9DMK, Dahlgren, VA
Replace "nobody" with my callsign for e-mail
http://www.qsl.net/w9dmk
http://zaffora/f2o.org/W9DMK/W9dmk.html


Cecil already defined the voltage and current at
the match point when he gave the characteristic
impedances of the two lines and the rate of
energy transfer through them. Knowing the voltage
and current, anyone can calculate
Pfwd2 and Prev2 using Pfwd2 = |(V+IZ0)/2sqrt(Z0)|^2 and
Prev = |(V-IZ0)/2sqrt(Z0)|^2, where Z0 is the characteristic
impedance of the second transmission line.
Cecil's ability to add powers together, which he did in
this instance, isn't anything unique, and doesn't
really teach anything about the general case.
In fact, for a quarter wave transformer, you can
do the following trick: compute the value of the
power as it just comes through the impedance discontinuity
for the first time and call it Pa. Call Rho^2 at the
load P. Then the power delivered to the load will be
Pa( 1 + P + P^2 + P^3 + P^4 ....) which looks the
same as if the power reflection coefficient looking
back toward the generator was 1 and the power at the
load was the result of the addition of an infinite
number of reflections. Such an interpretation, though,
can be shown to be absolutely wrong. Can anyone see why?
73,
Tom Donaly, KA6RUH

Tom Donaly wrote:
Cecil's ability to add powers together, which he did in
this instance, isn't anything unique, and doesn't
really teach anything about the general case.

I'm glad you agree, Tom. Other experts on this newsgroup
will argue with you as they have with me for four years
ever since Dr. Best posted his infamous Z0-match equation:

Ptot = 75w + 8.33w = 133.33w

to which I objected back then, only to have most of
the rest of the posters agree with Dr. Best. I was
dumbfounded to see so many otherwise knowledgable
engineers agree to a violation of the principle of
conservation of energy. I was told not to worry about
conservation of energy - that it takes care of itself.

In fact, for a quarter wave transformer, you can
do the following trick: compute the value of the
power as it just comes through the impedance discontinuity
for the first time and call it Pa. Call Rho^2 at the
load P. Then the power delivered to the load will be
Pa( 1 + P + P^2 + P^3 + P^4 ....) which looks the
same as if the power reflection coefficient looking
back toward the generator was 1 and the power at the
load was the result of the addition of an infinite
number of reflections. Such an interpretation, though,
can be shown to be absolutely wrong. Can anyone see why?

Destructive interference between the external reflection
at the match point and the internal reflection from the
load supplies additional constructive interference
energy to the forward wave in the quarter wave transformer.
You didn't include that constructive interference energy
above. Hint: That virtual power reflection coefficient looking
rearward into the match point doesn't reach 1 until steady-
state is reached (wrong premise above). The virtual power
reflection coefficient looking forward into the match point
also doesn't reach 0 until steady-state is reached. Those
two virtual power reflection coefficients actually start out
the same value and proceed in opposite directions during
the transient buildup to steady-state.
--
73, Cecil http://www.qsl.net/w5dxp


----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==----
http://www.newsfeeds.com The #1 Newsgroup Service in the World! 100,000 Newsgroups
---= East/West-Coast Server Farms - Total Privacy via Encryption =---

Cecil Moore wrote:
Tom Donaly wrote:

Cecil's ability to add powers together, which he did in
this instance, isn't anything unique, and doesn't
really teach anything about the general case.


I'm glad you agree, Tom. Other experts on this newsgroup
will argue with you as they have with me for four years
ever since Dr. Best posted his infamous Z0-match equation:

Ptot = 75w + 8.33w = 133.33w

to which I objected back then, only to have most of
the rest of the posters agree with Dr. Best. I was
dumbfounded to see so many otherwise knowledgable
engineers agree to a violation of the principle of
conservation of energy. I was told not to worry about
conservation of energy - that it takes care of itself.

In fact, for a quarter wave transformer, you can
do the following trick: compute the value of the
power as it just comes through the impedance discontinuity
for the first time and call it Pa. Call Rho^2 at the
load P. Then the power delivered to the load will be
Pa( 1 + P + P^2 + P^3 + P^4 ....) which looks the
same as if the power reflection coefficient looking
back toward the generator was 1 and the power at the
load was the result of the addition of an infinite
number of reflections. Such an interpretation, though,
can be shown to be absolutely wrong. Can anyone see why?


Destructive interference between the external reflection
at the match point and the internal reflection from the
load supplies additional constructive interference
energy to the forward wave in the quarter wave transformer.
You didn't include that constructive interference energy
above. Hint: That virtual power reflection coefficient looking
rearward into the match point doesn't reach 1 until steady-
state is reached (wrong premise above). The virtual power
reflection coefficient looking forward into the match point
also doesn't reach 0 until steady-state is reached. Those
two virtual power reflection coefficients actually start out
the same value and proceed in opposite directions during
the transient buildup to steady-state.

Hi Cecil,
you come up with the right answer, but is your
interpretation correct? Can you do the same thing in a
general sense? If there is no Z0 match between the two
transmission lines, does your method still work? The
little conundrum I posed is an example of a procedure
that will actually give the right answer, but the
interpretation I gave of how it works is wrong. Can you
be sure your method doesn't have the same flaw?
73,
Tom Donaly, KA6RUH
(P.S. The method of using V and I and the junction of
the two xmission lines to find the forward and reverse
powers on a transmission line doesn't prove the powers exist.
It works just as easily with a pair of resistors and is
more an algebraic stunt that works than anything else. It
does agree with you, however.)