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


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Cecil Moore wrote:

Who said powers can never be added?

Must have been someone who was unfamiliar with the expression 'figures
can lie and liars can figure'.
:-)

ac6xg

Jim Kelley wrote:

Cecil Moore wrote:
Who said powers can never be added?

Must have been someone who was unfamiliar with the expression 'figures
can lie and liars can figure'. :-)

You reckon Eugene Hecht was lying when he shows us how to
add two irradiances to obtain the total irradiance (power
per unit-area) in _Optics_? Adding EM wave powers during
interference is a well accepted way of handling EM wave
superposition in the field of optics. The bright constructive
interference rings contain more power than the dark destructive
interference rings. RF waves and light waves are both electro-
magnetic waves, just at different frequencies. Asserting that
RF waves obey a different set of laws of physics than do light
waves is naive ignorance at best.
--
73, Cecil http://www.qsl.net/w5dxp

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Cecil Moore wrote:

Jim Kelley wrote:

Cecil Moore wrote:

Who said powers can never be added?


Must have been someone who was unfamiliar with the expression 'figures
can lie and liars can figure'. :-)


You reckon Eugene Hecht was lying when he shows us how to
add two irradiances to obtain the total irradiance (power
per unit-area) in _Optics_?

Eugene Hecht doesn't have a dog in this fight, Cecil. But the quote is
a truism that applies in any case. Wrong numbers added correctly
produce a wrong number; correct numbers added incorrectly produce an
incorrect number; and in the special case, certain wrong numbers added
in a particular incorrect way can produce a desired result.

You take too great a liberty with the name Eugene Hecht. Among the
things which won't be found in any of Dr. Hecht's texts is a minus sign
in front of number expressing an irradiance. Nor will we find a
negative scalar quantity accompanied by the claim that the negative sign
indicates a change in direction, as you have done. Eugene Hecht also
did not claim that interference could be a cause for energy to reflect
or otherwise change direction, as you have done. Such claims are
blatently false.

Power and irradiance are derived and dependent quantities, not
fundamental independent quantities in nature. And although an
automobile moves at some speed, the scaler quantity itself is not
something which moves. Similarly, power and irradiance do not
physically propagate and they do not physically interact. 'They' do not
reflect, refract, diffract, disperse, interfere, or act upon other
'powers' or 'irradiances'. JC Maxwell and others observed that it is
electric and magnetic fields which propagate, interact with matter, and
add algebraically and vectorially. When fields physically interact with
matter, we can measure their effect and can quantify such things as
voltage, current, and heat, and hence calculate such things as power or
irradiance. But it is actually the fields themselves which
algebraically sum. Of course the interference equation accurately
expresses power and irradiance. The fact that power and irradiance
generally go as the square of the fields allows us to correctly make
certain additional mathematical assumptions. One must still be careful
not to mistake an effect for a cause. But it is the 2nd Amendment, the
internet, and the absence of peer review which afford men the freedom
and means to work equations and describe physical phenomena in any way
they like.

73, ac6xg

Jim Kelley wrote:

Jim, you have a habit of erecting strawmen somewhat like:
"I don't care what you say, the sun will rise tomorrow." For that
reason, I'm going to trim the parts of your posting with which I
agree and have never disagreed.

You take too great a liberty with the name Eugene Hecht. Among the
things which won't be found in any of Dr. Hecht's texts is a minus sign
in front of number expressing an irradiance.

Sure wish you would read the book before making such statements.

On the contrary, here's equation (9.16) representing total destructive
interference.

Imin = I1 + I2 - 2*SQRT(I1*I2) = 0

The third term is indeed a minus sign in front of a number expressing
irradiance. However, total average irradiance cannot be less than zero.
And for the record, I have never said total average power could be less
than zero but, like Hecht, I treat destructive interference energy as a
negative term and constructive interference energy as a positive term.

Dr. Stephen Best, VE9SRB, did the same thing for his "Wave Mechanics
of Transmission Lines, Part 3:" QEX article, Nov/Dec 2001. He said:

"When the voltages V1 and V2 are exactly 180 deg out of phase, the
total power can be determined as follows:"

"PFtotal = P1 + P2 - 2*SQRT(P1)*SQRT(P2)"

so if you don't like negative power terms, you should confront both
Eugene Hecht and Dr. Best.

Nor will we find a
negative scalar quantity accompanied by the claim that the negative sign
indicates a change in direction, as you have done.

On the contrary, in equation 9.16 above, according to Hecht, the
interference term is negative indicating "total destructive
interference", his words, not mine. Here's Hecht's quote from _Optics_.

“The principle of
conservation of energy makes it clear that if there is constructive
interference at one point, the ‘extra’ energy at that location must have
come from somewhere else. There must therefore be destructive
interference somewhere else."

Sorry, but a negative interference term denotes destructive
interference. A positive interference term denotes constructive
interference. In a transmission line with only two directions,
if destructive interference occurs in one direction, then
constructive interference must occur in the only other direction
in order to satisfy the conservation of energy principle.

A wave cancellation event in a transmission line implies an equal
constructive interference event in the opposite direction. Anything
else violates the laws of physics.

Similarly, power and irradiance do not
physically propagate and they do not physically interact.

On the contrary, they do physically interact for coherent waves as can
be inferred by the interference equations. Please reference Chapter 9
in _Optics_, by Hecht. The mathematical interaction of power and
irradiance is a *result* of superposition of coherent EM waves. That's
where the interference equations involving irradiance come from.

JC Maxwell and others observed that it is
electric and magnetic fields which propagate, interact with matter, and
add algebraically and vectorially.

And contain power equal to ExH. EM waves cannot exist without energy.
If EM waves interact, their energy components interact. Destructive
and constructive interference cannot occur without energy components
which follow the laws of physics.

Of course the interference equation accurately expresses power and
irradiance.

That is some progress on your part so there's hope. What you need to
realize is that those interference equations define what happens to
the energy at a match point in a transmission line. Dr. Best kicked
this discussion off by his QEX article. He just didn't realize that
the equations he published were virtually identical to the classical
optical interference equations.
--
73, Cecil http://www.qsl.net/w5dxp


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On Tue, 12 Jul 2005 22:48:00 -0500, Cecil Moore
wrote:

Nor will we find a
negative scalar quantity accompanied by the claim that the negative sign
indicates a change in direction, as you have done.

On the contrary, in equation 9.16 above, according to Hecht, the
interference term is negative indicating "total destructive
interference", his words, not mine. Here's Hecht's quote from _Optics_.

[Hecht rolls his eyes] Jim's point is won, absolutely nothing quoted
here explicitly states a change in direction. That is, as Jim points
out, the math follows the physics, it does not create the physics.
There is a vast gulf between being descriptive and being proscriptive.

Cecil Moore wrote:

"PFtotal = P1 + P2 - 2*SQRT(P1)*SQRT(P2)"

so if you don't like negative power terms, you should confront both
Eugene Hecht and Dr. Best.

Don't be foolish. Obviously neither of the P terms can be negative if
PFtotal is supposed to represent a real number. I have no issues to
confront with either of the gentlemen. It is where you diverge from
Hecht (and Maxwell, and Born and Wolfe, and Jackson) that I take issue.

Nor will we find a negative scalar quantity accompanied by the claim
that the negative sign indicates a change in direction, as you have done.


On the contrary, in equation 9.16 above, according to Hecht, the
interference term is negative indicating "total destructive
interference", his words, not mine. Here's Hecht's quote from _Optics_.

One statement does not contradict the other, as the subject in each
sentence is entirely different. Both statements are obviously true.

Similarly, power and irradiance do not physically propagate and they
do not physically interact.


On the contrary, they do physically interact for coherent waves as can
be inferred by the interference equations. Please reference Chapter 9
in _Optics_, by Hecht. The mathematical interaction of power and
irradiance is a *result* of superposition of coherent EM waves. That's
where the interference equations involving irradiance come from.

As I explained they come from the fact that the fields interact, and
that power and intensity (or irradiance) go as the square of the field.
Lets say the square of F1 (field 1) is proportional with P1, and the
square of F2 is proportional with P2. And lets say the mathematical
description of the way the two fields interact is as follows:
Ftotal = (F1 - F2)*(F1 - F2). (Looks kinda like modulation, but I
digress.) We can then write that as
Ftotal = F1^2 + F2^2 - 2*F1*F2. By substitution then,
PFtotal = P1 + P2 - 2*SQRT(P1*P2). And that's where your favorite
equation comes from.

73, ac6xg

Jim Kelley wrote:
"You take too great a liberty with the name Eugene Hecht. Among the
things which won`t be found in any of Dr. Hecht`s texts is a minus sign
in front of a number expressing an impedance."

Don`t know why not. It happens at radio frequencies all the time. Look
through Kraus` chapter on "Mutual Impedance of Other Configurations".

B. Whitfield Griffith, Jr. gives a practical example starting on page
427 of "Radio-Electronic Transmission Fundamentals". On page 429 he
writes:
"We immediately notice one disturbing fact: the resistance component of
tower 1 has come out to be a negative number. This does not mean the
computations are wrong; it simply means that this tower because of the
particular phasing and current relationships of this array, will be
absorbing more power than it will be radiating itself."

This is an old story with broadcast antenna arrays.

Best regards, Richard Harrison, KB5WZI

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