Analyzing Stub Matching with Reflection Coefficients
 

Gene Fuller wrote:
Where are the equations that describe this "delta-t" stuff that you keep
bringing up?

delta-t is a mathematical term, Gene, related in the limit
to the differential dt. If you have to ask, I'm sure you
wouldn't understand the explanation.

Given the following experiment with two signal generators
equipped with circulators and load resistors - the generators
are phased-locked to ensure coherency:

100W 50W
50 ohm---50 ohm line---+---291.4 ohm line---291.4 ohm
SGCL1 SGCL2

Here's the setup for determining s11 and s21 with
SGCL2 turned off.

100W
50 ohm---50 ohm line---+---291.4 ohm line---291.4 ohm load
SGCL1

Here are the results:

a1----|
|----s21(a1)
s11(a1)----|

THE EM WAVE, s11(a1), EXISTS, IS MEASURABLE, AND IS ALIVE
AND WELL.

Here's the setup for determining s12 and s22 with
SGCL1 turned off.

50W
50 ohm load---50 ohm line---+---291.4 ohm line---291.4 ohm
SGCL2

Here are the results:

|----s22(a2)
s12(a2)----|
|----a2

THE EM WAVE, s12(a2), EXISTS, IS MEASURABLE, AND IS ALIVE
AND WELL.

Now power off both signal generators and power them up
simultaneously. Let t=0 be the time when s11(a1) and
s12(a2) first exist. Superpose s11(a1) and s12(a2). How
long does it take to achieve b1 = s11(a1) + s12(a2) = 0?
I estimate that time to be delta-t which becomes dt
in the differential equation limit.

The principle of superposition allows us to observe that
s11(a1) and a12(a2) actually existed before they were
canceled.
--
73, Cecil http://www.w5dxp.com

Analyzing Stub Matching with Reflection Coefficients
 

Roy Lewallen wrote:
I think you've put your finger right on Cecil's conceptual problem.

Your conceptual problem is obviously that you don't
understand the difference between a DC circuit problem
and an RF distributed network problem.

Can you get DC current to flow in opposite directions
around a closed loop? Can you get RF current to flow
in opposite directions around a closed loop.

Please get real.
--
73, Cecil http://www.w5dxp.com

Analyzing Stub Matching with Reflection Coefficients
 

Gene Fuller wrote:
Where are the equations that describe this "delta-t" stuff that you keep
bringing up?

I made an error in my last reply and have canceled that reply.

delta-t is a mathematical term, Gene, related in the limit
to the differential dt. If you have to ask, I'm sure you
wouldn't understand the explanation.

Given the following experiment with two signal generators
equipped with circulators and load resistors - the generators
are phased-locked to ensure coherency:

100W 100W
50 ohm---50 ohm line---+---291.4 ohm line---291.4 ohm
SGCL1 SGCL2

Here's the setup for determining s11 and s21 with
SGCL2 turned off.

100W
50 ohm---50 ohm line---+---291.4 ohm line---291.4 ohm load
SGCL1

Here are the results:

a1----|
|----s21(a1)
s11(a1)----|

THE EM WAVE, s11(a1), EXISTS, IS MEASURABLE, AND IS ALIVE
AND WELL.

Here's the setup for determining s12 and s22 with
SGCL1 turned off.

100W
50 ohm load---50 ohm line---+---291.4 ohm line---291.4 ohm
SGCL2

Here are the results:

|----s22(a2)
s12(a2)----|
|----a2

THE EM WAVE, s12(a2), EXISTS, IS MEASURABLE, AND IS ALIVE
AND WELL.

Now power off both signal generators and power them up
simultaneously. Let t=0 be the time when s11(a1) and
s12(a2) first exist. Superpose s11(a1) and s12(a2). How
long does it take to achieve b1 = s11(a1) + s12(a2) = 0?
I estimate that time to be delta-t which becomes dt
in the differential equation limit.

The principle of superposition allows us to observe that
s11(a1) and a12(a2) actually existed before they were
canceled.
--
73, Cecil http://www.w5dxp.com

Analyzing Stub Matching with Reflection Coefficients
 

Gene Fuller wrote:
I am familiar with all sorts of weird and wonderful things that happen
on surfaces and really close to interfaces and discontinuities.

Here's a little brain teaser for you, Gene.

Given the following experiment with two signal generators
equipped with circulators and load resistors - the generators
are phased-locked to ensure coherency. The two feedlines
are of equal electrical lengths.

100W | 25W
50 ohm---50 ohm line---+---291.4 ohm line---291.4 ohm
SGCL1 Pfor1=100W-- | --Prev2=25W SGCL2
--Prev1 | Pfor2--

What is the magnitude of Prev1? Pfor2?
--
73, Cecil http://www.w5dxp.com

Analyzing Stub Matching with Reflection Coefficients
 

Roy Lewallen wrote:
I think you've put your finger right on Cecil's conceptual problem. When
we solve the battery example by superposition, we get the right answer,
zero current. But now let's put a resistor between the two batteries and
repeat the solution. When we "turn off" the left hand battery, we have a
lot of power being dissipated in that resistor. Using Cecil's view, we
would assign this to be the power associated with the current flowing to
the left. Then we "turn on" the left hand battery and "turn off" the
right hand battery. Again we have a lot of power being dissipated. This
would be the power associated with the current flowing to the right. The
problem comes in having to somehow manipulate these powers to get zero,
which is what we actually see. The mistake, as you continually point
out, is attributing a power to each current -- or each wave -- in the
first place.

Roy, the intensity equation from the field of optics actually
handles that situation perfectly. The batteries are hooked
up in opposition to each other. Mathematically, their voltages
are 180 degrees out of phase.

Let P1 be the power dissipated in the resistor when battery
#1 is on. Let P2 be the power dissipated in the resistor
when battery #2 is on. P1 = P2

When both batteries are on, the total power in the resistor is:

Ptot = P1 + P2 + SQRT(P1*P2)cos(180) = 0

The intensity equation gets the same total power as you do. If
it's wrong then you are wrong. If it's right then you are wrong.
Seems that you are wrong either way. :-)
--
73, Cecil http://www.w5dxp.com

Analyzing Stub Matching with Reflection Coefficients
 

On Apr 19, 11:57 pm, Cecil Moore wrote:
Keith Dysart wrote:
Minor changes to the generator for B can give you a very hot one
or one that is exactly the same temperature as the one with
the circulator.

You've been hammering people with the source of your choice
for days but choose to abandon it as soon as someone points out
a logical problem with it.

Yes indeed. But a key realization is that the behaviour on the line
can not depend on arcane details about the construction of the
generator. So when such statements are made, it is instructive
to explore the situation with a different generator. If the outcome
is different, then there is something wrong with the theory. Note
that the standard explanations only require that the impedance
of the source be known. It is not important how this impedance
is achieved. With the hypothesis you are proposing, the internal
details significantly change the behaviour on the line and require
that the line (or the wave) know the internals so that it can
decide to enter or reflect despite the impedance being the same.

The conditions on the lines are indistinguishable and yet you
claim one is reflecting and one is not. How did the line know
whether it should reflect or not? Or is the wave that knows?

One wave encounters a 50 ohm resistor and is dissipated.
The other wave encounters an infinite impedance and is
100% re-reflected - all in accordance with the wave
reflection model.

But how does it know, since the output (source) impedances are
the same for both cases? How does the 50 Ohm output
impedance of generator B become infinite while the 50 Ohm
output impedance of generator A does not? This is the key
issue with your explanation. In fact, there is no difference and
they behave the same. The reverse wave encounters 50 Ohms
and is not reflected in either case. It is the only sensible answer.

Actually Pnet is zero because of basic circuit theory and the
universally accepted understanding that P = VI.

Circuit theory completely falls apart with distributed
network problems. The currents at two points in the
closed loop are often flowing in opposite directions.

This is a misunderstanding. Look at the distributed inductors and
capacitors and make the measurements at a point that makes
sense. This is, of course, how a directional wattmeter works.
It measures the voltage and current at one place on the line.
And this is how a real wattmeter would work if it was placed
at the same point.

In circuits, we measure the power at a particular place.
I certainly want my power company to do this. If the voltage
or current is 0 at a particular place in the circuit then no
energy is flowing at that place in the circuit. If you are disputing
this, I contend that you do not accept that P = VI.

Plenty of energy is flowing - two equal magnitudes
in opposite directions equals zero net energy.

Correction to your P = VI based on DC, not AC/RF:

Net P = V*I*cos(theta) = Pfor - Pref

I think you have made this mistakte before. In my expression P=VI,
V and I are simultaneous instantaneous values and P is the
instantaneous power. If you want average you need to integrate
and divide.

Your expression applies only to sine waves and V and I are peak
voltages. Interestingly, the expression P=V*I*cos(theta) for sine
waves is always derived by starting with Pinst=Vinst*Iinst,
substituting the appropriate functions for Vinst and Inst, integrating
and dividing.

Forget V and I being 0. cos(theta) is always 0 for a
standing wave. There is ZERO net power anywhere in a
standing wave. (It's a lot like my bank account.)

True. But there IS energy flowing at every point that is not a voltage
minimum or maximum. At these points there is NEVER any energy
flowing.

Inventive. But it doesn't fly. P = VI or it doesn't.

Forward waves and reflected waves are completely independent
of each other and do NOT interact. Their powers do NOT
superpose. There is nothing but joules moving at the
speed of light in a transmission line. There's no net
energy but your zero energy assertions are just illusions.
EM waves are incompatible with zero energy.

Well that does leave you with a bit of a conundrum since on
an open ended line, zero energy is flowing at every quarter
wavelength point back from the load. Pinst=Vinst*Iinst.
Vinst or Iinst is for all times zero. Power must be zero for
all times.

At such a point a real instantaneous wattmeter would always
indicate zero. At other points a real wattmeter would show
energy flowing first one way then the other, though the average
would be zero. And the peak magnitude of the energy flow
depends on where you insert the real wattmeter on the line.
An instructive question: Where does the peak energy flow
occur? It is zero at every quarter wavelength point so it
must be highest somewhere else.

....Keith

Analyzing Stub Matching with Reflection Coefficients
 

On Apr 20, 12:23 am, Cecil Moore wrote:
Roy Lewallen wrote:
I think you've put your finger right on Cecil's conceptual problem.

Your conceptual problem is obviously that you don't
understand the difference between a DC circuit problem
and an RF distributed network problem.

Can you get DC current to flow in opposite directions
around a closed loop? Can you get RF current to flow
in opposite directions around a closed loop.

However when you draw the distributed elements, you find
there are many loops (an infinite number?) and each loop
behaves exactly as it does for those in a circuit built of
lumped elements. There is no magic at all. The same
theories work, though the calculus is a bit more tedious.

So the RF is not flowing in opposite directions around the
loop. You just have not drawn the appropriate loops.

....Keith

Analyzing Stub Matching with Reflection Coefficients
 

On Apr 20, 1:02 am, Cecil Moore wrote:
When both batteries are on, the total power in the resistor is:

Ptot = P1 + P2 + SQRT(P1*P2)cos(180) = 0

The intensity equation gets the same total power as you do. If
it's wrong then you are wrong. If it's right then you are wrong.
Seems that you are wrong either way. :-)

The difficulties with your explanations are not so much that they
give you the wrong numerical answers. In fact, they give you the
right numerical answers so often that they convince you of their
truth.

No, the problem is that they lead you to do other thinks wrong.

For example, in all your examples you now provide a circulator
because you think this is the only way to control reflections,
when in fact, all that is necessary is a source impedance that is
equivalent to the line impedance and there will be no reflection.

In other cases you claim that problems are insolvable because
insufficient information is provided about the generator. In fact,
all you need to do is use the source impedance to compute
the reflection coefficient and the problem is solved.

You claim that superposition does not work in a generator.

These examples are suboptimal outcomes that derive
from your explanations. Better explanations do not lead to
these limitations.

Of course, you keep providing examples where the numbers
work and these reinforce your beliefs. You need to temporarily
let go and explore the alternatives. You will find that the
alternatives work better, not because they necessarily
give you different numerical answers but because they free
you to solve more problems and provide more optimal solutions.

....Keith

Analyzing Stub Matching with Reflection Coefficients
 

Keith Dysart wrote:
But a key realization is that the behaviour on the line
can not depend on arcane details about the construction of the
generator.

Agreed, but the behavior on the line also doesn't depend
upon where the forward and reverse traveling wave came from.
All that is important is their existence. The reverse
traveling wave might come from a separate source and the
line doesn't know the difference.

So when such statements are made, it is instructive
to explore the situation with a different generator. If the outcome
is different, then there is something wrong with the theory.

No, if the outcome is different, something different is
happening while abiding by the wave reflection model rules.
Why are you surprised that changes occur when something is
changed?

Note
that the standard explanations only require that the impedance
of the source be known.

If the "standard explanations" were correct, this would
have been put to bed a long time ago.

But how does it know, since the output (source) impedances are
the same for both cases?

Conclusion: The output source impedance is NOT the impedance
encountered by the reflected wave. That's essentially what
Walter Maxwell's paper says.

I think you have made this mistakte before. In my expression P=VI,
V and I are simultaneous instantaneous values and P is the
instantaneous power. If you want average you need to integrate
and divide.

No, if you want to average sinusoidals , use RMS values, which
is the result of said integrating and dividing. Somebody else
already did all the work.

Your expression applies only to sine waves and V and I are peak
voltages.

I is a peak voltage??????? You are more confused than
I ever realized. :-)

Interestingly, the expression P=V*I*cos(theta) for sine
waves is always derived by starting with Pinst=Vinst*Iinst,

Interestingly, you forgot to say you were talking about
instantaneous power until now.

True. But there IS energy flowing at every point that is not a voltage
minimum or maximum. At these points there is NEVER any energy
flowing.

Every half cycle in every EM traveling wave, the instantaneous
power is zero. Would you therefore argue that traveling waves
cannot transfer any energy? Please get real.

Given two parallel water pipes carrying 100 gallons/minute
in opposite directions. I can see you arguing that there
is no NET flow of water. But I cannot see you arguing that
there is no flow of water at all when the pressure of just
one stream can knock you off your feet.

Well that does leave you with a bit of a conundrum since on
an open ended line, zero energy is flowing at every quarter
wavelength point back from the load.

The conundrum is all yours. Every half cycle, every EM
traveling wave is at a zero instantaneous power point.
If EM traveling waves can transfer energy while the
instantaneous power is zero every half cycle, superposing
two of them doesn't present a conundrum at all.

At such a point a real instantaneous wattmeter would always
indicate zero.

A real instantaneous wattmeter would always indicate zero
at the zero crossing of every EM traveling wave in the
universe so exactly how does the light from Alpha Centauri
ever make it to Earth when the instantaneous power is zero
every half cycle? Please get real.
--
73, Cecil http://www.w5dxp.com

Analyzing Stub Matching with Reflection Coefficients
 

Keith Dysart wrote:
For example, in all your examples you now provide a circulator
because you think this is the only way to control reflections,
when in fact, all that is necessary is a source impedance that is
equivalent to the line impedance and there will be no reflection.

Your above statement has been proved wrong in any number
of bench experiments.

In other cases you claim that problems are insolvable because
insufficient information is provided about the generator. In fact,
all you need to do is use the source impedance to compute
the reflection coefficient and the problem is solved.

Again, proved wrong by any number of bench experiments.

You claim that superposition does not work in a generator.

Not that it doesn't work - just that it is impossible
for a reflected wave to compete with an active dynamic
source. The source essentially ignores the reflected
wave like a fire hose ignores you trying to spit up
the hose against the flow.

You will find that the alternatives work better, ...

Sorry Keith, I stopped listening to you when you asserted
that I is voltage.
--
73, Cecil http://www.w5dxp.com