Re-Normalizing the Smith Chart (Changing the SWR into the same load)
 

From Pozar's Microwave Engineering (Pg. 606):

Reflection Coefficient looking into load = (Zl-Zo)/(Zl+Zo)

Where Zl is a purely real load impedance, and Zo is the
purely real characteristic impedance reference.

When you change Zo, you change the normalized center of the
Smith Chart, and therefore the Reflection Coefficient and SWR, looking
into the same load.



Check it out...


Slick

Among the posters on this topic, I'd be willing to bet a substantial sum
that at least Tom, Ian, Cecil, Bill, and I can recite that formula from
memory. In addition, there are a number of other regular and occasional
newsgroup posters in this category who've been wise enough to not having
posted on this thread, and some who have posted whom I don't know well
enough to put my money on. I'd be willing to further bet that Ian,
Cecil, Bill, and I could have done so at any time for at least the last
20 years. I omit Tom from this second list only because I haven't yet
met him in person and otherwise haven't gotten any hints of his age --
but I'll take a gamble and spot him 10 years at least. Furthermore, we
all know how to use it, and have done so countless times in the process
of designing systems that work.

I'm glad you've discovered this equation. Learn what it means and how to
use it, and you've taken a good first step toward understanding
transmission line phenomena.

Roy Lewallen, W7EL

Dr. Slick wrote:
From Pozar's Microwave Engineering (Pg. 606):

Reflection Coefficient looking into load = (Zl-Zo)/(Zl+Zo)

Where Zl is a purely real load impedance, and Zo is the
purely real characteristic impedance reference.

When you change Zo, you change the normalized center of the
Smith Chart, and therefore the Reflection Coefficient and SWR, looking
into the same load.



Check it out...


Slick

Dr. Slick wrote:

From Pozar's Microwave Engineering (Pg. 606):

Reflection Coefficient looking into load = (Zl-Zo)/(Zl+Zo)

Where Zl is a purely real load impedance, and Zo is the
purely real characteristic impedance reference.

When you change Zo, you change the normalized center of the
Smith Chart, and therefore the Reflection Coefficient and SWR, looking
into the same load.

Correct! What Roy is objecting to is the wording of an assertion of
yours that seemed to imply that changing the reference Z0 on a piece
of Smith Chart paper magically changes the physical Z0 of the transmission
line to a different value. As you have advised me, be careful of what you
say and how you say it.

I referred to "free space" the other day. The image I had in my own
mind was the free space halfway between here and Alpha Centauri. Others
had the image of "free space" as 0.001 WL away from a radiating antenna.
--
73, Cecil http://www.qsl.net/w5dxp



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"Dr. Slick" wrote
From Pozar's Microwave Engineering (Pg. 606):

Reflection Coefficient looking into load = (Zl-Zo)/(Zl+Zo)

Where Zl is a purely real load impedance, and Zo is the
purely real characteristic impedance reference.

================================

Either Pozar (who I've never heard of) is not quite correct or Dr Slick has
misquoted him or taken him out of context.

In fact, the equation is also true for complex values of Zl and Zo. The
angle of RC can lie in any of the 4 quadrants. Furthermore, the magnitude of
RC can exceed unity.

I offer no references in support of this statement. It is issued here
entirely on my own responsibility.
----
Reg, G4FGQ

There's nothing wrong with the formula or the context. It follows from a
straightforward derivation that begins with the ratio of reflected to
forward waves at the load, and results in satisfying the boundary
conditions and Kerchoff's voltage and current laws at the load. It holds
for any complex values of Zl and Z0. The resulting reflection
coefficient is of course complex, but it's often confused with its
magnitude or with the time domain reflection coefficient. Increasing
this confusion is that there's no standard notation for these terms, so
the complex value in one text might be denoted by the same character as
the magnitude in another text.

There's no problem with the reflection coefficient having any angle in
any of the four quadrants. However, I've frankly had trouble getting
around the notion of the magnitude of the reflection coefficient being
greater than one with a passive load. I know it sometimes happens with
active loads, and there are even Smith chart techniques to deal with it.
You'll find discussions of it in texts on microwave circuit design.

It looks like it's possible to get a reflection coefficient with
magnitude 1 any time Rl*R0 -X*X0. Reg recently posted some values of
Z0 for common coaxial cables that show the angle of Z0 approaching -45
degrees at low frequency. So it wouldn't be hard to envision a cable
with Z0 = 100 - j100 or thereabouts at some very low frequency. If we
were to terminate it with a pure inductor with 100 ohms reactance (Zl =
0 + j100), it looks like the reflection coefficient would be -1 + j2,
which has a magnitude of the square root of three, or about 1.73. What
does this mean? It means that the reflected wave has a greater magnitude
than the incident wave. I'm not sure there's anything wrong with this --
it's sort of like a resonant effect. It would have to be checked to make
sure that the law of conservation of energy isn't violated, and that
Kirchoff's laws are satisfied, but I'd be surprised if there were any
violations. The calculated SWR is negative, but that's pretty
meaningless considering we have a line with a huge amount of attenuation
per wavelength (in order to have such a highly reactive Z0). With that
kind of attenuation there's no danger of having an oscillator with no
power source.

I'd really like to hear from some of the folks who deal more frequently
with reflection coefficient than I do, to see if I'm on the right track,
or if there is some consideration that requires modification of the
equation for very lossy lines. I've got quite a few references that deal
with reflection coefficient. They all give the same formula without
qualifications, but none mentions the possibility of the magnitude
becoming greater than one.

Reg, you've got more experience with very lossy lines (in terms of loss
per wavelength, which is what counts here) than anyone else on this
group. What happens at the load if you terminate a 100 - j100 ohm Z0
line with 0 + j100 ohms?

Roy Lewallen, W7EL

Reg Edwards wrote:
"Dr. Slick" wrote

From Pozar's Microwave Engineering (Pg. 606):

Reflection Coefficient looking into load = (Zl-Zo)/(Zl+Zo)

Where Zl is a purely real load impedance, and Zo is the
purely real characteristic impedance reference.


================================

Either Pozar (who I've never heard of) is not quite correct or Dr Slick has
misquoted him or taken him out of context.

In fact, the equation is also true for complex values of Zl and Zo. The
angle of RC can lie in any of the 4 quadrants. Furthermore, the magnitude of
RC can exceed unity.

I offer no references in support of this statement. It is issued here
entirely on my own responsibility.
----
Reg, G4FGQ

Roy Lewallen wrote in message ...

There's no problem with the reflection coefficient having any angle in
any of the four quadrants. However, I've frankly had trouble getting
around the notion of the magnitude of the reflection coefficient being
greater than one with a passive load. I know it sometimes happens with
active loads, and there are even Smith chart techniques to deal with it.
You'll find discussions of it in texts on microwave circuit design.


[s11]**2 + [s21]**2 = 1

For a lossless passive two port network, where the brackets
indicate magnitude only.

If you find a passive network that reflects more voltage than it
receives, let us all know about your free energy device.




I'd really like to hear from some of the folks who deal more frequently
with reflection coefficient than I do, to see if I'm on the right track,
or if there is some consideration that requires modification of the
equation for very lossy lines. I've got quite a few references that deal
with reflection coefficient. They all give the same formula without
qualifications, but none mentions the possibility of the magnitude
becoming greater than one.



Reflection coefficients greater than unity, which go outside the
Smith, only happen with active devices, as you have mentioned above.
Stability circles are a related topic, as their centers are often
based outside the unity RC circle.


Slick

Roy, Chipman on page 138 of "Theory and Problems of Transmission Lines"
makes the statement

" The conclusion is somewhat surprising, though inescapable, that a
transmission line can be terminated with a reflection coefficient whose
magnitude is as great as 2.41 without there being any implication that the
power level of the reflected wave is greater than that of the incident wave.
Such a reflection coefficient can exist only on a line whose attention per
wavelength is high, so that even if the reflected wave is in some sense
large at the point of reflection, it remains so for only a small fraction of
a wavelength along the line away from that point . . . The large reflection
coefficients are obtained only when the reactance of the terminal load
impedance is of opposite sign to the reactance component of the
characteristic impedance."

Chipman makes these remarks after his derivation of the operation of lines
with complex characteristic impedance.


--
73/72, George
Amateur Radio W5YR - the Yellow Rose of Texas
Fairview, TX 30 mi NE of Dallas in Collin county EM13QE
"In the 57th year and it just keeps getting better!"






"Roy Lewallen" wrote in message
...
There's nothing wrong with the formula or the context. It follows from a
straightforward derivation that begins with the ratio of reflected to
forward waves at the load, and results in satisfying the boundary
conditions and Kerchoff's voltage and current laws at the load. It holds
for any complex values of Zl and Z0. The resulting reflection
coefficient is of course complex, but it's often confused with its
magnitude or with the time domain reflection coefficient. Increasing
this confusion is that there's no standard notation for these terms, so
the complex value in one text might be denoted by the same character as
the magnitude in another text.

There's no problem with the reflection coefficient having any angle in
any of the four quadrants. However, I've frankly had trouble getting
around the notion of the magnitude of the reflection coefficient being
greater than one with a passive load. I know it sometimes happens with
active loads, and there are even Smith chart techniques to deal with it.
You'll find discussions of it in texts on microwave circuit design.

It looks like it's possible to get a reflection coefficient with
magnitude 1 any time Rl*R0 -X*X0. Reg recently posted some values of
Z0 for common coaxial cables that show the angle of Z0 approaching -45
degrees at low frequency. So it wouldn't be hard to envision a cable
with Z0 = 100 - j100 or thereabouts at some very low frequency. If we
were to terminate it with a pure inductor with 100 ohms reactance (Zl =
0 + j100), it looks like the reflection coefficient would be -1 + j2,
which has a magnitude of the square root of three, or about 1.73. What
does this mean? It means that the reflected wave has a greater magnitude
than the incident wave. I'm not sure there's anything wrong with this --
it's sort of like a resonant effect. It would have to be checked to make
sure that the law of conservation of energy isn't violated, and that
Kirchoff's laws are satisfied, but I'd be surprised if there were any
violations. The calculated SWR is negative, but that's pretty
meaningless considering we have a line with a huge amount of attenuation
per wavelength (in order to have such a highly reactive Z0). With that
kind of attenuation there's no danger of having an oscillator with no
power source.

I'd really like to hear from some of the folks who deal more frequently
with reflection coefficient than I do, to see if I'm on the right track,
or if there is some consideration that requires modification of the
equation for very lossy lines. I've got quite a few references that deal
with reflection coefficient. They all give the same formula without
qualifications, but none mentions the possibility of the magnitude
becoming greater than one.

Reg, you've got more experience with very lossy lines (in terms of loss
per wavelength, which is what counts here) than anyone else on this
group. What happens at the load if you terminate a 100 - j100 ohm Z0
line with 0 + j100 ohms?

Roy Lewallen, W7EL

Reg Edwards wrote:
"Dr. Slick" wrote

From Pozar's Microwave Engineering (Pg. 606):

Reflection Coefficient looking into load = (Zl-Zo)/(Zl+Zo)

Where Zl is a purely real load impedance, and Zo is the
purely real characteristic impedance reference.


================================

Either Pozar (who I've never heard of) is not quite correct or Dr Slick
has
misquoted him or taken him out of context.

In fact, the equation is also true for complex values of Zl and Zo. The
angle of RC can lie in any of the 4 quadrants. Furthermore, the
magnitude of
RC can exceed unity.

I offer no references in support of this statement. It is issued here
entirely on my own responsibility.
----
Reg, G4FGQ

Thanks very much for that additional information about the consequences
of the magnitude of the reflection coefficient exceeding one. I couldn't
find it in any of the electromagnetics or transmission line books on my
shelf, which at last count include about 13 texts. Chipman, alas, isn't
among them. It confirms what I suspected, and provides further evidence
that the posted equation is universally correct.

While I'm mentioning books, I picked up a couple at Powell's Technical
Bookstore yesterday evening that look like real winners. They're
_Engineering Electromagnetics_ by Nathan Ida (2000), and
_Electromagnetic Fields, Energy, and Waves_ by Leonard M. Magid (1972).

The thing that attracted me to Ida was that he explains things in very
clear terms, then follows each section with a number of examples showing
how the principles are applied to real problems. And answers to all the
exercises (separate from the examples) are at the back of the book. This
is a pretty recent book and fairly expensive. I was lucky to have found
a used copy at a reduced price.

Magid has the most rigorous derivation of power and energy flow on
transmission lines I've seen, as well as other extensive transmission
line information. One conclusion that pricked my ears was that on a line
with a pure standing wave (e.g., a lossless line terminated with an open
or short circuit), ". . . power (and therefore, energy) is completely
trapped within each [lambda]/4 section of this lossless line, never able
to cross the zero-power points and thus constrained forever to rattle to
and fro within each quarter-wave section of this line." I had reached
this same conclusion some time ago, but realized I hadn't properly
evaluated the constant term when integrating power to find the energy.
But I didn't want to get into the endless shouting match going on in the
newsgroup, and dropped it before going back and fixing my derivation.
Hopefully some of the participants in power and energy discussions will
read Magid's analysis before resuming. I found this book used at a very
modest price.

Roy Lewallen, W7EL

George, W5YR wrote:
Roy, Chipman on page 138 of "Theory and Problems of Transmission Lines"
makes the statement

" The conclusion is somewhat surprising, though inescapable, that a
transmission line can be terminated with a reflection coefficient whose
magnitude is as great as 2.41 without there being any implication that the
power level of the reflected wave is greater than that of the incident wave.
Such a reflection coefficient can exist only on a line whose attention per
wavelength is high, so that even if the reflected wave is in some sense
large at the point of reflection, it remains so for only a small fraction of
a wavelength along the line away from that point . . . The large reflection
coefficients are obtained only when the reactance of the terminal load
impedance is of opposite sign to the reactance component of the
characteristic impedance."

Chipman makes these remarks after his derivation of the operation of lines
with complex characteristic impedance.


--
73/72, George
Amateur Radio W5YR - the Yellow Rose of Texas
Fairview, TX 30 mi NE of Dallas in Collin county EM13QE
"In the 57th year and it just keeps getting better!"

W5DXP wrote:

Roy Lewallen wrote:
Magid has the most rigorous derivation of power and energy flow on
transmission lines I've seen, as well as other extensive transmission
line information. One conclusion that pricked my ears was that on a
line with a pure standing wave (e.g., a lossless line terminated with
an open or short circuit), ". . . power (and therefore, energy) is
completely trapped within each [lambda]/4 section of this lossless
line, never able to cross the zero-power points and thus constrained
forever to rattle to and fro within each quarter-wave section of this
line."

He's obviously talking about net energy. There is no impedance
discontinuity
in a continuous piece of transmission line so there is nothing to cause
reflections at the zero-power points. Net energy doesn't cross the zero-
power points but equal forward energy and reflected energy must cross the
zero-power points. That is easy to prove by observing ghosting on a TV set
being fed by 1000 feet of ladder-line. If energy is completely trapped
within each 1/4WL section, ghosting would be impossible.

I forgot to add. At the "zero-power points", either voltage or current is
zero. All that means is that all the energy is contained in the opposite
field. If the voltage is zero, the H-field is at a maximum. If the current
is zero, the E-field is at a maximum. The energy still exists, just in one
field or the other. V*I*cos(theta) is NET energy. The power in the forward
wave is Vfwd*Ifwd and the power in the reflected wave is Vref*Iref. Those
values are constant all up and down the line for a lossless feedline.
--
73, Cecil http://www.qsl.net/w5dxp



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W5DXP wrote:

I forgot to add. At the "zero-power points", either voltage or current is
zero. All that means is that all the energy is contained in the opposite
field.

Not quite all. It also means that there is NO power since P = V x I.

There can be lots of energy present but none of it is flowing past
the zero voltage or zero current point; hence no power.

To believe that energy is flowing across a zero voltage or zero current
point requires the rejection of the view that instantaneous power is
equal to instantaneous voltage multiplied by instantaneous current.
Rejection of
P = V x I would have wide impacts on our understanding of electrical
power and energy flows. (My light bulb is not drawing any current but
there could still be energy flowing in it!?)
Think carefully before going there.

....Keith

On Sun, 17 Aug 2003 18:16:42 -0700, Roy Lewallen
wrote:

Interesting stuff snipped

Magid has the most rigorous derivation of power and energy flow on
transmission lines I've seen, as well as other extensive transmission
line information. One conclusion that pricked my ears was that on a line
with a pure standing wave (e.g., a lossless line terminated with an open
or short circuit), ". . . power (and therefore, energy) is completely
trapped within each [lambda]/4 section of this lossless line, never able
to cross the zero-power points and thus constrained forever to rattle to
and fro within each quarter-wave section of this line." I had reached
this same conclusion some time ago, but realized I hadn't properly
evaluated the constant term when integrating power to find the energy.
But I didn't want to get into the endless shouting match going on in the
newsgroup, and dropped it before going back and fixing my derivation.
Hopefully some of the participants in power and energy discussions will
read Magid's analysis before resuming. I found this book used at a very
modest price.


Roy:

Interesting point and I don't recall reading or hearing it elsewhere.

The following is dashed off without fully thinking it through, so no
warranty on its accuracy.

If you think of a sound wave (longitudinal transmission, of course) in
a lossless acoustic transmission line terminated with a short, the
individual air molecules within each 1/4 wave section are likewise
trapped since at the 1/4 wave points there is zero sound pressure.
This may be a useful analogy for the electromagnetic transverse
propagating T-line.

Jack K8ZOA

W5DXP wrote:

Roy Lewallen wrote:
Magid has the most rigorous derivation of power and energy flow on
transmission lines I've seen, as well as other extensive transmission
line information. One conclusion that pricked my ears was that on a line
with a pure standing wave (e.g., a lossless line terminated with an open
or short circuit), ". . . power (and therefore, energy) is completely
trapped within each [lambda]/4 section of this lossless line, never able
to cross the zero-power points and thus constrained forever to rattle to
and fro within each quarter-wave section of this line."

He's obviously talking about net energy.

Not obvious at all.

There is no impedance discontinuity
in a continuous piece of transmission line so there is nothing to cause
reflections at the zero-power points.

A mechanical analogue may help. Consider that executive toy: five steel
balls attached by string to a frame. Pull back the ball on one side and
let it go, when it strikes the second ball, the fifth ball swings up.
Reduce this toy to just two balls.

Pull one back, when it strikes the second, it swings out, then swings
back striking the first which then swings out. It is clear that there
is energy transfer between the balls.

Redo the experiment by pulling both balls back and letting them go at
the same time. After colliding, both balls bounce back. Were you to
place a very thin sheet of steel between the balls at the collision
point it would not move. Since Work is Force x Distance and the Distance
is zero there is no Work being done on the sheet so no energy can be
crossing it.

In the shorted or open transmission line (from Magid, above), the
analogue is two clumps of charge rushing towards each other and meeting
at a voltage maximum (current zero). No charge crosses this point
(obvious because the current is zero), but the charge coming from each
direction builds to a voltage maximum and bounces away again (since
like charge repels).

Net energy doesn't cross the zero-
power points but equal forward energy and reflected energy must cross the
zero-power points.

Not possible since NO energy crosses the zero voltage and zero current
points (unless you want to reject Pinst = Vist x Iinst).

That is easy to prove by observing ghosting on a TV set
being fed by 1000 feet of ladder-line. If energy is completely trapped
within each 1/4WL section, ghosting would be impossible.

This ghosting argument appears quite powerful along with the somewhat
similar observation that information can be sent in both directions
simultaneously on a phone line.

The difficulty I encountered, while trying to understand, is that
simultaneously holding the views that:
1) ghosting is caused by reflected energy flowing back along the line;
and
2) Pinst = Vinst x Iinst
required too much double think.

Item 2) seemed to be too universally applicable to let go, so a better
understanding of 1) was required.

Although it seems unrelated, it is worthwhile to consider how to send
information along a line without sending energy in the same direction.

For simplicity, consider an ideal transmission line of useful length
with
a matched Thevenin DC source on the left and a matched load connected
through a switch on the right. Initially, the switch is open and the
line is charged to V: the voltage of the voltage source. Observe that
there is no current flowing anywhere, hence no energy flowing and
therefore no power. This is entirely consistent with 2), above.

Close the switch. Charge starts flowing from the line through the load.
A negative voltage step begins to propagate backwards along
the line at the velocity of the line. When this voltage step reaches
the source, the line has entered a new energy state with constant
voltage V/2 across its length, a current of V/2/R flowing and
energy is flowing from the left to the right at V**2/R/4 Watts. This
power is dissipated in the load at the right.

Opening the switch will cause a positive voltage step to propagate to
the left and when it reaches the source, the line will have been
restored to its initial conditions with no energy flowing.

A detector at the source (monitoring voltage or current) can determine
if the switch is open or closed (after the voltage steps have finished
propagating), thus information can be transferred from right to left
while energy only flows from left to right.

This information is transferred by doing something that changes the
energy state on the line and waiting for the new energy state to
propagate along the line. It is important to note that the
propagation of the change in energy state is not the same as the
propagation of energy. They can, and often do, occur in different
directions at the same time.

And to return to the original question, this is the cause of ghosting;
it is the propagation of the change in energy state on the line that
results in ghosting. If the source was matched, then the line settles
to its final state in one round trip and no ghosting is observed.
When the source is not matched, it takes several round trips for the
line to settle and ghosting is what you see.

By the way, this is not quite Magid's situation since he was saying
the energy completely bounced back and forth only when the line was
open or shorted. In other situations with standing waves, some of the
energy is bouncing within the 1/4 wave sections, while other energy
is flowing forward.

....Keith

wrote:
W5DXP wrote:
I forgot to add. At the "zero-power points", either voltage or current is
zero. All that means is that all the energy is contained in the opposite
field.

Not quite all. It also means that there is NO power since P = V x I.

It means there is no NET power transfer. There are power flow vectors
in both directions that are equal in magnitude. Reference: _Fields_
and_Waves_in_Communications_Electronics_ by Ramo, Whinnery, & Van Duzer,
section 6.10, page 350, where they describe Pz-, the reflected wave
Poynting vector and Pz+, the forward wave Poynting vector.

There can be lots of energy present but none of it is flowing past
the zero voltage or zero current point; hence no power.

There is a forward power flow vector and a reflected power flow vector.
There is no net power flowing past any point. There are, however, equal
magnitude component constant power flow vectors flowing in both directions.

To believe that energy is flowing across a zero voltage or zero current
point requires the rejection of the view that instantaneous power is
equal to instantaneous voltage multiplied by instantaneous current.

No, it doesn't. It only requires acceptance of Ramo, Whinnery, & Van Duzer.

However, to reject energy flow across a zero voltage or zero current
point requires a confusion of cause and effect. Energy flow in both
directions is the *CAUSE* of the standing waves. You simply cannot
turn around and say that standing waves eliminate their own cause
but continue to exist anyway.

Rejection of
P = V x I would have wide impacts on our understanding of electrical
power and energy flows.

Nobody is rejecting it. If the lossless stub is one second long, it takes
two seconds of *POWER* to bring it to steady-state. If the stub contains
no moving energy, where did all those joules go? They cannot disappear or
stand still. If they are moving, power exists.
--
73, Cecil http://www.qsl.net/w5dxp



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Magid has the most rigorous derivation of power and energy flow on
transmission lines,
==========================
The following short question is adressed to all contributors to this
newsgroup who feel impelled to bolster their lack of self-confidence by
dragging in the chapter and verse of their favourite worshipped authors and
Gurus, most of whom nobody has ever heard of and highly unlikely ever to get
their hands on.

How do you know that?

W5DXP wrote:

wrote:
W5DXP wrote:
I forgot to add. At the "zero-power points", either voltage or current is
zero. All that means is that all the energy is contained in the opposite
field.

Not quite all. It also means that there is NO power since P = V x I.

It means there is no NET power transfer.

Do not be afraid to admit that you have changed the definition of P = V
x I
and therefore do not accept the standard definition.

When I learned Pinst = Vinst x Iinst there were no caveats about how
Pinst meant Pnet. Instantaneous energy is flowing or it is not. When
Pinst
is 0 for all time, then there is no energy flowing.

To satisfy your theory (and minimize double think), you have had to
change
this to Pnet is zero to allow these cancelling powers to flow. So be it.

There are power flow vectors
in both directions that are equal in magnitude. Reference: _Fields_
and_Waves_in_Communications_Electronics_ by Ramo, Whinnery, & Van Duzer,
section 6.10, page 350, where they describe Pz-, the reflected wave
Poynting vector and Pz+, the forward wave Poynting vector.

There can be lots of energy present but none of it is flowing past
the zero voltage or zero current point; hence no power.

There is a forward power flow vector and a reflected power flow vector.
There is no net power flowing past any point.

True, sort of. At the quarter wave points where voltage or current are
always
zero, there is no energy flowing. Period.

At other points, energy flows in one direction for a quarter cycle and
then in the other direction for the next quarter cycle, producing a net
of zero. A true instantenouse power meter (one which measures V and I
and
displays V x I) will easily demonstrate this. As a thought experiment,
move such a true power meter along a shorted or open line and think if
its indications in the time domain, then do the averages. It will be
quite instructive. Repeat for a line terminated in other than its
characteristic impedance.

By the way, since energy flows forward for a quarter cycle and backwards
for the next, the maximum distance travelled by this energy is one
quarter
wavelength on the line. It is not flowing all the way to the end of the
line and then back. There is not enough time for this to happen (on a
multi-wavelength line) since it changes direction every quarter cycle.

There are, however, equal
magnitude component constant power flow vectors flowing in both directions.

To believe that energy is flowing across a zero voltage or zero current
point requires the rejection of the view that instantaneous power is
equal to instantaneous voltage multiplied by instantaneous current.

No, it doesn't. It only requires acceptance of Ramo, Whinnery, & Van Duzer.

However, to reject energy flow across a zero voltage or zero current
point requires a confusion of cause and effect. Energy flow in both
directions is the *CAUSE* of the standing waves. You simply cannot
turn around and say that standing waves eliminate their own cause
but continue to exist anyway.

Not quite. Standing voltage and current waves (which are not waves in
the normal sense) can be observed on the line. They can be measured with
real voltage and current instruments; as can real energy flows with
a real (V x I) power meter (but not a 'Bird watt' meter which
is doing something quite different). It happens that if you assume the
existence of forward and reverse voltage and current waves, mathematical
functions can be derived that will produce the same distribution of
voltage and current as observed on the line. This is extraordinarily
convenient some analysis but does not mean that these assumed waves are
real.

A mechanical analogue would be to look at a guy wire on a pole. You can
analyze the forces as two vectors at 90 degrees (or any other angle of
convenience!), but never make the mistake of assuming that there are
actually two guy wires present. Just because it is mathematically
convenient to assume the existence of two vectors does not mean they
exist.

Rejection of
P = V x I would have wide impacts on our understanding of electrical
power and energy flows.

Nobody is rejecting it. If the lossless stub is one second long, it takes
two seconds of *POWER* to bring it to steady-state. If the stub contains
no moving energy, where did all those joules go?

This energy is indeed stored in the stub. None of it moves across zero
voltage
or current boundaries. Between these boundaries, it is indeed moving as
it changes from being stored in the capacitance and inductance of the
line.
This time variation of the location of the stored energy produces the
observed
voltages and currents on the line.

They cannot disappear

Absolutely. They do not disappear.

or stand still.

With AC excitation they do not stand still, but when similar analysis is
done
for a line excited with a DC source, the energy does indeed stand still.
An
open line stores the energy in the capacitance and a shorted line stores
it
in the inductance.

If they are moving, power exists.

Yes, but it never moves more than a quarter wavelength.

....Keith

W5DXP wrote in message ...

The transmission coefficient can certainly be greater than unity,
being (2*Z2)/(Z1+Z2).

From Pozar's Microwave Engineering (Pg. 606):

Reflection Coefficient looking into load = (Zl-Zo)/(Zl+Zo)

Where Zl is the load impedance, and Zo is the characteristic
impedance reference.

The reflection coefficient cannot be greater than unit.

How can you have an S21 that is greater than unity with a passive
network?


Slick

Keith wrote:
"At the quarter wave points where voltage and current are always zero,
there is no energy flowing. Period."

Yes there is energy flowing, and it is flowing in both directions if it
is flowing in one direction. Otherwise there would be no standing wave.

The waves flow right through each other so long as the impedance is
uniform. Energy flow is unabated at those points of illusory zeros.
Those zeros are not zeros at all in the forward and reverse waves. They
only appear as zeros when your sensor can`t separate the forward from
the reverse. Get a proper directional sensor and the forward wave is
sensed as full amplitude as is the reverse wave.

Best regards, Richard Harrison, KB5WZI

"Reg Edwards" wrote in message ...
Magid has the most rigorous derivation of power and energy flow on
transmission lines,
==========================
The following short question is adressed to all contributors to this
newsgroup who feel impelled to bolster their lack of self-confidence by
dragging in the chapter and verse of their favourite worshipped authors and
Gurus, most of whom nobody has ever heard of and highly unlikely ever to get
their hands on.

How do you know that?

Gee, Reg, since you took that out of context, it seems a bit unfair.
Roy wrote it, and after the comma was, "I've seen." I don't know it,
but I'm willing to take Roy at his word on the matter.

Cheers,
Tom

wrote:
In the shorted or open transmission line (from Magid, above), the
analogue is two clumps of charge rushing towards each other and meeting
at a voltage maximum (current zero). No charge crosses this point
(obvious because the current is zero), but the charge coming from each
direction builds to a voltage maximum and bounces away again (since
like charge repels).

Unfortunately, the analogy is not a good one. In a transmission line,
there must exist a discontinuity to cause a reversal of momentum of
the waves. No such discontinuity exists so there is nothing to reverse
the momentum of the forward and reflected waves.

Not possible since NO energy crosses the zero voltage and zero current
points (unless you want to reject Pinst = Vist x Iinst).

Ramo, Whinnery, and Van Duzer disagree. They say that the power
reflection coefficient is equal to the reflected Poynting vector
divided by the forward Poynting vector which in this case would
be unity.

The difficulty I encountered, while trying to understand, is that
simultaneously holding the views that:
1) ghosting is caused by reflected energy flowing back along the line;
and 2) Pinst = Vinst x Iinst required too much double think.

Life is tough all over. :-) What I am saying doesn't require double
think. :-)

Although it seems unrelated, it is worthwhile to consider how to send
information along a line without sending energy in the same direction.

Information transfer doesn't require energy? Methinks you are confusing
the carriers of the wave with the waves themselves. Zero carriers may
indeed cross the zero power point, but that does not prevent energy
from crossing the boundary. It just means that the energy crossing the
boundary must be equal in both directions.

One point. The power is not Vavg * Iavg. The power is Vavg*Iave*cos(theta)
All up and down a shorted lossless transmission line, theta is equal to
90 degrees. Around a voltage zero point, on one side the voltage is equal
to 0.00000001 volts and is 90 degrees out of phase with the current, i.e.
zero power. On the other side of the voltage zero point, the voltage is
equal to -0.00000001 volts and is 90 degrees out of phase with the current.
There are an infinite number of points where the voltage is 90 degrees out
of phase with the current. That's why the average power is always zero, not
because the average voltage is zero, but because the voltage is *ALWAYS* 90
degrees out of phase with the current in a lossless shorted feedline.

When the source is not matched, it takes several round trips for the
line to settle and ghosting is what you see.

Put an amplifier between the TV set and the feedline. Let the amplifier
have as high an impedance as possible (insulated gate FET). That will
cause reflections. Those reflections will still be there during steady-
state and observing the ghosting will prove that, during steady-state,
the ghosts have made one or more round trips back to the source where
they are re-reflected (assuming a mismatched source or a Z0-match
provided by a network).
--
73, Cecil http://www.qsl.net/w5dxp
"One thing I have learned in a long life: that all our science, measured
against reality, is primitive and childlike ..." Albert Einstein



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Reg Edwards wrote:
The following short question is adressed to all contributors to this
newsgroup who feel impelled to bolster their lack of self-confidence by
dragging in the chapter and verse of their favourite worshipped authors and
Gurus, most of whom nobody has ever heard of and highly unlikely ever to get
their hands on.

You never heard of Ramo, the 'R' in TRW? :-)
--
73, Cecil http://www.qsl.net/w5dxp



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It can be convenient to define a reference impedance for S-paramters
which is not the characteristic impedance of all the pieces of
transmission line in the system, and keep the same reference impedance
for all your work. For example, 50 ohms is commonly used, even though
the system contains sections of microstrip of various impedances.
But that's just a convenience for system analysis and design. If you
want to try to assign even a little physical significance to
reflection coefficient as used on a piece of line, you really should
be using the line's characteristic impedance as the reference
impedance. In addition, you should realize that it's going to make
sense only in a linear, time-invariant system with steady-state
excitation, with only one source of excitation (at a time). In
addition, it is of course a function of frequency, just as the line's
characteristic impedance is.

As others have noted, the magnitude of the reflection coefficient can
be greater than unity with a passive line and load. Don't try to read
too much physical significance into that, however.

Although the classic definition of (V)SWR involves knowing a voltage
maximum and a voltage minimum on a line, I much prefer a definition in
terms of forward and reverse voltages. That allows me to think about
SWR at a point on a uniform line, and realize that it will be
different at different points (because of line attenuation). In fact,
the _definition_ I use for reflection coefficient is Er/Ef, or
equivalently the ratio of electric fields (or magnetic fields)
associated with forward and reverse waves (which then applies also to
non-TEM waveguides). From that definition, it's straightforward to
determine that rho = (Zl-Zo)/(Zl+Zo). And in keeping with the idea
that you cannot have a voltage magnitude minimum less than zero, and
because I believe it's more practical than the classic definition to
have an SWR definition I can apply to any point on a line, my working
SWR defintion is SWR = (|Ef|+|Er|)/(|(|Ef|-|Er|)|). This will align
well with the usual formula that SWR = (1+|rho|)/(1-|rho|) when
|rho|=1, but it never gives you a negative SWR. If you can accept my
definition of SWR, we can talk about SWR. If you can't, then I just
won't talk with you about SWR, and limit the discussion to reflection
coefficient which we presumably would be able to agree on.

Cheers,
Tom

(Dr. Slick) wrote in message . com...
From Pozar's Microwave Engineering (Pg. 606):

Reflection Coefficient looking into load = (Zl-Zo)/(Zl+Zo)

Where Zl is a purely real load impedance, and Zo is the
purely real characteristic impedance reference.

When you change Zo, you change the normalized center of the
Smith Chart, and therefore the Reflection Coefficient and SWR, looking
into the same load.



Check it out...


Slick

wrote:
Do not be afraid to admit that you have changed the definition of P = V
x I and therefore do not accept the standard definition.

Well, I was taking 'x' as a multiplication sign. Did you mean it as
a cross product sign? In any case, your Vinst is *NET* voltage and
your Iinst is *NET* current. We know that the feedline Z0 forces
Vfwd*Ifwd to be a constant value for a lossless line. We know that
the feedline Z0 forces Vref*Iref to be a constant value for a lossless
line. Vfwd+Vref is the *NET* voltage. Ifwd+Iref is the *NET* current.
Of course, their product, in any form, is going to be *NET* power.

When I learned Pinst = Vinst x Iinst there were no caveats about how
Pinst meant Pnet. Instantaneous energy is flowing or it is not. When
Pinst is 0 for all time, then there is no energy flowing.

But RF energy cannot stand still so if it exists, it must necessarily
flow. If there is no energy flowing, then there is no RF. If there is
no RF, then your statements are irrelevant to this newsgroup. :-)

To satisfy your theory (and minimize double think), you have had to
change this to Pnet is zero to allow these cancelling powers to flow.
So be it.

Your Vinst is *NET* voltage equal to Vfwd+Vref. Your Iinst is *NET*
current equal to Ifwd+Iref. Of course, their product will be *NET*
power. It cannot be anything else.

True, sort of.

Ramo, Whinnery, and Van Duzer will be impressed that you "sort of"
agree with them. :-)

It is not flowing all the way to the end of the
line and then back. There is not enough time for this to happen (on a
multi-wavelength line) since it changes direction every quarter cycle.

Just as I suspected, you are confusing the carriers of the energy with
the energy itself in the waves which moves at the speed of light. A
quarter cycle of time is very, very large compared to the speed of light.

Since water molecules cannot travel at the speed of sound in the ocean,
I assume you would argue that the energy in a tsunami wave cannot travel
at 500 mph, right?

Not quite. Standing voltage and current waves (which are not waves in
the normal sense) can be observed on the line. They can be measured with
real voltage and current instruments; as can real energy flows with
a real (V x I) power meter (but not a 'Bird watt' meter which
is doing something quite different). It happens that if you assume the
existence of forward and reverse voltage and current waves, mathematical
functions can be derived that will produce the same distribution of
voltage and current as observed on the line. This is extraordinarily
convenient some analysis but does not mean that these assumed waves are
real.

So please amaze me with a model that produces standing waves without actual
forward and reflected waves (in a single source, single feedline, single load
system).

A mechanical analogue would be to look at a guy wire on a pole. You can
analyze the forces as two vectors at 90 degrees (or any other angle of
convenience!), but never make the mistake of assuming that there are
actually two guy wires present. Just because it is mathematically
convenient to assume the existence of two vectors does not mean they
exist.

Nobody is rejecting it. If the lossless stub is one second long, it takes
two seconds of *POWER* to bring it to steady-state. If the stub contains
no moving energy, where did all those joules go?

This energy is indeed stored in the stub. None of it moves across zero
voltage or current boundaries.

What exactly, keeps energy from crossing the boundary? Here is an example.

source-------------50 ohm coax--------------+----1/4WL stub-----open

What mechanism of physics keeps energy from crossing the '+' point? Note
that there is no physical impedance discontinuity at point '+'.

This is what I (and Ayn Rand) call "primacy of consciousness" type thinking.
If you believe it strongly enough, your math model will dictate reality.
Something about being able to move mountains with the faith of a grain of
mustard seed. Something about being able to change the SWR by changing the
normalization of a Smith Chart on a sheet of paper. OTOH, I believe in
"primacy of existence", where reality dictates my math models. They may
be wrong but are as close to reality as I can get.

All you have to do to convince me that you are right is explain exactly
how standing waves can be sustained without a forward wave and a reflected
wave (in a system with a single source, single feedline, and single load).
--
73, Cecil http://www.qsl.net/w5dxp
"One thing I have learned in a long life: that all our science, measured
against reality, is primitive and childlike ..." Albert Einstein



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Reg Edwards wrote:
Magid has the most rigorous derivation of power and energy flow on
transmission lines,

==========================
The following short question is adressed to all contributors to this
newsgroup who feel impelled to bolster their lack of self-confidence by
dragging in the chapter and verse of their favourite worshipped authors and
Gurus, most of whom nobody has ever heard of and highly unlikely ever to get
their hands on.

How do you know that?

I multiplied v(t) and i(t) in the forward and reverse waves and added
them as a function of position to get the instantaneous power at each
point along the line. Then I integrated to find the energy. As I
mentioned in the part of the posting you excluded from your quote, I
discovered that I hadn't evaluated the constant of integration.
Somewhere along the line, I got sidetracked, and didn't want to get
sucked into the interminable argument going on (which I see I've started
up again -- my sincere apology to all), so didn't go back and clean it
up. I had, however, reached the same conclusion as Magid, so apparently
the constant was zero, or didn't impact the results. Magid follows the
same process, although I haven't yet followed it through completely.

You've now heard of Magid, and you can very likely find a used copy on
the Internet for the price of a couple of bottles of mediocre Pinot Noir
in much less time than it would take to drink it (unless perhaps you're
a speed drinker). You could get your hands on one with even less effort
than I've taken -- I had to walk a few blocks, while you can do it all
from your easy chair, only having to rise and face the Sun when the
postman comes with your book.

Shoot, you can even get it from the same store where I got mine, if they
have another copy just now. http://www.powellsbooks.com.

Roy Lewallen, W7EL
Certified Reg's Old Wife, Nitpicker, Busy-Body, Lacker of
Self-Confidence, Worshipper of Authors and Gurus, and Other Notable
Distinctions and Honours which are Bestowed Almost Daily

Dr. Slick wrote:
Therefore, the reflected voltage can never be greater than the
input voltage for a passive network, and the reflection coefficient
can never be greater than 1 for such a case.

What must be realized is that the s-parameter reflection coefficients
are *PHYSICAL* reflection coefficients while 'rho' is not a physical
reflection coefficient. s11 and 'rho' do not usually have the same
value. There are physical constraints on s11. The s11 physical
constraints do not apply to 'rho'. s11 is the square root of the
ratio of Pref to Pfwd under special conditions. 'rho' is the square
root of the ratio of Pref to Pfwd under all conditions. There's a
big difference. s11 does not usually equal 'rho' anywhere except
at a one-port load.
--
73, Cecil http://www.qsl.net/w5dxp



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Roy Lewallen wrote in message ...
....
I omit Tom from this second list only because I haven't yet
met him in person and otherwise haven't gotten any hints of his age --
but I'll take a gamble and spot him 10 years at least.

Well, you can find archived postings from me from 10 years ago that I
think demonstrate that I understood how SWR meters actually work back
then. And you could probably find my name on a patent that would give
you a clue that I was at least starting to learn a little something
about transmission lines in 1969 or so, though I readily admit to not
worrying about "reflection coefficient" back then. Just hacked
through the raw, unabridged transmission line equations. I suppose
Reg would think that a better way to learn the stuff anyway. I had
bought the King, Mimno and Wing book back then, but didn't get around
to actually reading it till much, much later.

Cheers,
Tom

Roy Lewallen wrote:
You've now heard of Magid, ...

Unfortunately, every author and guru that I have ever encountered,
at some point, confuses cause and effect. I'm sorry I can't get over
to the Texas A&M library right now but Magid seems to believe that
standing waves can be sustained without a forward wave and a reflected
wave. Does he explain how that is possible? In all honesty, it is an easy
mistake to make. Even Ramo, Whinnery, and Van Duzer make the same
mistake. Not exactly a quote but: The reflection coefficient is caused
by the ratio of the reflected power to the forward power. Therefore,
the ratio of the reflected power to the forward power is caused by
the reflection coefficient. "Logic" like this seems to abound in
the field of transmission lines.
--
73, Cecil http://www.qsl.net/w5dxp



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Tom Bruhns wrote:
W5DXP wrote:
We know that the feedline Z0 forces
Vfwd*Ifwd to be a constant value for a lossless line. We know that
the feedline Z0 forces Vref*Iref to be a constant value for a lossless
line.

I trust I haven't taken that too far out of context. And...I hope you
meant that Zo = Vf/If = -Vr/Ir, or something equivalent. Surely the
product of Vf and If is independent of Zo.

Yep, sorry, brain fart. Should have been a '/' instead of a '*'.
The point was (is) that (Vfwd + Vref) can be zero but in a feedline
with reflections Vfwd and Vref cannot be zero. Same for the component
currents.
--
73, Cecil http://www.qsl.net/w5dxp



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"George, W5YR" wrote in message ...
Roy, Chipman on page 138 of "Theory and Problems of Transmission Lines"
makes the statement

" The conclusion is somewhat surprising, though inescapable, that a
transmission line can be terminated with a reflection coefficient whose
magnitude is as great as 2.41 without there being any implication that the
power level of the reflected wave is greater than that of the incident wave.


How did he set this up? How did he measure this excactly? How
can you get a reflection coefficient greater than one into a passive
network? I'd really like to know.


Slick

Dr. Slick wrote:
W5DXP wrote in message ...

What must be realized is that the s-parameter reflection coefficients
are *PHYSICAL* reflection coefficients while 'rho' is not a physical
reflection coefficient. s11 and 'rho' do not usually have the same
value. There are physical constraints on s11. The s11 physical
constraints do not apply to 'rho'. s11 is the square root of the
ratio of Pref to Pfwd under special conditions. 'rho' is the square
root of the ratio of Pref to Pfwd under all conditions. There's a
big difference. s11 does not usually equal 'rho' anywhere except
at a one-port load.

[s11]=rho, rho being just the magnitude of the s11.

I just told you that is not always true.

s11 is the reflection coefficient when a2=0

|rho| is the reflection coefficient when a2=a2

In a Z0-matched system with reflections, they are NOT equal.
--
73, Cecil http://www.qsl.net/w5dxp



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W5DXP wrote:

|rho| is the reflection coefficient when a2=a2

Prophetic!

Yes, Magid explains it thoroughly and well. His analysis includes
traveling waves from the beginning to the end. It's basically the same
analysis I've done myself.

What he *didn't* do was to throw the time-dependent information out the
window at the very beginning of the analysis, as you've done in all your
analyses involving only average power. That allows him to actually show
where the energy is at every instant of time. It looks to me like his
analysis is valid, so if your method leads to a different conclusion, I
feel there's good reason to doubt its validity.

Y'know, if every author seems to have it wrong, considering how many
really, really intelligent and learned people there are out there, and
how many times their work has been carefully reviewed, have you
considered for just a moment the possibility that maybe, just maybe,
they have it right and you have it wrong? But if you're absolutely
convinced you've discovered something that all those other folks are
mistaken about, why not write your own textbook? Or at least a paper or
two for the professional publications. The world will thank you for
straightening them out.

Roy Lewallen, W7EL

W5DXP wrote:
Roy Lewallen wrote:

You've now heard of Magid, ...


Unfortunately, every author and guru that I have ever encountered,
at some point, confuses cause and effect. I'm sorry I can't get over
to the Texas A&M library right now but Magid seems to believe that
standing waves can be sustained without a forward wave and a reflected
wave. Does he explain how that is possible? In all honesty, it is an easy
mistake to make. Even Ramo, Whinnery, and Van Duzer make the same
mistake. Not exactly a quote but: The reflection coefficient is caused
by the ratio of the reflected power to the forward power. Therefore,
the ratio of the reflected power to the forward power is caused by
the reflection coefficient. "Logic" like this seems to abound in
the field of transmission lines.

Thanks for the analogy.

One can mathematically and conceptually conceive two opposite-traveling
waves that add up to the observed standing wave, and that's fine. The
problem comes with assigning power or energy to the waves. Then you run
into the problem of how one wave got the energy over the barrier into
the pocket and the other wave took the same amount back out, without
transfering any air molecules across the barrier in the process.

The average power analysis looks to me something like this. Suppose you
have two batteries each with exactly 2 volts potential and zero internal
resistance, with a 2 ohm resistor connected between their positive
terminals. The negative terminals are connected together. You replace
the battery on the left with a short (turning it off), and observe that
the current through the resistor is one amp to the left. Then you hook
the left hand battery back up and turn the right hand battery "off" by
replacing it with a short. You observe that there's one amp now flowing
through the resistor to the right. Finally, turn both batteries "on" by
putting them both in place. You can use superposition to conclude,
correctly, that there's zero current flowing in the resistor.

But it's silly to insist that there's a forward two watt "power wave"
flowing to the right, and another two watt wave flowing to the left. You
subtract one from the other and, sure enough, get zero. But are the
"power waves" real? Studying and analyzing these imaginary waves is
surely a lot more interesting than simply looking at the circuit and
noting that the "boring" (as Cecil calls it) net power is zero. But
aren't you studying ghosts?

Even more risky is adding the things. This time hook two one volt
batteries in series with the 2 ohm resistor and energize one at a time.
With the upper one on and the lower one "off" (replaced with a short)
you get 1/2 amp. You've got a "power wave" of I^2 * R = 1/2 watt. Turn
the lower one on and the upper one "off", and you get another "power
wave" of 1/2 watt, in the same direction. Turn them both on, and you
have a power flow of, um, 2 watts. Welcome to the new math.

Roy Lewallen, W7EL

Jack Smith wrote:

Roy:

Interesting point and I don't recall reading or hearing it elsewhere.

The following is dashed off without fully thinking it through, so no
warranty on its accuracy.

If you think of a sound wave (longitudinal transmission, of course) in
a lossless acoustic transmission line terminated with a short, the
individual air molecules within each 1/4 wave section are likewise
trapped since at the 1/4 wave points there is zero sound pressure.
This may be a useful analogy for the electromagnetic transverse
propagating T-line.

Jack K8ZOA

Roy Lewallen wrote:
Y'know, if every author seems to have it wrong, considering how many
really, really intelligent and learned people there are out there, and
how many times their work has been carefully reviewed, have you
considered for just a moment the possibility that maybe, just maybe,
they have it right and you have it wrong?

"It" is not always the same thing. What I am saying is that nobody's
perfect, nobody knows everything, and 1000 years from now, most of what
you and I believe today will be obsolete.

For instance, Ramo & Whinnery use the square root of the ratio of reflected
power to forward power to define the reflection coefficient. Shortly
thereafter, they use the reflection coefficient as the cause of that
power ratio. So in essence, they are saying that the reflection
coefficient causes itself.

Another example from this newsgroup. After standing waves have been caused
by the superposition of forward waves and reflected waves, they somehow
become self-sustaining without the forward waves and reflected waves.
--
73, Cecil http://www.qsl.net/w5dxp



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Roy Lewallen wrote:
One can mathematically and conceptually conceive two opposite-traveling
waves that add up to the observed standing wave, and that's fine. The
problem comes with assigning power or energy to the waves.

Proposing waves that exist without energy seems more problematical
to me. And nobody has come up with an explanation of how standing
waves can exist without forward waves and reflected waves in a
single source, single feedline, single load system. Seems to me,
that is an absolutely necessary first step in proposing self-
sustaining standing waves.

The average power analysis looks to me something like this. Suppose you
have two batteries ...

In a typical ham installation, there is only one source and all system
energy is supplied by that single source. Please limit your examples
to one source to adhere to reality. Having to change the example
away from reality to multiple sources to make a point is a weakness,
not a strength, in the argument. In a typical ham installation, instant
energy is NOT available since the source is nanoseconds away. Any
energy needed instantaneously must already be there.
--
73, Cecil http://www.qsl.net/w5dxp



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Richard Harrison wrote:

Keith wrote:
"At the quarter wave points where voltage and current are always zero,
there is no energy flowing. Period."

Yes there is energy flowing, and it is flowing in both directions if it
is flowing in one direction. Otherwise there would be no standing wave.

This view that there is energy flowing in both directions at the same
time
leads to some strange conclusions.

The directional wattmeter uses the following expressions to compute its
displayed value:
Vf = (V + IR)/2
Vr = (V - IR)/2
where V and I are the instantaneous voltage and current at the same
point
on the line and then computes power from:
Pf = average(Vf**2/R)
Pr = average(Vr**2/R)
Appropriate scaling (which we'll ignore for simplicity) is needed
depending on the characteristics of the voltage and current sensors.

Let's apply these expressions to some simple examples.

Connect a length of 50 Ohm transmission line to a 9 Volt battery and
wait
for the transient to die:
Vf = (9 + 0)/2 - 4.5 V
Vr = (9 - 0)/2 - 4.5 V
Pf = 0.405 Watt
Pr = 0.405 Watt
Disconnect the battery and the capacitance of the line remains charged
to
9 Volts. Are you really quite comfortable with the notion that this
line,
charged to 9 Volts, with 0 current everywhere has 0.405 Watts flowing
forward canceling at all points the 0.405 Watts flowing in reverse?

Others have stated that lamp cord has an impedance of about 100 Ohms.
Assume this to be correct.
With a table lamp plugged in but the lamp turned off:
Vf = (120 + 0)/2 - 60 V
Vr = (120 - 0)/2 - 60 V
Pf = 36 Watt
Pr = 36 Watt
Again, are you comfortable with this conclusion?

How about phone lines (phone on hook) at 600 Ohms and 48 Volts:
Vf = 24 V
Vr = 24 V
Pf = 0.96 W
Pr = 0.96 W
In a 400 pair cable there is 384 Watts flowing in the forward and
reverse direction. From a hundred thousand line central office we
get 960 Kilowatts forward and reverse, when the current is zero,
everywhere. Comfort level?

With the forward and reverse power view, to completely understand
the behaviour of the circuit, we need to know Z0 or we can not
compute these forward and reverse powers which seem to be
fundamental.

Consider a flashlight with the lamp off: the wire twists and turns
and has a very non-constant Z0. While computationally tedious, it
is possible to determine Pf and Pr at every point along the wire.
Isn't it necessary to do this to obtain a complete understanding
of what is happening in the circuit, if Pf and Pr are flowing?
If not, why not? The necessity for doing this is the logical
conclusion of the Pf and Pr approach.

Presented with a wire charged to high static voltage, what is Pf and
Pr? Presented with a sphere charge to a high static voltage, what
is the Pf and Pr?
Presented with a simple length of wire, how much power is flowing?

Similarly for a capacitor. The plates of a capacitor must have power
reflecting in all directions even when it is just charged to a
constant voltage (according to the Pf and Pr theory).

Given these bizarre results it would seem wise to limit the application
of the watts indicated by a directional 'watt'meter to those things
which are proper and not assume that it necessarily means that there
is real power flowing.

....Keith

How can you get a reflection coefficient greater than one
into a passive network? I'd really like to know.

=====================================

Dear Dr Slick, it's very easy.

Take a real, long telephone line with Zo = 300 - j250 ohms at 1000 Hz.

Load it with a real resistor of 10 ohms in series with a real inductance of
40 millihenrys.

The inductance has a reactance of 250 ohms at 1000 Hz.

If you agree with the following formula,

Magnitude of Reflection Coefficient of the load, ZL, relative to line
impedance

= ( ZL - Zo ) / ( ZL + Zo ) = 1.865 which exceeds unity,

and has an angle of -59.9 degrees.

The resulting standing waves may also be calculated.

Are you happy now ?
---
Reg, G4FGQ

Keith wrote:
"Connect a length of 50 Ohm transmission line to a 9 Volt battery and
wait for the transient to die."

The transient is the only part of your proposal for which the 50-ohm
surge impedance has meaning. That`s why it`s called the "surge"
impedance.

The 50 ohms is Zo and equals the sq.rt. of Z/Y, or in a perfect line
this is sq. rt. of L/C.

L and C require an alternatng current (changing, not static) to produce
reactance, so the d-c calculations are meaningless to Zo.

Best regards, Richard Hsarrison, KB5WZI

W5DXP wrote in message ...

[s11]=rho, rho being just the magnitude of the s11.

I just told you that is not always true.

s11 is the reflection coefficient when a2=0

|rho| is the reflection coefficient when a2=a2

In a Z0-matched system with reflections, they are NOT equal.


You're right, but we are talking about a one-port network, the
antenna and transmission line.


Slick