what happens to reflected energy ?
 

Roy Lewallen wrote in
:


Adding a directional wattmeter to the mix raises an interesting issue.

Roy,

I suspect that the directional wattmeter is in a large part responsible
to the strong committment to the view the Pf and Pr are separate tangible
quantities, and that they can be dealt with independently.

You will recall a raging argument some years ago about the possible range
of rho, and insistence by some practical practitioners that rho could
never be greater than unity because it would mean that reflected power
was greater than forward power, obviously impossible, then citing
experience with a directional wattmeter (calibrated for V/I being purely
real) as evidence. Of course, telephony line designers know about this
effect and do not have directional wattmeters to confuse their thinking.

You example goes on to point out the absurdity of application of that
kind of thinking.

Miguel has asked why I keep quoting the term 'reflected power' etc, it is
because it is a deceptive label that misrepresents the quantity.

Roy has explained, and I have explained in the articles I referenced, the
assumptions that underly the concept that Pf=Vf*If, and its limitations.

There are many ham texts and articles that try to explain some concepts
and deliver to the user a language and model that encourages extension to
invalid application.

Miguel, in a steady state scenario with a source at one end of a TEM line
and a load at the other, the power (rate of flow of energy) at any point
is given by the product of instantaneous voltage and current. This is a
time varying quantity which can be integrated over time to obtain the
average power. At some points on the line, the instantaneous power may be
negative for some parts of the cycle, which indicates that over a cycle,
some energy is exchanged to and fro across that point, but the average
power will always be from source to load, and that quantity will decrease
from source to load as accounted for by line loss elements, though in the
general case, not purely exponentially as sometimes believed.

Owen

what happens to reflected energy ?
 

On Jun 17, 4:45*pm, Roy Lewallen wrote:
Recalling my earlier example of a 100 watt transmitter connected to a
half wavelength 200 ohm transmission line and then to a 50 ohm resistive
load, the "forward power" on the line was 156.25 watts and the "reverse
power" or "reflected power" 56.25 watts. The subject of this topic asks
where the "reflected energy" goes. Does it go back into the transmitter?

------Z01=50------+------Z02=200------load

Pfor1 = 100w, Pref1 = 0w
Pfor2 = 156.25w, Pref2 = 56.25w
rho^2 = 0.36, rho = 0.6

That's obviously a Z0-match to 50 ohms. There are, as usual, four
wavefront components:

Pfor1(rho^2) = 36w, the amount of forward power on the 50 ohm line
that is reflected back toward the source.

Pfor1(1-rho^2) = 64w, the amount of forward power on the 50 ohm line
that is transmitted through the impedance discontinuity toward the
load.

Pref2(rho^2) = 20.25w, the amount of reflected power on the 200 ohm
line that is re-reflected back toward the load by the impedance
discontinuity.

Pref2(1-rho^2) = 36w, the amount of reflected power on the 200 ohm
line that is transmitted through the impedance discontinuity toward
the source.

Pref1 = Pfor1(rho^2) + Pref2(1-rho^2) - total destructive interference

Pref1 = 36w + 36w - 2*SQRT(36w*36w) = 0

Pfor2 = Pfor1(1-rho^2) + Pref2(rho^2) + constructive interference

Pfor2 = 64w + 20.25w + 2*SQRT(64w*20.25w)

Pfor2 = 64w + 20.25w + 72w = 156.25w

Note that the 72 watts of total destructive interference energy toward
the source has changed directions and joined the forward wave toward
the load as constructive interference.

No need to haul out a wattmeter or SWR meter. A pencil, paper, and
calculator is all one needs PLUS an understanding of the rules of
physics governing energy flow at a Z0-match.

Actually, you didn't even need to tell me what the Pfor2 and Pref2
values are. As you can see from the calculations, they are easy to
calculate even when they are unknown.
--
73, Cecil, w5dxp.com

what happens to reflected energy ?
 

On Jun 17, 5:53*pm, K1TTT wrote:
On Jun 17, 10:31*am, Keith Dysart wrote:





On Jun 14, 7:04 pm, K1TTT wrote:

On Jun 14, 12:09 pm, Keith Dysart wrote:
Choose an arbitrary point on the line and observe the voltage. The
observed voltage will be a function of time. Let us call it V(t).
At the same point on the line, observe the current and call it I(t)..

if you can choose an arbitrary point, i can choose an arbitrary
time... call it a VERY long time after the load is disconnected.

That works for me, as long as the time is at least twice the
propagation
time down the line.

The energy flowing (i.e. power) at this point on the line will
be P(t)=V(t)I(t). If the voltage is measured in volts and the current
in amps, then the power will be in joules/s or watts.
We can also compute the average energy flow (power) over some
interval by integrating P(t) over the interval and dividing by
the length of the interval. Call this Pavg. For repetitive
signals it is convenient to use one period as the interval.

Note that V(t) and I(t) are arbitrary functions. The analysis
applies regardless of whether V(t) is DC, a sinusoid, a step,
a pulse, etc.; any arbitrary function.

To provide some examples, consider the following simple setup:
1) 50 ohm, Thevenin equivalent circuit provides 100 V in to an
* *open circuit.

this is the case we were discussing... the others are irrelevant at
this point

2) The generator is connected to a length of 50 ohm ideal
* *transmission line. We will stick with ideal lines for the
* *moment, because, until ideal lines are understood, there
* *is no hope of understanding more complex models.
3) The ideal line is terminated in a 50 ohm resistor.

snip the experiments you do not consider relevent

Now let us remove the terminating resistor. The line is
now open circuit.

OK, now we are back to where i had objections above.

Again, set the generator to DC. After one round trip
time, the observations at any point on the line will
be:
* V(t)=100V
* I(t)=0A
* P(t)=0W
* Pavg=0W

ok, p(t)=0w, i=0a, i connect my directional wattmeter at some very
long time after the load is disconnected so the transients can be
ignored and i measure 0w forever.

Not correct. See the equations for a directional wattmeter below.

there is no power flowing in the line.

I agree with this, though a directional wattmeter indicates
otherwise. See below.

Now it is time to introduce forward and reflected waves to
the discussion. Forward an reflected waves are simply a
mathematical transformation of V(t) and I(t) to another
representation which can make solving problems simpler.

The transformation begins with assuming that there is
a forward wave defined by Vf(t)=If(t)*Z0, a reflected
wave defined by Vr(t)=Ir(t)*Z0 and that at any point
on the line V(t)=Vf(t)+Vr(t) and I(t)=If(t)-Ir(t).

Since we are considering an ideal line, Z0 simplifies
to R0.

Simple algebra yields:
* Vf(t)=(V(t)+R0*I(t))/2
* Vr(t)=(V(t)-R0*I(t))/2
* If(t)=Vf(t)/R0
* Ir(t)=Vr(t)/R0

The above equations are used in a directional wattmeter.

Now these algebraically transformed expressions have all
the capabilities of the original so they work for DC,
60 HZ, RF, sinusoids, steps, pulses, you name it.

hmmm, they work for all waveforms you say... even dc?? *ok, open
circuit load, z0=50, dc source of 100v with 50ohm thevenin
resistance...

You have made some arithmetic errors below. I have re-arranged
your prose a bit to allow correction, then comment.

well, maybe i missed something... lets look at the voltages:

Vf(t)=(100+50*2)/2 = 100
Vr(t)=(100-50*2)/2 = 0

You made a substitution error here. On the transmission line
with a constant DC voltage:
* V(t)=100V
* I(t)=0A

substituting in to
* Vf(t)=(V(t)+R0*I(t))/2
* Vr(t)=(V(t)-R0*I(t))/2

yields
* Vf(t)=50V
* Vr(t)=50V

and
* If(t) = 50/50 = 1a
* Ir(t) = 50/50 = 1a

Recalling that the definition of forward and reflected is
* V(t)=Vf(t)+Vr(t)
* I(t)=If(t)-Ir(t)
we can check our arithmetic
* V(t)= 50+50 = 100V
* I(t)= 1-1 * = 0A
Exactly as measured.
If you are disagree with the results, please point out
my error in the definitions, derivations or arithmetic.

except i can measure If and Ir separately and can see they are both
zero. *

How would you measure them? The standard definition of forward
and reflected waves are as I provided above. I suppose you could
construct your own definitions, but such a new definition would
be confusing and unlikely to have the nice properties of the
standard definitions.

this is where your argument about putting a resistor between
two batteries falls apart... there is a length of cable with a finite
time delay in the picture here. *so it is possible to actually watch
the waves reflect back and forth and see what really happens. *in the
case where the line is open at the far end we can measure the voltage
at both ends of the line and see the constant 100v, so where is the
voltage difference needed to support the forward and reverse current
waves? *

Well, as I suggested, the forward and reflected wave are somewhat
fictitious. They are very useful as intermediate results in solving
problems but, as you note above, quite problematic if one starts
to assign too much reality to them.

That is why they are quite analogous to the two currents in the two
battery example.

you can't have current without a voltage difference. *

You can, though, have an arbitrary current as an intermediate
result in solving a problem. With the two battery example, reduce
the resistor to 0 and an infinitely large current is computed in
both directions, though we know in reality that no current is
actually flowing.

plus i
can substitute in a lossy line and measure the line heating and
measure no i^2r loss that would exist if there were currents flowing
in the line, so there are none.

I certainly agree that there is no real current flowing, but the
fictititious If and Ir are indeed present (or pick a better word
if you like).

try this case, which is something we do at work... take a 14ga piece
of copper wire maybe 50' above ground, its characteristic impedance is
probably a few hundred ohms, and put 100kv on it... how much power is
it dissipating from your If and Ir heating losses?? *lets say 200 ohms
and 100kv gives 100,000/200a=500a, shouldn't take long to burn up that
wire should it? *and yet it never does... so why doesn't it have your
circulating currents

I really like this example. I thought of using something similar once
to convince Cecil, but never did. He is unreedemable. But your
example aptly demonstrates why If and Ir are not real, though still
useful because If-Ir is the actual current flowing.

Just as an aside, your comment above about requiring a voltage to
cause a current to flow is not quite correct. An ideal conductor has
zero resistance, so current can flow without voltage in an ideal
conductor. Strangely, this can even be achieved in the real world
with superconductors.

So I offer another example for your consideration. The experiment
is the same as above (100V, 50 ohm, ideal source and conductors)
but this time short the end of the transmission line. After waiting
for the line to settle, we find 0 volts everywhere and 2A flowing.
Recalling the coversion equations:
Vf(t)=(V(t)+R0*I(t))/2
Vr(t)=(V(t)-R0*I(t))/2
If(t)=Vf(t)/R0
Ir(t)=Vr(t)/R0

we obtain:
Vf(t)=(0+100)/2 = 50V
Vr(t)=(0-100)/2 = -50V
If(t)=1A
Ir(t)=-1A

Substituting to verify
V(t)=50+(-50)=0V
I(t)=1-(-1)=2A

Exactly as measured.

I will caution again, do not assign too much reality to these
forward and reflected waves; very useful for solving problems,
but trouble if carried too far.

....Keith

what happens to reflected energy ?
 

On Jun 17, 7:00*pm, Keith Dysart wrote:
An ideal conductor has
zero resistance, so current can flow without voltage in an ideal
conductor.

Quoting "Fields and Waves ...", by Ramo and Whinnery: "A perfect
conductor is usually understood to be a material in which there is no
electric field at any frequency. Maxwell's equations ensure that there
is then also no time-varying magnetic field in the perfect conductor."
How does one go from zero current to a non-zero current if the
magnetic field is prohibited from varying (changing) with time?
--
73, Cecil, w5dxp.com

what happens to reflected energy ?
 

On Jun 17, 7:00*pm, Keith Dysart wrote:
I will caution again, do not assign too much reality to these
forward and reflected waves; very useful for solving problems,
but trouble if carried too far.

I'm afraid you are too late with that advice. Optical physicists have
assigned reality to forward and reflected waves inside an
interferometer and have tracked all of the energy including the energy
in canceled wavefronts.

http://www.teachspin.com/instruments...eriments.shtml

Scroll down to, "Using Dielectric Beamsplitters to find the "missing
energy" in destructive interference" It says: "... when interference
is destructive at the standard output, it is constructive at the non-
standard output."

In RF terms, when interference is destructive at a Z0-match, it is
constructive in the direction of the load.
--
73, Cecil, w5dxp.com

what happens to reflected energy ?
 

On 17 jun, 21:30, Cecil Moore wrote:
On Jun 17, 7:00*pm, Keith Dysart wrote:

An ideal conductor has
zero resistance, so current can flow without voltage in an ideal
conductor.

Quoting "Fields and Waves ...", by Ramo and Whinnery: "A perfect
conductor is usually understood to be a material in which there is no
electric field at any frequency. Maxwell's equations ensure that there
is then also no time-varying magnetic field in the perfect conductor."
How does one go from zero current to a non-zero current if the
magnetic field is prohibited from varying (changing) with time?
--
73, Cecil, w5dxp.com

An ideal conductor has
zero resistance, so current can flow without voltage in an ideal
conductor.

Inertia first Newton law applied to the free electron lttle balls
perhaps?
Now, if at first of experiment little balls was at rest, how do they
set in movement without a force? Tell us.
......
Keith, OM, if you do not make the rope experiment, make this another
simple one = get from Radio Shack a long, long lossles TL, (with
vf=1, why not?), 6*10^8 meters long it is good, open or short ended
(it does not matter). Connect it to your 100 W rig, key for a one
second the TX full CW power, inmediately disconnect the TL and touch
the connector with your fingers. Just count: tree, two, one, ¡zero!.
If after "zero" you still with the connector in your fingers without
blink, then reflected waves really have not too much reality... have
some ointment for burns, may be the boys are right.) (sorry, do not
be angry with me I am practicing translate some creole humor to
english :D )

73 Miguel LU6ETJ

what happens to reflected energy ?
 

On 6/6/2010 9:22 AM, JC wrote:
Basic question (at least for me) for a very poor antenna matching :
-100 w reach the antenna and 50 w are radiated.
- 50 w are "reflected", what is their fate ?
Are they definitely lost for radiation and just heat the line, the final....
JC




It would be really really REALLY effing great if you dorks would take
this to a suitable forum. This has nothing to do with antennas.

And it is an unproductive infinite length, in case you bunch have missed
that point. But I'm guessing that given how intelligent you all are,
you already figured that out, and are ignoring it. No one is changing
their position, and they never will.

And yes, I have the thread set to "ignore". Which everyone should.

I will NOT see your responses.

tom
K0TAR

what happens to reflected energy ?
 

On 17 jun, 18:45, Roy Lewallen wrote:
Owen Duffy wrote:

While Zo of transmission lines might not be purely real, if the sampler
element is calibrated for V/I being real, then the power is given by
'forward power' - 'reflected power'. This is true even if the
calibration impedance is different to the transmission line in which the
measurements are made, though significant departure will impact
measurement uncertainty.

For example, if I insert a 50 ohm Bird 43 in a 75 ohm line at 1.8MHz,
and measure Pf=150W and Pr=50W, then the power is 150-50=100W. *The
insertion VSWR due to the Bird 43 is trivial in this case, so it hardly
disturbs the thing being measured. Each of the power measurements is of
little value alone, no inference can be made (in this case) of the
actual line VSWR, but the difference of the Pf and Pr readings does give
the power at that point.

I discuss this in my article entitled "http://vk1od.net/blog/?p=1004" at
http://vk1od.net/blog/?p=1004.

Owen

Adding a directional wattmeter to the mix raises an interesting issue.

Recalling my earlier example of a 100 watt transmitter connected to a
half wavelength 200 ohm transmission line and then to a 50 ohm resistive
load, the "forward power" on the line was 156.25 watts and the "reverse
power" or "reflected power" 56.25 watts. The subject of this topic asks
where the "reflected energy" goes. Does it go back into the transmitter?
Well, let's put a 50 ohm directional wattmeter between the transmitter
and the line. It measures 100 watts forward and 0 watts reverse. So what
happened to the "reverse power" on the transmission line? If it was
going back into the transmitter, we've now fixed the problem just by
adding the wattmeter. In fact, we could replace the wattmeter with a
piece of 50 ohm transmission line of any length, as short or long as we
want, and the "forward power" on that line will be 100 watts and
"reverse power" zero. So we've protected the transmitter from this
horrible, damaging "reverse power" just by adding a couple of inches of
50 ohm line -- or a directional wattmeter. Is this cool or what?

Or we can put the wattmeter at the far end of the line and read 100
watts forward and zero watts reverse, eliminating "reverse power" at the
load -- although we've lost 56.25 watts of "forward power". Or put it at
the center of the line, where we'd read a whopping 451.6 watts forward
and 351.6 reverse(*).

These are the "forward power" and "reverse power" on the short 50 ohm
transmission line (or lumped equivalent) inside the wattmeter. So we can
create "forward power" and "reverse power" just by moving the wattmeter
around. And most importantly, we can use it to isolate the transmitter
from that bad "reverse power". What a powerful tool!

I should point out that the example doesn't even mention, or need to
mention, the transmitter output impedance. The entire analysis holds for
any transmitter impedance, whether "dissipative", "non-dissipative",
linear, or nonlinear. Wherever the "forward power" and "reverse power"
come from and do to, it doesn't depend on any particular value or kind
of transmitter impedance.

(*) This brings up a great idea. Drop down to Radio Shack
or HRO and pick up one of those circulator things that separates forward
and reverse power. Put it into the middle of the line where you have
451.6 watts of "forward power" and 351.6 watts of "reverse power". Take
the 351.6 watts of "reverse power", rectify it, send it to an inverter,
and use it to run the transmitter -- you'll have plenty, even with poor
efficiency, and you can unplug the transmitter from the mains. There
should even be enough power left over to run your cooler and keep a
six-pack cold. Then work DX with your 451.6 watts of forward power while
you enjoy a cool one. Goodbye to electric bills! Hello to DX, free
power, and a tall cool one!

Roy Lewallen, W7EL- Ocultar texto de la cita -

- Mostrar texto de la cita -

Dear Roy:
This is a brilliant piece of work, it clarify so much misconceptions
(or extension of certain concepts beyond its scope). However the
question of "why" persist floating somehow.
Without leaving conventional approachs I think in the article cited in
the "Where does it go?" thread there are some helpful and valuable
hints: http://www.ittc.ku.edu/~jstiles/622/...es_package.pdf

Page 25 = "The precise values of Vo+, Io+ Vo+ and Io- are therefore
determined by satisfying the boundary conditions applied at each end
ofthe transmission line.
Page 28 = Although Vo+- and Io+- are determined by boundary
conditions
(i.e., what’s connected to either end of the transmission line),
the ratio V+- / V+- is determined by the parameters of the
transmission line only ( R, L, G, C).

What it is your opinion about?

73 - Miguel Ghezzi - LU6ETJ

what happens to reflected energy ?
 

lu6etj wrote:

Dear Roy:
This is a brilliant piece of work, it clarify so much misconceptions
(or extension of certain concepts beyond its scope). However the
question of "why" persist floating somehow.
Without leaving conventional approachs I think in the article cited in
the "Where does it go?" thread there are some helpful and valuable
hints: http://www.ittc.ku.edu/~jstiles/622/...es_package.pdf

Page 25 = "The precise values of Vo+, Io+ Vo+ and Io- are therefore
determined by satisfying the boundary conditions applied at each end
ofthe transmission line.
Page 28 = Although Vo+- and Io+- are determined by boundary
conditions
(i.e., what’s connected to either end of the transmission line),
the ratio V+- / V+- is determined by the parameters of the
transmission line only ( R, L, G, C).

What it is your opinion about?

73 - Miguel Ghezzi - LU6ETJ

Hello Miguel,

The part of the paper you quote is very standard transmission line
analysis which can be found in any textbook on the subject -- although I
find this treatment to be better written than most. Nothing I have ever
posted contradicts, or intends to contradict, the very well known and
established theory which has been found to be correct and useful for
over a hundred years. What I find silly is the idea that there are waves
of average power bouncing back and forth, needing to "go" somewhere, and
the specious arguments put forth in desperate attempts to explain where
these elusive waves "go" and how they supposedly behave. (But
interestingly, people seem much less concerned about where that extra
"forward power" comes from.) It's very much like a pseudo-scientific
exposition on the nature of ghosts or arguments about the effects of
various astrological signs.

As you've hopefully seen from my few recent postings, it's painfully
simple to show that most of the conceptions about these power waves are
ridiculous. But one can't change someone's religion by means of rational
arguments, and it's a fool's errand to try. So my postings aren't
directed to the power-wave zealots but rather to the few people who are
still open to reason. I see you're one of them, and I'm glad you've
giving the topic some careful thought.

Roy Lewallen, W7EL

what happens to reflected energy ?
 

On Jun 17, 10:57*pm, lu6etj wrote:
On 17 jun, 21:30, Cecil Moore wrote:





On Jun 17, 7:00*pm, Keith Dysart wrote:

An ideal conductor has
zero resistance, so current can flow without voltage in an ideal
conductor.

Quoting "Fields and Waves ...", by Ramo and Whinnery: "A perfect
conductor is usually understood to be a material in which there is no
electric field at any frequency. Maxwell's equations ensure that there
is then also no time-varying magnetic field in the perfect conductor."
How does one go from zero current to a non-zero current if the
magnetic field is prohibited from varying (changing) with time?
--
73, Cecil, w5dxp.com
An ideal conductor has
zero resistance, so current can flow without voltage in an ideal
conductor.

Inertia first Newton law applied to the free electron lttle balls
perhaps?
Now, if at first of experiment little balls was at rest, how do they
set in movement without a force? Tell us.
.....
Keith, OM, *if you do not make the rope experiment, *make this another
simple one = *get from Radio Shack a long, long lossles TL, (with
vf=1, why not?), 6*10^8 meters long it is good, open or short ended
(it does not matter). Connect it to your 100 W rig, key for a one
second the TX full CW power, inmediately disconnect the TL and touch
the connector with your fingers. Just count: tree, two, one, ¡zero!.
If after "zero" you still with the connector in your fingers without
blink, then reflected waves really have not too much reality... have
some ointment for burns, may be the boys are right.) (sorry, do not
be angry with me I am practicing translate some creole humor to
english :D *)

73 Miguel LU6ETJ- Hide quoted text -

- Show quoted text -

Good day Miguel,

In the example you provide, there would be real energy flowing and
it would be clearly shown by
P(t)=V(t)*I(t)
P(t) will be negative because the energy is flowing in the reverse
direction.

Computing Pf and Pr would reveal
Pf(t)=0
Pr(t)=-P(t)

So Pf(t)-Pr(t) = P(t) as it must.

I see no conflict with anything that I have written previously.

There are many examples where one can observe energy flow in a
reflected wave.

It just takes one counter-example to demonstrate that this is not
always
the case and that one should not assign *too* much reality to such
waves.

What this neans is that when someone asks "what happesn to reflected
energy?", the first question you have to answer is "Is this a
situation
where the computed reflected power represents something real?"
because
if it does not, the original question is moot.

....Keith