On Jun 10, 5:52 pm, K1TTT wrote:
On Jun 9, 2:04 am, Keith Dysart wrote:

Exactly. And yet, were you to attach a DC-couple directional
wattmeter to the line, it would indicate 50W forward and 50W
reflected (using your corrected numbers).

only as you monitored the initial transient. once the transient has
passed there is no power flowing in the line.

I agree completely with the second sentence. But the first is in
error. A directional wattmeter will indeed show 50W forward and 50W
reflected on a line with constant DC voltage and 0 current. I’ll
start by providing support for your second sentence and then get
back to reflected power.

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).

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.
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.

Set the generator to DC:
V(t)=50V, wherever it is measured on the line
I(t)=1A, wherever it is measured on the line
P(t)=50W, wherever it is measured on the line
Pavg=50W
No disagreement, I hope.

Set the generator to generate a sinusoid:
V(t)=50sin(wt)
I(t)=1sin(wt)
P(t)=25+25sin(2wt)
Pavg=25W
describes the observations at any point on the line.
Measurements at different points will have a different
phase relationship to each due to the delay in the line.
Note carefully the power function. The power at any point
on the line varies from 0 to 50W with a sinusoidal
pattern at twice the frequency of the voltage sinusoid.
The power is always positive; that is, energy is always
flowing towards the load.

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

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

The story for a sinusoid is more interesting...
The observations now depend on the location on the
line where the measurements are performed.
At a current 0:
V(t)=100sin(Wt)
I(t)=0A
P(t)=0W
Pavg=0W
At a voltage 0:
V(t)=0V
I(t)=2sin(wt)
P(t)=0W
Pavg=0W
Halfway between the current 0 at the load and the
preceding voltage 0 (that is, 45 degrees back
from the load:
V(t)=70.7sin(wt)
I(t)=1.414cos(wt)
P(t)=50sin(2wt)
Pavg=0W
This tells us that for the first quarter cycle of the
voltage (V(t)), energy is flowing towards the load,
with a peak of 50W, for the next quarter cycle, energy
flows towards the source (peak –50W), and so on, for
an average of 0, as expected.

For a sinusoid, when the load is other than an open
or a short, energy flows forward for a greater
period than it flows backwards which results in a
net transfer of energy towards the load.

At the location of current or voltage minima on
the line, energy flow is either 0 or flows towards the
load.

All of the above was described without reference to forward
or reflected voltages, currents or power. It is the basic
circuit theory description of what happens between two
networks. It works for DC, 60 HZ, RF, sinusoids, steps,
pulses, you name it. If there is any disagreement with the
above, please do not read further until it is resolved.

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

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.

Some people look at the expression If(t)=Vf(t)/R0 and
think it looks a lot like a travelling wave so they decide
to compute the energy being moved by this wave:
Pf(t)=Vf(t)*Vf(t)/R0
Pr(t)=Vr(t)*Vr(t)/R0

So let us explore the results when applied to the DC examples from
above.

Terminated with 50 ohms:
Vf(t)=50V
If(t)=1A
Vr(t)=0V
Ir(t)=0A

Pf(t)=50W
Pr(t)=0W

This all looks good, no reflected wave and the appropriate
amount of energy is being transported in the forward wave.

Let us try the open ended line:
Vf(t)=50V
If(t)=1A
Vr(t)=50V
Ir(t)=1A

which yield
V(t)=Vf(t)+Vr(t)=100V
I(t)=If(t)-Ir(t)=0A

which is in agreement with the observations. But there is
definitely a forward voltage wave and a reflected voltage
wave. Carrying on to compute powers:
Pf(t)=50W
Pr(t)=50W
and
P(t)=Pf(t)-Pr(t)=0W

So there you have it. The interpretation that the forward
and reflected wave are real and transport energy leads to
the conclusion that on a line charged to a constant DC
voltage, energy is flowing in the forward direction and
energy is flowing in the reflected direction.

Because it is obvious that there is no energy flowing on
the line, this interpretation makes me quite uncomfortable.

Now there are several ways out of this dilemma.

Some say they are only interested in RF (and optics), and
that nothing can be learned from DC, steps, pulses or other
functions. They simply ignore the data and continue with
their beliefs.

Some claim that DC is special, but looking at the equations,
there is no reason to expect the analysis to collapse with
DC.

The best choice is to give up on the interpretation that the
forward and reflected voltage waves necessarily transport
energy. Consider them to be a figment of the analysis technique;
very convenient, but not necessarily representing anything real.

As an added bonus, giving up on this interpretation will remove
the need to search for “where the reflected power goes” and
free the mind for investment in activities that might yield
meaningful results.

....Keith

On Jun 14, 12:09*pm, Keith Dysart wrote:
On Jun 10, 5:52 pm, K1TTT wrote:

On Jun 9, 2:04 am, Keith Dysart wrote:

Exactly. And yet, were you to attach a DC-couple directional
wattmeter to the line, it would indicate 50W forward and 50W
reflected (using your corrected numbers).

now remember, this is AFTER you disconnected the load resistor.


only as you monitored the initial transient. *once the transient has
passed there is no power flowing in the line.

I agree completely with the second sentence. But the first is in
error. A directional wattmeter will indeed show 50W forward and 50W
reflected on a line with constant DC voltage and 0 current. I’ll
start by providing support for your second sentence and then get
back to reflected power.


no, it won't... as i will explain below.

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.

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.

Set the generator to DC:
* V(t)=50V, wherever it is measured on the line
* I(t)=1A, wherever it is measured on the line
* P(t)=50W, wherever it is measured on the line
* Pavg=50W
No disagreement, I hope.

only if this is a case WITH a load... this is not what we were
discussing above.



Set the generator to generate a sinusoid:
* V(t)=50sin(wt)
* I(t)=1sin(wt)
* P(t)=25+25sin(2wt)
* Pavg=25W
describes the observations at any point on the line.
Measurements at different points will have a different
phase relationship to each due to the delay in the line.
Note carefully the power function. The power at any point
on the line varies from 0 to 50W with a sinusoidal
pattern at twice the frequency of the voltage sinusoid.
The power is always positive; that is, energy is always
flowing towards the load.

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. there is no power flowing in the
line.


The story for a sinusoid is more interesting...
The observations now depend on the location on the
line where the measurements are performed.
At a current 0:
* V(t)=100sin(Wt)
* I(t)=0A
* P(t)=0W
* Pavg=0W
At a voltage 0:
* V(t)=0V
* I(t)=2sin(wt)
* P(t)=0W
* Pavg=0W
Halfway between the current 0 at the load and the
preceding voltage 0 (that is, 45 degrees back
from the load:
* V(t)=70.7sin(wt)
* I(t)=1.414cos(wt)
* P(t)=50sin(2wt)
* Pavg=0W
This tells us that for the first quarter cycle of the
voltage (V(t)), energy is flowing towards the load,
with a peak of 50W, for the next quarter cycle, energy
flows towards the source (peak –50W), and so on, for
an average of 0, as expected.

For a sinusoid, when the load is other than an open
or a short, energy flows forward for a greater
period than it flows backwards which results in a
net transfer of energy towards the load.

At the location of current or voltage minima on
the line, energy flow is either 0 or flows towards the
load.

All of the above was described without reference to forward
or reflected voltages, currents or power. It is the basic
circuit theory description of what happens between two
networks. It works for DC, 60 HZ, RF, sinusoids, steps,
pulses, you name it. If there is any disagreement with the
above, please do not read further until it is resolved.

so you have derived voltages and currents at a couple of special
locations where superposition causes maximum or minimum voltages or
currents... this of course ONLY works after sinusoidal steady state
has been achieved, so your cases of connecting and disconnecting the
load must be thrown out. and how do you find a maximum or minimum
when the load is matched to the line impedance? standing waves are
always a trap... step back from the light and come back to the truth
and learn how to properly account for forward and reflected waves with
voltages or currents and all will be revealed.



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

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...

If(t) = 100/50 = 2a
Ir(t) = 100/50 = 2a

so there is a reflected 2a current you say? where does that go when
it gets back to the source? we know that when the 100v source is open
circuited as it must be in this case that it develops 100v across its
terminals... sure doesn't seem like any way that 2a can go back into
the source. it can't be reflected since the source impedance matches
the line impedance. so where does it go? maybe those simple
algebraic expressions have to be reconsidered a bit.

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

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

hmm, you do seem to get the right constant voltage... but it kind of
messes up your power flow since Vr=0 there can be no power in that
reflected wave, so it really doesn't make it much of a wave.

but wait, the equations must be consistent so

Ir(t)=0/50

So there really is no reflected current... but there is forward
current? so where does that forward current go if there is no load?
it doesn't reflect back if Ir=0... does it just build up at the open
end? going to be lots of trons sitting there after a while.





Some people look at the expression If(t)=Vf(t)/R0 and
think it looks a lot like a travelling wave so they decide
to compute the energy being moved by this wave:
* Pf(t)=Vf(t)*Vf(t)/R0
* Pr(t)=Vr(t)*Vr(t)/R0

So let us explore the results when applied to the DC examples from
above.

Terminated with 50 ohms:
* Vf(t)=50V
* If(t)=1A
* Vr(t)=0V
* Ir(t)=0A

* Pf(t)=50W
* Pr(t)=0W

This all looks good, no reflected wave and the appropriate
amount of energy is being transported in the forward wave.

Let us try the open ended line:
* Vf(t)=50V
* If(t)=1A
* Vr(t)=50V
* Ir(t)=1A

which yield
* V(t)=Vf(t)+Vr(t)=100V
* I(t)=If(t)-Ir(t)=0A

which is in agreement with the observations. But there is
definitely a forward voltage wave and a reflected voltage
wave. Carrying on to compute powers:

sorry, you can't have a 'voltage wave' without a 'current wave'...
just doesn't work i'm afraid. to have one you must have the other.

* Pf(t)=50W
* Pr(t)=50W
and
* P(t)=Pf(t)-Pr(t)=0W

So there you have it. The interpretation that the forward
and reflected wave are real and transport energy leads to
the conclusion that on a line charged to a constant DC
voltage, energy is flowing in the forward direction and
energy is flowing in the reflected direction.

obviously incorrect, by reductio ad absurdum if i remember the latin
correctly.


Because it is obvious that there is no energy flowing on
the line, this interpretation makes me quite uncomfortable.

Now there are several ways out of this dilemma.

Some say they are only interested in RF (and optics), and
that nothing can be learned from DC, steps, pulses or other
functions. They simply ignore the data and continue with
their beliefs.

Some claim that DC is special, but looking at the equations,
there is no reason to expect the analysis to collapse with
DC.

The best choice is to give up on the interpretation that the
forward and reflected voltage waves necessarily transport
energy. Consider them to be a figment of the analysis technique;
very convenient, but not necessarily representing anything real.

As an added bonus, giving up on this interpretation will remove
the need to search for “where the reflected power goes” and
free the mind for investment in activities that might yield
meaningful results.

...Keith

On Jun 14, 11:04*pm, K1TTT wrote:
read more »

no, i've read enough thank you.

I think there are 2 basic things you are missing... V and I are
functions not only of time, but of distance along the line, both for
the forward and reflected waves. looking at them at specific points
can lead to incorrect assumptions and over simplified conclusions.
Also, there are very good reasons why the sinusoidal steady state
condition is used to simplify the equations they way they are normally
shown, and why your over simplifications won't hack it in the long
run. to do the full analysis of a distributed system (except in very
special cases) for an arbitrary waveform requires summations of
reflections both ways on the line over many reflection periods.. the
results of this for step functions can be seen with simple tdr's on
lines with multiple discontinuities where you can see the reflections
ringing down.

Keith Dysart wrote:

. . .

So there you have it. The interpretation that the forward
and reflected wave are real and transport energy leads to
the conclusion that on a line charged to a constant DC
voltage, energy is flowing in the forward direction and
energy is flowing in the reflected direction.

Because it is obvious that there is no energy flowing on
the line, this interpretation makes me quite uncomfortable.

Now there are several ways out of this dilemma.

Some say they are only interested in RF (and optics), and
that nothing can be learned from DC, steps, pulses or other
functions. They simply ignore the data and continue with
their beliefs.

Some claim that DC is special, but looking at the equations,
there is no reason to expect the analysis to collapse with
DC.

The best choice is to give up on the interpretation that the
forward and reflected voltage waves necessarily transport
energy. Consider them to be a figment of the analysis technique;
very convenient, but not necessarily representing anything real.

As an added bonus, giving up on this interpretation will remove
the need to search for “where the reflected power goes” and
free the mind for investment in activities that might yield
meaningful results.

...Keith

Well done, Keith. Thanks.

Roy Lewallen, W7EL

On Jun 16, 4:11*pm, Roy Lewallen wrote:
Well done, Keith. Thanks.

Two ignorant people patting each other on the back. If DC methods
worked on distributed networks, there would never have been any need
to invent distributed network analysis - yet it was invented because
DC methods failed miserably for distributed network analysis. How does
a 1/4WL impedance transformer perform at DC? How does a DC analysis
work for light waves? Good grief!
--
73, Cecil, w5dxp.com

On Jun 16, 5:26*pm, Cecil Moore wrote:
On Jun 16, 4:11*pm, Roy Lewallen wrote:

Well done, Keith. Thanks.

Two ignorant people patting each other on the back. If DC methods
worked on distributed networks, there would never have been any need
to invent distributed network analysis - yet it was invented because
DC methods failed miserably for distributed network analysis. How does
a 1/4WL impedance transformer perform at DC? How does a DC analysis
work for light waves? Good grief!

Ahhh. Your standard answer. You are an RF dude and there is nothing
that can possibly be learned from simple DC circuits.
Especially when the results challenge your beliefs.

Still, I observe that you have discovered no errors in the exposition.
Which is good.

....Keith

On Jun 17, 5:38*am, Keith Dysart wrote:
You are an RF dude and there is nothing
that can possibly be learned from simple DC circuits.

We studied DC circuits in EE101. Later we were forced to expand our
knowledge into distributed networks because the DC circuit model
failed at RF frequencies.

Still, I observe that you have discovered no errors in the exposition.

I have not wasted my time time trying to discover any errors. In a
nutshell, what new knowledge have you presented?
--
73, Cecil, w5dxp.com

On Jun 17, 9:12*am, Cecil Moore wrote:
On Jun 17, 5:38*am, Keith Dysart wrote:

You are an RF dude and there is nothing
that can possibly be learned from simple DC circuits.

We studied DC circuits in EE101. Later we were forced to expand our
knowledge into distributed networks because the DC circuit model
failed at RF frequencies.

One does, however, usually require that the more sophisticated model
continue to provide the correct answer for the simpler problems that
could be solved in another way.

And it does here as well, though the answers make some uncomfortable.

Still, I observe that you have discovered no errors in the exposition.

I have not wasted my time time trying to discover any errors. In a
nutshell, what new knowledge have you presented?

Well, at least you continue to read the posts, so perhaps there is
some slim hope, but I am not holding my breath.

As for me, I have learned a lot from your postings since I first
encountered them circa 1994. Back then, I was not so knowledgeable
about transmission lines and I was trying to decide who was right.
You posted well thought out, convincing arguments and I found myself
switching from side to side. Reflected waves heat finals, no they
don't, yes they do, ... but eventually I learned. Then I continued
to read your posts, visited your web site and studied your arguments.
The challenge was to discover where you had gone wrong. Occasionally
I would think up a suitable counter example and offer it to you, but,
even then, you refused to look at examples that might threaten your
beliefs. So I no longer post with any expectation that you might be
persuaded.

Mostly now I post to ensure that a view other than yours is voiced
to help prevent those who are not yet sure from being sucked in to
your vortex.

Anyhow, I suggest you read my examples and find the flaws, for that
is the surest way to convince me that I am wrong. And don't just
post other examples that support your position, for it only takes
one counter-example to disprove a hypothesis, no matter how many
examples are in agreement. And don't reject simple DC examples
because
they lead to uncomfortable answers for it is by examining the
examples
that do not support your hypotheses that you will learn, not by
sticking
just with the ones that do.

....Keith

On Jun 18, 7:00*pm, Keith Dysart wrote:
Reflected waves heat finals, no they
don't, yes they do, ... but eventually I learned.

Hopefully, you learned that rail arguments are rarely valid and the
truth usually lies somewhere in between. Sometimes reflected waves
heat finals and sometimes they don't. It all depends on what kind and
how much interference exists between the forward wave and reflected
wave at the source impedance and (apparently) how much of that
impedance is dissipative or non-dissipative. Roy's food-for-thought
article doesn't take interference into account at all and therefore
arrives at magic conclusions divorced from reality.

Anyhow, I suggest you read my examples and find the flaws, for that
is the surest way to convince me that I am wrong.

I have pointed out many of the flaws in your examples. The way you
parse the energy violates the laws of physics. Maxwell's equations
only work on a function that contains classic EM traveling waves.
Standing waves and DC steady-state examples don't meet the necessary
boundary conditions. In fact, it can be proved that EM waves do not
existent during DC steady-state. The fact that an instrument designed
to detect EM waves malfunctions during DC steady-state is not
unexpected. Even an SWR meter calibrated for a different Z0 than the
one being used will indicate false results.

You really need to learn to recognize when your experiment and your
math model are contradictory, when your concepts violate the laws of
physics, and when you have chosen the wrong measuring equipment. You
remind me of myself when I was 14 years old and tried to measure the
screen voltage on a vacuum tube using a Simpson multimeter. It wasn't
anywhere near the voltage specified in the manual so I wasted my money
by buying a new tube the next time my family made a trip to Houston.

You would report those same measurement results as proof of a strange,
magical happening.
--
73, Cecil, w5dxp.com