Keith Dysart wrote:

You are certainly correct; many people believe in reflected power,
though I've
always found that to be a poor basis for my own beliefs.

You have also provided the classic example where the numerology works
and 'reverse power' offers a tidy explanation. I am sure this neat
example is
the basis for many people's belief.

Belief has nothing to do with it. Observations suffice. If you have any
doubt that EM radiation can be reflected, just look in a mirror. Did you
need to 'believe' you saw your reflection to confirm your observation?

What drove me to look at alternate explanations for these kinds of
examples
was that the 'reverse power' explanation fails miserably when the
power
gets back to the generator.

Red Herring. There is no 'failure' in the accounting of reflected power.
The observations and numbers work out quite nicely.

On Mar 21, 1:32 pm, Dan Bloomquist wrote:

Belief has nothing to do with it. Observations suffice. If you have any
doubt that EM radiation can be reflected, just look in a mirror. Did you
need to 'believe' you saw your reflection to confirm your observation?

I have yet to question the reflection of EM radiation, just the
existence of
"reverse power" in transmission lines.

Red Herring. There is no 'failure' in the accounting of reflected power.
The observations and numbers work out quite nicely.

A simple example that I can never make add up is a 50 Watt generator
with a 50 ohm output impedance, driving a 50 ohm line which is open at
the end. Using the "reverse power" explanation, 50 W of "forward
power"
from the generator is reflected at the open end, providing 50 W of
"reverse
power". Since the generator is matched to the line there is no
reflection
when this "reverse power" reaches the generator so it disappears into
the generator. If this is truly power, it must go somewhere else, be
dissipated, transformed into some other form or stored (based on the
conservation of energy principle). Where did it go?

Most correspondents agree that what happens depends on the design
of generator; dissipation either increases, decreases or stays the
same (compared to when the line was terminated in 50 Ohms and the
power going down the line is dissipated in the termination). This
does
not make an easy explanation for where that supposedly real power
goes. Of course, if it is not real power, then there is no issue,
which
leads one back to looking for explanations other than "reverse power".

....Keith

Keith Dysart wrote:

On Mar 21, 1:32 pm, Dan Bloomquist wrote:


Belief has nothing to do with it. Observations suffice. If you have any
doubt that EM radiation can be reflected, just look in a mirror. Did you
need to 'believe' you saw your reflection to confirm your observation?


I have yet to question the reflection of EM radiation, just the
existence of
"reverse power" in transmission lines.

Use of a TDR makes for a valid observation on the line.

Red Herring. There is no 'failure' in the accounting of reflected power.
The observations and numbers work out quite nicely.

A simple example that I can never make add up is a 50 Watt generator
with a 50 ohm output impedance, driving a 50 ohm line which is open at
the end. Using the "reverse power" explanation, 50 W of "forward
power"
from the generator is reflected at the open end, providing 50 W of
"reverse
power". Since the generator is matched to the line there is no
reflection
when this "reverse power" reaches the generator so it disappears into
the generator....

I explained this last week, albeit for a different reason. I'll paste:
***
Hi Richard,
He says it in the last sentence. But here is an example. Take a 50 ohm
thevinin source. Power off, it looks like 50 ohms back into it. Take a
second thevinin source to represent a reflection and drive 5 volts into
the first source. Now set your first source 180 degrees to the
reflection and drive forward 5 volts.

(s)-----/\/\/\--------(c)-----/\/\/\--------(r)

(s)source (c)connection (r)reflection. With (s) 180 degrees out of phase
from (r), (r) will see a short at (c). It is because of the power
generated at the source that the impedance into it can look purely
reactive. And, you can use 5 ohms with 1 volt at the source, (c) will
still look like a short to (r). The source resistance doesn't matter as
long as a 'match' is made.

And for the same reason, why the 50 ohm line doesn't look like 50 ohms
is because of reflected power. Drive an open quarter wave line and it
looks like a short because the reflected voltage is 180 degrees out from
the source.
***

If this is truly power, it must go somewhere else, be
dissipated, transformed into some other form or stored (based on the
conservation of energy principle). Where did it go?

The energy is sitting on the line. It didn't disappear. See all the
posts I made last week.


Most correspondents agree that what happens depends on the design
of generator;

Actually, it doesn't. The exception is minor and pointed out in the case
of a non linear source.

dissipation either increases, decreases or stays the
same (compared to when the line was terminated in 50 Ohms and the
power going down the line is dissipated in the termination). This
does
not make an easy explanation for where that supposedly real power
goes. Of course, if it is not real power, then there is no issue,
which
leads one back to looking for explanations other than "reverse power".

The mistake is in assuming the energy must disappear. It doesn't.

...Keith

Best, Dan.

On Mar 21, 4:25 pm, Dan Bloomquist wrote:
Keith Dysart wrote:
A simple example that I can never make add up is a 50 Watt generator
with a 50 ohm output impedance, driving a 50 ohm line which is open at
the end. Using the "reverse power" explanation, 50 W of "forward
power"
from the generator is reflected at the open end, providing 50 W of
"reverse
power". Since the generator is matched to the line there is no
reflection
when this "reverse power" reaches the generator so it disappears into
the generator....

I explained this last week, albeit for a different reason. I'll paste:
***
Hi Richard,
He says it in the last sentence. But here is an example. Take a 50 ohm
thevinin source. Power off, it looks like 50 ohms back into it. Take a
second thevinin source to represent a reflection and drive 5 volts into
the first source. Now set your first source 180 degrees to the
reflection and drive forward 5 volts.

(s)-----/\/\/\--------(c)-----/\/\/\--------(r)

(s)source (c)connection (r)reflection. With (s) 180 degrees out of phase
from (r), (r) will see a short at (c). It is because of the power
generated at the source that the impedance into it can look purely
reactive. And, you can use 5 ohms with 1 volt at the source, (c) will
still look like a short to (r). The source resistance doesn't matter as
long as a 'match' is made.

And for the same reason, why the 50 ohm line doesn't look like 50 ohms
is because of reflected power. Drive an open quarter wave line and it
looks like a short because the reflected voltage is 180 degrees out from
the source.
***

So you don't like my example? Well I can use yours then.

While you use the word 'power', the real analysis in your example is
all
done with volts. This is excellent and helps demonstrate my point that
'reverse power' is not needed as an explanation. We can carry on from
the analysis you have done and compute some powers. The real power at
(c)
is 0 Watts (the voltage is 0 at all times so using P=VI, the power
must
be zero). Assuming that when you say "drive 5 volts", you mean that
the
voltage source in the Thevenin equivalent generator is set to 10 V,
the 'forward power' at (c) is 0.5 W and the 'reverse power' is 0.5 W.
When subtracted, these produce the expected result of 0 W which agrees
with the actual computed power. All is well. (And yes, the Bird works
for determining this result).

Now consider someone who believes in the reality of 'forward' and
'reflected power'. There is 0.5 W of 'forward power' which reaches
the generator at the right. Since the impedance of this generator is
the
same as the characteristic impedance of the line (or the left
generator
in this example because there is no line), there is no reflection
so the 'power' must go into the generator. Similarly with no
reflection,
the 'reverse power' goes into the generator. Where does this power go?
If 'reverse power' is real, then it needs to be accounted. Since there
are many examples for which this accounting can not be done
it leads to the inescapable conclusion that 'reverse power' does not
really exist. (Please though, this does not mean that forward and
reverse voltages and currents do not exist. The fundamental error
being committed is that it is not valid to multiply the voltages and
currents computed using superposition to produce powers that actually
represent flowing energy).

Any reader is invited to prove me wrong by providing an accounting for
the 'reverse power' when it reaches the generator whose output
impedance
matches the characteristic impedance of the line (i.e. no reflection).

....Keith

Keith Dysart wrote:
Any reader is invited to prove me wrong by providing an accounting for
the 'reverse power' when it reaches the generator whose output
impedance
matches the characteristic impedance of the line (i.e. no reflection).

Destructive interference at the generator output
terminal is all the accounting one needs.

Here's a mental aid to help see what happens to
the reflected energy. Assume the output impedance of
the generator and the characteristic impedance of the
line are Z1. Assume THE GENERATOR SEES Z2 AS A LOAD.
Z2 is a general case impedance and could have any
value depending upon the load ZLoad which is not
equal to Z1. The load is mismatched and therefore
Pref1 is not zero.

Gen---Z1 t-line----ZloadZ1
Pfor1--
--Pref1

You say reflected energy flows into the generator.
I'll show you why it does NOT and that total
destructive interference is responsible for
eliminating reflections at the generator.

Add any length of lossless transmission line
with a characteristic impedance of Z2 at the output
of the generator (nothing changes) so it looks like:

Gen---Z2 t-line---+---Z1 t-line----loadZ1
Pfor2-- Pfor1--
--Pref2=0 --Pref1

Pfor1 and Pref1 are the same values as above.

The Z2 t-line has NO reflected waves because it
is terminated in its characteristic impedance Z2.
The Generator continues to see Z2 as a load. One
has to admit that nothing at the Generator output
has changed. THE GENERATOR SEES EXACTLY THE SAME
LOAD IMPEDANCE IN BOTH CASES. NOTHING HAS CHANGED
IN THE SYSTEM EXCEPT OUR ABILITY TO ANALYZE IT.

Now a simple energy analysis will indicate what
happens to the reflected waves on the Z1 t-line. Since
Pref2=0, the system is Z0-matched to Z2 at point '+'.
Total destructive interference is what happens to the
reflected waves on the Z2 t-line. All of the reflected
energy is re-reflected back toward the load at the Z2-
match point '+'.

NONE OF THE REFLECTED ENERGY REACHES THE GENERATOR SO
NONE OF THE REFLECTED ENERGY CAN BE ABSORBED BY THE
GENERATOR IMPEDANCE.

So there you have it - a detailed mathematically based
reason why the reflected energy does not make it into
the generator in your example. To fully understand why
zero reflected energy flows in the generator, one needs
to understand superposition along with destructive and
constructive interference. This is covered in more detail
in my WorldRadio Energy Analysis article at:

http://www.w5dxp.com/energy.htm

I would recommend that everyone interested in this subject
read the "Optics", by Hecht chapters on superposition and
interference of EM waves. It is the best treatment that
I have seen in writing.
--
73, Cecil http://www.w5dxp.com

Keith Dysart wrote:

On Mar 21, 4:25 pm, Dan Bloomquist wrote:

Keith Dysart wrote:

A simple example that I can never make add up is a 50 Watt generator
with a 50 ohm output impedance, driving a 50 ohm line which is open at
the end. Using the "reverse power" explanation, 50 W of "forward
power"
from the generator is reflected at the open end, providing 50 W of
"reverse
power". Since the generator is matched to the line there is no
reflection
when this "reverse power" reaches the generator so it disappears into
the generator....

So you don't like my example?

Your example assumes that the reflected power will see the 50 ohms of
the generator. And I had shown you a condition where it will see a short
no mater what the 'output' impedance of the generator.

While you use the word 'power', the real analysis in your example is
all
done with volts. This is excellent and helps demonstrate my point that
'reverse power' is not needed as an explanation. We can carry on from
the analysis you have done and compute some powers. The real power at
(c)
is 0 Watts (the voltage is 0 at all times so using P=VI, the power
must
be zero).

It is a short. It is just that simple. If you have any kind of lab, take
a long piece of coax and drive it with a pulse and watch with a scope.
Find out for yourself that energy will reflect off a short.

Assuming that when you say "drive 5 volts", you mean that
the
voltage source in the Thevenin equivalent generator is set to 10 V,
the 'forward power' at (c) is 0.5 W and the 'reverse power' is 0.5 W.
When subtracted, these produce the expected result of 0 W which agrees
with the actual computed power. All is well. (And yes, the Bird works
for determining this result).

Now consider someone who believes in the reality of 'forward' and
'reflected power'. There is 0.5 W of 'forward power' which reaches
the generator at the right. Since the impedance of this generator is
the
same as the characteristic impedance of the line (or the left
generator
in this example because there is no line), there is no reflection
so the 'power' must go into the generator.

No. (c) is a short. It is just that simple. The power does not flow past
it 'into the other generator'. If you have any doubt that power is
reflected from a short you have not made the observation. And if you
have not made an observation that contradicts this, you can not make the
claim of a contradiction.

snip assumptions based on a false premise

Any reader is invited to prove me wrong by providing an accounting for
the 'reverse power' when it reaches the generator whose output
impedance
matches the characteristic impedance of the line (i.e. no reflection).

If you stick to the premise that the reflected power sees the generator
impedance it can't be 'proven wrong'. I can say that gravity doesn't
exist all day long but that won't make it so. The observation will
'prove' me wrong.

On Mar 23, 11:05 am, Dan Bloomquist wrote:
Your example assumes that the reflected power will see the 50 ohms of
the generator. And I had shown you a condition where it will see a short
no mater what the 'output' impedance of the generator.


Truly, we have found the root of the disconnect.

Kindly compute the reflection coefficient at the connection where
a 50 Ohm line is driving a 75 Ohm line.

For your convenience, recall that RC = (Z2-Z1)/(Z2+Z1).

Now redo the same, except that the 75 Ohm line is one-half wavelength
long terminated in a short.

If you get the same answer, then you will see why the reflected
voltage in the example does not encounter a discontinuity at the
entrance to the generator and is therefore not re-reflected.

If you get a different answer, then some study of reflection
coefficient
is in order.

....Keith

Keith Dysart wrote:
Truly, we have found the root of the disconnect.

Truly, we have but it is not what you think. A
source doesn't obey the passive reflection rules.
The V/I ratio encountered within a source is
*active*, not passive. Active V/I ratios can and
do cause reflections.
--
73, Cecil http://www.w5dxp.com

Keith Dysart wrote:
I have yet to question the reflection of EM radiation, just the
existence of "reverse power" in transmission lines.

Then you are certainly engaging in the proverbial Red Herring.

A simple example that I can never make add up is a 50 Watt generator
with a 50 ohm output impedance, driving a 50 ohm line which is open at
the end. Using the "reverse power" explanation, 50 W of "forward
power"
from the generator is reflected at the open end, providing 50 W of
"reverse
power". Since the generator is matched to the line there is no
reflection
when this "reverse power" reaches the generator so it disappears into
the generator. If this is truly power, it must go somewhere else, be
dissipated, transformed into some other form or stored (based on the
conservation of energy principle). Where did it go?

This is a lot like the 1/2WL W7EL example in his food for
thought articles. The generator is *NOT* matched to the line
as it sees an open circuit and cannot continue to stuff 50
watts into the open circuit. The generator is as mismatched
as it can possibly be. The reflected wave also sees that open
circuit and is 100% reflected. Since the generator is not
delivering any power and there is a forward power and a
reflected power, the reflected power is supplying the
forward power. Anything else violates the conservation
of energy principle.

Most correspondents agree that what happens depends on the design
of generator; dissipation either increases, decreases or stays the
same (compared to when the line was terminated in 50 Ohms and the
power going down the line is dissipated in the termination). This
does
not make an easy explanation for where that supposedly real power
goes. Of course, if it is not real power, then there is no issue,
which
leads one back to looking for explanations other than "reverse power".

Any level of interference is possible depending upon the
phase angle between the forward E-field and the reflected
E-field. All this is explained in "Optics" by Hecht which
some people have apparently avoided reading/understanding.
Optical physicists solved this problem a century ago. They
don't have the luxury of dealing with voltages and currents
and are forced to deal with power densities. You should
try trodding their paths and enlightening yourself.
--
73, Cecil http://www.w5dxp.com

On Mar 22, 12:20 am, Cecil Moore wrote:

This is a lot like the 1/2WL W7EL example in his food for
thought articles. The generator is *NOT* matched to the line
as it sees an open circuit and cannot continue to stuff 50
watts into the open circuit. The generator is as mismatched
as it can possibly be. The reflected wave also sees that open
circuit and is 100% reflected. Since the generator is not
delivering any power and there is a forward power and a
reflected power, the reflected power is supplying the
forward power. Anything else violates the conservation
of energy principle.

I suggest that you have quite mixed up your impedances here
thus rendering all further analysis invalid.

At any point in the system there are 4 impedances.

There is the characteristic impedance looking left and the
characteristic impedance looking right. For systems in
sinusoidal steady-state, there is also the effective impedance
looking left and the effective impedance looking right.

The characteristic impedance is dependant only the elements of
the system. It does not depend on the length of the line or
the frequency of excitation. At changes in the characteristic
impedance, reflections occur. This impedance can be used for
transient as well as steady-state analysis of a system.

The effective impedance is dependant on the characteristic
impedances of the components of the system, as well as line
length and excitation frequency. It can only be used for
steady-state analysis. It does not cause reflections.

Until these impedances are kept straight in discussions, there
is no hope for correctness.

....Keith