On 22 Mar 2007 20:08:59 -0700, "Dr. Honeydew"
wrote:

What sort of output impedance does this generator have?

With the "accurate RF power meter" determination of 1W, it appears to
be a precision source (of 1W). The output impedance IS the 50 Ohm
resistance you applied to it. This loading at the output terminals is
a conventional usage of a precision source.

73's
Richard Clark, KB7QHC

Dr. Honeydew wrote:
"We have a new signal generator here at the Labs, but it came without
documentation,"

Terman elaborates on Thevenin`s theorem on page 74 of his 1955 opus.
From Fig. 3-19 (b) on page 75, I would suggest loading the generator
with resistance until its open-circuit voltage is cut in half. That
resistance should equal the internal impedance of the generator which
you have already determined is resistive. Provided you don`t maintain
that sort of overload too long on your 1-watt source, you should do no
damage to it.

Best regards, Richard Harrison, KB5WZI

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

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

"Dr. Honeydew" wrote
Oh, and we thought to do another experiment. While the generator was
operating, we sent a short burst of RF down a 50 ohm transmission line
to the generator's output. Because the power measurements seem to say
the generator's output resistance is very low, we thought we would see
a return pulse, delayed by the round-trip time down the 15 meters of
high-quality cable. But we didn't see any echo. What could be going
on? We double-checked everything, and even looked into the line next
to the generator with a high impedance probe. We can see the burst
going in there, but still nothing comes back.
____________

Below is a link leading to a documented, real-world RF pulse measurement to
think about.

The reflection results from an antenna mismatch at the far end of ~1,600
feet of air-dielectric transmission line. A directional coupler and
calibrated, step attenuator allow for accurate measurement of the weak
return pulse amplitude in the presence of the much greater incident pulse.

Note that the reflection is (barely) seen even in the incident display,
time-aligned with its location in the reflected display.

Reflections DO exist in such systems, but measuring them requires
appropriate instrumentation and methodology.

http://i62.photobucket.com/albums/h8...easurement.gif

RF

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 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.

No, I may have confused you but I did not mix up the
impedances. What the generator and reflected wave "sees"
above refers to the effective or virtual impedance, the
ratio of V/I designated by Z1, Z2, etc. The characteristic
impedances are designated by Z01, Z02, etc.

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.

Yes, nothing I said disagrees with that. I will try to
draw an ASCII block diagram from now on.

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.

Yes, nothing I said disagrees with that.

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.

Yes, nothing I said disagrees with that.

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

OK, I will insert the impedances for you in what you quoted
above.

This is a lot like the 1/2WL W7EL example in his food for
thought articles.

50wGen---1/2WL open-circuit stub---Z=infinity
Pfor=50w--
--Pref=50w

The generator is *NOT* matched to the load as it sees an
open circuit (effective Z=infinity) 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 (Z=infinity) and is 100% reflected.
The reflected wave encounters the source wave which puts the
voltages in phase and the currents 180 degrees out of phase.
That's what happens at an open circuit.

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, i.e. being 100% re-reflected. The
re-reflection is associated with total destructive interference
toward the generator and total constructive interference
toward the load. Anything else violates the conservation of
energy principle.
--
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 22, 10:43 pm, Richard Clark wrote:
On 22 Mar 2007 20:08:59 -0700, "Dr. Honeydew"
wrote:

What sort of output impedance does this generator have?

With the "accurate RF power meter" determination of 1W, it appears to
be a precision source (of 1W). The output impedance IS the 50 Ohm
resistance you applied to it. This loading at the output terminals is
a conventional usage of a precision source.

73's
Richard Clark, KB7QHC


Thank you, Mr. Clark. But I'm having trouble reconciling what you
wrote with what Mr. Harrison posted about the output impedance being
equal to the conjugate of the load which dissipates the most power.
Clearly the 40 ohm load dissipates more power than the 50 ohm load, so
we don't see how your answer and Mr. Harrison's posting can both be
correct.

From the labs,
Dr. Honeydew

On Mar 20, 3:37 pm, Walter Maxwell wrote:
One of the issues discussed in this thread that Owen originated concerned whether or not reflected power
enters the power amp and dissipates as heat in the plates of the amp. Some of the posters apparently are
unable to appreciate that the reflected power does not cause heating of the amp, unless the reflected power
detunes the amp and the amp is left detuned from resonance, which of course is not the correct manner of
operating the amp.

In the last post of the original thread I presented the details of an experiment I performed (one of many
using the same procedure) on a Kenwood TS-830S transceiver that proves how and why reflected power in no way
causes heating of the amp when the amp is properly adjusted in the presence of the reflected power.

Usually, such a presentation as in the last post in that thread evokes a great deal of response, as for
example, Art Unwin's. So I'm somewhat surprised, and a little disappointed that my post has resulted in total
silence. Have my efforts in helping to solve the problem gone for naught?

Walt, W2DU

I have followed this thread with interest There are some who seem to
dispute the existance of reflected power and its ability to do damage.
My answear is that reflected power can certainly do so any one want to
run a completly detuned antenna to prove me wrong may do so at their
expence and risk. as to it causing heat in the amp of course not the
extra work that the amp has to do to "push" the signal to the antenna
does. if a semiconductive device is over worked then it will produce
heat.
The old thermonic valves are far more tolerant than semiconductors.
mike M0DMD