Jim - NN7K wrote in news:C2LFh.5953$re4.1319
@newssvr12.news.prodigy.net:

And, in fact, the "Reflected power" is
"Re-reflected" from the Source, Back to
the Load, if memory serves (minus loss's
accumulated from the first "Reflection"
of power, if memory serves! Jim NN7K

All this consideration of re-reflected, re-re-reflected, re-re-re-re-
reflected energy is something that is appropriate to analysing how the
steady state is established, and yes I know it takes an infinite time,
but it establishes subtantially quite quickly. The transient converges to
the steady state.

To find the steady state solution, take a short cut, bypass the transient
analysis and jump straight to the converged situation. In the steady
state... The complex ratio of Vf to Vr at the load end is entirely
determined by the constraints of a transmission line of Zo and the
passive load. The complex ratio of Vf to Vr at the line input can be
determined from the load end conditions by applying the line transmission
formulas with the complex propagation constant. Knowing Vf and Vr at the
input of the line and Zo, the complex V/I ratio (the equivalent load)
that the loaded line input will enforce can be calculated. The power
delivered by the source can be then found by finding the voltage or
current it will supply into the equivalent load.

This might seem long winded, but it is a sight easier than solving some
thousands of iterations of reflection, re-reflection and so on.

Owen

On Fri, 02 Mar 2007 00:31:55 GMT, Owen Duffy wrote:

The misunderstandings will frustrate your analysis, so try again.

Hi Owen,

Is it noteworthy that I found a blistering hot resistor for exactly
the conditions you set forth? Was I balanced in my reply to note the
alternative did the opposite? Even with my gaff of missing the
halfwave description, was the discussion incomplete in noting there
being a spectrum of responses?

Given I come to exactly the same analysis, same solution, same
conclusion (with more explanation, perhaps in that I do demonstrate
the reflected energy is absorbed/dissipated/what-have-you in the
source resistor) as your posting timestamped 2033 hours my time - what
exactly do I need to try again? Did the intervening 4 hours between
this post and the second one of yours find some catharsis?

Could you give me a dope slap instead of a hint about this frustration
I seem to be suffering? Like, should I take an aspirin or a syringe
of morphine?

73's
Richard Clark, KB7QHC

Richard Clark wrote in
:

On Fri, 02 Mar 2007 00:31:55 GMT, Owen Duffy wrote:

The misunderstandings will frustrate your analysis, so try again.

Hi Owen,

Is it noteworthy that I found a blistering hot resistor for exactly
the conditions you set forth? Was I balanced in my reply to note the
alternative did the opposite? Even with my gaff of missing the
halfwave description, was the discussion incomplete in noting there
being a spectrum of responses?

Richard, I must admit I didn't read the rest of your post when you stated
that you didn't know the line length and requested that info.

I have now read it.

I make the comment that just because the situation exists where the Volts
and Current from the source are the same as the Volts and Current into
the line (they have to be don't they), that does not imply matching in
the Jacobi Maximum Power Transfer Theoram sense.


Owen

Owen Duffy wrote in
:

As an exercise, think of a generator that has a Thevenin equivalent of
some voltage V and a series impedance of R+j0, connected to a half
wave of lossless transmission line where Zo=R. To give a numerical
example, lets make V=100 and R=50, so Vr=50 and "reflected power"=50.
How much of the "reflected power" is dissipated in the generator. In
this case, the generator dissipates less heat than were it terminated
in 50 ohms.

Ok, the solution:

Lets examine the matched load scenario for a start, the 100V generator
with 50 ohms internal resistance and a matched 50 ohm load. The current
is 100/(50+50) or 1A, the power in the load is 1^2*50 or 50W, the power
dissipated in the source is 1^2*50 or 50W.

No lets look at the scenario with the o/c half wave lossless line
attached to the generator. In the steady state, current from the
generator is zero, dissipation in the generator is 0^2*50 or 0, voltage
at the generator terminals and at the o/c (load end) of the line is 100V.
At the o/c load end, the complex reflection coefficient is 1, so Vf=50V,
Vr=50V and "reflected power"=50^2/50 or 50W.

But, wait a minute, there is 50W of "reflected power" on the line, the
line is matched to the source, and there is zero dissipation in the
source, less than when it has a matched load.

Don't take anything above to mean that I represent that a simple linear
model is a good representation of a transmitter PA.

This simple example that shows that existence of "reflected power" on a
transmission line does not necessarily result in some or all of the
"reflected power" being dissipated in the generator.

I will leave it to Cecil to take to confuse this simple example with some
photon based complication.

Owen

Ya know, all the bits sent on to the great bit bucket in the sky
during this 'discussion' could have been avoided by simply reading
Walt Maxwell's "Reflections" until you understand it...
denny / k8do

On 2 Mar 2007 05:27:13 -0800, "Denny" wrote:



Ya know, all the bits sent on to the great bit bucket in the sky
during this 'discussion' could have been avoided by simply reading
Walt Maxwell's "Reflections" until you understand it...
denny / k8do

Hi Denny,

I am to presume you are adding bits to the bit bucket then? Your own
answer to any of Owen's examples would have trumped hoary advice.

73's
Richard Clark, KB7QHC

Owen Duffy wrote:
This simple example that shows that existence of "reflected power" on a
transmission line does not necessarily result in some or all of the
"reflected power" being dissipated in the generator.

I will leave it to Cecil to take to confuse this simple example with some
photon based complication.

No photons necessary, Owen. You are using a Thevenin
equivalent source. Paraphrasing Ramo et.al of "Fields
and Waves ..." fame: No valid conclusions can be
automatically drawn from the calculation of power
dissipation inside a Thevenin equivalent source.
(Sorry, I don't have the book with me for the exact
quote.)

For a lot of real-world sources, double the voltage
with zero current output would be very bad news.
--
73, Cecil http://www.w5dxp.com

Cecil Moore wrote:
Owen Duffy wrote:
This simple example that shows that existence of "reflected power" on
a transmission line does not necessarily result in some or all of the
"reflected power" being dissipated in the generator.

I will leave it to Cecil to take to confuse this simple example with
some photon based complication.

No photons necessary, Owen. You are using a Thevenin
equivalent source.

What is the dissipation in the generator using a Norton
source?
--
73, Cecil http://www.w5dxp.com

Cecil, W5DXP wroote:
"What is the dissipation in the generator using a Norton source?"

According to page 76 of Terman`s 1955 opus:
"Alternatively, a load impedance may be matched to a source of power in
such a way as to make the power delivered to the load a maximum (the
available power of the power source). This is accomplished by making the
load impedance the conjugate of the generator impedance as defined by
Thevenin`s theorem."

On page 75 Terman labels a Norton diagram: "Equivalent Arrangement".

Equivalence means the open-circuit voltage and short-circuit current are
the same whichever diagram represents the power source.

The power dissipated in the source under matched conditions depends not
on the diagramatic representation, but upon how much of the internal
resistance of the source behaves as a resistor does, and how much is
"dissipationless resistance". It`s real, but makes no heat. If it were
fictional, final amplifiers would be limited to 50% efficiency. We all
know many R-F amplifiers have discontinuous input power
which allows efficiencies much in excess of 50%.

Best regards, Richard Harrison. KB5WZI

Cecil Moore wrote in news:fuWFh.3131$M65.1761
@newssvr21.news.prodigy.net:

Cecil Moore wrote:
Owen Duffy wrote:
This simple example that shows that existence of "reflected power" on
a transmission line does not necessarily result in some or all of the
"reflected power" being dissipated in the generator.

I will leave it to Cecil to take to confuse this simple example with
some photon based complication.

No photons necessary, Owen. You are using a Thevenin
equivalent source.

What is the dissipation in the generator using a Norton
source?

The existence of cases that show that dissipation in the source does not
necessarily increase due to VSWR on the transmission line does not
support the assertion that "VSWR causes reflected power that is
dissipated in the source". One sound case is enough to disprove the
generality.

Sure, transforming the source to a Norton equivalent would produce an
answer, and in this case a different answer for what happens inside the
generator. That Norton equivalent source with half wave s/c line will
also produce zero dissipation in the source.

Cecil, it appears your motive is to create confusion to divert attention
from the cases that are inconsistent with the assertion that "VSWR causes
reflected power that is dissipated in the source".

I have no difficulty with the statement "a transmitter is usually
specified to work over a limited range of load impedances (often
specified as a maximum VSWR at the transmitter terminals), the user
should expect it works properly over that range and should understand
that operation outside of that range may expose it to voltages or
currents (consequent heat), that may cause permanent damage". This advice
can be given to a six hour ham without telling them any lies, but
imparting the knowledge that they need to operate safely.

The explanation expressed / supported by some here that "we use ATUs to
cause total re-reflection of power "reflected" from the antenna so
protecting the PA" is a nonsense explanation of how the ATU protects the
PA from the effects of a poor load.


Owen