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

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

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

On Fri, 02 Mar 2007 06:16:15 GMT, Owen Duffy wrote:

that does not imply matching in
the Jacobi Maximum Power Transfer Theoram sense.

-um, OK-

I find negative propositions a bit oblique, what DOES it imply?

73's
Richard Clark, KB7QHC

Richard Clark wrote in
:

On Fri, 02 Mar 2007 06:16:15 GMT, Owen Duffy wrote:

that does not imply matching in
the Jacobi Maximum Power Transfer Theoram sense.

-um, OK-

I find negative propositions a bit oblique, what DOES it imply?

I was responding to your words:

"I will at this point re quote Chipman to roughly this scenario (being
more general, he didn't specify the reflection).

"At the signal source end of the line ... none of the power
reflected by the terminal load impedance is re-reflected on
returning to the input end of the line."
The ellipsis reveals that the source Z matches the line Z."

I disagree that the conditions that exist in the steady state at the
source end of the line imply in the general case, matching in the Jacobi
Maximum Power Transfer Theoram sense.

Owen

Owen Duffy wrote:
To find the steady state solution, take a short cut, bypass the transient
analysis and jump straight to the converged situation.

That certainly works but in the process, many people
have ignored and/or forgotten the total energy content
of a transmission line with reflections is greater
than the net energy being transferred to the load.

A one microsecond long transmission line transferring
100 watts to a matched load (Pfor=100w, Pref=0w)
contains 100 microjoules of energy during steady-state.

A one microsecond long transmission line transferring
100 watts to a mismatched load (Pfor=200w, Pref=100w)
contains 300 microjoules of energy during steady-state.

The transmission line always contains exactly the right
amount of energy to support the forward and reflected
waves. The most logical conclusion is that forward and
reflected waves continue to exist during steady-state
and those EM waves cannot stand still.

Forward and reflected waves during steady-state are
the building blocks of the standing wave which couldn't
exist without them. Forward and reflected waves contain
energy levels which can be easily calculated. The extra
energy, more than the matched flat case, is what causes
the extra losses due to SWR.
--
73, Cecil http://www.w5dxp.com

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

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