It's true that even a simple DC battery circuit can be described in
terms of incident and reflected power, SWR, and reflection
coefficients, but that's very unrealistic when trying to conceptually
understand what's going on.

I suppose a point that I'm making is that understanding how to match a
source with a load doesn't have much to do with reflections and
standing waves although, again, it can certainly be described
(unrealistically) that way.

Take the case of a transmitter and an antenna connected together with
a two-port black box between them, and that black box happened to
contain a transmission line. That unseen transmission line has
standing waves and reflections (assuming a mismatched condition),
losses, etc. all contained within the box. However, the transmitter
only sees a steady state complex impedance when looking into the box
(at a single frequency). Obviously there's some conjugate matching
that needs to take place for maximum power transfer, but there are no
reflections or standing waves involved outside the box -- unless
transmission line stubs are used for matching.

In short, transmission lines have reflections and standing waves, and
as a "black box" they affect how the load is seen by the source. But
extending the power reflection concept outside of that black box only
confuses things, even though it works mathmatically.

Al

Cecil Moore wrote in message ...
alhearn wrote:
Herein lies one of the big problems with the "reflection" definition,
conceptually.

That's why I often resort to a signal generator with a circulator/load
to illustrate my point. That signal generator *is* a constant power
source.

Therefore, what is commonly called "reflected power" is power that
never leaves the transmitter and is dissipated as heat by the
transmitter's internal 50 ohm impedance (if the transmitter's design
doesn't prematurely shut down first).

You can mount an argument that if the source doesn't see its
source impedance, then there is a reflection at that internal
mismatch. But that's not what is commonly called reflected power.

When we talk about reflected power on this newsgroup, we are usually
referring to the forward power rejected by a mismatch between the
transmission line Z0 and the antenna impedance (associated with mismatch
loss). In a typical ham radio antenna system, the "lost" reflected power
is forced to engage in destructive interference at the tuner and thus
joins the forward power wave.

alhearn wrote:
Take the case of a transmitter and an antenna connected together with
a two-port black box between them, and that black box happened to
contain a transmission line. That unseen transmission line has
standing waves and reflections (assuming a mismatched condition),
losses, etc. all contained within the box. However, the transmitter
only sees a steady state complex impedance when looking into the box
(at a single frequency).

That's a steady-state shortcut which assumes pure sine waves that
don't exist in reality. Please don't confuse steady-state shortcuts
with reality. Noise and modulation cause the "steady-state complex
impedance" not to be steady-state at all. Many will say it's close
enough, but one cannot understand reflections by assuming an un-
varying steady-state.

In a TV system with ghosting due to reflections, the unvarying steady-
state condition doesn't exist. In fact, when you assume steady-state
conditions, you eliminate ghosting, at least in your own mind. In
reality, steady-state doesn't really exist because of random noise
and unpredictable modulation.
--
73, Cecil http://www.qsl.net/w5dxp



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"Cecil Moore" wrote in message
...
That's a steady-state shortcut which assumes pure sine waves that
don't exist in reality. Please don't confuse steady-state shortcuts
with reality. Noise and modulation cause the "steady-state complex
impedance" not to be steady-state at all. Many will say it's close
enough, but one cannot understand reflections by assuming an un-
varying steady-state.

In a TV system with ghosting due to reflections, the unvarying steady-
state condition doesn't exist. In fact, when you assume steady-state
conditions, you eliminate ghosting, at least in your own mind. In
reality, steady-state doesn't really exist because of random noise
and unpredictable modulation.

The more you talk about 'steady-state-, the more apparent it becomes that
you don't really understand what it means.

73, Jim AC6XG

Concept below

However this is not an accurate model of a transmitter.

For an example, take an old Heathkit DX-100 generating a measured 180 watts
of CW RF into a matched 50 ohm load. To do this, it does NOT also dissipate
180 watts of RF into some "virtual" internal RF load in the DX-100. In
fact, the PAs and power supply in the DX-100 could not produce a total RF
output power of 360 watts without exceeding their ratings.

The dissipation in the PA is essentially related only the DC to RF
conversion efficiency of the PA, which in this case probably is about 75%,
max (Class C). So a PA input power of about 240 watts DC is required to
produce 180 watts of RF output power. The other 60 watts of plate input
power is converted to heat by the PA tube anodes.

The entire RF output generated by the PA stage is applied virtually 100% to
the output connector. How much of that is absorbed by the load connected
there is a function of load SWR and system losses.

- RF
________________

"alhearn" wrote
If you leave out the complex part of impedences for the moment and
think of 100 volt generator that has a 50 ohm internal impedance
driving a 50 ohm load, current is 1 amp and the power dissipated by
the load is 50 watts. There is also 50 watts dissipated by the
generator's internal impedance, for a total of 100 watts dissipated by
the entire system. Therefore, the "available" power for this generator
is 50 watts.

"Richard Fry" wrote in message
.........................
................................
Concept below

However this is not an accurate model of a transmitter.

For an example, take an old Heathkit DX-100 generating a measured 180
watts
of CW RF into a matched 50 ohm load. To do this, it does NOT also
dissipate
180 watts of RF into some "virtual" internal RF load in the DX-100. In
fact, the PAs and power supply in the DX-100 could not produce a total RF
output power of 360 watts without exceeding their ratings.

The dissipation in the PA is essentially related only the DC to RF
conversion efficiency of the PA, which in this case probably is about 75%,
max (Class C). So a PA input power of about 240 watts DC is required to
produce 180 watts of RF output power. The other 60 watts of plate input
power is converted to heat by the PA tube anodes.

The entire RF output generated by the PA stage is applied virtually 100%
to
the output connector. How much of that is absorbed by the load connected
there is a function of load SWR and system losses.

- RF
There is a Motorola ap note that agrees with what Richard is saying. To
paraphrase it, if the the DX100 had an output impedance of 50 Ohms, then the
overall efficiency would be 37.5%.
.....................................

"Tam/WB2TT" wrote in message
...

"Richard Fry" wrote in message
.........................
...............................
Concept below

However this is not an accurate model of a transmitter.

For an example, take an old Heathkit DX-100 generating a measured 180
watts
of CW RF into a matched 50 ohm load. To do this, it does NOT also
dissipate
180 watts of RF into some "virtual" internal RF load in the DX-100. In
fact, the PAs and power supply in the DX-100 could not produce a total
RF
output power of 360 watts without exceeding their ratings.

The dissipation in the PA is essentially related only the DC to RF
conversion efficiency of the PA, which in this case probably is about
75%,
max (Class C). So a PA input power of about 240 watts DC is required to
produce 180 watts of RF output power. The other 60 watts of plate input
power is converted to heat by the PA tube anodes.

The entire RF output generated by the PA stage is applied virtually 100%
to
the output connector. How much of that is absorbed by the load
connected
there is a function of load SWR and system losses.

- RF
There is a Motorola ap note that agrees with what Richard is saying. To
paraphrase it, if the the DX100 had an output impedance of 50 Ohms, then
the
overall efficiency would be 37.5%.

Unfortunately I can't read all the digressions in the thread. I skim by
author...

This is an interesting twist, Tam. I think if this were the case, then
there would be MORE power dissipated in the Tx than Mr. Fry is saying -
making the situation worse. By that, I mean, getting further from what is
going on. I think this goes in the wrong direction. I believe the flaw is
believing that the Rs=RL must exist for the transmitter.
--
Steve N, K,9;d, c. i My email has no u's.

On Mon, 24 May 2004 16:40:44 -0500, "Steve Nosko"
wrote:
This is an interesting twist, Tam. I think if this were the case, then
there would be MORE power dissipated in the Tx

Hi Steve,

And why would that be?

73's
Richard Clark, KB7QHC

"Steve Nosko" wrote in message
...

"Tam/WB2TT" wrote in message
...

"Richard Fry" wrote in message
.........................
...............................
Concept below

However this is not an accurate model of a transmitter.

For an example, take an old Heathkit DX-100 generating a measured 180
watts
of CW RF into a matched 50 ohm load. To do this, it does NOT also
dissipate
180 watts of RF into some "virtual" internal RF load in the DX-100.
In
fact, the PAs and power supply in the DX-100 could not produce a total
RF
output power of 360 watts without exceeding their ratings.

The dissipation in the PA is essentially related only the DC to RF
conversion efficiency of the PA, which in this case probably is about
75%,
max (Class C). So a PA input power of about 240 watts DC is required
to
produce 180 watts of RF output power. The other 60 watts of plate
input
power is converted to heat by the PA tube anodes.

The entire RF output generated by the PA stage is applied virtually
100%
to
the output connector. How much of that is absorbed by the load
connected
there is a function of load SWR and system losses.

- RF
There is a Motorola ap note that agrees with what Richard is saying. To
paraphrase it, if the the DX100 had an output impedance of 50 Ohms, then
the
overall efficiency would be 37.5%.

Unfortunately I can't read all the digressions in the thread. I skim
by
author...

This is an interesting twist, Tam. I think if this were the case,
then
there would be MORE power dissipated in the Tx than Mr. Fry is saying -
making the situation worse. By that, I mean, getting further from what is
going on. I think this goes in the wrong direction. I believe the flaw
is
believing that the Rs=RL must exist for the transmitter.

That is what I am saying. The efficiency goes from 75% to 37.5%; so, there
is more power dissipated in the TX.

Tam/WB2TT
--
Steve N, K,9;d, c. i My email has no u's.

"Richard Fry" wrote in message
...
Concept below

However this is not an accurate model of a transmitter.

For an example, take an old Heathkit DX-100 generating a measured 180
watts
of CW RF into a matched 50 ohm load. To do this, it does NOT also
dissipate
180 watts of RF into some "virtual" internal RF load in the DX-100. In
fact, the PAs and power supply in the DX-100 could not produce a total RF
output power of 360 watts without exceeding their ratings.

The dissipation in the PA is essentially related only the DC to RF
conversion efficiency of the PA, which in this case probably is about 75%,
max (Class C). So a PA input power of about 240 watts DC is required to
produce 180 watts of RF output power. The other 60 watts of plate input
power is converted to heat by the PA tube anodes.

The entire RF output generated by the PA stage is applied virtually 100%
to
the output connector. How much of that is absorbed by the load connected
there is a function of load SWR and system losses.

Finally someone who is getting closer to the flaw in the way many
mis-interpret the maximum power transfer theorm concept.
--
Steve N, K,9;d, c. i My email has no u's.

On Sun, 23 May 2004 02:07:13 GMT, "Henry Kolesnik"
wrote:

I know that any power not dissipated by an antenna is reflected back to the
transmitter. Then the transmitter "reflects" this reflection back to
antenna, ad nauseum until its all gone. I also know that a short or an open
is required to reflect power and I'm searching for which it is, an open or a
short. I'm inclined to think it's a virtual open but I'm at a loss to
understand that and I wonder if someone has a good explanation or analogy
and some math wouldn't hurt.
tnx
Hank WD5JFR


Hi Hank,

Round Two.

A short or an open certainly reflects. However, as you have observed
through your question, so does a poorly matched antenna; thus you must
agree that it presents neither a short nor an open. As such,
reflection is not confined to these two conditions.

We can display a condition where we have a 2:1 mismatch. This is
fairly commonplace as a consideration, if not as a reality. Here, the
reflected power is less than the total applied (some 12% if dead
reckoning is correct). That is, if the antenna presents a 2:1
mismatch to the power applied to it, nearly 90% of that power will
proceed to be radiated with a trivial portion returned to the source
(or the tuner). If we boost that mismatch to 10:1, that increases the
reflection substantially (let's call it 90% in this spirit of dead
reckoning) and naturally less is radiated. The math is quite as
simple as a balance ledger.

10:1 for a 50 Ohm source would mean either the load presents a Z of
500 Ohms, or 5 Ohms. From this you can see that we are approaching
either an open (hi-Z) or a short (lo-Z) and either perform the same
job of reflection - short of total reflection. As such, there is no
distinction to the power whether it encounters either, the reflection
ensues by virtue of the simple mismatch, not by the literal condition.
The computation of mismatch defines how reflective the interface is.

73's
Richard Clark, KB7QHC