Thanks guys -- some really great posts here -- it will take me quite a
while to digest all this!!

Thanks Again,

-Bill

"billcalley" wrote in message
oups.com...
We are all told that VSWR doesn't matter when using low loss
transmission lines, since the RF energy will travel from the
transmitter up to the mismatched antenna, where a certain amount of
this RF energy will reflect back towards the transmitter; after which
the RF will then reflect back up to the antenna -- where the energy is
eventually radiated after bouncing back and forth between the
transmitter and antenna. I understand the concept, but what I don't
quite understand is why the reflected RF energy isn't simply absorbed
by the 50 ohm output of the transmitter after the first reflection?
For the RF to bounce back and forth, wouldn't the transmitter's
impedance have to be very, very high (or low) when the reflected RF
energy hit its output stages? I know I'm missing something vital
here...

Thanks!

-Bill



Your mistake is that you assume the output of the tx is 50 ohms, in the case
you stated the transmitter must be matched to the impedance it sees looking
into the transmission line.

On Mon, 12 Mar 2007 11:00:04 GMT, Jan Panteltje
wrote:

Also normally, there is a pi type filter (to prevent harmonics), between
amplifier and antenna.
This filter _WILL_ match the antenna to the output impedance of the
transmitter, so _even_ if the transmitter output impedance is very
very low (low voltage high current output stage for example), the reflected
power will be nicely converted to match the transmitter, and heat up the
output amp, with its possible destruction as result.

Hi Jan,

Actually, there is a transformer there in the typical Ham transmitter
(and probably in every general purpose power source) that typically
transforms the native Z to the output Z. This is a step up for solid
state, and step down for tubes. In the solid state rigs, it is a
literal transformer feeding the 1-2 Ohms through a 5:1 winding ratio
to a switched bank of low pass filters. This stuff is mud ordinary.

As for the reflected energy, depending upon the phase it will either
combine destructively (heat) or constructively (cool) in the extremes.
There are, of course, 179 degrees of variation between these extremes
before they repeat themselves again. Cooling, of course, is something
of a misnomer as nothing useful is happening (poor power transfer) so
perhaps the terms should be destructive through uselessly benign.

73's
Richard Clark, KB7QHC

I understand the concept, but what I don't
quite understand is why the reflected RF energy isn't simply absorbed
by the 50 ohm output of the transmitter after the first reflection?

In my opinion the simplest way to answer your question is that you are assuming
that the transmitter is equivalent to a 50 ohm load, which is not true because
the transmitter is instead equivalent to the series of a 50 ohm load and a
voltage generator.

A simple DC example grossly clarifies thre issue: connect a 12V battery to the
series of a 50-ohm load and another 12V battery. How much current flows through
the load? Naught (assuming the correct polarity). So no power is dissipated in
it.

73

Tony I0JX

In message .com,
billcalley writes
We are all told that VSWR doesn't matter when using low loss
transmission lines, since the RF energy will travel from the
transmitter up to the mismatched antenna, where a certain amount of
this RF energy will reflect back towards the transmitter; after which
the RF will then reflect back up to the antenna -- where the energy is
eventually radiated after bouncing back and forth between the
transmitter and antenna. I understand the concept, but what I don't
quite understand is why the reflected RF energy isn't simply absorbed
by the 50 ohm output of the transmitter after the first reflection?
For the RF to bounce back and forth, wouldn't the transmitter's
impedance have to be very, very high (or low) when the reflected RF
energy hit its output stages? I know I'm missing something vital
here...

It matters when it changes suddenly, like mine did recently on my 70MHz
beam, when one of the elements came off in a gale.

Brian GM4DIJ
--
Brian Howie

billcalley wrote:

We are all told that VSWR doesn't matter when using low loss
transmission lines, since the RF energy will travel from the
transmitter up to the mismatched antenna, where a certain amount of
this RF energy will reflect back towards the transmitter; after which
the RF will then reflect back up to the antenna -- where the energy is
eventually radiated after bouncing back and forth between the
transmitter and antenna.

As pointed out, VSWR does matter. A lot of bouncing means you heat the
transmission line with the power instead of radiating the power.
'Doesn't matter', really means it can be tolerated if need be.

I understand the concept, but what I don't
quite understand is why the reflected RF energy isn't simply absorbed
by the 50 ohm output of the transmitter after the first reflection?
For the RF to bounce back and forth, wouldn't the transmitter's
impedance have to be very, very high (or low) when the reflected RF
energy hit its output stages? I know I'm missing something vital
here...

Here is what you are missing. In the case of the output, (real/resistive
component of the transmitter), seeing the reflected wave, it is _not_
reflecting that power back up the transmission line as you think it is.
It would go back to that real impedance and heat the transmitter. Here
is what is done with a miss match in the real world.

trans-output - match - line - antenna

The 'match' is where the magic happens. All the energy coming down the
line that got reflected from the antenna 'sees' the 'trans-output -
match' as a perfect reflector and gets bounced back[*]. On the other
side of the match is the trans-output. There the trans-output sees a
perfect impedance, (technically, the conjugate of the trans-output), so
that all the power travels through the match toward the antenna.

The magic is that when the match is tuned, both of the above conditions
are satisfied.

*The reflected wave sees a purely reactive reflector not just because of
the network but also because of the output power of the transmitter.
Without transmitter power the impedance as seen from the load will
dramatically change.

Best, Dan.

On Mon, 12 Mar 2007 13:37:00 -0400, "Jimmie D"
wrote:

Your mistake is that you assume the output of the tx is 50 ohms,

Hi Jimmie,

At the risk of yet another, non-quantitative reply I will repeat:
a question that has NEVER been answered by those who know what the
transmitter output Z ISN'T:
"What Z is it?"

in the case
you stated the transmitter must be matched to the impedance it sees looking
into the transmission line.

THAT is true, and it brings us to the point of all this energy
sloshing around until the antenna finally dissipates it out into the
Æther. It is the reflection off the mismatch of the tuner (the
mismatch seen by the antenna as source to the line going back) that
prevents energy from presenting any destructive results to the source
- the whole point of using a tuner in the first place.

73's
Richard Clark, KB7QHC

"Richard Clark" wrote in message
...
On Mon, 12 Mar 2007 13:37:00 -0400, "Jimmie D"
wrote:

Your mistake is that you assume the output of the tx is 50 ohms,

Hi Jimmie,

At the risk of yet another, non-quantitative reply I will repeat:
a question that has NEVER been answered by those who know what the
transmitter output Z ISN'T:
"What Z is it?"

in the case
you stated the transmitter must be matched to the impedance it sees
looking
into the transmission line.

THAT is true, and it brings us to the point of all this energy
sloshing around until the antenna finally dissipates it out into the
Æther. It is the reflection off the mismatch of the tuner (the
mismatch seen by the antenna as source to the line going back) that
prevents energy from presenting any destructive results to the source
- the whole point of using a tuner in the first place.

73's
Richard Clark, KB7QHC

Correct but I just want to remember that the purpose of the tuner is to
match the impedance of the transmitter to the impedance of the antenna/
transmission line.The standing waves can be viewed as a reflect voltage, a
reflect current or as a reflected impedance. Besides I thought there had
been enough quanitative analysis of the question and was hoping a simple
answer may be enough to turn on the light bulb for the OP. If he still
wanted to know more I figure he would ask.

On Mon, 12 Mar 2007 19:30:59 +0100, "Antonio Vernucci"
wrote:

A simple DC example grossly clarifies thre issue: connect a 12V battery to the
series of a 50-ohm load and another 12V battery. How much current flows through
the load? Naught (assuming the correct polarity). So no power is dissipated in
it.

Hi Tony,

Turn the second battery over. Double the power is dissipated in it.

Phase, you can't live with it, you can't live without it.

73's
Richard Clark, KB7QHC

A simple DC example grossly clarifies thre issue: connect a 12V battery to the
series of a 50-ohm load and another 12V battery. How much current flows
through
the load? Naught (assuming the correct polarity). So no power is dissipated in
it.

Hi Tony,

Turn the second battery over. Double the power is dissipated in it.

Phase, you can't live with it, you can't live without it.

73's
Richard Clark, KB7QHC

Of course. Mine was just a DC example to illustrate things in a simple manner.

When the transmitter is properly tuned, the phase relationship is such that the
reflected wave does not get dissipated at all into the 50 ohm output of the
transmitter, and is then reflected back to the antenna

Tony I0JX