On 7/5/2015 12:21 PM, Jeff Liebermann wrote:
On Sat, 04 Jul 2015 22:37:41 -0400, Jerry Stuckle
wrote:

On 7/4/2015 9:43 PM, Jeff Liebermann wrote:
Let's see if I understand you correctly. You claim that with a power
amplifier (source) output impedance that is perfectly matched to the
coax cable, but not necessarily the load (antenna), any reflected
power from the load (antenna) is bounced back to the load (antenna) by
the perfectly matched source (power amp). Is that what you're saying?

With a perfect matching network and a perfect feedline (which is what we
are discussing), that is true.

Ok, so we agree on that part.

Yet, when I have a perfectly matched load (antenna), all the power it
is fed is radiated and nothing is reflected. You can't have it both
ways because the reflected power from the load (antenna), becomes the
incident power going towards the source (power amp). Matched and
mismatched loads do NOT act differently depending on the direction of
travel. If you claim were true, then transmitting into a matched
antenna or dummy load would reflect all the power back towards the
transmitter.

Nothing wrong with it at all

Ah, but there's plenty wrong with your view. At the load (antenna)
end of the coax, we both agree that with a perfect match, perfect
coax, and perfect load, there's is no reflection. Yet when you look
at the other end of the same coax, the same perfectly matched coax
(S21 back into the PA) suddenly decides to reflect any RF that might
be returned from a load (antenna) mismatch. It seems rather odd that
RF would act differently at opposing ends of the coax cable. In the
forward direction, a matched load either absorbs or radiate. In the
reverse direction, a matched load changes its mind and decides to
reflect? I don't think so.


That's because you don't understand a perfect matching network or
feedline - which is what we are discussing. Your knowledge of theory is
sadly lacking.

There's another problem with your view of how VSWR works. If I
transmitted into an open or short circuit load (antenna), all the
forward RF would be reflected back to the source (PA). That would
mean that the PA will need to protect itself from over voltage or over
current using the traditional VSWR protection circuit. Yet, if the PA
were perfectly matched into a perfect coax cable, all that reflected
RF would bounce off the PA and back to the load (antenna). The PA
would not see any of that RF, and there would not be any need for a
VSWR protection circuit. I don't think so.


Once again, you're not discussing the same subject.

Personal experience with blowing up finals has demonstrated to my
satisfaction that a perfectly matched PA is quite capable of being
blown up by transmitting into an open or short with no VSWR
protection. Presumably, the damage was done by the reflected RF
(causing over voltage or over current in the PA) which would not be
present in your scheme of things, with a perfectly matched PA and
coax.


See above.

Another problem is IMD (intermodulation) products produced in the
power amplifier (PA). This is not a major problem with HF radios, but
is a serious problem with mountain top repeater sites. The antennas
on the towers tend to be rather close together. RF from a nearby
transmitter can couple into adjacent antennas, travel down the coax to
the PA, mix with the transmit signal in the PA, get amplified by the
PA, and get re-radiated by the antenna. The effect is typically
blocked by cavity filters and one-way isolators or circulators. The
problem here is that if the perfectly matched PA really did reflect
anything coming down from the antenna back to the antenna, there would
be no need for such IMD protection. The RF from the adjacent antenna
would simply bounce back towards the antenna and be re-radiated
without any mixing taking place. It would be a wonderful world if it
worked that way, but it obviously does not.


Once again, see above.

There are other problems, but they require math to explain, which
requires more time than I have available right now.



Because it's obvious you don't understand the theoretical aspects of the
system. Sure, they don't exist in the real world. But they provide a
simplified system for a start to the math; real world deficiencies can
then be added to the math to define the real-world aspects.

But you don't have the background to understand the theoretical aspects
we are discussing. I suggest you quit showing your ignorance.

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Jerry, AI0K

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