In message , Roger Hayter
writes
Ian Jackson wrote:

In message , rickman
writes
On 7/3/2015 2:50 AM, Jeff wrote:

Are you suggesting that the conjugate match will reflect back to the
antenna 100% of the original reflected wave from the antenna?


Yes, it must.

For example with an external ATU that provides a conjugate match it is
clearly the case that if a 1:1 VSWR is achieved then no reflected power
reaches the TX. (as shown on an SWR meter between the Tx and ATU.)

I am very certain that this assumption is not correct. I wish I had
the math to back me up. The only total reflection I am aware of is an
open circuit which of course absorbs no power at all.

Here is a point. The VSWR only shows no power being sent back to the
txmt output. That does not mean no power is absorbed from the
reflected wave by the matching circuit. I believe the example you gave
was Z of 1063 -j0. Isn't that a real impedance of 1063 ohms which is
equivalent to a resistor? Resistors dissipate power don't they?

I have to admit that I am, to some extent, confused.

Maybe it helps to look at the situation from the point of view that the
matching circuit doesn't 'know' that there is a reflected wave. All it
sees is the impedance looking into the sending end of the coax - and
this is whatever is on the antenna end, transformed by the length of
coax. The load the matching unit sees could be replaced with the same
physical values of L, C and R, so there IS nowhere for a reflected wave
to exist.

Provided the TX sees a 50 ohm load when looking into the input of the
matcher, there will be no theoretical losses. However, a real-life
matcher WILL have loss, and so will the coax. Also, the coax will have a
loss greater than when it is matched, mainly because of the 'I -squared
R' (literal) hot-spots.

Surely it *is* the reflected wave that mediates the transformation of
the aerial impedance to what is seen at the transmitter end? The
transmitter sees the vector sum of all the waves traversing the
transmission line at that point. Or else how would it "know" what was
happening at the other end?

I guess that until reflections are received back from the far end of the
coax, the transmitter will see the 50 ohms Zo (surge impedance) of the
coax. But once things have settled worn, the transmitter neither knows
nor cares what's at the far end. All it knows is that the load presented
to it isn't what it ought to be. But insert a matcher, and it will be as
happy as Larry. The system will work fine, but will suffer the penalty
of the additional SWR losses on the coax, and those of the matcher.
Provide these are not unacceptable, the benefit is that all the matching
can be done in the comfort of shack.
--
Ian