On 10/5/2015 4:02 AM, rickman wrote:
On 10/5/2015 4:17 AM, Ian Jackson wrote:
In message , rickman
writes



You keep saying that the 1:1 match between the TX and the ATU prevents
any power from being sent to the TX which is not true. You are
confusing the power from the TX which is not reflected and the power
reflected from the antenna which passes through the ATU to the TX.

If the SWR meter between the TX output indicates a 1:1 SWR, then there
can be NO power travelling between the ATU input and the TX output - ie
there IS no reflected power. QED, surely?

If you ignore the losses in the ATU, all the power that the mismatched
antenna reflects, and that makes it back to the ATU output, MUST be
re-reflected by the ATU output impedance, and head off back towards the
antenna. This is because the reflected signal cannot heat up a lossless
ATU, and the SWR meter says it isn't coming back through the ATU. It
simply has nowhere to go except back down the coax.

You saying something is true or imagining a SWR reading is not the same
as understanding what is going on. What SWR reading are you imagining?
Can you explain this in terms of the circuit analysis? The ATU
consists of what circuit? The TX has some source impedance, what would
that be? I don't think you can design an ATU circuit that will isolate
the real source impedance of the TX from the reflected wave from the
antenna.


Okay, Rick, here ya go...

Using the antenna info given by Wayne (20-j130) and his transmission
line, I find the following:

Using characteristics of the line I found on the 'Net, I see that the
velocity factor is .66 and the loss is about 0.7dB/100 feet
(insignificant in this case, but I feel I must mention it).

This should transform the antenna to about 6.453+j52.544 ohms at the far
end (don't forget the velocity factor). I used a Smith chart but if you
wish to verify using the Telegraphers Equations, be my guest.

What does it take to make this impedance look like 50+j0 ohms?

One solution is a shunt capacitance at the far end of the cable of
140.5pF and then a series capacitance of 80.9pF to the transmitter. That
would be a shunt Xc of about 80.9 ohms and a series Xc of about -138.64
ohms respectively.

Please verify that this will present a 50 ohm load to the transmitter.
If so, the transmitter will see a 1:1 SWR and no power will be returned
from the matching network.