"Cecil Moore" wrote in message
.. .
Dave wrote:

"Cecil Moore" wrote in message
Also assume that the impedance bump caused by the insertion of the
Bird is negligible, i.e. the same net voltage and net current exists
whether the Bird is in or out of the circuit.

you can not assume this, especially if you consider the actual results of
the experiment. the bird is a 50 ohm transmission line segment and it is
seeing a 50 ohm load. the 50 ohm load plus 1/2 wave of anything is 50
ohms. so as far as the bird knows there is no reflected power and the
real world reading is correct.

Remove the Bird, reconnect the two pieces of 75 ohm coax, and I'll
bet you will measure 70.7 volts and 1.414 amps at that point, 1/2
WL back from the 50 ohm load. If so, the Bird is NOT changing the
conditions when it is inserted. And we know there has to be reflected
power all up and down the 75 ohm coax. There is no place in the system
on the load side of the tuner where reflected power is zero. The
reflected power exists and the Bird reports a bogus reading for it.

With or without the Bird in the circuit the ratio of net voltage
to net current at the measurement point, 1/2WL away from the 50
ohm load, will be 50 ohms according to transmission line theory.
That's all the Bird is seeing - a V/I ratio of 50 which is irrelevant
to SWR unless Z0 is known and Z0 is known to be 75 ohms, NOT 50 ohms.

The ratio of net voltage to net current is All THE BIRD EVER SEES.
"As far as the bird knows ..." it is embedded in a 50 ohm environment.
But the Bird is not all-knowing. In this case, it is embedded in a
75 ohm environment and is giving bogus readings because all it sees
(samples) is a net voltage to net current whose ratio is 50 ohms.

It is what the Bird doesn't know that is important. The Bird doesn't
know that it is not embedded in a 50 ohm environment, and in its
ignorance of that fact, reports bogus results.

The actual SWR on a lossless line doesn't change. Yet, in another
posting, I showed that moving the Bird 1/4WL closer to the load
caused a reported SWR change by the Bird from 1:1 to 2.25:1. How
could both results possibly be right?
--
73, Cecil http://www.qsl.net/w5dxp

obviously at 1/4 wave from the load the impedance seen by the bird is not 50
ohms. you just can't ignore that internally the bird is a 50 ohm line, even
though it is only a couple inches long it is 50 ohms, put a 50 ohm load on
it and it will read no reflected power. this is just how it works, you
can't wish it to be anything else, especially when the actual measurements
prove me correct. you may try to assume away the bird and its
characteristic impedance, but it just won't go away as evidenced by the real
world measurements. you have to realize that the bird is not measuring what
is going on outside it's case, the only thing it can measure is the voltage
and current in its little 50 ohm internal world. so look at it from the
meter's point of view... it looks out the load side and sees 50 ohms, it
can't know that it is really a 75 ohm line, in fact it can't know there is
any line there at all, it just sees a 50 ohm load, therefore no reflection
into the meter, and no reflected power reading. qed.