"Jerry Stuckle" wrote in message
...
On 7/2/2015 1:56 PM, Ralph Mowery wrote:
"Jerry Stuckle" wrote in message
...

You are correct in that if a 75 ohm bridge is used, the indicated SWR
would be 1:1, because everything from that point on is 75 ohms.
However, the mismatch (and reflection) occurs on the transmitter side of
the bridge, not the antenna side. So the bridge will never see it. But
an accurate bridge will show lower power output due to the mismatch.

A mismatch is a mismatch, no matter where in the system it occurs. And
any mismatch will cause less than 100% power to be transferred. The
rest is reflected.

Just look at the specs of any amateur transceiver. They show an
impedance of 50 ohms. So a load of 50 ohms provides for maximum power
transfer; any other impedance causes a mismatch.

--
The real impedance of the transmitter is not 50 ohms. It is whatever the
device is used in the final stage and the poewr level. For a 100 watt
transmitter it is in the thousand ohm range and for solid state devices it
is very low. The matching circuit is often fixed to be 50 ohms,but could be
made for most any impedance. The older tube circuits were adjustable by the
user for a range of somewhat bleow 50 ohms to around 200 ohms. Could be
more or less depending on the design.

The mismatch you are counting on for a 50 ohm transmitter and a 75 ohm
feedline and 75 ohm antenna is in the tuned circuits/matching circuit in the
transmitter. Whatever power comes out of the transmitter will make it to
the antenna minus the loss of the coax, but not additional loss due to swr.
The power comming out of a 50 ohm transmitter will be less due to mismatch,
but not because of swr of the antenna system which is 1:1.