rickman wrote:
On 7/8/2015 8:34 PM, wrote:
rickman wrote:
On 7/8/2015 3:38 PM, John S wrote:
On 7/8/2015 10:48 AM, rickman wrote:
On 7/8/2015 10:09 AM, John S wrote:
On 7/2/2015 1:38 PM, rickman wrote:
On 7/2/2015 1:56 PM, Ralph Mowery wrote:
"Jerry Stuckle" wrote in message
...

Try this - connect the output of an HF transmitter to an SWR bridge.
Now connect a piece of 75 ohm coax such as RG-59 to the output of the
SWR meter, and connect that to a 75 ohm resistive load. Do you think
the SWR bridge will show a 1:1 SWR? Not a chance. It will be 1.5:1.



What you have described is a case of using the wrong swr bridge. You
are
trying to use a 50 ohm bridge on a 75 ohm system. If a 75 ohm
bridge is
used it will show a 1:1 SWR.

The real SWR is 1:1. With a 75 ohm line and 75 ohm load there is no
reflected power.

My knowledge of antenna systems is limited, but I do know that this is
correct, there will be no reflection from the antenna.

If there is no reflections from the antenna, how can there be a loss in
the source end? There is NO power returned according to your own
statement.

I don't see any contradiction. The power comes from the source through
the source impedance. The source impedance will create a loss, no?


If the
transmitter output is 50 ohms there will be a loss in this matching
that
will result in less power being delivered to the feed line, but that
will not result in reflections in the feed line.

Why? What causes the loss? The transmitter output resistance? So that
would mean that one can never achieve more that 50% efficiency at the
transmitter's OUTPUT! And that would mean that a 1000W transmitter is
dissipating 500 watts under the BEST circumstances. Good luck on getting
that to work to your satisfaction.

Maybe "loss" isn't the right term then. The output of a 50 ohm source
driving a 75 ohm load will deliver 4% less power into the load than when
driving a 50 ohm load. That comes to -0.177 dB. Is there any part of
that you disagree with?

All of it. Let's say you have a 1A source and it has a 50 ohm impedance
in series with its output. With a 50 ohm load it will provide 50W to the
load. With a 75 ohm load it will provide 75W to the load. The only
difference is that the 50 ohm load will cause the source voltage (before
the series impedance) to be 100V while the 75 ohm load will require 112V
(before the series impedance). If the series impedance is 0 +/- j75
ohms, it will have no power loss. If the series impedance is 50 + j0 it
will have a 50W loss.

I was referring to a voltage source.


Instead of arguing about it, one can download QUCS for free which
will simulate the whole thing and one can see what really happens.

Download QUCS for your operating system:

http://qucs.sourceforge.net/

Generate a model consisting of a voltage source with a series resistance
of a few Ohms to simulate a solid state source or a much higher
resistance to simulate a vacuum tube source. Chose a convienient
frequency for the source.

Go to: http://home.sandiego.edu/~ekim/e194r.../matcher2.html
to calculate an impedance matching network to match the resistance
you've chosen to 50 Ohms.

Put the matching circuit in the model.

Add a transmission line to the model.

Terminate the transmission line with a 50 Ohms resistor.

Add a fixed frequency AC simulation at the desired frequency.

Change various parameters to your heart's content to see what happens.

Change the matching network such that the output of your transmitter
is no longer 50 Ohms and see what happens.

When the QUCS output disagrees with your beliefs, you can argue with
the program.

Uh, I already did a simulation using LTspice. No one even commented on
the simulation as I recall. Besides, I don't really need a simulation
to measure the power delivered to a load. That is a *very* simple
circuit to calculate in a few seconds. But first we have to agree on
what we are discussing.

Does LTspice do transmission lines?

I'm not going to retrace this entire conversation, but someone said a
matching network which presents a complex impedance with a non-zero real
part and a zero imaginary part would reflect 100% of the wave which had
been reflected from the antenna back to the transmitter output. I was
trying to nail down this example so I could do some calculations on it.
If you want to do that I am happy to work on the problem. If not,
that's fine too.

Nope, I know what happens.

BTW, my response was not directed at any particular person other
than those that do not understand the conditions for maximum
power transfer given a fixed source.



--
Jim Pennino