July 5th 15, 04:35 PM
posted to rec.radio.amateur.antenna
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First recorded activity by RadioBanter: May 2011
Posts: 550
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An antenna question--43 ft vertical
On 7/5/2015 10:02 AM, Jerry Stuckle wrote:
On 7/5/2015 9:23 AM, Ian Jackson wrote:
In message ,
writes
Wayne wrote:
"Jeff Liebermann" wrote in message
...
On Sat, 04 Jul 2015 19:04:01 -0400, Jerry Stuckle
wrote:
Think of it this way, without the math. On the transmitter side of
the
network, the match is 1:1, with nothing reflected back to the
transmitter.
So you have a signal coming back from the antenna. You have a perfect
matching network, which means nothing is lost in the network. The
feedline is perfect, so there is no loss in it. The only place for
the
signal to go is back to the antenna.
Wikipedia says that if the source is matched to the line, any
reflections that come back are absorbed, not reflected back to the
antenna:
https://en.wikipedia.org/wiki/Impedance_matching
"If the source impedance matches the line, reflections
from the load end will be absorbed at the source end.
If the transmission line is not matched at both ends
reflections from the load will be re-reflected at the
source and re-re-reflected at the load end ad infinitum,
losing energy on each transit of the transmission line."
Well, I looked at that section of the writeup.
And, I have no idea what the hell they are talking about.
Looks like a good section for a knowledgeable person to edit.
If the termination matches the line impedance, there is no reflection.
Both the antenna and the source are terminations.
This is a bit difficult to visualze with an RF transmitter, but is
more easily seen with pulses.
Being essentially a simple soul, that's how I sometimes try to work out
what happening.
You'll be better off if you killfile the troll. You'll get a lot less
bad information and your life will become much easier.
The wikipedia entry is correct as written.
In the real world, the output of an amateur transmitter will seldom
be exactly 50 Ohms unless there is an adjustable network of some
sort.
I've always understood that the resistive part of a TX output impedance
was usually less than 50 ohms.
If a transmitter output impedance WAS 50 ohms, I would have thought that
the efficiency of the output stage could never exceed 50% (and aren't
class-C PAs supposed to be around 66.%?). Also, as much power would be
dissipated in the PA stage as in the load.
Fixed output amateur transmitters are a nominal 50 ohms. It can vary,
but that is due to normal variances in components, and the difference
can be ignored in real life.
But output impedance has little to do with efficiency. A Class C
amplifier can run 90%+ efficiency. It's output may be anything, i.e.
high with tubes but low with solid state. But the output impedance can
be converted to 50 ohms or any other reasonable impedance through a
matching network.
A perfect matching network will have no loss, so everything the
transmitter puts out will go through the matching network. Of course,
nothing is perfect, so there will be some loss. But the amount of loss
in a 1:1 match will not be significant.
Also, the amplifier generates the power; in a perfect world, 100% of
that power is transferred to the load. The transmitter doesn't
dissipate 1/2 of the power and the load the other 1/2. It's not like
having two resistors in a circuit where each will dissipate 1/2 of the
power.
BTW - the resistive part of the impedance is not the same as resistance.
For a simple case - take a series circuit of a capacitor, an inductor
and a 50 ohm resistor. At the resonant frequency, the impedance will
be 50 +j0 (50 ohms from the resistor, capacitive and inductive
reactances cancel). But the DC resistance is infinity. Again, a simple
example, but it shows a point.
I encourage all of you to read Walter Maxwell's (W2DU) book "Reflections
III". It will explain everything about this.
Everything you are discussing has been put to bed.
Cheers.
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