Dr. Slick wrote:
Roy Lewallen wrote in message ...
So, in theory, if you have achieved a perfect match with your
antenna, you will have matched the impedance to the 377 Ohms of free
space, you will not have reflections at the matching point, and the
energy will radiate in whatever pattern you have designed for.
No, an antenna doesn't "match" the impedance of free space. The input
impedance of an antenna is the ratio of V to I. The impedance of free
space is the ratio of the E field to the H field of a plane wave. They
both happen to have units of ohms, but they're different things and
there's no "matching" going on. If you apply 100 watts to an antenna,
resonant or not, 100 watts will be radiated, less loss, regardless of
the antenna's input impedance.
I disagree on this point. You can think of an antenna as a type
of tranformer, from 50 Ohms to 377 Ohms of free space. A tuned
antenna is "matching" 50 to 377 ohms, and will therefore have no
Thinking of it doesn't make it so. All the energy applied to the antenna
is radiated, less loss, regardless of the antenna's feedpoint impedance.
How does that fit into your model.
Have you heard of something called mis-match loss? If the
antenna's input impedance is not matched to the tranmission line (or
the final PA), then the radiated power will be significantly less than
I have indeed, and have posted several times about this
often-misunderstood and misused term. You can find the postings by going
and searching this group for postings by
me containing "mismatch loss".
The funny thing about this, is that you cannot say that the 50
Ohms in the center of the chart is a "resistive" 50 Ohms, as there is
very little real resistance in the average antenna.
Sure you can. You're confusing resistance with resistors. Resistance is
a dimension, like length. There are lots of things which have dimensions
of resistance but aren't resistors, like transresistance, characteristic
resistance of a transmission line, or radiation resistance to name just
a few. People who have a shaky understanding of basic electric circuit
theory seem to have trouble dealing with this, but it becomes easier to
deal with as you learn more about basic electricity. A mechanical
example is torque and work, which have the same dimensions (force times
distance) but are definitely different things.
ok, so perhaps the way to think of it is: when an antenna is
matched, the I and V curves
will be in phase (no reactance), and the
product of I*V (integrated) will be the power transmitted.
The average power radiated is always the real part of V*I(conjugate). If
V and I are in phase, this is simply equal to V*I. But what does this
have to do with the confusion between a resistor and resistance?
50 Ohms is really what people call the "radiation" resistance, which
is something of a misnomer again, because this is trying to equate the
successful impedance matching and subsequent non-reflected EM
radiation with a truly real resistance like an ideal dummy load.
Sorry, that doesn't make a whole lot of sense. Yes, it's called the
radiation resistance, but it's not a misnomer at all. (I suppose it
would be if you called it a "radiation resistor", but nobody I know of
has ever called it that.) If you calculate the power "consumed" by this
resistance (that is, the power flow into it), it's the power being
radiated. If the radiation and loss resistance of an antenna were zero,
no energy would flow into it and consequently none would be radiated.
(If the radiation resistance was zero and the loss resistance wasn't,
then energy would flow into the antenna but none would be radiated -- it
would all be dissipated as heat.)
If both the radiation and loss resistance were zero, i would
expect an ideal short, and therefore full -180 degree reflections.
You're assuming that the antenna is fed with a transmission line, but
I suppose what this all means is that if you have a matched
antenna, it's V and I curves
will be IN PHASE and will have the exact
same RMS values as if you had a truly resistive dummy load instead.
Yes and no. If you're feeding the antenna with a 450 ohm line, it's
matched to the line only if its impedance is 450 + j0 ohms, so it looks
like a 450 ohm dummy load, not a 50 ohm one. On the other hand, you can
feed a 50 + j0 ohm antenna with a half wavelength of 450 ohm line, and
get a perfect 50 ohm match to a transmitter at the input end of the
line, while running a 9:1 SWR on the the transmission line. Then either
the antenna or the input of the line looks like a 50 ohm dummy load.
Therefore, you can consider the center of the Smith Chart (or the
entire real (non-reactive) impedance line) as a real resistance like
an ideal dummy load.
Do you agree with this statement Roy?
Your use of "non-reflected EM radiation" seems to imply that the
radiation from an antenna is somehow bounced back from space if the
antenna feedpoint impedance is reactive. That's one of the rather
bizarre and very wrong conclusions you could draw from the mistaken idea
that the antenna "matched" the characteristic impedance of free space.
I think i'm correct to think of antennas as impedance matching
50 Ohms to 377 Ohms.
I like to think of 'em as sort of potato guns, launching RF potato
photons into the aether. But that doesn't make them potato guns.
Feel free to think of them any way you like, as long as you consistently
get the right answer.
Now tell me, why can't you just make some 377 ohm transmission line
(easy to make) open circuited, and dispense with the antenna altogether?
When you figure out the answer to that one, you might begin to see the
error with your mental model.
. . .
I believe i understand the Chart better than you think, Roy,
enough to know that you do know what you are talking about.
I still think it's ok to consider antennas as impedance
tranformers, but you have brought up some very good points.
Ok by me. I'll be waiting for your patented no-antenna 377 ohm feedline.
Roy Lewallen, W7EL