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#1
July 15th 03, 06:32 AM
 Dr. Slick Posts: n/a
50 Ohms "Real Resistive" impedance a Misnomer?

Hello,

I'd like to start a discussion (or light a brush-fire, as the
case may be for this NG!), about what a 50 Ohm impedance match really
means.

On our trusty Smith Chart, assuming it is normalized to 50 Ohms,
the center is considered and labeled as the "real resistive" 50 match
point. In fact, the entire middle horizontal line is the "real" part
of the impedance.

I'm sure many of you have read the popular description/model of a
transmission line as an infinite chain of alternating series inductors
with shunt capacitors, with the resulting characteristic impedance as
Z=(L/C)**1/2, where the L and C are distributed inductances and
capacitances.

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.

Ohms in the center of the chart is a "resistive" 50 Ohms, as there is
very little real resistance in the average antenna. This "resistive"
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

Of course, it's well known that a truly real resistive 50 Ohm
dummy load should appear exactly like a properly matched antenna to
the transmitter.

Why do i ask all this? Well, if you believe that complex
impedance measurements (series equivalent) by MFJ antenna analyzers
are not completely inaccurate, then it appears that two 1/4 watt 100
Ohm resistors in parallel (lead lengths short) are a much more
consistent 50 Ohms over the VHF band than almost all the higher power

Problem is, the high power dummy loads will vary from 52 to 45
"real" ohms depending on the frequency, with the "real" part of the
impedance getting lower with increasing frequency, so it doesn't seem
to be a "skin effect". The spread gets much worse when you put a 3'
jumper coax in between, and even more worse when you add a power/swr
meter. Then the "real" Ohms will be from 65 to 35 ohms, with the max
and mins not correlating with frequency at all, and the stray
reactances will be much more too, but just as varied with frequency.
So much for "50 ohm" jumper cables! I suppose they are as close as
they can get them for a particular price.

My theory is that the "real" part of the impedance is mainly the
truly resistive 50 ohms of the dummy load at low frequencies around 10
MHz or so...but as you go up in frequency, the parasitics of the dummy
load and the coax jumper cable will cause "radiation" resistance to be
mixed in with this truly real 50 ohms, giving us readings all over the
map.

What do you folks think?

Dr. Slick

#2
July 15th 03, 07:52 AM
 Helmut Wabnig Posts: n/a

#3
July 15th 03, 09:20 AM
 Roy Lewallen Posts: n/a

Dr. Slick wrote:
Hello,

I'd like to start a discussion (or light a brush-fire, as the
case may be for this NG!), about what a 50 Ohm impedance match really
means.

On our trusty Smith Chart, assuming it is normalized to 50 Ohms,
the center is considered and labeled as the "real resistive" 50 match
point. In fact, the entire middle horizontal line is the "real" part
of the impedance.

I'm sure many of you have read the popular description/model of a
transmission line as an infinite chain of alternating series inductors
with shunt capacitors, with the resulting characteristic impedance as
Z=(L/C)**1/2, where the L and C are distributed inductances and
capacitances.

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.

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
example is torque and work, which have the same dimensions (force times
distance) but are definitely different things.

This "resistive"
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

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.)

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.

Of course, it's well known that a truly real resistive 50 Ohm
dummy load should appear exactly like a properly matched antenna to
the transmitter.

At one frequency.

Why do i ask all this? Well, if you believe that complex
impedance measurements (series equivalent) by MFJ antenna analyzers
are not completely inaccurate, then it appears that two 1/4 watt 100
Ohm resistors in parallel (lead lengths short) are a much more
consistent 50 Ohms over the VHF band than almost all the higher power

No surprise. It's much harder to make something that's physically big
have a consistent impedance at high frequencies than for something
physically small, simply because stray inductances and capacitances are
both larger for large objects. Let that be a lesson.

Problem is, the high power dummy loads will vary from 52 to 45
"real" ohms depending on the frequency, with the "real" part of the
impedance getting lower with increasing frequency, so it doesn't seem
to be a "skin effect". The spread gets much worse when you put a 3'
jumper coax in between, and even more worse when you add a power/swr
meter. Then the "real" Ohms will be from 65 to 35 ohms, with the max
and mins not correlating with frequency at all, and the stray
reactances will be much more too, but just as varied with frequency.
So much for "50 ohm" jumper cables! I suppose they are as close as
they can get them for a particular price.

It's no trick at all to tranform a real resistance to a different value
of real resistance using only purely reactive L and C components. It's
done all the time. An L network, with two components, is the simplest
circuit which can pull off this magical trick. Just pick a point on the
real axis of that Smith chart of yours and follow reactance lines around
-- first XL, then XC, or vice-versa, until you end up back on the real
axis again -- at a different resistance value. The amount of XL and XC
you transit along the way are in fact the values you'd need to make an L
network to do the transformation. As for the transmission line, start
at, say, 45 ohms, then go in a circle around the center, reading off R
and X values as you go. Those are the values of R and X you can get with
a 50 ohm transmission terminated with a 45 + j0 ohm load. Of course, "50
ohm" lines are often quite a ways off -- I've measured them at up to 62
ohms or so. And you're right, you can get better ones if it's worth a
lot to you. Oh, also notice that if you start on the real axis anywhere
but the center and go around a half circle, representing 90 degrees of
lossless transmission line, you end up at a different place on the real
axis. Presto! You've pulled off a transformation of a purely real
impedance with a lossless transmission line. Cool, huh?

My theory is that the "real" part of the impedance is mainly the
truly resistive 50 ohms of the dummy load at low frequencies around 10
MHz or so...but as you go up in frequency, the parasitics of the dummy
load and the coax jumper cable will cause "radiation" resistance to be
mixed in with this truly real 50 ohms, giving us readings all over the
map.

Although radiation will cause an increase in terminal resistance
(remember, it accounts for the radiated power), it's not at all
necessary in order to cause the dummy load resistance (real part of the
impedance) to vary. The stray L and C can do that all by themselves,

What do you folks think?

I think you'd benefit a lot from learning how to do some basic
operations with a Smith chart. It would broaden your horizons a lot.

Roy Lewallen, W7EL

#4
July 15th 03, 12:31 PM
 W5DXP Posts: n/a

Dr. Slick wrote:
Ohms in the center of the chart is a "resistive" 50 Ohms, as there is
very little real resistance in the average antenna.

From the IEEE Dictionary: "resistance (1)(B) The real part of impedance."

Apparently, all the resistance in the average antenna is real. :-)
--
73, Cecil http://www.qsl.net/w5dxp
"One thing I have learned in a long life: that all our science, measured
against reality, is primitive and childlike ..." Albert Einstein

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#5
July 15th 03, 03:05 PM
 W5DXP Posts: n/a

Dr. Slick wrote:
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
100 watts.

Not if there's an antenna tuner (Z0-match) in the circuit. The

100W XMTR--50 ohm feedline--+--1/2WL 150 ohm feedline--50 ohm load

All of the reflected energy is re-routed back toward the load at
the '+' Z0-match point through re-reflection and wave cancellation.
--
73, Cecil http://www.qsl.net/w5dxp

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#6
July 15th 03, 04:22 PM
 Richard Harrison Posts: n/a

Dr. Slick wrote:
the center of the chart is a "resistive" 50 ohms as there is very little
real resistance in the average antenna."

Resistance is defined as real. That is, current is instantaneously
proportional to the voltage.

Any efficient antenna has a high ratio of radiation resistance to loss
resistance.

Resistance is the ratio of in-phase voltage to current accepted by an
from the antenna. They are often represented by an equivalent circuit of
two resistors in series.

Dr, Frederick Emmons Terman says of radiation resistance:
"This is the resistance that, when inserted in series with the antenna,
will consume the same amount of power as is actually radiated. ---it is
customary to refer the radiation resistance to a current maximum in the
case of an ungrounded antenna, and to the base of the antenna when the
antenna is grounded."

Best regards, Richard Harrison, KB5WZI

#10
July 15th 03, 07:23 PM
 Roy Lewallen Posts: n/a

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
reflections, ideally.

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
100 watts.

often-misunderstood and misused term. You can find the postings by going
to http://www.groups.google.com and searching this group for postings by
me containing "mismatch loss".

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
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

what 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?

This "resistive"

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

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
that's ok.

I suppose what this all means is that if you have a matched
antenna, it's V and I curves

what curves?

will be IN PHASE and will have the exact

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

Do you agree with this statement Roy?

Yes.

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
transformers.

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

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

. . .

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

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