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.

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. 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
radiation with a truly real resistance like an ideal dummy load.

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
dummy loads we have tested.

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

On 14 Jul 2003 22:32:12 -0700, (Dr. Slick) wrote:

.............
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?

Where do I get a "good" dummy load?

w.

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.

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.

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
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.)

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.

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
dummy loads we have tested.

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,
without any radiation at all.


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

Dr. Slick wrote:
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.

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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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
following will radiate most of the mismatch loss from the load.

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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Dr. Slick wrote:
"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."

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
antenna. Part is made by loss in the antenna. part is made by radiation
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

On 15 Jul 2003 07:03:28 -0700, (Dr. Slick) wrote:

A strange mix of notions, some valid, others fanciful interpretation.

What do you mean by this? Can you explain yourself more? Or am i
confusing you?

Confused is a good word.

You might not, others would, and there is nothing "funny" going on in
the first place. As to "real" resistance, that is merely semantics
and does not illuminate just what you want to talk about.

I think it's "funny" that you can't give me your opinion on this.
I don't think you have one.

Confused Opinion being sought? Enough is provided.

This "resistive"
50 Ohms is really what people call the "radiation" resistance, which
is something of a misnomer again,

It is only a misnomer that you brought to the table. There is, by and
large, no antenna that is by nature 50Ohms in the first place, but
there are many antennas that have been cobbled together to present
50Ohms - not the same thing; but also hardly distinctive either,
except to the semanticist. Radiation resistance is NOT a function of
drive point impedance even if the two share the same value for some
obscure design.

This paragraph was written by someone who doesn't know what they
are saying AT ALL.
You clearly are way more confused than even me.

Again you have reversed the attribution. I am not the one seeking
"opinion."

"We?"

You presume MFJ is not completely inaccurate? That is a wordy long
way around the barn. And this is a function of "if you believe?" Why
not simply state what YOU believe and leave out the tea leaf reading?

Why not simply stop pretending you know what the hell i'm talking
about when you clearly don't!

An "MFJ is not completely inaccurate?" Now there is clear writing. I
think my statement was rather explicit. But if you need it restated,
I think you said it well enough yourself.


Theories abound and I have a 50Ohm dummy load that is fairly flat from
DC to 1GHz (confirmed by calibration to be less than 1:1.2 anywhere)
that I bought for $50. True, it is actually 52Ohms (the actual
standard of post WW2 Electronics), but its frequency characteristics
are just fine. More so than your experience in either obtaining
accuracy, or confirming it, or both.

How you make the leap to "radiation" resistance to make up the
difference in your observations is inventive, but not compelling; and
certainly lacks considerable discussion.

Discussion which you have not illuminated at all. I don't have a
high opinion of most HAM people, and you certainly fit the bill.

You cannot distinguish the specification of a Dummy Load being flat
over the frequency range of DC to 1MHz to within less than 1:1.2?
What opinions are you seeking of HAM people, the quality of bacon?

But some Hams like Roy actually know what they are talking about.
You certainly do not.

Now there is opinion I can recognize. ;-)

So, to respond directly to the subject:
50 Ohms "Real Resistive" impedance a Misnomer?
Yes, as the nominative you employ.

73's
Richard Clark, KB7QHC

Stay out of discussions that go over your head.

Slick

Thanks but no thanks, OM. You certainly have a crappy dummy load you
may have confirmed through haphazard testing that you then rejected
with opinion and speculative theory. That quality was evident
throughout.

You also have a low threshold for your avowed mission to
light a brush-fire
As always, you have complete control in what you choose to write, and
you chose opinion over technical discussion. As far as opinion or
brush fires goes, you don't seem well suited for that either.

73's
Richard Clark, KB7QHC

On Tue, 15 Jul 2003 10:22:02 -0500 (CDT),
(Richard Harrison) wrote:

Dr. Slick wrote:
"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."


Hi Richard,

This is seems to be a point of convergence for the derivation of many
fascinating and strange theories of fancy and speculation.

You point out in the remainder of your posting about the combination
of resistances (giving particular care to describe in terms of phase)
and yet many posters here fail to account for those same assortment of
R's available from real life.

One notable offering of measuring antenna (or load) impedance involved
the use of a thermometer to which I asked "what is the Z for a 1
degree rise?" I was not surprised to find no answer forthcoming even
when the premise was sound. Such is the shortfall of speculation in
the face of analytical enquiry. There is no corresponding shortfall
of opinion draped in the mantle of citations unfortunately.

These attempts to separate "real" resistance from other resistances
are challenged with volumes of formulaic recitation, and the absolute
resistance (yet another meaning) to merely stepping out into the field
with an OhmMeter (much less that same thermometer). Clearly,
resolution of these imponderables is not a target for some scribblers.

For many years there has been this effete distinction of there being
dissipative and non dissipative resistance (perhaps the basic, or
elemental concern of this thread; yet through hazy writing that agenda
remains elusive). The transmitter cannot possibly separate the two.
It thus remains for the target audience to resolve, but even they
cannot either unless the incident of gaining or losing this additional
resistance occurs in a short enough interval to allow its perception
(notably expressed in dB). If the researchers refuse to do some field
work, it will always remain among their mysteries of the sacrament.

73's
Richard Clark, KB7QHC

(Dr. Slick) wrote in message . com...
....
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
dummy loads we have tested.

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?

With just a bit of experience in this area, I think you need to find a
way to control your experiments more carefully so you KNOW what's
going on, to a better approximation. I'm quite sure it's possible to
build fairly high power dummy loads--at least high enough power to
handle legal amateur transmitter outputs--that present much better
matches than you describe. Our most usual problem in calibrating
precision instruments at high frequencies is in finding cables which
are really close to 50 ohms and are also practical for routine tests.
Some of the other parts such as power splitters present problems, too.
But we've gotten pretty good at figuring out just where the errors
creep in, through a combination of analysis and experiments. For
sure, our work builds on a tremendous amount of work that has gone on
in the past in the area of precision RF measurements. We currently
work at frequencies from DC to several GHz, and worry in the GHz range
about errors equivalent to an ohm or so...less down at 100MHz and
below.

But for ham applications, do you really care about all this? There's
probably not a need to, if you're just trying to get power to an
antenna, but if you care because you simply want to learn how to build
a precision system and make precision measurements, that's fine too.
You should be able to find lots of info on the web about that...try,
for example, to find Hewlett-Packard/Agilent ap notes on RF
measurements and calibrations.

Cheers,
Tom

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.

I have indeed, and have posted several times about this
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".




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

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
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
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
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?

Yes.


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
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
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