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

On Tue, 15 Jul 2003 01:20:02 -0700, Roy Lewallen
wrote:


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.
Roy Lewallen, W7EL

Roy;
Hope you don't mind if I ask you a couple of questions about your
last sentence.
Given a 75 ohm dipole fed 100 watts with 75 0hm coax. Assume no
losses, now if the antenna's input impedance is changed to say 50 ohms
(non-reactive). You would have some loss with the mismatch between
the coax and antenna?
The transmitter would see 50 ohms?

Dilon Earl wrote:
On Tue, 15 Jul 2003 01:20:02 -0700, Roy Lewallen
wrote:


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.
Roy Lewallen, W7EL


Roy;
Hope you don't mind if I ask you a couple of questions about your
last sentence.
Given a 75 ohm dipole fed 100 watts with 75 0hm coax. Assume no
losses, now if the antenna's input impedance is changed to say 50 ohms
(non-reactive). You would have some loss with the mismatch between
the coax and antenna?

Mismatch does not cause loss (that is, conversion of electrical energy
to heat), except that the loss of a lossy transmission line will
increase (very slightly unless initial loss is great) when SWR is
elevated. In this case, the line SWR would be 1.5:1, which would not
cause a significant amount of extra loss even if the cable were very
lossy when matched. So the answer is no.

The transmitter would see 50 ohms?


The transmitter could see any of a variety of impedances, depending on
the length of the 75 ohm transmission line. Only if the line were an
exact multiple of an electrical half wavelength would the transmitter
see 50 ohms, resistive. If the line were an odd number of quarter
wavelengths, the transmitter would see 112.5 ohms, resistive. At all
other lengths, the transmitter would see a complex (partly resistive and
partly reactive) load.

There is a term called "mismatch loss", which is widely misunderstood in
the amateur community because of its name. It doesn't really represent
loss at all, but a signal reduction for other reasons. I've explained
this before in this newsgroup, so if you're interested, you should be
able to find my earlier postings via http://www.groups.google.com.

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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W5DXP wrote in message ...
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. :-)
--


You misunderstand my point, Cecil.

If the antenna is tuned correctly, the "radiation" resistance is a
real 50 Ohms, with the V and I waveforms IN PHASE. But, my point is
that you can take a DC measurement anywhere on the ideal lossless
antenna and you will never see 50 Ohms anywhere, only shorts.

Ideally, the antenna never heats up, and has no resistive losses.
This is not to say that it won't have a "radiation" resistance of 50
Ohms at the resonant, tuned frequency. And to a transmitter at the
resonant frequency, this will electrically appear to be the same thing
as an ideal dummy load of 50 Ohms.

Roy's message has clarified a few things.


Slick

Dr. Slick wrote:

W5DXP wrote:
Apparently, all the resistance in the average antenna is real. :-)

You misunderstand my point, Cecil.

Actually, it was a play on words using your definition of "real" Vs
the IEEE Dictionary and mathematical definition of "real". The Devil
made me do it but I did provide a smiley face.
--
73, Cecil http://www.qsl.net/w5dxp



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Dr. Slick wrote:
"But, my point is that you can take a DC measurement anywhere on the
ideal lossless antenna and you will never see 50 Ohms anywhere, only
shorts."

True. D-C resistance of a lossless wire is zero. But, apply 50 watts r-f
power to an antenna adjusted to present 50-ohms resistance at a
particular frequency.

What happens? Our loading, adjusted for 50 watts output, produces 50
volts at an in-phase current of 50 amps.

The 50 watts is being radiated. As far as the transmitter is concerned,
the power might just as well be feeding a dummy load.

There are several reasons a d-c ohmmeter doesn`t read radiation
resistance. Once the ohmmeter`s d-c has charged the antenna, current
flow stops. D-C doesn`t radiate. A d-c ohmmeter doesn`t even measure the
right loss resistance value. Some of the loss resistance in a true
antenna with actual losses, comes from skin effect at radio frequencies,
where the conduction is forced to the outside of the conductors,
decreasing their effective cross-section and increasing their loss. R-F
also has the ability to induce eddy currents in surrounding conductors
and to agitate molecular movement in surrounding insulators. R-F thus
increases loss over that measured by passing d-c through the antenna.

Radiation resistance can be readily measured by using an r-f bridge
instrument. The bridge and antenna are excited by a generator operating
at the same frequency the transmitter will use. The bridge will indicate
reactance in the antenna, if it is not resonant. The null detector for
the bridge is typically a good radio receiver.

Radiation resistance is real though it does not heat the antenna. Loss
resistance is real though it does heat the antenna and its surroundings.

A resistance is a volt to amp ratio in which amps are in-phase with the
volts.

A resistor is a special type of resistance in which the electrical
energy applied to the resistance is converted into heat.

Best regards, Richard Harrison, KB5WZI

Richard Harrison wrote:
What happens? Our loading, adjusted for 50 watts output, produces 50
volts at an in-phase current of 50 amps.

Hmmmm, 50 watts in, 2500 watts out. How much will you take for
that antenna, Richard? :-)
--
73, Cecil http://www.qsl.net/w5dxp



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