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50 Ohms "Real Resistive" impedance a Misnomer?
Hello,
I'd like to start a discussion (or light a brushfire, 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 nonreflected 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 
#2





#3




Dr. Slick wrote:
Hello, I'd like to start a discussion (or light a brushfire, 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 nonreflected 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 "nonreflected 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 viceversa, 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 
#4




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 = Posted via Newsfeeds.Com, Uncensored Usenet News = http://www.newsfeeds.com  The #1 Newsgroup Service in the World! == Over 80,000 Newsgroups  16 Different Servers! = 
#5




Dr. Slick wrote:
Have you heard of something called mismatch 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 (Z0match) in the circuit. The following will radiate most of the mismatch loss from the load. 100W XMTR50 ohm feedline+1/2WL 150 ohm feedline50 ohm load All of the reflected energy is rerouted back toward the load at the '+' Z0match point through rereflection and wave cancellation.  73, Cecil http://www.qsl.net/w5dxp = Posted via Newsfeeds.Com, Uncensored Usenet News = http://www.newsfeeds.com  The #1 Newsgroup Service in the World! == Over 80,000 Newsgroups  16 Different Servers! = 
#6




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 inphase 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 
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#10




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 mismatch 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 oftenmisunderstood 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 nonreflected 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 (nonreactive) impedance line) as a real resistance like an ideal dummy load. Do you agree with this statement Roy? Yes. Your use of "nonreflected 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 noantenna 377 ohm feedline. Roy Lewallen, W7EL 
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