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Pi network question
Hi Group,
I have a pi-network, or at least I believe it is a pi-network in my antenna tuner. It has two caps to ground and one in series inductor. If I tune my SWR for 1:1 and go back and measure each value of the capacitors and inductor, is there a equation or computer program I can use these measured values to obtain my complex impedance at the input of the coax cable? de KJ4UO |
Arn't you going around your hindparts to get to your elbow?
Connect an MFJ259B or similar and measure at the coax. Once these valuses are measured at the frequency of interest, you can determine from the tuners specs if you will be able to achieve a match. If you can already achieve a match isn't determining the value simply a neat mental exercise? wrote in message ups.com... Hi Group, I have a pi-network, or at least I believe it is a pi-network in my antenna tuner. It has two caps to ground and one in series inductor. If I tune my SWR for 1:1 and go back and measure each value of the capacitors and inductor, is there a equation or computer program I can use these measured values to obtain my complex impedance at the input of the coax cable? de KJ4UO |
On 31 Jul 2005 21:01:19 -0700, "
wrote: Hi Group, I have a pi-network, or at least I believe it is a pi-network in my antenna tuner. It has two caps to ground and one in series inductor. If I tune my SWR for 1:1 and go back and measure each value of the capacitors and inductor, is there a equation or computer program I can use these measured values to obtain my complex impedance at the input of the coax cable? You ought to be able to do it from first principles, the computation is not that difficult. Some of the issues include your estimate of the value of the capacitors, the value of inductance, and the Q of the inductor. You could try to form a value for those by substitution of some known capacitors and "matching up" a known load, working back to estimate C, L, then use those values to solve the unknown case. The accumulated error might render the answer of little value. One program that springs to mind that solves a pi network is in the Hamcalc suite, but IIRC, it treats all components as ideal (ie lossless). Reg Edwards has another that you could play with, and IIRC, it does allow you to specify unloaded Q of the inductor. Have a play, you have nothing to lose and everything to learn! Owen -- |
On 31 Jul 2005 21:01:19 -0700, "
wrote: I have a pi-network, or at least I believe it is a pi-network in my antenna tuner. It has two caps to ground and one in series inductor. If I tune my SWR for 1:1 and go back and measure each value of the capacitors and inductor, is there a equation or computer program I can use these measured values to obtain my complex impedance at the input of the coax cable? Hi OM, Good time to learn the Smith Chart. The components in that network are simply arcs moving the presented Z to the transformed Z (50Ohms). Unfortunately this is easier said than done - or understood. Like all skills, time and practice are necessary. As for understanding, the Smith Chart will bring that quicker than any software - which simply dumps an answer in your lap, the quality or actual utility of which is as much a mystery as the problem you wanted to solve. There are a multitude of network settings that will satisfy the matching and software or equations will do little more than say "yup, that setting does exactly as you observed." Not very filling, but when you do the same thing on the Chart, you can see where you are going before you get there. This means you also have the choice of choosing the best path (shortest). It would be a rare piece of code that could do that - from a mountain of formulas. 73's Richard Clark, KB7QHC |
On 31 Jul 2005 21:01:19 -0700, "
wrote: Hi Group, I have a pi-network, or at least I believe it is a pi-network in my antenna tuner. It has two caps to ground and one in series inductor. If I tune my SWR for 1:1 and go back and measure each value of the capacitors and inductor, is there a equation or computer program I can use these measured values to obtain my complex impedance at the input of the coax cable? de KJ4UO I would caution you in regard to measuring the values of the C's and the L of your pi network. If you actually have an RF bridge or other test equipment suitable for accurately making such measurements, you are indeed lucky. Even with such equipment, you should realize that you will have to disconnect certain components from one another in the tuner in order to actually make the measurements on each of the three components. You are also probably going to have to remove some of those components entirely in order to make the measurements with reasonably short leads. For anything below 10 MHz, the lead length will not be a significant factor. Assuming that you are successful in making the measurements with good accuracy, the necessary calculations require a good knowledge of complex numbers and complex arithmetic - much of which is adequately explained in the Amateur Radio Handbook. My suggestion at that point would be to simply post your results, including the frequency of measurement, to this forum and request that one of us make the calculations for you. I'm sure many, including myself, would be glad to do that for you. It's the quickest way, by far. Bob, W9DMK, Dahlgren, VA Replace "nobody" with my callsign for e-mail w9dmkatcrosslinkdotnet http://www.qsl.net/w9dmk http://zaffora/f2o.org/W9DMK/W9dmk.html |
If he has an RF bridge, couldn't he simply terminate the tuner's input
with 50 ohms resistive and measure the impedance at the tuner's output? Well, the conjugate, anyway. I'm trying to imagine the benefit of doing it that way, rather than simply measuring the impedance at the transmission line directly. Chuck, NT3G W9DMK (Robert Lay) wrote: On 31 Jul 2005 21:01:19 -0700, " wrote: Hi Group, I have a pi-network, or at least I believe it is a pi-network in my antenna tuner. It has two caps to ground and one in series inductor. If I tune my SWR for 1:1 and go back and measure each value of the capacitors and inductor, is there a equation or computer program I can use these measured values to obtain my complex impedance at the input of the coax cable? de KJ4UO I would caution you in regard to measuring the values of the C's and the L of your pi network. If you actually have an RF bridge or other test equipment suitable for accurately making such measurements, you are indeed lucky. Even with such equipment, you should realize that you will have to disconnect certain components from one another in the tuner in order to actually make the measurements on each of the three components. You are also probably going to have to remove some of those components entirely in order to make the measurements with reasonably short leads. For anything below 10 MHz, the lead length will not be a significant factor. Assuming that you are successful in making the measurements with good accuracy, the necessary calculations require a good knowledge of complex numbers and complex arithmetic - much of which is adequately explained in the Amateur Radio Handbook. My suggestion at that point would be to simply post your results, including the frequency of measurement, to this forum and request that one of us make the calculations for you. I'm sure many, including myself, would be glad to do that for you. It's the quickest way, by far. Bob, W9DMK, Dahlgren, VA Replace "nobody" with my callsign for e-mail w9dmkatcrosslinkdotnet http://www.qsl.net/w9dmk http://zaffora/f2o.org/W9DMK/W9dmk.html |
I have an MFJ-269 which I can measure the input to the coax and get a
direct impedance measurement. Then I would put the tuner in line with the MFJ-269 as the source and adjust the tuner until I see 50 + j0 on the MFJ. I would follow by measuring the pi-network components also using the MFJ as it will measure values at the desired RF frequency. I would keep the signal frequency low so that stray inductance and capacitance do not affect the measurement. What I would like to do is see by using the values I obtain, calculate the cable load and compare with MFJ direct measurement. I will review the handbook to see if there is a standard equation knowing the values and assume that the Q of the inductor is greater than 10. All in the fun of the hobby. de KJ4UO |
That's a worthwhile educational exercise.
You might find a single equation, but it gets a bit messy. However, it's easily done in a few steps, if you're comfortable with complex arithmetic. I use an HP48GX calculator for this sort of thing, since it readily and directly handles complex numbers. Call the capacitor on the input side of the network C1 and the on the output side C2, and the inductor L. First calculate the reactance of each of them at the frequency of interest, Xc1, Xl, and Xc2. You'll find formulas for those in the Handbook. Note that Xc1 and Xc2 will be negative and Xl will be positive. First calculate the parallel combination of the load impedance (Zl) and Xc2. That will be the impedance seen looking toward the load from the output side of the inductor: Za = Zl || Xc2 = (Rl + jXl) || (jXc2), where Xc2 is negative. The combined impedance of two impedances Z1 and Z2 in parallel are 1 / ((1/Z1) + (1/Z2)) = (Z1 * Z2) / (Z1 + Z2). Now add Xl to your result to find the Z looking into the input side of the inductor: Zb = Za + jXl. (If the inductor has appreciable loss, use Zl = Rl + jXl instead of just jXl.) Finally, calculate the parallel combination of that impedance and the impedance of the input capacitor to find the impedance looking into the network: Zin = Zb || jXc1, remembering that Xc1 will be negative. You can of course combine all this into one equation, but it gets pretty big. I prefer to do it in steps, one of the reasons being that I understand exactly what I'm doing at each step rather than just dumping numbers into an equation and hoping that what comes out is right. A Smith Chart gives you an even better feel for what's going on, and this would be a good opportunity to get acquainted with that valuable tool. You might solve the problem both arithmetically and by using the Smith Chart and compare results. If you aren't comfortable with complex arithmetic, the equations get considerably more complicated since you'll have to deal with the resistance and reactance separately. If that's the case, I second the advice already given that you either solve it using a Smith Chart or consider just letting one of the other folks on this group do the calculation for you. Roy Lewallen, W7EL wrote: I have an MFJ-269 which I can measure the input to the coax and get a direct impedance measurement. Then I would put the tuner in line with the MFJ-269 as the source and adjust the tuner until I see 50 + j0 on the MFJ. I would follow by measuring the pi-network components also using the MFJ as it will measure values at the desired RF frequency. I would keep the signal frequency low so that stray inductance and capacitance do not affect the measurement. What I would like to do is see by using the values I obtain, calculate the cable load and compare with MFJ direct measurement. I will review the handbook to see if there is a standard equation knowing the values and assume that the Q of the inductor is greater than 10. All in the fun of the hobby. de KJ4UO |
On Mon, 01 Aug 2005 12:21:59 GMT, chuck wrote:
If he has an RF bridge, couldn't he simply terminate the tuner's input with 50 ohms resistive and measure the impedance at the tuner's output? Well, the conjugate, anyway. I'm trying to imagine the benefit of doing it that way, rather than simply measuring the impedance at the transmission line directly. Chuck, NT3G I will break my own rule of not responding to responders and answer your question - otherwise you would assume that I was ignoring you. If I might answer a question with a question - why do you think that a tuner would give up its settings so easily? It's a an interesting speculation. First, let me clarify what I mean by a Tuner's input terminal and its output terminal. Considering that a tuner handles transmitted power in one direction only, we should refer to its "input" side as the port that connects to the transmitter and its "output" port as the one connecting to the antenna or feedline. That agrees with what I interpret your post as suggesting. Therefore, a 50 ohm termination at the "input" side "should" produce a measured impedance at the "output" port that is, as you say, the complex conjugate of the impedance seen looking into the transmission line. I haven't done a rigorous analysis of that configuration, but it certainly seems reasonable and correct. It would, indeed be an optimum way of inferring the impedance that the original poster is trying to measure. I see nothing wrong with your suggestion and would recommend it. I also see no reason to do any of that as it would be just as easy to measure the transmission line input impedance directly - as you said. It's amazing how much good stuff comes out of the woodwork when people are interested in their hobby and interested in learning. Thanks, Bob, W9DMK, Dahlgren, VA Replace "nobody" with my callsign for e-mail w9dmkatcrosslinkdotnet http://www.qsl.net/w9dmk http://zaffora/f2o.org/W9DMK/W9dmk.html |
If the tuner can cope, there's not much interest in what the input
impedance of the transmission line is. What IS of interest is what impedance terminates the remote end of the transmission line. Usually this is the antenna. And usually this is impractical to measure directly. From the input impedance of the transmission line, Program ZL_ZIN computes the input impedance of the antenna (or whatever may be at the other end ). It is also necessary to know the length of the line and its Zo. Download in a few seconds program ZL_ZIN from website below and run immediately, ---- .................................................. .......... Regards from Reg, G4FGQ For Free Radio Design Software go to http://www.btinternet.com/~g4fgq.regp .................................................. .......... |
With sufficient accuracy, knowing its maximum value, it is possible to estimate the value of an air-spaced variable capacitor setting just by looking at it. Similarly, the value of a coil can be calculated by counting the number of turns and guessing its length and diameter. The calculation is simple enough. I've done it dozens of times. It gets you into the right ballpark. Which is quite accurate enough for amateur radio purposes. There's no need for rocket science. I can't understand the abysmal state of education in simple arithmetic in our primary schools. We should import the kids who inhabit the sewers of Rio de Janerio and other American cities and employ them as arithmetic teachers. It would be more economic than culling by armed police. ---- Reg, G4FGQ |
On Mon, 01 Aug 2005 21:09:16 GMT, (Robert
Lay) wrote: On Mon, 01 Aug 2005 12:21:59 GMT, chuck wrote: If he has an RF bridge, couldn't he simply terminate the tuner's input with 50 ohms resistive and measure the impedance at the tuner's output? Well, the conjugate, anyway. I'm trying to imagine the benefit of doing it that way, rather than simply measuring the impedance at the transmission line directly. Chuck, NT3G I will break my own rule of not responding to responders and answer your question - otherwise you would assume that I was ignoring you. If I might answer a question with a question - why do you think that a tuner would give up its settings so easily? It's a an interesting speculation. First, let me clarify what I mean by a Tuner's input terminal and its output terminal. Considering that a tuner handles transmitted power in one direction only, we should refer to its "input" side as the port that connects to the transmitter and its "output" port as the one connecting to the antenna or feedline. That agrees with what I interpret your post as suggesting. Therefore, a 50 ohm termination at the "input" side "should" produce a measured impedance at the "output" port that is, as you say, the complex conjugate of the impedance seen looking into the transmission line. I haven't done a rigorous analysis of that configuration, but it certainly seems reasonable and correct. It would, indeed be an optimum way of inferring the impedance that the original poster is trying to measure. I see nothing wrong with your suggestion and would recommend it. Let's pretend that the antenna (load) Z = 49 +j0. Let's also pretend that the "tuner" consists of a series 1 ohm resistor. Looking into the input of the "tuner" we measure 50 +j0. Pretty good, huh? Terminating the input of the tuner with 50 +j0 and looking from the other end we measure 51 +j0. Whoops. Best have a lossless tuner to play this game. I also see no reason to do any of that as it would be just as easy to measure the transmission line input impedance directly - as you said. It's amazing how much good stuff comes out of the woodwork when people are interested in their hobby and interested in learning. Thanks, Bob, W9DMK, Dahlgren, VA Replace "nobody" with my callsign for e-mail w9dmkatcrosslinkdotnet http://www.qsl.net/w9dmk http://zaffora/f2o.org/W9DMK/W9dmk.html |
"Reg Edwards" wrote in message ... With sufficient accuracy, knowing its maximum value, it is possible to estimate the value of an air-spaced variable capacitor setting just by looking at it. Similarly, the value of a coil can be calculated by counting the number of turns and guessing its length and diameter. The calculation is simple enough. I've done it dozens of times. It gets you into the right ballpark. Which is quite accurate enough for amateur radio purposes. There's no need for rocket science. I can't understand the abysmal state of education in simple arithmetic in our primary schools. We should import the kids who inhabit the sewers of Rio de Janerio and other American cities and employ them as arithmetic teachers. It would be more economic than culling by armed police. ---- Reg, G4FGQ I credit the California and Oklahoma Public Schools for my mathematical illiteracy. Went to school in Bakersfield in 1947, got straight A cause the math they taught in the 9th grade was the math I learned in Oklahoma in the 4th grade. Returned to Oklahoma in the middle of the school year and was placed in a 2nd semester Algebra class. Teacher was a coach who spent most of his class time working with (on?) a female student whom he later married. Even if I had been truly interested I'd never have been able to catch up. I was passed on even though I was totally incompetent in Algebra.( this was before social passing became common in the public schools)They should have adjusted my schedule and had me start at the beginning Algebra semester even if it required some class juggling the following year. Perhaps they thought they were doing me a favor, or maybe they just didn't realize how retro the California school math was. Harold KD5SAK |
Good point, Wes. A reminder that a 1:1 swr at the input of a lossy tuner
is not proof of a perfect match at the output! Chuck NT3G Wes Stewart wrote: Let's pretend that the antenna (load) Z = 49 +j0. Let's also pretend that the "tuner" consists of a series 1 ohm resistor. Looking into the input of the "tuner" we measure 50 +j0. Pretty good, huh? Terminating the input of the tuner with 50 +j0 and looking from the other end we measure 51 +j0. Whoops. Best have a lossless tuner to play this game. |
The difference between teaching reading and writing and teaching
arithmetic is that practical uses for reading and writing are immediately apparent to children. Whereas teachers of arithmetic, and then maths, fail to show children the many interesting and practical uses of the subject. It's lack of imagination on the part of teachers - and laziness. They have no interest in the subject themselves. They are themselves victims of poor teaching and of the system which selects them to be teachers. In the UK it can be traced back to ignorance on the part of Ministers of Education. It is a great pity children are not given the opportunity to appreciate the beauty in Mathematics. In England, first year engineering university students have to spend the first 6 months being taught what they should have learned at the age of 14. Including how to read and write. Just prior to a General Election (of Members of Parliament) in the UK I once asked a prospective MP what seven nines were. He didn't know! Yet there was the possibility he could end up as the Chancellor of the Exchequer or even as a Minister of Education. As the meeting was being held in a public house the prospective MP changed the subject and bought me a pint of beer. Presumably in a futile attempt at bribery for a vote. Eventually he lost his deposit (of £500) for failing to collect the necessary small minimum number of votes. ---- Reg. |
Reg Edwards wrote: Just prior to a General Election (of Members of Parliament) in the UK I once asked a prospective MP what seven nines were. He didn't know! He probably thought you were using Cockney rhyming slang, and wasn't familiar with the expression 'seven nines'. The correct answer to your querry was of course: extremely pure. :-) ac6xg |
Reg Edwards wrote:
Just prior to a General Election (of Members of Parliament) in the UK I once asked a prospective MP what seven nines were. He didn't know! Interesting that the same things have very different meanings. In my current business 7 nines would mean roughly 3.16 seconds. Can you tell me why? tom K0TAR |
the square root of seven nines is 3162.27...
if the figure above were microseconds would be ~3.162 seconds. John "Tom Ring" wrote in message . .. Reg Edwards wrote: Just prior to a General Election (of Members of Parliament) in the UK I once asked a prospective MP what seven nines were. He didn't know! Interesting that the same things have very different meanings. In my current business 7 nines would mean roughly 3.16 seconds. Can you tell me why? tom K0TAR |
.... microseconds = milliseconds...
.... even ... frown John "John Smith" wrote in message ... the square root of seven nines is 3162.27... if the figure above were microseconds would be ~3.162 seconds. John "Tom Ring" wrote in message . .. Reg Edwards wrote: Just prior to a General Election (of Members of Parliament) in the UK I once asked a prospective MP what seven nines were. He didn't know! Interesting that the same things have very different meanings. In my current business 7 nines would mean roughly 3.16 seconds. Can you tell me why? tom K0TAR |
On 2005-08-02, Tom Ring wrote:
In my current business 7 nines would mean roughly 3.16 seconds. Can you tell me why? That's how much downtime you get each year if you have 99.9999% uptime. Of course that's *unplanned* downtime. ;-) -- Ben Jackson http://www.ben.com/ |
one more "9" My calculator says that 3.16 sec per year (365.25 days) is an
up percent of 99.99990 73 Mac N8TT - back from Perry -- J. Mc Laughlin; Michigan U.S.A. Home: "Ben Jackson" wrote in message ... On 2005-08-02, Tom Ring wrote: In my current business 7 nines would mean roughly 3.16 seconds. Can you tell me why? That's how much downtime you get each year if you have 99.9999% uptime. Of course that's *unplanned* downtime. ;-) -- Ben Jackson http://www.ben.com/ |
KJ4UO asked: If I tune my SWR for 1:1 and go back and measure each value of the capacitors and inductor, is there a equation or computer program I can use these measured values to obtain my complex impedance at the input of the coax cable? Jim Tonne, WB6BLD, wrote a program in Visual Basic that does precisely that for all common tuner configurations. It is called Revload and can be downloaded for free from: http://www.tonnesoftware.com/revload.html You may want to check out his other ham-radio related programs, listed on his main page: http://www.tonnesoftware.com/ As others have pointed out, this is not a particularly accurate way of measuring impedance, but it should be adequate for "ballpark" estimates. Jim, K7JEB |
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