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Zo and Ro
Not having much else to do at present I thought I would make a comment on Zo and Ro of transmission lines. For entertainment and educational value, of course, if you like that sort of thing. The condition for which Zo of a transmission line is always purely resistive (Zo = Ro) is extremely simple. It is - G = C * R / L where G is shunt conductance, C is shunt capacitance, R is series resistance, L is series inductance, all per unit length of line. Which applies to any line length, at any frequency from DC to UHF. It is a shortcoming of the Smith Chart, with Zo always equal to Ro, that it does not make you aware of this and can lead you up the garden path if you are not careful. As has recently occurred on this newsgroup. Don't get me wrong. I'm not against Smith Charts. They are graphically educational within their limitations. ---- Reg, G4FGQ |
Reg Edwards wrote:
The condition for which Zo of a transmission line is always purely resistive (Zo = Ro) is extremely simple. It is - G = C * R / L Wonder why Ramo and Whinnery say that's an approximation for low-loss lines? If the R+jwL angle is equal to the G+jwC angle, doesn't that make Z0 purely resistive? -- 73, Cecil http://www.qsl.net/w5dxp |
On Mon, 22 Nov 2004 22:08:29 +0000 (UTC), "Reg Edwards"
wrote: Not having much else to do at present I thought I would make a comment on Zo and Ro of transmission lines. For entertainment and educational value, of course, if you like that sort of thing. The condition for which Zo of a transmission line is always purely resistive (Zo = Ro) is extremely simple. It is - G = C * R / L where G is shunt conductance, C is shunt capacitance, R is series resistance, L is series inductance, all per unit length of line. Which applies to any line length, at any frequency from DC to UHF. It is a shortcoming of the Smith Chart, with Zo always equal to Ro, that it does not make you aware of this and can lead you up the garden path if you are not careful. As has recently occurred on this newsgroup. Don't get me wrong. I'm not against Smith Charts. They are graphically educational within their limitations. ---- Reg, G4FGQ Dear Reg, You say that it is a shortcoming of the Smith Chart that Zo equals Ro. However, I think that is either a misunderstanding or just misleading. The Smith Chart only constrains the normalizing quantity to be purely resistive - not the characteristic impedance of a particular transmission line being shown on that chart. My program, SmartSmith, for example, allows the user to specify both an Ro and an Xo term for all transmission line sections. When it's all said and done, the Smith Chart only implements the transmission line equation (as shown on pages 24-10 and 27-29 in the 17th Edition of The ARRL Antenna Book). With my respects and best wishes, Bob, W9DMK, Dahlgren, VA http://www.qsl.net/w9dmk |
Cecil,
Try it. I believe you will find that your equality requirement on angles reduces to precisely the simple equation offer by Reg. 73, Gene W4SZ Cecil Moore wrote: Reg Edwards wrote: The condition for which Zo of a transmission line is always purely resistive (Zo = Ro) is extremely simple. It is - G = C * R / L Wonder why Ramo and Whinnery say that's an approximation for low-loss lines? If the R+jwL angle is equal to the G+jwC angle, doesn't that make Z0 purely resistive? -- 73, Cecil http://www.qsl.net/w5dxp |
Gene Fuller wrote:
Try it. I believe you will find that your equality requirement on angles reduces to precisely the simple equation offer by Reg. Exactly! That's why I wonder why Ramo and Whinnery said it's an approximation. Wonder why Ramo and Whinnery say that's an approximation for low-loss lines? If the R+jwL angle is equal to the G+jwC angle, doesn't that make Z0 purely resistive? -- 73, Cecil http://www.qsl.net/w5dxp |
Cecil,
They were undoubtedly confused by their models, and they could not deal with reality. 73, Gene W4SZ Cecil Moore wrote: Gene Fuller wrote: Try it. I believe you will find that your equality requirement on angles reduces to precisely the simple equation offer by Reg. Exactly! That's why I wonder why Ramo and Whinnery said it's an approximation. Wonder why Ramo and Whinnery say that's an approximation for low-loss lines? If the R+jwL angle is equal to the G+jwC angle, doesn't that make Z0 purely resistive? -- 73, Cecil http://www.qsl.net/w5dxp |
Gene Fuller wrote:
They were undoubtedly confused by their models, and they could not deal with reality. Thanks Gene, I really appreciate it when you contribute something techincal. -- 73, Cecil http://www.qsl.net/w5dxp |
Gene Fuller wrote: Cecil, They were undoubtedly confused by their models, and they could not deal with reality. 73, Gene W4SZ I guess one could infer that that if G / C R / L, and R + jwL G + jwC, then perhaps there are losses. I would only add that there are probably also small currents in shunt distributed along the line. 73, ac6xg Cecil Moore wrote: Gene Fuller wrote: Try it. I believe you will find that your equality requirement on angles reduces to precisely the simple equation offer by Reg. Exactly! That's why I wonder why Ramo and Whinnery said it's an approximation. Wonder why Ramo and Whinnery say that's an approximation for low-loss lines? If the R+jwL angle is equal to the G+jwC angle, doesn't that make Z0 purely resistive? -- 73, Cecil http://www.qsl.net/w5dxp |
For those with a mathematical bent, alternatively stated, for the angle of Zo to be zero, or for Zo to be purely resistive - L * G = C * R or L / C = R / G which relationship is derived from - Zo = Squareroot( Z / Y ) Where line series impedance Z = R + j * Omega * L and line shunt admittance Y = G + j * Omega * C provided the angle of Z is equal to the angle of Y. Which makes Zo = Ro + j * ZERO QED ------ Reg, G4FGQ |
Cecil,
Do you s'pose that if the equality is perfect for zero-loss lines then maybe it is an useful approximation for low-loss lines? Do you really think R&W were proposing that this simple relationship is more appropriate for low loss lines than for zero loss lines? 73, Gene W4SZ Cecil Moore wrote: Gene Fuller wrote: Try it. I believe you will find that your equality requirement on angles reduces to precisely the simple equation offer by Reg. Exactly! That's why I wonder why Ramo and Whinnery said it's an approximation. Wonder why Ramo and Whinnery say that's an approximation for low-loss lines? If the R+jwL angle is equal to the G+jwC angle, doesn't that make Z0 purely resistive? -- 73, Cecil http://www.qsl.net/w5dxp |
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Jim Kelley wrote:
I guess one could infer that that if G / C R / L, and R + jwL G + jwC, then perhaps there are losses. I would only add that there are probably also small currents in shunt distributed along the line. If R 0 and G 0, then there are losses. The only time a line is lossless is when R = G = 0 which, according to Reg, is only in my wet dreams about circles on Smith Charts. :-) For real world transmission lines at HF, (usually) R/Z0 G*Z0. When I was a member of the high speed cable group at Intel, I remember test leads designed for R/Z0=G*Z0 but they were expensive special order devices. We apparently are more successful at designing very good dielectrics than in finding an economically feasible conductor with a couple of magnitudes less resistance than copper. Thus our ordinary transmission lines have a lot more series resistance than shunt conductance, especially open-wire transmission lines in free space. :-) -- 73, Cecil http://www.qsl.net/w5dxp |
Gene Fuller wrote:
Do you s'pose that if the equality is perfect for zero-loss lines then maybe it is an useful approximation for low-loss lines? Do you really think R&W were proposing that this simple relationship is more appropriate for low loss lines than for zero loss lines? Nope, exactly the opposite. Apparently, they were proposing that this simple relationship doesn't hold for highly lossy lines. Chipman also has something to say about highly lossy lines. -- 73, Cecil http://www.qsl.net/w5dxp |
Dr. Slick wrote:
Well, that is pretty entertaining and interesting, i will admit. However, the result isn't very practical... On the contrary, the result is extremely practical and isn't very ideal, just like real-world physics. :-) -- 73, Cecil http://www.qsl.net/w5dxp |
Wes Stewart wrote in message . ..
On 23 Nov 2004 07:36:11 -0800, (Dr. Slick) wrote: [snip] | Since i trust Les Besser more than |other people, Heh heh. Reminds me... I took his video taped course, "RF/Microwave Transistor Amplifier Design". I was watching one tape at home while the XYL was making dinner. She could hear the audio and finally asked me to turn it off since it was making *her* fall asleep. [g]. Really smart guy, but not a very dynamic speaker. Hehe! Yeah, also not exactly the sexiest guy on the planet, is he! I have his full RF Fund. course One and Two on tape, like 12 tapes in all. We should have a Les Besser Get together sometime. Wooohaaa! Fun! Slick |
Cecil Moore wrote in message ...
Dr. Slick wrote: Well, that is pretty entertaining and interesting, i will admit. However, the result isn't very practical... On the contrary, the result is extremely practical and isn't very ideal, just like real-world physics. :-) When was the last time you used "G = C * R / L" for anything? Slick |
Dr. Slick wrote:
When was the last time you used "G = C * R / L" for anything? Yesterday. |
Cecil Moore wrote:
Dr. Slick wrote: When was the last time you used "G = C * R / L" for anything? Yesterday. Just curious, Cecil. What were you using a distortionless line for? Tom Donaly, KA6RUH |
Tom Donaly wrote:
Cecil Moore wrote: Dr. Slick wrote: When was the last time you used "G = C * R / L" for anything? Yesterday. Just curious, Cecil. What were you using a distortionless line for? Methinks you jumped to conclusions. Slick didn't ask when was the last time I used a distortionless line. He asked when was the last time I used the equation "G=C*R/L" for anything. I'm using it right now. I used it yesterday to refresh my memory about distortionless lines which I did use quite often before I retired from Intel in 1998. If I presently held a job in the cable modem group, I suppose I would still be using them. -- 73, Cecil http://www.qsl.net/w5dxp |
Cecil Moore wrote:
Tom Donaly wrote: Cecil Moore wrote: Dr. Slick wrote: When was the last time you used "G = C * R / L" for anything? Yesterday. Just curious, Cecil. What were you using a distortionless line for? Methinks you jumped to conclusions. Slick didn't ask when was the last time I used a distortionless line. He asked when was the last time I used the equation "G=C*R/L" for anything. I'm using it right now. I used it yesterday to refresh my memory about distortionless lines which I did use quite often before I retired from Intel in 1998. If I presently held a job in the cable modem group, I suppose I would still be using them. -- 73, Cecil http://www.qsl.net/w5dxp I'm surprised the golden ears croud hasn't discovered distortionless lines as a means of separating the gullible from their gold. 73, Tom Donaly, KA6RUH |
Cecil Moore wrote in message ...
Dr. Slick wrote: When was the last time you used "G = C * R / L" for anything? Yesterday. Oh really? Please explain in detail? S. |
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Cecil Moore wrote in message ...
Dr. Slick wrote: When was the last time you used "G = C * R / L" for anything? Yesterday. Ok, kids, can you say: "BULL****"? Slick |
Dear Reg,
The Smith Chart only constrains the normalizing quantity to be purely resistive - not the characteristic impedance of a particular transmission line being shown on that chart. My program, SmartSmith, for example, allows the user to specify both an Ro and an Xo term for all transmission line sections. Bob, W9DMK, Dahlgren, VA =================================== Hi Bob, But your program is not a Smith Chart. It's probably better than a Smith Chart. ---- Reg. |
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Bob, your program can probably calculate the input impedance, Rin + j*Xin,
of a line having Zo = Ro + j*Xo, with given attenuation Alpha dB, and given phase-shift Beta radians, with a terminating impedance Rt + j*Xt. Which is a commonly needed quantity on the way to calculating the ultimate, all-important, single number, transmission efficiency. But can you do it with nothing at hand except a Smith Chart? ---- Yours, Reg. |
On Fri, 26 Nov 2004 22:05:13 +0000 (UTC), "Reg Edwards"
wrote: But can you do it with nothing at hand except a Smith Chart? Turn the chart over and write the math on the back - GEEZ |
Turn the chart over and write the math on the back - GEEZ
=========================== Can you do it yourself within 3 months? It will take that long to dig out the formulae. I've just realised this is the first occasion I've ever been at the bottom end Z of a newsgroup message listing. You can't get any lower! ---- Yours, Punchinello. |
On Sat, 27 Nov 2004 01:02:37 +0000 (UTC), "Reg Edwards"
wrote: Turn the chart over and write the math on the back - GEEZ Can you do it yourself within 3 months? It will take that long to dig out the formulae. Turn the chart back over and save yourself the time. Any more effort will burn more calories than any transmission line. I've just realised this is the first occasion I've ever been at the bottom end Z of a newsgroup message listing. You can't get any lower! You can if you start the thread and no one comes to the party. It's been said that if you haven't failed - you aren't trying hard enough. 73's Richard Clark, KB7QHC |
On Fri, 26 Nov 2004 22:05:13 +0000 (UTC), "Reg Edwards"
wrote: Bob, your program can probably calculate the input impedance, Rin + j*Xin, of a line having Zo = Ro + j*Xo, with given attenuation Alpha dB, and given phase-shift Beta radians, with a terminating impedance Rt + j*Xt. Which is a commonly needed quantity on the way to calculating the ultimate, all-important, single number, transmission efficiency. But can you do it with nothing at hand except a Smith Chart? And the answer is... an unequivocal NO. However, just to satisfy my curiousity, exactly which of your beautiful, zipped up Pascal programs will do that for me? Bob, W9DMK, Dahlgren, VA http://www.qsl.net/w9dmk |
Bob, your program can probably calculate the input impedance, Rin + j*Xin, of a line having Zo = Ro + j*Xo, with given attenuation Alpha dB, and given phase-shift Beta radians, with a terminating impedance Rt + j*Xt. Which is a commonly needed quantity on the way to calculating the ultimate, all-important, single number, transmission efficiency. =============================================== Bob asked - However, just to satisfy my curiousity, exactly which of your beautiful, zipped up Pascal programs will do that for me? =============================================== Bob, I thought nobody would ever ask. But they've been available from my website for years. Look at Programs - RJELINE2 RJELINE3 RJELINE4 COAXPAIR COAXRATE and give yourself a few practical examples. There is a one-line description after each program's name on the download page in my website. The above programs are dedicated to transmission lines. Input data includes one or two physical dimensions which avoids restriction to particular type-of-line numbers. Users are given a free hand to design cables to their own specifications. Nevertheless, they are practical in nature and simple to use. There are other programs which incorporate the same calculations but which are not explicitly apparent to the user. They use exact classical transmission line formulae and so are as accurate as the input data over the stated frequency ranges. Usually from power frequencies up to UHF. They take skin effect and the increase in inductance at low frequencies and other subtle factors such as conductor proximity effect in twin-lines in their stride. They are good enough for the highest precision engineering applications. I have not disclosed the source code to prevent it falling into the hands of argumentative vandals, so-called guru's, and technically ignorant old-wives who would ruin the programs' reputation, not forgetting mine, for RELIABILITY. Reliability is Quality versus Time. For references I quote my only tutors - Ohm, Ampere and Volta. When considering transmission lines you can check your's and other programs against mine for accuracy with confidence. You will discover the effects of both your known and other, unsuspected approximations. Readers should bear in mind I'm not getting paid for this. Bob, I'm on MontGras, Chilean, Reserve Merlot, tonight. Nuff said. ---- .................................................. .......... Regards from Reg, G4FGQ For Free Radio Design Software go to http://www.btinternet.com/~g4fgq.regp .................................................. .......... |
On Sun, 28 Nov 2004 02:03:57 +0000 (UTC), "Reg Edwards"
wrote: Bob, I'm on MontGras, Chilean, Reserve Merlot, tonight. Nuff said. There's nothing that can top vintage Reg. Tnx, I'll visit the site and pick up my free samples. Tschuss! Bob, W9DMK, Dahlgren, VA http://www.qsl.net/w9dmk |
Who the heck are Ramo and Whinnery. Never heard of them! Presumably, because you refer to them, they are or were people who make or made a living out of re-iterating old wive's tales in book-form. It was obvious I introduced G = C * R / L simply to show that a line's Zo can be purely resistive even when it is NOT lossless. It can have any loss you like. Apparently you have not yet grasped the idea. And, despite what R and W or YOU may have to say on the subject, it is an exact expression at all frequencies from DC to almost infinity. My only references are Ohm, Ampere and Volta who I'm sure you have heard of. But no hard feelings. ;o) Tonight I'm on Chilean, dry, 2004, MontGras, Reserve Chardonnay. I didn't choose it myself. My loving daughter does my shopping. But it's quite a pleasant, refreshing plonk. ---- Regards, Reg. ============================================ "Cecil Moore" wrote Reg Edwards wrote: The condition for which Zo of a transmission line is always purely resistive (Zo = Ro) is extremely simple. It is - G = C * R / L Wonder why Ramo and Whinnery say that's an approximation for low-loss lines? If the R+jwL angle is equal to the G+jwC angle, doesn't that make Z0 purely resistive? -- 73, Cecil |
"Reg Edwards" wrote in message ... Who the heck are Ramo and Whinnery. Never heard of them! Presumably, because you refer to them, they are or were people who make or made a living out of re-iterating old wive's tales in book-form. It was obvious I introduced G = C * R / L simply to show that a line's Zo can be purely resistive even when it is NOT lossless. It can have any loss you like. Apparently you have not yet grasped the idea. And, despite what R and W or YOU may have to say on the subject, it is an exact expression at all frequencies from DC to almost infinity. My only references are Ohm, Ampere and Volta who I'm sure you have heard of. But no hard feelings. ;o) Tonight I'm on Chilean, dry, 2004, MontGras, Reserve Chardonnay. I didn't choose it myself. My loving daughter does my shopping. But it's quite a pleasant, refreshing plonk. ---- Regards, Reg. ============================================ "Cecil Moore" wrote Reg Edwards wrote: The condition for which Zo of a transmission line is always purely resistive (Zo = Ro) is extremely simple. It is - G = C * R / L Wonder why Ramo and Whinnery say that's an approximation for low-loss lines? If the R+jwL angle is equal to the G+jwC angle, doesn't that make Z0 purely resistive? -- 73, Cecil Fields & Waves in Communication Electronics, by S. Ramo, J.R. Whinnery, and T. Van Duzer, Wiley, 3rd edition, 1994 107 proof Baker's for me 73, H. |
Reg Edwards wrote:
Who the heck are Ramo and Whinnery. Never heard of them! Presumably, because you refer to them, they are or were people who make or made a living out of re-iterating old wive's tales in book-form. It was obvious I introduced G = C * R / L simply to show that a line's Zo can be purely resistive even when it is NOT lossless. It can have any loss you like. Apparently you have not yet grasped the idea. Ramo and Whinnery are the authors of my 50's college textbook on fields and waves. Of course it could be a misprint, but they say your above formula is an approximation that is good for low-loss lines. Apparently, something additional happens for high-loss lines. Chipman seems to agree with Ramo and Whinnery when he introduces some additional interference terms (discussed some time ago on this newsgroup). At the time, I didn't realize the additional terms were interference terms but the impedance of the load apparently somehow interacts with the characteristic impedance of the high-loss transmission line to upset the ideal relationships in your equation above. -- 73, Cecil http://www.qsl.net/w5dxp |
Cecil Moore wrote:
Reg Edwards wrote: It was obvious I introduced G = C * R / L simply to show that a line's Zo can be purely resistive even when it is NOT lossless. It can have any loss you like. Ramo and Whinnery are the authors of my 50's college textbook on fields and waves. Of course it could be a misprint, but they say your above formula is an approximation that is good for low-loss lines. Certainly good at HF and UHF when the skin depth is likely to be a small fraction of the conductor radius. Apparently, something additional happens for high-loss lines. Not so much high loss, as low frequency. Both L and R are frequency dependent assuming normal (non superconducting) metallic conductors. G and C may have a frequency dependency depending on the dielectric characteristics. Once the frequency is high enough so that the current can be considered to flow only on the skin of the conductor, the effective AC resistance is proportional to the square root of the frequency and the inductance is constant. At frequencies below the above defined 'critical frequency', the internal inductance must be considered as well as the complicated frequency dependence of resistance. Chipman seems to agree with Ramo and Whinnery when he introduces some additional interference terms (discussed some time ago on this newsgroup). Yep. At the time, I didn't realize the additional terms were interference terms but the impedance of the load apparently somehow interacts with the characteristic impedance of the high-loss transmission line to upset the ideal relationships in your equation above. The relationship is correct for all frequencies and standing wave ratios as long as the correct frequency dependent values of transmission line parameters are used. The wave equation still describes the relationship between current and voltage. The additional 'interference' terms appear when calculating the energy distribution and loss characteristics. bart wb6hqk |
Bart Rowlett wrote:
The relationship is correct for all frequencies and standing wave ratios as long as the correct frequency dependent values of transmission line parameters are used. Here's the equation that Ramo and Whinnery says is an approximation for low-loss lines. Z0 = SQRT(L/C)[1 + j(G/2wC - R/2wL)] If G = C * R / L then Z0 = SQRT(L/C) So why did Ramo and Whinnery say it is an approximation for low-loss lines? -- 73, Cecil http://www.qsl.net/w5dxp |
On Sun, 28 Nov 2004 23:01:04 -0600, Cecil Moore
wrote: Here's the equation that Ramo and Whinnery says is an approximation for low-loss lines. So you keep saying. Is it that difficult to find their exact solution for any lines? |
Richard Clark wrote:
Cecil Moore wrote: Here's the equation that Ramo and Whinnery says is an approximation for low-loss lines. So you keep saying. This is the first time I have posted the equation. Is it that difficult to find their exact solution for any lines? Maybe the math is easier for the approximation? -- 73, Cecil http://www.qsl.net/w5dxp |
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