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Calculating loss on a mismatched line
Apparently, Michaels described in "Technical Correspondence" in QST NOV 1997 a method for calculating loss on a mismatched line. I don't have the article, and haven't been able to find it on the net, so I am working from references to it that I have seen, mainly in Usenet. Apparently, the method involves calculation of a factor (let's call it MR) as MR=|(Zl-Zo*)/(Zl+Zo)|, and the line loss between two points is given by 10*log((1-MR1**2)/(1-MR2**2)) where MR1 and MR2 are the values for MR at points 1 and 2. I have compared the results of this on lines with Xo0 and high VSWR, and the results are identical to calcuating the loss by subtraction of Real(VI*) at point 1 from Real(VI*) at point 2. Has anyone a link to, or a reference to the derivation of the formula? Owen -- |
On Wed, 06 Jul 2005 22:07:41 GMT, Owen wrote:
Apparently, the method involves calculation of a factor (let's call it MR) as MR=|(Zl-Zo*)/(Zl+Zo)|, and the line loss between two points is given by 10*log((1-MR1**2)/(1-MR2**2)) where MR1 and MR2 are the values for MR at points 1 and 2. Sorry, that gives the additional loss "due to SWR" which has to be added to the matched line loss to obtain the total loss between points 1 and 2. Zl in the formulas is the impedance (V/I) at the point where MR is calculated. Owen -- |
"Owen" wrote in message ... On Wed, 06 Jul 2005 22:07:41 GMT, Owen wrote: Apparently, the method involves calculation of a factor (let's call it MR) as MR=|(Zl-Zo*)/(Zl+Zo)|, and the line loss between two points is given by 10*log((1-MR1**2)/(1-MR2**2)) where MR1 and MR2 are the values for MR at points 1 and 2. Sorry, that gives the additional loss "due to SWR" which has to be added to the matched line loss to obtain the total loss between points 1 and 2. Zl in the formulas is the impedance (V/I) at the point where MR is calculated. Owen Well, Owen, you might take a look at Appendix 8 of Reflections 2, which can be found on my web site at w2du.com. Click on 'Read Appendices from Reflections 2', the click on Appendix 8. I think what you'll see there will be of interest. Walt, W2DU |
On Wed, 6 Jul 2005 18:32:50 -0400, "Walter Maxwell"
wrote: Well, Owen, you might take a look at Appendix 8 of Reflections 2, which can be found on my web site at w2du.com. Click on 'Read Appendices from Reflections 2', the click on Appendix 8. I think what you'll see there will be of interest. Thanks Walt. I had a look at it, and although it doesn't state as much, isn't it correct only for distortionless lines (Xo=0)? Owen -- |
Owen,
Your formula is too short to be anything but an approximation. It may be a very good approximation. On the other hand it may contain exactly the same errors as whatever you may have checked it against. The exact formula is exceedingly involved and occupies about half a dozen lines of source code in new program SWRARGUE which by coincidence I have just placed in my website. You can check your formula against my program. Let us know how you get on. ---- Reg, G4FGQ --------------------------------------------------------------- "Owen" wrote in message ... Apparently, Michaels described in "Technical Correspondence" in QST NOV 1997 a method for calculating loss on a mismatched line. I don't have the article, and haven't been able to find it on the net, so I am working from references to it that I have seen, mainly in Usenet. Apparently, the method involves calculation of a factor (let's call it MR) as MR=|(Zl-Zo*)/(Zl+Zo)|, and the line loss between two points is given by 10*log((1-MR1**2)/(1-MR2**2)) where MR1 and MR2 are the values for MR at points 1 and 2. I have compared the results of this on lines with Xo0 and high VSWR, and the results are identical to calcuating the loss by subtraction of Real(VI*) at point 1 from Real(VI*) at point 2. Has anyone a link to, or a reference to the derivation of the formula? Owen -- |
On Wed, 6 Jul 2005 23:48:53 +0000 (UTC), "Reg Edwards"
wrote: Owen, Your formula is too short to be anything but an approximation. It may be a very good approximation. On the other hand it may contain exactly the same errors as whatever you may have checked it against. The exact formula is exceedingly involved and occupies about half a dozen lines of source code in new program SWRARGUE which by coincidence I have just placed in my website. I have calculated the loss using P=Real(V*I*) at the two points, and it is long winded. Michaels approach produces the same result, and the coding is more elegant, probably faster to calculate. He is a SK, so I can't ask him, but in the hope that it is well known, someone might know of the derivation. You can check your formula against my program. Let us know how you get on. Your calculator does not allow complex Zo does it? Doesn't that mean it assumes a distortionless line. I am calculating loss in the general case. Thanks... Owen -- |
"Owen" wrote in message ... On Wed, 6 Jul 2005 18:32:50 -0400, "Walter Maxwell" wrote: Well, Owen, you might take a look at Appendix 8 of Reflections 2, which can be found on my web site at w2du.com. Click on 'Read Appendices from Reflections 2', the click on Appendix 8. I think what you'll see there will be of interest. Thanks Walt. I had a look at it, and although it doesn't state as much, isn't it correct only for distortionless lines (Xo=0)? Owen As I understand it, Owen, a line has to be lossless for Xo to be 0, while distortionless lines have loss but have equal series R and shunt G. All lines that I've measured have a small negative X, that would be zero if the line were lossless. So I'd have to say that the material in Appendix is is correct for standard lines. Distortionless lines are normally found only in long-distance phone lines used at voice frequencies, not RF. Or am I missing something? Walt,W2DU |
Owen:
Right of you to be cautious of that wise and crafty old brit. Not only has he survived in this den of treacherous and particularly vicious hams--he has THRIVED--this is quite suspicious in itself! However, most often you will find there is true wisdom in his words and programs... and it is of a highly practical nature. He is a welcomed resource here... John "Owen" wrote in message ... On Wed, 6 Jul 2005 23:48:53 +0000 (UTC), "Reg Edwards" wrote: Owen, Your formula is too short to be anything but an approximation. It may be a very good approximation. On the other hand it may contain exactly the same errors as whatever you may have checked it against. The exact formula is exceedingly involved and occupies about half a dozen lines of source code in new program SWRARGUE which by coincidence I have just placed in my website. I have calculated the loss using P=Real(V*I*) at the two points, and it is long winded. Michaels approach produces the same result, and the coding is more elegant, probably faster to calculate. He is a SK, so I can't ask him, but in the hope that it is well known, someone might know of the derivation. You can check your formula against my program. Let us know how you get on. Your calculator does not allow complex Zo does it? Doesn't that mean it assumes a distortionless line. I am calculating loss in the general case. Thanks... Owen -- |
On Wed, 06 Jul 2005 22:27:08 GMT, Owen wrote:
On Wed, 06 Jul 2005 22:07:41 GMT, Owen wrote: Apparently, the method involves calculation of a factor (let's call it MR) as MR=|(Zl-Zo*)/(Zl+Zo)|, and the line loss between two points is given by 10*log((1-MR1**2)/(1-MR2**2)) where MR1 and MR2 are the values for MR at points 1 and 2. Sorry, that gives the additional loss "due to SWR" which has to be added to the matched line loss to obtain the total loss between points 1 and 2. Zl in the formulas is the impedance (V/I) at the point where MR is calculated. I think "Computation of Impedance and Efficiency of Transmission Line with High Standing-Wave Ratio", W.W. Macalpine, Transactions of the AIEE, vol. 72, pp 334-339; July, 1953 might be of interest to you. The same reference is used as the basis for the section "Transformation of Impedance on Lines With Highh SWR" in the ITT handbook. |
On Wed, 6 Jul 2005 20:06:13 -0400, "Walter Maxwell"
wrote: As I understand it, Owen, a line has to be lossless for Xo to be 0, while distortionless lines have loss but have equal series R and shunt G. All Walt, I learnt that a distortionless line is one where attenuation and phase velocity are constant for all frequencies, and that requires that R/XL=G/XC in the RLGC model of a lines characteristics, and the result is that Zo is purely real. A lossless line is a special case of a distortionless line. lines that I've measured have a small negative X, that would be zero if the line were lossless. So I'd have to say that the material in Appendix is is correct for standard lines. Distortionless lines are normally found only in long-distance phone lines used at voice frequencies, not RF. Or am I missing something? If you like, I am saying your approach is valid for lossless lines, it is also valid for all distortionless lines, but I think it is not accurate for lines in the general case because it isn't correct if Xo!=0. Owen -- |
My program assumes line Zo is purely resistive. But at HF it is
practically impossible to tell the difference in line loss between an angle of 0 degrees and 1 degree which is the range of Zo angles involved. I have the exact formula for any line but its far too complicated to write here. I didn't use it in my program as it would have meant another input value for no useful purpose. ---- Reg ---------------------------------------------------------- "Owen" wrote in message ... On Wed, 6 Jul 2005 23:48:53 +0000 (UTC), "Reg Edwards" wrote: Owen, Your formula is too short to be anything but an approximation. It may be a very good approximation. On the other hand it may contain exactly the same errors as whatever you may have checked it against. The exact formula is exceedingly involved and occupies about half a dozen lines of source code in new program SWRARGUE which by coincidence I have just placed in my website. I have calculated the loss using P=Real(V*I*) at the two points, and it is long winded. Michaels approach produces the same result, and the coding is more elegant, probably faster to calculate. He is a SK, so I can't ask him, but in the hope that it is well known, someone might know of the derivation. You can check your formula against my program. Let us know how you get on. Your calculator does not allow complex Zo does it? Doesn't that mean it assumes a distortionless line. I am calculating loss in the general case. Thanks... Owen -- |
On Wed, 06 Jul 2005 17:17:17 -0700, Wes Stewart
wrote: I think "Computation of Impedance and Efficiency of Transmission Line with High Standing-Wave Ratio", W.W. Macalpine, Transactions of the AIEE, vol. 72, pp 334-339; July, 1953 might be of interest to you. The same reference is used as the basis for the section "Transformation of Impedance on Lines With Highh SWR" in the ITT handbook. Yes, I saw (and I think, understand) the formulas under the heading "Power and Efficiency" in my ed #6 of the ITT H'book. Not wanting to express the algorithm in pseudo code (AKA ugly ASCII math) here, 'cos it *will* be unreadable, I have pasted the formula I have worked up, and a plot of it in a particular scenario at http://www.vk1od.net/temp/LineLoss.htm . Given, that to support this, there is are functions for Gamma(load) and Gamma(x), it is a whole bunch of code, and Michaels approach is attractively simple, and it produces exactly the same answer on this scenario. Owen -- |
"Owen" wrote in message ... On Wed, 6 Jul 2005 20:06:13 -0400, "Walter Maxwell" wrote: As I understand it, Owen, a line has to be lossless for Xo to be 0, while distortionless lines have loss but have equal series R and shunt G. All Walt, I learnt that a distortionless line is one where attenuation and phase velocity are constant for all frequencies, and that requires that R/XL=G/XC in the RLGC model of a lines characteristics, and the result is that Zo is purely real. A lossless line is a special case of a distortionless line. lines that I've measured have a small negative X, that would be zero if the line were lossless. So I'd have to say that the material in Appendix is is correct for standard lines. Distortionless lines are normally found only in long-distance phone lines used at voice frequencies, not RF. Or am I missing something? If you like, I am saying your approach is valid for lossless lines, it is also valid for all distortionless lines, but I think it is not accurate for lines in the general case because it isn't correct if Xo!=0. Owen Owen, if X = 0 there is no attenuation, but you're saying my material is invalid if X is not 0? I'm sorry, but I'm confused. Walt, W2DU |
On Wed, 6 Jul 2005 21:23:52 -0400, "Walter Maxwell"
wrote: If you like, I am saying your approach is valid for lossless lines, it is also valid for all distortionless lines, but I think it is not accurate for lines in the general case because it isn't correct if Xo!=0. Owen Owen, if X = 0 there is no attenuation, but you're saying my material is invalid if X is not 0? I'm sorry, but I'm confused. Walt, I don't think that Xo=0 implies zero attenuation, but it is true that if the line has zero attenuation that Xo=0. I think the method in your appendix is true when Xo=0, and an approximation where Xo!=0. I have just added a function (DLoss) to calculate loss as per your appendix to the graphic at http://www.vk1od.net/temp/LineLoss.htm , and it can be seen that in this (perhaps extreme) case, it is not a very good approximation. I also added the same calcs for f=100MHz and Zo is much closer to real, and it can be seen the approximation is much closer. Owen -- |
"Owen" wrote in message
... Apparently, Michaels described in "Technical Correspondence" in QST NOV 1997 a method for calculating loss on a mismatched line. I don't have the article, and haven't been able to find it on the net, so I am working from references to it that I have seen, mainly in Usenet. Apparently, the method involves calculation of a factor (let's call it MR) as MR=|(Zl-Zo*)/(Zl+Zo)|, and the line loss between two points is given by 10*log((1-MR1**2)/(1-MR2**2)) where MR1 and MR2 are the values for MR at points 1 and 2. I have compared the results of this on lines with Xo0 and high VSWR, and the results are identical to calcuating the loss by subtraction of Real(VI*) at point 1 from Real(VI*) at point 2. Has anyone a link to, or a reference to the derivation of the formula? Owen In my first edition of "Paul and Nasar", p 335, dealing with lossy transmission lines; there is a footnote as follows: For the interested reader, a very thorough discussion of the derivation the transmission-line equations is given in R. B. Adler, L. J. Chu, and R. M. Fano , "Electromagnetic Energy Transmission and Radiation", Wiley, New York, 1960, chap. 9. I have never seen the reference, so cannot comment, but may be available in a university library. Wonder if that is the Fano in "Fano's limit". 73, Frank |
"Owen" wrote in message ... On Wed, 6 Jul 2005 21:23:52 -0400, "Walter Maxwell" wrote: If you like, I am saying your approach is valid for lossless lines, it is also valid for all distortionless lines, but I think it is not accurate for lines in the general case because it isn't correct if Xo!=0. Owen Owen, if X = 0 there is no attenuation, but you're saying my material is invalid if X is not 0? I'm sorry, but I'm confused. Walt, I don't think that Xo=0 implies zero attenuation, but it is true that if the line has zero attenuation that Xo=0. I think the method in your appendix is true when Xo=0, and an approximation where Xo!=0. I have just added a function (DLoss) to calculate loss as per your appendix to the graphic at http://www.vk1od.net/temp/LineLoss.htm , and it can be seen that in this (perhaps extreme) case, it is not a very good approximation. I also added the same calcs for f=100MHz and Zo is much closer to real, and it can be seen the approximation is much closer. Owen I'll have to study your math, Owen, to let it sink in. I don't get it on the first glance. Walt |
On Wed, 6 Jul 2005 23:05:03 -0400, "Walter Maxwell"
wrote: I have just added a function (DLoss) to calculate loss as per your appendix to the graphic at http://www.vk1od.net/temp/LineLoss.htm , and it can be seen that in this (perhaps extreme) case, it is not a very good approximation. I also added the same calcs for f=100MHz and Zo is much closer to real, and it can be seen the approximation is much closer. Owen I'll have to study your math, Owen, to let it sink in. I don't get it on the first glance. Sorry all, The function for DLoss on the web page was for the additional loss due to SWR. I have now added a term to the DLoss function to include the matched line loss and updated the web page. The outcome hasn't changed in that the two methods of calculating loss only agree if Zo is real. Owen -- |
Owen wrote:
"Has anuypne a link to or reference to derivation of the formula?" My 19th edition of the ARRL Antenna Book treats additional power loss due to SWR on page 24-10. Best regards, Richard Harrison, KB5WZI |
"Richard Harrison" wrote in message ... Owen wrote: "Has anuypne a link to or reference to derivation of the formula?" My 19th edition of the ARRL Antenna Book treats additional power loss due to SWR on page 24-10. Best regards, Richard Harrison, KB5WZI ======================================== On this subject the only hope of discovering the exact loss on a line with SWR is to do an exact complex mathematical analyis for yourseslf and hope you don't make any mistakes. For starters, the calculated curves shown in ARRL publications over the years, as plagiarised by the RSGB, are in error. But they are near enough not to lose any sleep about. Why the importance attatched to this line feature I can't imagine. --- Reg. |
On Wed, 6 Jul 2005 23:48:53 +0000 (UTC), "Reg Edwards"
wrote: Owen, Your formula is too short to be anything but an approximation. It may be a very good approximation. On the other hand it may contain exactly the same errors as whatever you may have checked it against. The exact formula is exceedingly involved and occupies about half a dozen lines of source code in new program SWRARGUE which by coincidence I have just placed in my website. Dear Reg, I do not believe in coincidence (that's how we knew when the 2nd airplane flew into the World Trade Center, that it was not an accident.) Bob, W9DMK, Dahlgren, VA Replace "nobody" with my callsign for e-mail http://www.qsl.net/w9dmk http://zaffora/f2o.org/W9DMK/W9dmk.html |
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"Owen" wrote in message ... On Thu, 7 Jul 2005 11:23:24 -0500, (Richard Harrison) wrote: Owen wrote: "Has anuypne a link to or reference to derivation of the formula?" My 19th edition of the ARRL Antenna Book treats additional power loss due to SWR on page 24-10. Richard, I have an 18th edition, and the formula it gives does not factor in the angle of the reflection, and so is an approximation. I don't know if that applies to the 19th edition. Back to my original post, I was not in search of a formula for an exact solution, nor a reason to want an exact solution, but rather whether anyone has seen the derivation of Michaels formula quoted in QST Nov 97. Thanks for your help. Owen Well, Owen, if you believe the expressions I presented in Reflections 2 are approximate, then why do I get the correct answers? Walt, W2DU |
On Thu, 7 Jul 2005 18:46:36 -0400, "Walter Maxwell"
wrote: Well, Owen, if you believe the expressions I presented in Reflections 2 are approximate, then why do I get the correct answers? It seems to me that the method requires that rho is not greater than one (otherwise the denominator (1-rho2**2) becomes negative, which is a nonsense). This hints that it does not apply in the general case where rho *can* be greater than 1, and is therefore probably limited to cases of distorionless line (Xo=0). To avoid publishing "ugly" maths here, I have put a page up at http://www.vk1od.net/temp/reflection.htm with a bunch of expressions for conditions on the modelled line, including functions for power flow at an arbitrary point, Loss calculated from powerflow at two points and loss based on your loss formula + matched line loss. The graphs show the loss from point x to the load, x is 0 at the load and negative toward the source. The algorithms produce quite different results. If I ignore Xo (ie force Zo to be real), then both algorithms produce the same results. Have I made a mistake in the maths, or in modelling the scenario? Owen -- |
"Owen" wrote in message ... On Thu, 7 Jul 2005 18:46:36 -0400, "Walter Maxwell" wrote: Well, Owen, if you believe the expressions I presented in Reflections 2 are approximate, then why do I get the correct answers? It seems to me that the method requires that rho is not greater than one (otherwise the denominator (1-rho2**2) becomes negative, which is a nonsense). This hints that it does not apply in the general case where rho *can* be greater than 1, and is therefore probably limited to cases of distorionless line (Xo=0). To avoid publishing "ugly" maths here, I have put a page up at http://www.vk1od.net/temp/reflection.htm with a bunch of expressions for conditions on the modelled line, including functions for power flow at an arbitrary point, Loss calculated from powerflow at two points and loss based on your loss formula + matched line loss. The graphs show the loss from point x to the load, x is 0 at the load and negative toward the source. The algorithms produce quite different results. If I ignore Xo (ie force Zo to be real), then both algorithms produce the same results. Have I made a mistake in the maths, or in modelling the scenario? Owen Thanks for responding, Owen, but I'm going to be otherwise occupied until Saturday, so the fact that I don't respond immediately doesn't mean that I'm ignoring you. Walt |
On Wed, 6 Jul 2005 21:23:52 -0400, "Walter Maxwell"
wrote: If you like, I am saying your approach is valid for lossless lines, it is also valid for all distortionless lines, but I think it is not accurate for lines in the general case because it isn't correct if Xo!=0. Owen Owen, if X = 0 there is no attenuation, but you're saying my material is invalid if X is not 0? I'm sorry, but I'm confused. Walt, it has just occurred to me that I am using the "actual" Zo, not the nominal Zo, and I think your rho calc is based on the nominal Zo, as it will be measured with an instrument presumably calibrated for nominal Zo. I have compared the loss calculated by your method (with rho based on nominal Zo, Zo=Ro+j0) and my method and they are very similar (though not the same). I have added a function to calculate the loss using your formula based on nominal Zo and plotted it, along with the difference to the power based loss calc. They are at http://www.vk1od.net/temp/reflection.htm . If your method is based on nominal Zo, rather than the actual Zo, it is likely to be an approximation, though on this example, it is pretty close and probably is quite adequate for most practical lines at HF and above. (The error increases as frequency is reduced (Zo departs more from nominal Zo).) Having resolved the apparent inconsistency... I am still in search of a derivation of the Michaels formula. Owen -- |
Owen, I tried to send this as a reply to you, but your email address was
rejected, so I had to send this to the group. My response appears below your Walt "Owen" wrote in message ... On Wed, 6 Jul 2005 21:23:52 -0400, "Walter Maxwell" wrote: If you like, I am saying your approach is valid for lossless lines, it is also valid for all distortionless lines, but I think it is not accurate for lines in the general case because it isn't correct if Xo!=0. Owen Owen, if X = 0 there is no attenuation, but you're saying my material is invalid if X is not 0? I'm sorry, but I'm confused. Walt, it has just occurred to me that I am using the "actual" Zo, not the nominal Zo, and I think your rho calc is based on the nominal Zo, as it will be measured with an instrument presumably calibrated for nominal Zo. I have compared the loss calculated by your method (with rho based on nominal Zo, Zo=Ro+j0) and my method and they are very similar (though not the same). I have added a function to calculate the loss using your formula based on nominal Zo and plotted it, along with the difference to the power based loss calc. They are at http://www.vk1od.net/temp/reflection.htm . If your method is based on nominal Zo, rather than the actual Zo, it is likely to be an approximation, though on this example, it is pretty close and probably is quite adequate for most practical lines at HF and above. (The error increases as frequency is reduced (Zo departs more from nominal Zo).) Having resolved the apparent inconsistency... I am still in search of a derivation of the Michaels formula. Owen ----- Original Message ----- From: "Owen" Newsgroups: rec.radio.amateur.antenna Sent: Friday, July 08, 2005 6:08 PM Subject: Calculating loss on a mismatched line Hi Owen, I'm trying to understand your Mathcad presentations, but I've run into some roadblocks concerning terminology, some of which I'm not familiar with. I confess my questions prove my ignorance, but that's ok if one's trying to learn. However, I was using nominal Zo. First, Xo!=0. I don't know what this means. Second, what does MML stand for in English? Third, in 'functions for V, I, Z, etc at z'. Where is 'z'? I cannot find any reference to it. Fourth, 'exp'. Exponent? If so, of what? e? Fifth, I understand 'x' as distance along the line from the termination, but what is 'y'? Sixth, what is AppLoss? Approximate? Apparent? Applied? Seventh, 'DLoss'. What is 'D'? Dielectric? Again, what is the 'y' term? An ordinate value? Eighth, in the LineLoss(x,y) = 10log... the identical right-hand terms in both numerator and denominator, the identical functions of 'e^^ x e^^. what is the meaning of the bar above the second appearance of 'e'? And above gamma(x)? I want to understand your math presentation, Owen, especially when I see that Loss(x,0 - W2DUloss(x,0) is so small I want to understand what makes the difference. So I'd appreciate it if you'd set me straight on the points I made above. Walt |
On Sat, 9 Jul 2005 23:17:18 -0400, "Walter Maxwell"
wrote: I'm trying to understand your Mathcad presentations, but I've run into some roadblocks concerning terminology, some of which I'm not familiar with. I confess my questions prove my ignorance, but that's ok if one's trying to learn. However, I was using nominal Zo. Not at all, you are far more eminent that I on this topic, and I appreciate your review. I am learning from all this. Apologies for the difficulty in understanding my notation. Some of it breaks into psuedo programming code. First, Xo!=0. I don't know what this means. Not equals. Second, what does MML stand for in English? MLL? Matched Line Loss (dB/m) Third, in 'functions for V, I, Z, etc at z'. Where is 'z'? I cannot find any reference to it. These quantities are a function of z, where z is a position on the line. The convention that I have used for displacement is that it is negative towards the generator. When it matters, displacement is in metres. The z is just used in definition of some functions in Matchcad (where you see :=), I have used x for position variable in the graphs. Fourth, 'exp'. Exponent? If so, of what? e? exp(x) is e to the power of x (For clarity, I shouldn't have written it that way, it works, but Mathcad understands the meaning of e superscript x as e to the power of x, as you will see in some of the expressions, and it is easier to read.) Fifth, I understand 'x' as distance along the line from the termination, but what is 'y'? In some of the functions, I have written them to calculate some quantity between two arbitrary points x and y. They are used in the definition of fuctions (where you see :=). Most of the graphs use 0 for y so they are plotted wrt the load position Sixth, what is AppLoss? Approximate? Apparent? Applied? Approximate Loss, and it was incorrectly based on Zo rather than nominal Zo. Seventh, 'DLoss'. What is 'D'? Dielectric? Again, what is the 'y' term? An ordinate value? DLoss was equivalent to AppLoss. Eighth, in the LineLoss(x,y) = 10log... the identical right-hand terms in both numerator and denominator, the identical functions of 'e^^ x e^^. what is the meaning of the bar above the second appearance of 'e'? And above gamma(x)? The bar above the variable is the complex conjugate operator. I want to understand your math presentation, Owen, especially when I see that Loss(x,0 - W2DUloss(x,0) is so small I want to understand what makes the difference. So I'd appreciate it if you'd set me straight on the points I made above. Walt, in the models at http://www.vk1od.net/temp/LineLoss.htm , I now know why there is such a gap between DLoss and LineLoss. You will recognise AppLoss / DLoss is your Appendix 8 expression, but my rho function was based on the modelled complex value of Zo (characteristic impedance), not the nominal value of Zo. In the second lot at http://www.vk1od.net/temp/reflection.htm , AppLoss is equivalent to DLoss and it is based on nominal Zo, W2DULoss you will see calculates the rho term (though not identified) using nominal Ro. Comparing the results with loss calcuated from P(x)/P(y) (the ratio of the real power at points x and y), the conclusion is that using your expression with actual Zo is not at all accurate, using it with nominal Zo is very close. If I force Zo to be real for all modelling, the results of all methods is exactly the same (within rounding errors of the order of 10 to the power of -14) Some of your questions are just about the Mathcad notation (though that is not too dissimilar to normal handwritten math notation), but some of it is my expression and usage. Again my apologies for confusing with too little explanation. I appreciate your review and comments Walt. Owen -- |
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"Owen" wrote in message ... On Sat, 9 Jul 2005 23:17:18 -0400, "Walter Maxwell" wrote: I'm trying to understand your Mathcad presentations, but I've run into some roadblocks concerning terminology, some of which I'm not familiar with. I confess my questions prove my ignorance, but that's ok if one's trying to learn. However, I was using nominal Zo. Not at all, you are far more eminent that I on this topic, and I appreciate your review. I am learning from all this. Apologies for the difficulty in understanding my notation. Some of it breaks into psuedo programming code. First, Xo!=0. I don't know what this means. Not equals. Second, what does MML stand for in English? MLL? Matched Line Loss (dB/m) Third, in 'functions for V, I, Z, etc at z'. Where is 'z'? I cannot find any reference to it. These quantities are a function of z, where z is a position on the line. The convention that I have used for displacement is that it is negative towards the generator. When it matters, displacement is in metres. The z is just used in definition of some functions in Matchcad (where you see :=), I have used x for position variable in the graphs. Fourth, 'exp'. Exponent? If so, of what? e? exp(x) is e to the power of x (For clarity, I shouldn't have written it that way, it works, but Mathcad understands the meaning of e superscript x as e to the power of x, as you will see in some of the expressions, and it is easier to read.) Fifth, I understand 'x' as distance along the line from the termination, but what is 'y'? In some of the functions, I have written them to calculate some quantity between two arbitrary points x and y. They are used in the definition of fuctions (where you see :=). Most of the graphs use 0 for y so they are plotted wrt the load position Sixth, what is AppLoss? Approximate? Apparent? Applied? Approximate Loss, and it was incorrectly based on Zo rather than nominal Zo. Seventh, 'DLoss'. What is 'D'? Dielectric? Again, what is the 'y' term? An ordinate value? DLoss was equivalent to AppLoss. Eighth, in the LineLoss(x,y) = 10log... the identical right-hand terms in both numerator and denominator, the identical functions of 'e^^ x e^^. what is the meaning of the bar above the second appearance of 'e'? And above gamma(x)? The bar above the variable is the complex conjugate operator. I want to understand your math presentation, Owen, especially when I see that Loss(x,0 - W2DUloss(x,0) is so small I want to understand what makes the difference. So I'd appreciate it if you'd set me straight on the points I made above. Walt, in the models at http://www.vk1od.net/temp/LineLoss.htm , I now know why there is such a gap between DLoss and LineLoss. You will recognise AppLoss / DLoss is your Appendix 8 expression, but my rho function was based on the modelled complex value of Zo (characteristic impedance), not the nominal value of Zo. In the second lot at http://www.vk1od.net/temp/reflection.htm , AppLoss is equivalent to DLoss and it is based on nominal Zo, W2DULoss you will see calculates the rho term (though not identified) using nominal Ro. Comparing the results with loss calcuated from P(x)/P(y) (the ratio of the real power at points x and y), the conclusion is that using your expression with actual Zo is not at all accurate, using it with nominal Zo is very close. If I force Zo to be real for all modelling, the results of all methods is exactly the same (within rounding errors of the order of 10 to the power of -14) Some of your questions are just about the Mathcad notation (though that is not too dissimilar to normal handwritten math notation), but some of it is my expression and usage. Again my apologies for confusing with too little explanation. I appreciate your review and comments Walt. Owen Thank you, Owen, for kicking aside the roadblocks preventing me from understanding your math presentation. I get it now, and realize I was a knothead for being confused. It is now perfectly clear why one can't get the true answer using my expressions for calculating loss on the line when using only the nominal Zo and not the actual Zo when there is loss. I should have known that intuitively, and why it escaped me is puzzling. Studying your math approach let me see the light, and for that I thank you. And thank you also for taking the time to teach me. Walt PS--I note from your telephone numbers in your email to me that you are not located in the US. Also, the name Duffy sounds somewhat British. Are you in the UK? ---------------------------------------------------------------------------- ---- |
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