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