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" 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

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" 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
...
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
--

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
--