Jeff wrote:
What utter rot. The TDR indicates a discontinuity that is IN ADDITION to
the one that gives you your result. Ignoring that discontinuity will
certainly give you the wrong answer, regardless if which domain you are
working in!!!

When line loss is given in watts, the discontinuity
at the source certainly has an effect.

When line loss is given in dB (10x log of a ratio),
the discontinuity at the source has NO effect.

The discontinuity at the source has NO effect on
the RATIO of two powers.
--
73, Cecil http://www.w5dxp.com

Jeff wrote:
With a 50 ohm source and 200 ohm load the loss is calculated as is 3.93dB
and VSWR is 3.98:1
Changing to a 100 ohm source the loss is calculated as is 4.78dB and VSWR
is 3.41:1

Looks like your losses and VSWR are taken on the wrong
side of the source resistor.

source R
source V----x--/\/\/\/\/\/\--y----T-line---load

Loss and VSWR should be calculated at 'y', not at 'x'.
We are interested in the VSWR and losses *on the T-line*.
Please change your reference point from 'x' to 'y'.
--
73, Cecil http://www.w5dxp.com

Jeff wrote:
Your application of the above equations neglects the first discontinuity.
The first discontinuity (inside the source) doesn't
have any effect on the SWR on the transmission line.

The discontinuity is NOT "inside the source", it is at the source to coax
interface, and as such effects the VSWR that the source sees.

There is usually a piece of coax running from the
source connector back to a filter. I would suggest
that the discontinuity that you are talking about
is indeed some distance "inside the source".

But we users don't care or measure what VSWR the source
sees. Only the PA designer worries about such. We users
only care and measure the VSWR *ON* the transmission line.
--
73, Cecil http://www.w5dxp.com

Owen Duffy wrote:
"Jeff" wrote in
e.com:

If you used a TDR, for example, to look at the set-up you would see 2
points of discontinuity, firstly at the 100 ohm source to 50 ohm cable
interface, and secondly at the cable to 200 ohm load. BOTH of these
discontinuities add to the overall mismatch as seen by the 100 ohm
load.

Your TDR does not work in the steady state frequency domain space, and is
misleading you.

In the steady state, the (complex) ratio of forward voltage to reflected
voltage is determined solely by the load impedance and characteristic
impedance of the line.

In crude terms, during establishement of steady state, you can view that
a load end reflected wave which is then partially reflected at a
mismatched source end, will reach the load end and be reflected in the
same ratio as the earlier passes. The subsequent round trips as steady
state is approached do not change the (complex) ratio of forward voltage
to reflected voltage in the steady state.

I know you have support here for the assertion that source end mismatch
affects VSWR in the steady state, but you won't find it in reputable text
books.

Owen and Cecil are right: the source (transmitter) has no effect
whatever on the VSWR on the line.

That isn't just an assertion - it is part of the bedrock transmission
line theory. Owen referred to "reputable textbooks", one of which would
surely be 'Theory and Problems of Transmission Lines' by R A Chipman
[1]. This book gains a lot of its reputation from its very complete
mathematical development of the theory, showing all the detailed
working.

Chipman treats standing wave patterns in two different ways: first by
assuming the final steady-state conditions, and then in much more detail
by considering multiple reflections between the load and the source.
Given a sufficient number of reflections, the multiple-reflection model
converges on exactly the same results as the steady-state analysis -
just as it does in the physical world.

VSWR on the line is determined by the ratio |Vmax|/|Vmin|. The complex
impedance that the source sees at the input terminals of the line is the
ratio V/I at that point (where V and I are both vector quantities which
include phase information). An alternative way of calculating either
VSWR or Zin is through the ratio Vforward/Vreflected (again vector
quantities).

All of these approaches are alternative pathways through the same body
of theory. They are all consistent with one another, and there is no
contradiction between any of them.

You will notice that all these standing wave relationships involve
ratios. Chipman's detailed analysis confirms that these ratios are
determined EXCLUSIVELY by the properties of the line and the load -
never the source.

The source properties do determine the magnitudes of all of the
individual voltages and currents - but when you change the source
properties (output voltage and/or impedance) all the individual voltages
and currents on the line and at the load are changed by the same factor.
So when you take the ratio, the source properties cancel right out
again.

All this confirms that, if you sweat out the math in all the different
levels of detail that Chipman did, the source (transmitter) still has no
effect whatever on the VSWR on the line.




[1] Out of print, but well worth searching for: ISBN 0-07-010747-5.
The web bookstores currently have eight copies on offer, at a range of
prices.



--

73 from Ian GM3SEK

Owen and Cecil are right: the source (transmitter) has no effect whatever
on the VSWR on the line.

That isn't just an assertion - it is part of the bedrock transmission line
theory. Owen referred to "reputable textbooks", one of which would surely
be 'Theory and Problems of Transmission Lines' by R A Chipman [1]. This
book gains a lot of its reputation from its very complete mathematical
development of the theory, showing all the detailed working.

I am sorry but you are not correct, I have not read Chipman so I cannot
comment on his analysis or your interpretation of his results, but my
understanding , practical experiments and CAD analysis would lead me to
disagree.

If we take the situation where the source is matched (50ohms) to the 5.35
wavelength transmission line (lossless to simplify things) with a 100ohm
load, I agree that the vswr is 4:1, unchanging with frequency.

Plotted on a Smith Chart when swept against frequency this gives a circle
centred on 1 (50ohms) with a radius of 4. i.e. on a constant VSWR circle.

Now if we change the source impedance to 100ohms and repeat the same sweep
and re-plot, keeping the chart normalized to 50 ohms, the circle moves on
the resistance axis, still with a radius of 4 and now passing though 2 (100
ohms) resistive. The centre moves to about 0.6 (30ohms). It then becomes
obvious that the locus of the circle is NOT a constant VSWR against
frequency.

You will come to the same conclusion if you normalize the chart to 100 ohms,
the new source impedance and re-plot.

The coax is acting as an impedance transformer, causing a shift along the
resistance axis.

Looking at it another way, the vswr changes sinusoidally with frequency, in
our example, between 2:1 and 8:1. (The same as the Smith chart plot with a
circle of radius 4 centred at about 0.6).

73
Jeff

Cecil Moore wrote:
Jim Kelley wrote:
Most engineers equate the units of power to power, i.e.
joules/sec = watts and so does the IEEE dictionary.

I can't speak for most engineers, but I think the first time I saw it
was in high school physics, and of course later in engineering school.
That was about 35 years ago. I think of it a fundamental concept -
one that I happen to understand very well. Not unlike the
relationship between Joules and electron-volts.

But I
am content to assert that the joules in the joules per
second of a reflected wave is real energy. Do you disagree?

I don't agree that the terms power and energy become interchangeable
by virtue of the fact that their units can both be expressed with the
word Joule in them. One can find himself making unrealistic
predictions if he is not precise in his application of the ideas which
underlie these terms.

73, ac6xg

Jeff wrote:
You will come to the same conclusion if you normalize the chart to 100 ohms,
the new source impedance and re-plot.

The Z0 of the transmission line has not changed to 100
ohms so normalizing the chart to 100 ohms is not valid.
--
73, Cecil, http://www.qsl.net/w5dxp

" You will come to the same conclusion if you normalize the chart to 100
ohms,
the new source impedance and re-plot.

The Z0 of the transmission line has not changed to 100
ohms so normalizing the chart to 100 ohms is not valid.
--
73, Cecil, http://www.qsl.net/w5dxp

It is just as valid as using 50 ohms, and the result is the same, a changing
vswr.

I see you have not commented on the main point of my post, that being that
the smith chart shows a changing vswr when you change the source impedance.

Hint: transmission line transformers would not work if the vswr did not
change.

73
Jeff

On Thu, 01 Mar 2007 09:49:52 GMT, Owen Duffy wrote:

Reminding you that your question was "What is the loss in the line?",
check your own post.

Hi Owen,

Can you offer why I should? Well, I suppose not or you would have.

However, I am one to never turn aside a suggestion and I did review
everything (except my own quote - I've repeated it enough, haven't I?)
and I will respond to that review within the body of this text.

Well, you posted an answer, not a solution. It wouldn't have been your
solution anyway, because it looks like it is copied straight out of a
book.

Does it being someone else's solution make any difference to the
outcome? Owen, your comment reveals a prejudice by implication.
Copying it right out has removed any issue of authority has it not? It
has also removed any issue of accuracy too - if you accept that
authority. Ultimately, having copied it out makes for the best
resolution. Having copied it out, and offering the citation, gives us
both access to the chain of evidence. Did I withhold or otherwise
linger with the citation? You asked for my solution and I immediately
offered both.

Ironically, does your suggestion that
It wouldn't have been your solution anyway
mean you would suspect I would have come up with a different answer?
That is, ascribing to me the quality of being able to get it right
instead? That would be generous, thank you. However, it appears I
fell short of that mark (and may have been the intent of your
elliptical pat on the back).

Looking at Reference Data for Engineers, Sixth Edition, p24-12, Example 3
(which is the same as the problem you posed), they give the answer as
3.27dB.

A simple review of example four distinctly reveals the details to the
problem I posed; example three contains only some of them. Example 3
is a subset of example 4 (as that example dwells on at great length).

However, example 4 does have one notable difference, it asks:
"What is mismatch loss between the generator and the line?"
for which the answer is:
"1.62 dB"

Ah, the devil is in the details. Continuing from example 4:
"The transducer loss is found by using the results
of 3 and 4 in (4). This is
1.27 + 2.00 + 1.62 = 4.9 decibels"

I am happy that my answer rounded to 3.3dB is correct.

Congratulations. You may note in my earlier correspondence I allowed
exactly that.

The source resistance has no influence over the line loss at all.

Upon review of my own reference (not the recommendation you offer
above) I must concur. I was trapped by what I have already described
as being the classic confusion between systems of match and loss. My
solution was not for the loss in the line, but for transducer loss,
and specifically for the inclusion of mismatch loss within the
transducer loss. All caloric, but mis-ascribed to the line loss.

You posed this problem as difficult and one that no one has ever got
right. No wonder, you have a different answer to the the book!

Well, in fact my answer conforms exactly to the book. The problem is
not one of inexactitude, it is of poor referencing. As to the matter
of no one else having ever got it right, no one even consulted the
book - even partially. You can count yourself among a population of
one and hashing it through served us well.

73's
Richard Clark, KB7QHC

Richard Clark wrote:

On Wed, 28 Feb 2007 13:55:47 -0800, Jim Kelley
wrote:


What I meant was, in what way were you able to attribute and apportion
this heat to its various sources? What evidence were you able to
obtain to show reflected energy re-entering the source output? What
component in the system in fact dissipated the reflected energy? How
were you able to determine the exact source and amount of energy at
any given location within the source? Or did you just presume that
you understood the underlying mechanisms?


Hi Jim,

This knowledge arrived by many avenues.

But primarily, it seems, by speculation. I know how to measure heat,
Richard. What I am asking, and what you have thus far been unable to
answer (which is as I suspected), is how is it that you were able to
ascertain that this heat energy was caused by energy that was
reflected from the load rather than having come directly from the
power supply within the source? How is it that this electromagnetic
energy is so easily reflected from a load, but is utterly immune to
reflection when it encounters the output of a source? I think it's
been fairly well established that the output impedance of these things
is far from 50 ohms. Why should reflected energy not be, at least in
some part, re-reflected back toward the load?

Someone who alleges to be so familiar with load lines should be able
to contend with an increase in dissipation against a mismatched load
without having to explain it as 're-absorbed' reflected energy.

Inductive logic leads us to this junction as the principle target of
reflected power (the signal path is symmetric, after all).

Speculation could also lead to that juction.

Experience
has supported this logic.

It could be experience coupled with misattributed fact. Possible?

73, Jim AC6XG