Peter O. Brackett wrote:
My eyes glazed over and I nearly fell asleep and had to stop following after
a couple of screens of what seemed to turn into gibberish before my eyes.

Not your fault mind you, it's mine.

What helps for me is to print it out. I evolved looking at printed
pages, not computer screens. I can scribble the correct math
operators down so I don't have to remember what e** means.

I just don't get the point of all of your wonderful efforts!

Roy may have explained Richard's data. In any case, it's good
to know that we cannot use the simplified wave reflection model
on very lossy lines. It appears that a lossy line doesn't yield
a smooth spiral on a Smith Chart.
--
73, Cecil http://www.qsl.net/w5dxp



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It obviously doesn't bother you that your new "forward power" isn't the
product of forward voltage and current, or that the new "reverse power"
isn't the product of reverse voltage and current. But then I guess we
shouldn't be surprised.

Roy Lewallen, W7EL

Cecil Moore wrote:
Roy Lewallen wrote:

Well, shucks, that makes it easy.


Just being logical. There are only two directions in a transmission
line, forward and reverse. If all the waves are coherent, all forward
waves superpose to one wave and all reverse waves superpose to one
other wave. Your net forward power is greater than your net reflected
power by the net amount of power accepted by the load. This happens
locally at the load no matter what is happening elsewhere in the
transmission line.

Cecil Moore wrote:

Seems to me, all the terms with a '+' sign would be forward power, by
definition, and all the terms with a '-' sign would be reflected power,
by definition. I don't see any violation of the conservation of energy
principle. The power equation balances.

I've given the equation. With that and a spreadsheet or plotting program
(or graph paper) of your choice, you can have the plot in minutes. Note
that x is the distance from the load.

Roy Lewallen, W7EL

Cecil Moore wrote:

. . .
What does a plot of that extra power look like up
and down the line?

Note also that if the equation for total average power is used to find
average power at any point along the line, fE1 becomes the forward
voltage at the observation point, not necessarily the forward voltage at
the input end of the line. So to calculate the total power as a function
of position along the line, it's probably best to use a voltage at a
fixed point, such as fE2, the forward voltage at the load, in its place.
Make the substitution

|fE1|^2 = |fE2|^2 * exp(2 * ax)

for |fE1|^2 to make the power equation more usable for this purpose.

Roy Lewallen, W7EL

Roy Lewallen wrote:
I've given the equation. With that and a spreadsheet or plotting program
(or graph paper) of your choice, you can have the plot in minutes. Note
that x is the distance from the load.

Roy Lewallen, W7EL

Cecil Moore wrote:

. . .

What does a plot of that extra power look like up
and down the line?

As derived, the equation for total average power at any point along the
line contains the term |fE1|^2, which is the square of the magnitude of
the forward voltage at that point. It might be more useful to replace
fE1 with a value which doesn't vary with position along the line, such
as fE2, the forward voltage at the load. The substitution is:

|fE1|^2 = |fE2|^2 * exp(2 * ax)

which gives the alternate formula

P1 = (|fE2|^2 / |Z0|) * (exp(2 * ax) * (1 - rho^2 * exp(-4 * ax)) *
cos(delta) + rho * (2 * sin(delta) * sin(2 * bx - 2 * psi)))

And, subtituting values for "forward power" and "reverse power":

P1 = fP - rP + (|fE2|^2 / |Z0|) * rho * 2 * sin(delta) * sin(2 *
bx - 2 * psi).

Roy Lewallen, W7EL


Roy Lewallen wrote:
Here's the calculation of total average power P1 at any point on a
transmission line. The point on the line is called point 1, and the
location of the load is called point 2. The distance between them is x.

. . .

= (|fE1|^2 / |Z0|) * ((1 - rho^2 * exp(-4 * ax)) * cos(delta) + rho *
exp(-2 * ax) * (2 * sin(delta) * sin(2 * bx - 2 * psi)))

Subtituting values for "forward power" and "reverse power", we have:

P1 = fP - rP + (|fE1|^2 / |Z0|) * rho * exp(-2 * ax) * 2 *
sin(delta) * sin(2 * bx - 2 * psi).
. . .

Cecil:

[snip]
In any case, it's good
to know that we cannot use the simplified wave reflection model
on very lossy lines. It appears that a lossy line doesn't yield
a smooth spiral on a Smith Chart.
--
73, Cecil http://www.qsl.net/w5dxp
[snip]

Amen brother and... heh, heh... especially for broad band signals.

Smith Charts are for mono-chomatic signals. Most tough transmission
problems
are broad band and the Smith Chart yeilds no useful insight in those
problems.

A widely applied practical example is the transmission of bi-directional
broad band digital subscriber loop (DSL) signals over telephone twisted
pair.

Telephone twisted pair is very lossy... at the "standard" 18,000 foot
length you can barely tell what is connected on the other end, short
or open. In fact it might just as well be "semi-infinite"! The longest
spans we have built chips for were up to 47,000 feet of #24 AWG
full duplex data transmission at the basic rate with digital echo
cancellation
on both ends using trellis coded pulse amplitude modulation.

I can assure you that 47,000 feetof #24 AWG definitely has a complex
and lossy Zo! I had quite a few big "spools" of such cable in my lab
for the beta tests!

The big problem with such designs is not maximum power transfer,
rather it is hearing the remote end in the presence of the local transmitter
blasting away on the same pair as the receiver [talker echo] and so one
needs to "image match" the transmitter to eliminate as much talker echo
as possible and just take whatever power reaches the receiver at
the other end. Of course you have some control over the spectrum
of the power that reaches the other end by "pre-coding" at the
transmitter, still the optimum strategy at the transmitter is to get
an "image match". i.e. make the generator internal impedance
as close to the complex Zo as you can make it!

And... in those problems you need to differentiate two forward waves
and two reflected waves. Heh, heh... hard to do that using just the
two symbols Vfwd and Vref or V_+ and V_-, you need symbols
for at least two each... Say Vfwd_1 and Vfwd_2 and Vref_1 and
Vref_2, etc... messy to say the least!

For this reason I much prefer the Scattering Formalism symbols
"a" for incident and "b" for reflected, a1 for indicdent on port 1
and a2 for incident on port 2, then b1, b2, etc...

Sometime, when I get some free time from my current
consulting gig, I'll prepare a short example for the group of the
problems inherent in full duplex signalling over complex Zo lines in
situations where the "best" Engineering solution is "image match"
not "conjugate match". ;-)

--
Peter K1PO
Indialantic By-the-Sea, FL.

Reg:

[snip]
Peter, the stage is now set to introduce Eigenvectors,
Eigenvalues and Sylvesters theorem for square matrices.
;o)
---
Reg
[snip]

Ahhhh... Sylvester, I knew him well!

Roy even fusses at me and insists that the transformation matrix M between
v, i and a, b include a factor of 1/2 as:

a = v/2 + Ri/2 = 1/2 [v + Ri]
b = v/2 - Ri/2 = 1/2 [v - Ri]

i.e. M is:

|1/2 R/2|
|1/2 -R/2|

so that the M is chosen for the voltage and current values to match the
values of v and i found
in solutions of the Telegraphists Equation.

That's fine, but with that simple multiplier common to all elements of the
linear combination of electricals to make the waves, it just don't matter
since as you know, when you form rho = b/a the factor of 1/2 just drops out.

I suppose introducing the "outer product" of the two vectors B = [b1, b2]
and [A^-1]'[ = [1/a1, 1/a2]'
as Bx[A^-1]' = S to form the two by two Scattering Matrix and showing that
the factor of 1/2 disappears there as well would be far far too much for
this group to assimilate!

I can hear Oliver rolling over in his grave!

Long live Sylvester!

;-)

--
Peter K1PO
Indialantic By-the-Sea, FL.

Roy Lewallen wrote:
It obviously doesn't bother you that your new "forward power" isn't the
product of forward voltage and current, or that the new "reverse power"
isn't the product of reverse voltage and current. But then I guess we
shouldn't be surprised.

I guess I will quote Aristotle on that one, Roy. A thing is what it is -
It's not something else. Two directions are all that exist, forward and
backwards. Do I need to publish a binary truth table?
--
73, Cecil http://www.qsl.net/w5dxp



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"Peter O. Brackett" wrote in message
link.net...
Sometime, when I get some free time from my current
consulting gig, I'll prepare a short example for the group of the
problems inherent in full duplex signalling over complex Zo lines in
situations where the "best" Engineering solution is "image match"
not "conjugate match". ;-)

--
Peter K1PO
Indialantic By-the-Sea, FL.


Peter,

You have just answered the question that people have been arguing about the
last few weeks. I hope you have time to describe this in more detail.
Tam/WB2TT

Peter O. Brackett wrote:
A widely applied practical example is the transmission of bi-directional
broad band digital subscriber loop (DSL) signals over telephone twisted
pair.
Peter K1PO
Indialantic By-the-Sea, FL.

As Peter notes,
telephone transmission line impedance is always complex.
The parameters R, L, G & C per unit length
(series resistance, series inductance, shunt conductance,
shunt capacitance) are NOT CONSTANT with frequency,
or temperature. So the cable impedance is not
constant either.
Signal spectra extend from nearly DC (a few kHz)
up to 12MHz or more - many octaves.
Even over voiceband, 400Hz to 2800 Hz,
the cable impedance changes a _lot_.

Lengths vary from several feet to 10s of kft.
There are often/usually open-ended shunt cable sections,
a.k.a. bridged taps, along the cable.
Other things, like series lumped loading coils (inductors),
may appear if not removed from longer cables.
Signals at the DSL receiver ends are umpteen dB below
the transmitter signal levels on the same pair of wires,
and can be in the same band if separate to-the-customer
and to-the-network bands are not used.

Smith Charts, as much as I like them for ham purposes,
are of no help.

This subject is addressed in T1.417-2001, Issue 1
"Spectrum Management For Loop Transmission Systems"
January, 2001
Developed by Sub-Committee T1E1.4
which develops the xDSL standards (DSL, HDSL,
ADSL, VDSL,...) for North America.

Annex B of T11.417 deals with the modeling
of cables for such cases: formulas, RLCG vs. freq.
and other data for common AWG and metric cables...

Software packages are available offline and online.

http://net3.argreenhouse.com:8080/dsl-test/index.htm
(A free registration is needed.)

(The other 200+ pages are left for the reader.)

The latest working draft of Issue 2 is available free*
as document T1E1.4/2003-002R3 from

http://www.t1.org/filemgr/filesearch.taf

Do a "Simple Search" for filename 3e140023

When the "Results of Simple Search" page appears,
click on the blue full name T1E1.4/2003-002R3
under the Contributions column.

When the next page appears,
click on the blue 3e140023 after "File Prefix"
to finally download the document. (2.1MB)

I just tried this procedure to be sure it works.

* The official Issue 1 is over US$300.

In 1995 I was the first editor and wrote the first draft
of what became T1.417. Much of what I wrote is intact
word-for-word as the first half of Annex B
(to my amazment) - the general descriptive part
before the nitty-gritty models and numbers.

Have fun.

There will be an exam.

Cheers, 73,

Ron McConnell
Retired Secretary T1E1.4


N 40º 46' 57.9" W 74º 41' 21.9"
FN20ps77GU46 [FN20ps77GV75]

http://home.earthlink.net/~rcmcc