A Feedline Energy Analysis
Cecil A. Moore (W5DXP)
on
December 12, 2005
View comments about this article!
First published in WorldRadio, Oct. 2005, and reproduced with permission.
An Energy Analysis at an Impedance Discontinuity in an RF Transmission Line, Part I
Where Does the Power Go? [1] There continue to be many differing responses to the question within
the amateur radio community and, so far, no one has presented the facts of the physics of power as
understood from the field of optics. Those facts from optics have been known and understood for
decades and are consistent with the laws of physics and the equations governing the behavior of RF
transmission lines. Light and RF waves are both composed of electromagnetic energy. Most of the
following information comes from "Optics" [2]. In the field of optics, irradiance is the same thing as
power in an RF transmission line if the cross sectional area of the transmission line is taken into
account. Irradiance has the dimensions of energy per unit area per unit time. If the light beam of a
particular laser occupies the same cross sectional area as a particular coaxial RF transmission line then
the irradiance of the laser beam is comparable to the RF power in the transmission line. The 1/4
wavelength thin-film deposited on glass to obtain a non-reflective surface performs in a virtually
identical way to a 1/4 wavelength series matching section in a transmission line. Single-source RF
energy in a transmission line and laser light are both coherent electromagnetic energy waves that obey
the laws of superposition, interference, conservation of energy, and conservation of momentum.
My Historical Perspective
My first memories of the answer to "Where does the power go?" are articles published in QST written
by Walter Maxwell, W2DU, some quarter of a century ago. Mr. Maxwell later compiled the
information into a book titled, "Reflections", which quickly became the bible for Amateur Radio
applications involving stub matching, transmission lines, and forward and reflected energy flow. Mr.
Maxwell coined the terms, "virtual short" and "virtual open", as a shorthand description of what
rearward-traveling reflected energy encounters at a match point in a transmission line resulting in
100% re-reflection. He also explained the function of destructive wave interference and constructive
wave interference in achieving a match point on a transmission line [8] which is what a large part of
this article is about.
Sometime after the publication of Reflections, some people questioned the validity of Mr. Maxwell's
concepts. In particular, Dr. Steven Best, VE9SRB, took Mr. Maxwell to task in a series of articles
published in QEX [3]. Simply put, Dr. Best disagreed with Mr. Maxwell that reflected power is 100%
re-reflected in a matched system. Before publication of his Part 3 QEX article, Dr. Best sent up trial
balloons for his ideas on the usenet newsgroup, rec.radio.amateur.antenna. My opinion was that Dr.
Best's future article contained numerous errors which were pointed out to him. However, the article as
published still contained the alleged errors. My determination to resolve the conflicts between the
concepts presented by Walter Maxwell and the ones presented by Dr. Best culminated in this present
article. The conclusions will be presented first with the technical details to follow in Part II and
Part III.
In a nutshell, Walter Maxwell's "virtual short" is a two step process. The reflected wave from the load
encounters the impedance discontinuity at the match point. A re-reflection occurs that equals the
incident reflected power multiplied by the power reflection coefficient at the match point (the square
of the voltage reflection coefficient). This re-reflected energy joins the forward wave traveling toward
the load. That first energy re-reflection is not the only energy that joins the forward wave. That fact is
what Dr. Best missed in his article. Interference of any kind was never mentioned in Dr. Best's QEX
article.
The part of the reflected wave that is not re-reflected is transmitted back through the impedance
discontinuity at the match point and attempts to flow toward the source. We know the reflected energy
doesn't make it to the source in a matched system, so where does it go? The answer is mentioned in
"Reflections II" [8]. What Mr. Maxwell is describing is wave cancellation due to total destructive
interference between two reflected waves. The first wave is the part of the source forward wave that is
initially reflected back toward the source from the match point. The second wave is the part of the
reflected wave from the load that is transmitted through the match point toward the source. These
waves are equal in magnitude and opposite in phase so, as Mr. Maxwell asserts in Reflections II, they
cancel to zero at the match point thus eliminating reflections between the match point and the source..
The canceling of these two waves to zero is the second step in Mr. Maxwell's virtual short process of
100% reflection.
Voltages can cancel and currents can cancel but energy cannot cancel. What happens to the energy that
existed in the waves before they were cancelled? Since we know that all the energy in a matched
system winds up flowing toward the load, the answer is a no-brainer. There are only two directions in
a transmission line. If energy that was previously flowing toward the source isn't flowing toward the
source anymore, it must necessarily be flowing toward the load. The conclusion is inescapable. Not
only is 100% of the reflected energy re-reflected at the match point, but wave cancellation is the cause
of part of that re-reflection. This is a well understood phenomenon in the field of optics [9] but not
well understood in the field of RF engineering.
An RF engineer will usually say there are three things that can cause 100% reflection. Those are a
short-circuit, an open-circuit, or a purely reactive impedance. This is true at a load. But at an
impedance discontinuity with waves incident from both directions, to that list of three, we can add a
fourth, namely wave cancellation due to total destructive interference. In general:
The destructive interference energy resulting from wave cancellation at an impedance
discontinuity becomes an equal magnitude of constructive interference in the opposite direction.
Since there are only two directions in a transmission line, wave cancellation is the equivalent of
an energy reflection. 100% wave cancellation means 100% energy reflection. [9]
References are included in this Part I of the series and will not be repeated in Parts II and III.
In Part II, the general case qualitative analysis will be presented.
I would like to thank Mr. Robert E. Lay, W9DMK, for his substantial contributions to this article.
References
[1] Bloom, Jon, Where Does the Power Go?, "QEX", Dec. 1994
[2] Hecht, Eugene, "Optics", Fourth Edition, (c)Aug. 2001, Addison-Wesley, ISBN 0805385665
[3] Best, Steven R., Wave Mechanics of Transmission Lines, Part 3, "QEX", Nov/Dec 2001
[4] "Interference term", "Optics", Eugene Hecht, Fourth Edition
Section 7.1 The Addition of Waves of the Same Frequency
It follows ... that the resultant flux density is not simply the sum of the component flux densities;
there is an additional contribution 2*E01*E02*cos(a2-a1), known as the interference term.
("a" replaces the Greek letter Alpha and INTERFERENCE TERM is emphasized.)
Section 9.1 General Considerations
The 'interference term' becomes I12 = 2*SQRT[(I1)(I2)]*cos(s) [where I is irradiance (power)]
('SQRT' replaces the square root sign and "s" replaces the Greek letter Sigma.)
[5] "S-Parameter Techniques", Hewlett Packard Application Note 95-1, available on the web.
The S-Parameter normalized voltage equations are:
b1 = (s11)(a1) + (s12)(a2) and b2 = (s21)(a1) + (s22)(a2)
The squares of all those terms are related to power as explained in the application note. It is left as an
exercise for the reader to square both sides of both equations above and observe that the resulting
equations contain the interference term that agrees with Eq 1 and Eq 2 in the body of this paper.
[6] "Optics", Eugene Hecht, Fourth Edition
Section 3.3 Energy and Momentum, "One of the most significant properties of the electromagnetic
wave is that it transports energy and momentum." [Note from W5DXP: Energy and momentum must
be conserved. The direction of the energy and momentum associated with reflected waves must be
reversed for a match to occur.]
Section 4.11 Photons, Waves and Probability, "The principle of conservation of energy makes it clear
that if there is constructive interference at one point, the 'extra' energy at that location must have come
from somewhere else. There must therefore be destructive interference somewhere else. "If two or
more electromagnetic waves arrive at point P out-of-phase and cancel, 'What does that mean as far as
their energy is concerned?' Energy can be distributed, but it doesn't cancel out."
Section 7.1 The Addition of Waves of the Same Frequency, "The superposition of coherent waves
generally has the effect of altering the spatial distribution of the energy but not the total amount (of
energy) present."
[7] "Optics", Eugene Hecht, Fourth Edition
Section 9.1 General Considerations, "A maximum irradiance (power) is obtained when cos(s) = 1. ...
In this case of total constructive interference, the phase difference between the two waves is an
integer multiple of 2*Pi, and the disturbances are in-phase. ... A minimum irradiance (power) results
when the waves are 180 degrees out-of-phase, ... cos(s) = -1, ... and is referred to as total destructive
interference." ("s" replaces the Greek letter Sigma and TOTAL DESTRUCTIVE INTERFERENCE is emphasized.)
[8] Maxwell, Walter, "Reflections II", (c) 2001 Worldradio Books, ISBN 0-9705206-0-3
page 4-3, "The destructive wave interference between these two complementary waves ...
causes a complete cancellation of energy flow in the direction toward the generator. Conversely,
the constructive wave interference produces an energy maximum in the direction toward the load, ..."
page 23-9, "Consequently, all corresponding voltage and current phasors are 180 degrees out of phase
at the matching point. ... With equal magnitudes and opposite phase at the same point (point A, the
matching point), the sum of the two (reflected) waves is zero."
[9] Quotes from two web pages from the field of optical engineering:
www.mellesgriot.com/products/optics/oc_2_1.htm
"Clearly, if the wavelength of the incident light and the thickness of the film are such that a phase
difference exists between reflections of p, then reflected wavefronts interfere destructively, and overall
reflected intensity is a minimum. If the two reflections are of equal amplitude, then this amplitude (and
hence intensity) minimum will be zero." (Referring to 1/4 wavelength thin films.)
"In the absence of absorption or scatter, the principle of conservation of energy indicates all 'lost'
reflected intensity will appear as enhanced intensity in the transmitted beam. The sum of the reflected
and transmitted beam intensities is always equal to the incident intensity. This important fact has been
confirmed experimentally."
http://micro.magnet.fsu.edu/primer/java/scienceopticsu/interference/waveinteractions/index.html
"... when two waves of equal amplitude and wavelength that are 180-degrees ... out of phase with
each other meet, they are not actually annihilated, ... All of the photon energy present in these waves
must somehow be recovered or redistributed in a new direction, according to the law of energy
conservation ... Instead, upon meeting, the photons are redistributed to regions that permit constructive
interference, so the effect should be considered as a redistribution of light waves and photon energy
rather than the spontaneous construction or destruction of light."
Note from W5DXP: In an RF transmission line, since there are only two possible directions, the only
"regions that permit constructive interference" at an impedance discontinuity is the opposite direction
from the direction of destructive interference.
Cecil A. Moore, w5dxp@arrl.net
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A Feedline Energy Analysis
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by WD0CT on December 12, 2005
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Been a long time since I had the original qsts with Maxwell's articles. I do have a print out of Reflections though. I'm looking forward to the discussion.
Does reflection 'gain' really exist?
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RE: A Feedline Energy Analysis
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by W5DXP on December 12, 2005
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Part II was published in the Dec. 2005 issue of "WorldRadio" and will be made available here as soon as I receive my "WorldRadio" Jan. 2006 issue around Dec. 15 which will contain Part III. Part III will be made available here as soon as I receive my Feb. 2006 issue. Simultaneous with my submission here, the articles will appear on my web page:
http://www.qsl.net/w5dxp/energy.htm
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RE: A Feedline Energy Analysis
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by N6AJR on December 12, 2005
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I find the biggest thing in antennas, swr and all that is that the antenna, feed, tuner, and radio output make up a system. If the antenna is x=5 and y=6 and the system needs to be x=10 and y=10 to be resonant ( the best kind of antenna) then between the feed and tuner options you want to add 5 x's and 4 y's. this allows the antenna to see what it wants, , x=10 and y=10. You did not actually change the antenna, but you did change the system.
The other way to do this is to buy a steppir, and physically change the elements until x=10 and y=10 and most all of your signal goes out the end of the pipe.
( you still have a little loss in the feed line, due to resistance in the wire etc, but the steppir accounts for all that, along with metal in the near field, ground conductivity, and such. but the steppir can be adjusted to a "perfect match " in each situation, to make the most of your output.)
Good ole' Kurt Sterba in the World ZRadio Magazine covers this often and well..
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A Feedline Energy Analysis
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by W2DU on December 12, 2005
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In rsponse to WD0CT's question, does reflection gain really exist?
Yes it does. The power reflected at an antenna mismatch is call reflection 'loss'. However, at the matching network (antenna tuner), the reflected power is re-reflected back to the antenna. The process by which the reflected power is re-reflected back up the line to the load is called reflection 'gain'. The term appears in engineering texts. The result is that the only portion of the reflected power that is lost is in the small attenuation in the tuner and the attenuation of the feed line.
Walter Maxwell, W2DU
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RE: A Feedline Energy Analysis
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by W9AC on December 12, 2005
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Walt and Cecil, you've both done some excellent work on making the subject matter understandable.
I have to ask the question:
Did Dr. Best finally concede at the conclusion of the QEX articles?
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RE: A Feedline Energy Analysis
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by W4CNG on December 12, 2005
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The answer as to where it went is HEAT. Dissipation into the atmosphere. You can match most any transmitter into almost any antenna, and the device in the middle gets to dissipate the HEAT. There are lots of folks that do not understand these issues and will go to great extents to explain the issues. Fancy formulas not withstanding, it is all I2/R losses at the transition point of impediances.
Steve W4CNG
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RE: A Feedline Energy Analysis
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by W5DXP on December 12, 2005
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If I remember correctly, Dr. Best said (in an email to me) something to the effect that his article needed to be expanded but I haven't seen his "expanded" material. That doesn't mean it doesn't exist.
--
73, Cecil, W5DXP
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RE: A Feedline Energy Analysis
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by W5DXP on December 12, 2005
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>W4CNG wrote: Fancy formulas not withstanding, it is all I2/R losses at the transition point of impediances.<
An impedance discontinuity *point* is pretty small. If there were a lot of I^2*R losses at that *point*, it would get mighty hot but it doesn't.
--
73, Cecil, W5DXP
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DEDs are DEAD
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by KA4KOE on December 12, 2005
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Thanks to the fine folks at Eham, you will not see any future bonafide DEDs from me. Note that DED 20 was quickly pulled as it was not written by me, and no permission was asked of me in the first place by its author.
Any DEDs you see that are not written by myself, WA6BFH, WA2JJH, K4JSR, or KC8VWM (good friends), are done so without my permission, and are in violation of creative commons rights.
I am undergoing radiation treatments at the current time for tumors on my neck, and frankly have not had the wherewithal to write any more DEDs.
The DEDs are DEAD.
Thank you. It was a pleasure doing these articles for you fine folks. I have gotten many emails from you, and I do appreciate the kind thoughts.
73 and Goodbye
PHILIP NEIDLINGER, PE
Electrical Engineer
DWG, Inc.
KA4KOE
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RE: DEDs are DEAD
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by W9PMZ on December 13, 2005
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I somewhat expected that this post would come in concert with the removal of the posting about, the railway engineer (I appologize that I can remember his name).
I enjoy reading about radio's heritage and the men and women who made it come true.
So dude take your dude's and your copyright and, well just go away. I wonder if this dude will go DeForrest?
I hope that the poster of this article reposts the article without the copyright violation. I further hope that other authors will come forward interesting tales of the creators of radio.
73,
Carl - W9PMZ
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RE: A Feedline Energy Analysis
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by K5UJ on December 13, 2005
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For years, I took it as a given (one of those things you know is true, but don't know how you know, and never think about it) that on 80 and 160 m., the frequencies are so low, that just about any coaxial feed will do the job with minimal loss. 300 feet of 8X on 160? No problem. 200 feet of 58U? Fine. I used to wonder why broadcasters, even at 250 w. stations, would invest in runs of fat heliax for medium wave RF. Then last year, I started using a modelling program (TLW) to come up with a transmatch circuit. I began playing around with it, as it could show me predicted losses for different kinds of feeds, of various lengths, depending on the frequency and load impedence. I was astonished to find that, even on 160, with a 50j0 load, a 300' run of rg213 had loss far greater than I would have ever guessed. the point is that if you can afford to do it, and have long runs (over 100') it pays to invest in good feedline.
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RE: DEDs are DEAD
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by KT8K on December 13, 2005
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Good luck, Philip! The DEDs have been great. Thanks so much for entertaining and educational writings, and hope to C U around here for a long time to come.
73 de kt8k - Tim
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A Feedline Energy Analysis
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by W2DU on December 13, 2005
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In answer to W4AC's question, did Steve Best concede after the QEX articles? No John, he did not.
Prior to those articles Steve continuously disputed my position on re-reflection, so he told me he was going to write a definitive article for QEX that would prove him right and prove me wrong. The result was his 3-part article in QEX.
All three parts of his article contained serious errors of fact, which prompted me to respond with a rebutting article in QEX. A slightly modified version of this article is repeated as Chapter 25 in the third edition of Reflections, which is in production. A preview of Chapter 25 is available for downloading from my web page at www.w2du.com.
I have not heard from Steve since my rebuttal article was published in QEX, and he has not appeared on the newsgroup rraa since then either, where before he was a constant poster. I wonder why (g).
Walt Maxwell, W2DU
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RE: DEDs are DEAD
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by W9PMZ on December 13, 2005
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Are DEDs copyright protected?
From the US gov copyright web site;
"http://www.copyright.gov/circs/circ1.html#wccc"
--------------------------------------------------------------------------------
WHAT IS NOT PROTECTED BY COPYRIGHT?
Several categories of material are generally not eligible for federal copyright protection. These include among others:
Works that have not been fixed in a tangible form of expression (for example, choreographic works that have not been notated or recorded, or improvisational speeches or performances that have not been written or recorded)
Titles, names, short phrases, and slogans; familiar symbols or designs; mere variations of typographic ornamentation, lettering, or coloring; mere listings of ingredients or contents
Ideas, procedures, methods, systems, processes, concepts, principles, discoveries, or devices, as distinguished from a description, explanation, or illustration
Works consisting entirely of information that is common property and containing no original authorship (for example: standard calendars, height and weight charts, tape measures and rulers, and lists or tables taken from public documents or other common sources)
DED is a short phrase and I submit not copyrightable. Repost the railway engineer!!!
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RE: A Feedline Energy Analysis
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by WF7A on December 13, 2005
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Forgive my density, but I'm still having a little wrapping my head around the concept of wave cancellation and still having energy left over. Maybe it's the analogy I've constructed in my head that's causing me problems:
Say we sample the instantaneous voltage on an ac line. The positive alternation on one side is at 110v and at the same moment--on the other side--there's a negative alternation of 110v. With our trusty o'scope we'd measure 0v since the positive and negative waves cancel (no difference of potential), but each side still has its own energy level.
Am I close or way off on this? I'm trying to think of it as a particle of matter meeting antimatter: they both would cancel upon meeting each other, yet release a lot of energy.
Second, I imagine the wave cancellation effect happens to sound waves as well as RF and optics, yes? So, anything from just above DC to daylight would react the same way.
Lastly, and forgive my gushing, but it's a genuine honor to see you here, Walt--I'm an admirer of your work.
Cheers,
Rich
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RE: DEDs are DEAD
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by KC9AGG on December 13, 2005
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hodgkin's lymphoma ? pretty treatable nowadays---good luck OM.
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RE: A Feedline Energy Analysis
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by W5DXP on December 13, 2005
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>WF7A wrote: Forgive my density, but I'm still having a little wrapping my head around the concept of wave cancellation and still having energy left over.<
EM fields can be canceled, voltages can be canceled, currents can be canceled, but energy cannot be cancelled. That pesky conservation of energy principle prohibits the cancellation of energy. Please read the optics web page references at the end of my article.
>Am I close or way off on this? I'm trying to think of it as a particle of matter meeting antimatter: they both would cancel upon meeting each other, yet release a lot of energy.<
Again, read the optics web page references. EM wave cancellation is NOT annihilation. It is merely a redistribution of energy. Two EM waves cancel when their amplitudes are equal, their phases are opposite, and they occupy the same path. "Optics", by Hecht is the best reference on superposition and interference that I know of.
>Second, I imagine the wave cancellation effect happens to sound waves as well as RF and optics, yes?<
Yes, but by a completely different mechanism. Sound is a physical longitudinal wave requiring a medium. An EM wave is a transverse field wave and requires no medium, i.e. it can propagate in empty space. There are ExH watts in an EM wave where 'x' is the cross-product.
--
73, Cecil, W5DXP
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RE: A Feedline Energy Analysis
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by WD0CT on December 13, 2005
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"In rsponse to WD0CT's question, does reflection gain really exist?
Yes it does. The power reflected at an antenna mismatch is call reflection 'loss'. However, at the matching network (antenna tuner), the reflected power is re-reflected back to the antenna. The process by which the reflected power is re-reflected back up the line to the load is called reflection 'gain'. The term appears in engineering texts. The result is that the only portion of the reflected power that is lost is in the small attenuation in the tuner and the attenuation of the feed line.
Walter Maxwell, W2DU"
Thanks Walt. Am I right to conclude that reflection gain equals reflection loss minus any line and tuner attenuation?
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RE: A Feedline Energy Analysis
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by KD2BD on December 13, 2005
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WF7A wrote:
> Forgive my density, but I'm still having a little
> wrapping my head around the concept of wave
> cancellation and still having energy left over.
> Maybe it's the analogy I've constructed in my head
> that's causing me problems:
>
> Say we sample the instantaneous voltage on an ac
> line. The positive alternation on one side is at
> 110v and at the same moment--on the other
> side--there's a negative alternation of 110v.
> With our trusty o'scope we'd measure 0v since the
> positive and negative waves cancel (no difference
> of potential), but each side still has its own
> energy level.
Hi Rich.
The 110 volt line voltage you mention is 110 volts RMS, or about 311 volts peak-to-peak. Connecting a 'scope across the line will measure 311 volts peak-to-peak (Don't try this without an isolation transformer!).
Since we're talking about an AC signal, the polarity and instantaneous peak voltages are constantly changing. And, yes, there is a brief moment in time in every cycle when the voltage will be exactly zero volts.
If we could AVERAGE the AC signal over time (through integration, or similar technique), we would read ZERO volts for a symmetrical AC waveform carrying no DC component.
These are just several interpretations of an AC voltage. We typically deal with RMS and peak-to-peak values in electronics. Multimeters typically read RMS, while scopes provide peak-to-peak readings (or readings between ANY points along the waveform you wish to measure).
But at no time does energy (or voltages) "cancel" in AC, unless they are summed in some manner.
Another point of confusion occurs when we measure the voltage along a transmission line delivering RF power to a load with a high VSWR. In this example, we might find a location along the line where the RF voltage is zero (RMS or peak-to-peak).
*** This does NOT mean that there is a cancellation of energy, or power, or voltage along the line! ***
This reading is simply the result of our measuring the RF voltage along the line using a non-directive measuring device.
Since RF is actually travelling in BOTH directions along the line (since reflections are taking place in this example), the zero volt reading is the result of a voltage cancellation that is taking place, not within the transmission line, but within the diode detector in the RF probe we've placed across the line!
Since the RF probe cannot distinguish between the incident and reflected voltages travelling along the line, and each differ in magnitude and phase at the point of measurement, the probe gives us the arithmetic sum of everything it sees. If we pick the right point along the line, it might just be zero volts.
To get a better picture of what's happening on the transmission line, a directional wattmeter (or voltmeter) is required. A good-quality directional probe will show the ratio of forward to reflected power to be constant along the transmission line, regardless of where the measurement is made (neglecting line losses, of course).
73, de John, KD2BD
http://kd2bd.ham.org/
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RE: A Feedline Energy Analysis
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by WF7A on December 13, 2005
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Thank you, Cecil, and for your erudite reply, John.
Maybe I'm suffering from Stollmayer's Law of Conservation of Cellular Momentum: if you don't try stretching your mind a little bit to learn something your brain cells just sit there and die off.
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A Feedline Energy Analysis
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by N0AH on December 13, 2005
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Copyright on Eham? There are no copyrights on Eham. hi Just be glad people are discussing the subject. I've never seen anyone put in so many notations and credits. And what happens? Very technical stuff. I too thank you for making this theory(s) a bit easier to understand.
73 Paul
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A Feedline Energy Analysis
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by VE3WGO on December 13, 2005
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I think readers might understand the concept of where the energy goes if a bit more information was presented.
Maybe the concept of re-reflection is clouding the concept of where does the power go? Answer: heating the line.
One key fact that is necessary to know is that at any point along a transmission line, the energy flowing along it is constant from one end to the other, except for the small amount of energy that is lost through imperfections in the dielectric and conductors in the form of heat. Thus, it can be understood that except for this small loss (often called the transmission line loss), the energy flowing "into" a line is the same as the energy flowing "out" the other end. This applies in all conditions of source, line, and load impedance.
When a wave travelling on a line encounters a load of an impedance different from the line, there will be a reflection from the load, back toward the source. When this happens, the refected wave and the original ("incident") wave will coexist on the transmission line. If we want to measure these waves we can use a variety of devices to do so. One such device is a directional coupler, which can seperate these waves and allows us to measure either the incedent or the reflected wave, depending on the direction we connect the coupler to the line. Another way to measure these waves, although not seperately, is with a slotted line, which is a device made of a short section of a transmission line that has been opened up with a narrow slot through which a small probe can be adjusted to sample the signal along the transmission line.
Now, back to the waves themselves. When a wave is reflected by an impedance change (a bad load or a discontinuity) on a transmission line, the incident and reflected waves will pass each other as they travel in opposite directions on that line. If we try measure the voltage on the line with a simple device such as a voltmeter, or a probe in a slotted line, we can't seperately measure these these waves, we can only measure their combined effect. We will observe that the voltage will not be constant along the line, and will form peaks and valleys as the result of constructive and destructive interference between the travelling waves. These peaks and valleys will be stationary (and a half-wavelength apart*), and the resulting envelope is known as a Standing Wave. The Ratio of these peaks and valleys is the Standing Wave Ratio (the Ham's second most measured quantity after Frequency).
One key thing to understand...since the power flowing along a transmission line is always constant everywhere along that line (except for the pesky transmission line loss which is small), then we know that any point where the voltage is high the current is low. Conversely, where the voltage is low the current is high. That's right...the current at points along the line also varies in the same way as the voltage does, but at different places on the line. The ratio of the voltage to current Standing Wave magnitudes and their phase differences would be seen to vary along the line, and it is this key fact that allows us to select an point on the line using appropriate tools such as a Smith Chart or its software equivalent, to add a matching device and try to reduce the SWR on the line.
Re-reflections. Think for a moment, and you'll be able to grasp where this controversial and confusing notion of re-reflection comes about....
With a signal source driving a transmission line that is connected to a mismatched load, there will be a portion of the incident wave that is reflected from the load back toward the source. If we imagine that the source is also not perfectly matched to the line, then there will be a another reflection of that wave from the source back toward the load again (it has therefore been reflected twice by now). Before trying to use math to examine this situation, just consider this...that tired little wave has already been reflected twice (once on the first trip toward the load, and again on the return trip back to the source, and now it wants to go back toward the load yet agan...), and passed through this transmission line twice and suffered line losses twice in the process. That's where the power went....heating the line...twice. And it will keep on doing it, being reflected at each end, and losing more energy in the line on each round trip until is dissipates completely.
So therein lies a nonmathematical attempt to explain where the power goes.
Just for your infotainment...up until about 15 years ago we still had high power microwave transmission systems all across the continent, carrying all of our telephone and data signals (AT&T, MCI, Bell Canada, CNCP, GTE, etc). Frequencies were in the 3.5 to 11 GHz range. These systems generally used waveguide antenna feeders instead of coaxial transmission lines. In some systems there was no load failure protection other than an isolator at the transmitter port. So when an antenna got damaged by ice or whatever, the SWR would of course skyrocket on the waveguide feeder, and with several hundred watts of power into these guides, the current peaks would be huge, that is until the TWT PA would back off or shut down. Thre were incidents when the waveguides would get hot spots every inch or so from these current peaks, and burn the grey paint coating black. A graphic lesson that SWR can even be "seen" in some situations!
I'd like to respectfully disagree with W5DXP, (in a friendly manner of course) as he distinguishes between wave propagation effects in electromagnetics and sound. They are in fact the same in virtually all respects insofar as the transfer of energy is a physical process, and both electromagnetic waves and mechanoacoustic waves behave identically. They both reflect, diffract, diverge, refract, and transmit in identical manners and under the same laws. No matter that one is transverse and the other longitudinal, nor that one needs a physical medium while the other can do without one. Dielectrics, impedance, density, inverse square law, doppler, radiators, apertures, ...it's all the same fun. While I was an EE studying electromagnetics and (James Clerk) Maxwell's equations, my roommate was an ME studying Fluid Dynamics. Same formulas (different variable names), same physics, same laws, same fear and panic in the midterms, same problems sets with only the names changed.
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RE: A Feedline Energy Analysis
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by W9PMZ on December 14, 2005
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Re-refelections.
Let's summarize the system:
Source Impedance = 50 Ohms
Coax Impedance = 50 Ohms
Load Impedance = open, or infinite
So the transmitted wave traveling down the transmission line, the coax, encounters the open line. At this time the wave is reflected back towards the source. When the wave encounters the source, does it re-reflect? Probably some of it; but since it is reflected into a matched condition (the source impedance matches the line impedance) wouldn't most of the reflection be terminated in the source?
73,
Carl - W9PMZ
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RE: A Feedline Energy Analysis
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by W5DXP on December 14, 2005
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>ve3wgo wrote: ... both electromagnetic waves and mechanoacoustic waves behave identically.<
Hi Ed: Good posting until you went overboard with that exclusive "identical" statement. Words like "identical", "always", "never", "perfect", etc. make more sentences false than true.
Think about it: If those two types of waves "behaved identically", we would be able to hear the sun with our naked ears. What I said was that the wave cancellation mechanism is different for EM waves (in free space) Vs sound waves (in air). That statement can easily be proven to be true. I suspect we have a "half-empty" Vs "half-full" argument going. :-)
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73, Cecil, W5DXP
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RE: A Feedline Energy Analysis
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by W5DXP on December 14, 2005
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> w9pmz wrote: When the (reflected) wave encounters the source, does it re-reflect?<
In a perfectly matched system, where reflected energy is not allowed to reach the source, the reflected energy is 100% re-reflected. In the average amateur radio system, this is accomplished by the antenna tuner. Total destructive interference toward the source causes total constructive interference toward the load. This will be explained in Parts II and III of my article.
When the system is not matched by a tuner (or other method), the picture is not as clear. Reflected energy flows into the source and superposition/interference occurs. What happens can be deduced from the laws of physics but cannot be measured directly.
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73, Cecil, W5DXP
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RE: A Feedline Energy Analysis
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by W9AC on December 14, 2005
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As long as we're trying to strive for clarity, wouldn't the term "match point" as used in the article be better replaced with "feed point" -- the point where the transmission line interfaces with the radiator?
The term "match point" implies that the feed point is always "matched." And unless the characteristic Z of the line is equal to the radiator terminal Z, there may not be a match.
Paul, W9AC
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RE: A Feedline Energy Analysis
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by W9AC on December 14, 2005
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">In a perfectly matched system, where reflected energy is not allowed to reach the source, the reflected energy is 100% re-reflected."
I really hate to nit-pik and sound like an old griping lady, but things can easily turn on the use or mis-use of a single written term.
If the system is perfectly matched (source Z, line Z, and radiator terminal Z are all equal), why is there any reflection at all at any point on the line? I think what you mean is the following:
"In a perfectly TUNED system, where a reflected wave is not allowed to reach the source because of an intervening network (e.g., ATU), the reflected wave is 100% re-reflected at the terminals of that network."
The use of the term TUNE is used to describe 100% system reactance cancellation, not necessarily a pure line Zo to feed match. In such case where there is (i) a line-to-radiator feed point Z mis-match; and (ii) a network inserted at some point on the transmission line in which system reactance is perfectly cancelled (tuned), a condition is created whereby the line is MATCHED between the Tx and network, and the line is TUNED between the network and radiator, resulting in near 100% transmission efficiency when using a loss-less line and network.
Paul, W9AC
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RE: A Feedline Energy Analysis
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by W9PMZ on December 14, 2005
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"In a perfectly matched system, where reflected energy is not allowed to reach the source, the reflected energy is 100% re-reflected. "
OK, now I'm confused.
What I think learned is that in a perfectly matched system, where;
Zs = Ztl = Zl,
Zs -> impedance of source
Ztl -> impedance of transmission line
Zl -> impedance of load.
But ignore an impedance matching device for a moment; if we look into the source we see a voltage generator with Thevenin equivalent impedance. Why wouldn’t that reflected wave be 100% terminated in the Thevenin equivalent impedance; in a perfectly matched system?
73,
Carl - W9PMZ
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RE: A Feedline Energy Analysis
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by W9AC on December 14, 2005
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>"But ignore an impedance matching device for a moment; if we look into the source we see a voltage generator with Thevenin equivalent impedance. Why wouldn’t that reflected wave be 100% terminated in the Thevenin equivalent impedance; in a perfectly matched system?"
What reflected wave?
I have to go back to the mechanics of a "perfectly matched system." Why is any reflection created whatsoever in such a system where the source Z, line Z, and radiator terminal Z are equal?
Paul, W9AC
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RE: A Feedline Energy Analysis
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by W9PMZ on December 14, 2005
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"In a perfectly matched system, where reflected energy is not allowed to reach the source, the reflected energy is 100% re-reflected. "
Because, the statement here is that in a 100% matched system it is re-refected. I agree with you (W9AC). I'm trying to understand this statement from W5DXP.
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RE: A Feedline Energy Analysis
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by W5DXP on December 14, 2005
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Instead of five replies, I'm going to try one summary.
Semantics strikes again. By "perfectly matched system", I mean "a system in which a perfect Z0-match exists". Since it is confusing, I will avoid the words "perfectly matched system" from now on. Given that Z0 is 50 ohms for most modern transceivers, The Z0-match point would be the point at which reflections are eliminated on 50 ohm coax, usually at the input terminal of a tuner such that there are no reflections between the tuner and the source. I apologize for any confusion about definitions. My technical vocabulary is half a century old and sometimes obsolete. Here's an example of a Z0-matched system that we can use in our discussion.
50 ohm XMTR--50 ohm coax--+--1/4WL 75 ohm coax--112.5 ohm load
A system where no reflections exist anywhere in the system is known as a "flat" system. The above system is "flat" from the XMTR to the '+' 50 ohm Z0-match point.
Note that in a lossless perfectly tuned system, a perfect conjugate match exists everywhere in the system if a Z0-match is achieved and Zsource = Z0. There's no limitations on reflected energy existing in a conjugately matched system. If the source impedance has reactance, then reflected waves are actually flowing in the source even though the system is perfectly conjugately matched.
> w9pmz wrote: But ignore an impedance matching device for a moment; if we look into the source we see a voltage generator with Thevenin equivalent impedance. Why wouldn't that reflected wave be 100% terminated in the Thevenin equivalent impedance; in a perfectly matched system?<
The Thevenin equivalent model prohibits us from determining what happens inside the source and is limited to determining only what happens outside the source. We could substitute a signal generator with a circulator and load resistor in which case all reflected power would be dissipated in the circulator load resistor. But your question is confusing to me. In a perfectly matched system, seems that no reflected energy would reach the source so none would be dissipated there.
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73, Cecil, W5DXP
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RE: A Feedline Energy Analysis
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by W5DXP on December 14, 2005
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> w9ac wrote:
Paul, your call brings back pleasant memories of when I operated from Texas A&M's w5ac during the '50s.
> The term "match point" implies that the feed point is always "matched." And unless the characteristic Z of the line is equal to the radiator terminal Z, there may not be a match.<
I touched on this subject in my other reply. A "perfectly matched *system*" doesn't prohibit the existence of reflected waves within the *system*. In an ideal system, if maximum available power is being delivered to the load, no matter how mismatched the load is to the feedline, the *system* is perfectly matched.
The term I use for a system in which no mismatch exists anywhere, is a "flat" system.
But it's good to get the semantic kinks ironed out and I will avoid the use of "perfectly matched" from now on.
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73, Cecil, W5DXP
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RE: A Feedline Energy Analysis
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by W9AC on December 14, 2005
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>"But it's good to get the semantic kinks ironed out and I will avoid the use of "perfectly matched" from now on."
Probably a good percentage of misunderstandings on this subject occur because two or more people are thinking alike but calling things in common something different!
The association between voltage, current, power, incident wave, reflected wave, reflection coeffient, matching, tuning, losses, VSWR, return loss and even the word "system" is so complex that we almost need to set defintions as part of a preamble before engaging in a discussison!
Nice work on the clarification, Cecil. I can now rest peacefully tonight.
Paul, W9AC
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RE: A Feedline Energy Analysis
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by W9PMZ on December 14, 2005
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"The Thevenin equivalent model prohibits us from determining what happens inside the source and is limited to determining only what happens outside the source."
In testing wireless local loop amplifiers, 3.4GHz 40W amplifiers (you see them ocassionally on eBay), we turned on the amp, without drive, and used a network analyzer to measure the S Parameters looking into the output of the amplifier. Perhaps I've confused terms, but shouldn't you be able to relate S22 (looking into the output) to an equivalent impedance as a function of frequency? In the passband the output has a high return loss, meaning that the passband is matched. And conversly the return loss is low outside the passband, meaning that outside the passband it is not matched. Wouldn't this help describe the Thevenin equivalent model of impedance?
Regardless of the output match, if you have a matching device in-line (such as an ATU), shouldn't you be able to use a network analyzer and examine S22 (at the matching device's output) at the frequency in question and predict the amount of reflection based on S22?
73,
Carl - W9PMZ
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RE: A Feedline Energy Analysis
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by W5DXP on December 14, 2005
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> w9pmz wrote: Regardless of the output match, if you have a matching device in-line (such as an ATU), shouldn't you be able to use a network analyzer and examine S22 (at the matching device's output) at the frequency in question and predict the amount of reflection based on S22?<
Therein lies a big problem as yet unresolved in a 15 year long argument between Warren Bruene, w5oly, and Walter Maxwell, w2du. Measuring s22 requires replacing the source impedance with its equivalent lumped source impedance when, in the typical amateur radio transmitter, the value of the source impedance is the center of the argument. The s-parameters are small signal, i.e. linear, parameters. How does a class-C amp fit into the linear equations? How does a push-pull class-AB fit into the linear equations?
Of course, there are devices like signal generators and IC's where s22 can be reliably measured. You can certainly come up with a real-world amp that looks like a Thevenin equivalent but it won't resemble at all an 813 running class-C. (813 is showing my age, but Ameritron still sells 811's.)
Truth is, I cannot answer your question. Perhaps Walt, w2du, will jump in here and explain his take on the subject. If we assume the 50 ohm Z0-match is a clue to the source impedance, what you say above can certainly be done. But the ARRL will disagree with you (and refuse to publish any articles you might come up with to support that view).
Here's a question for you. What effect can the source impedance possibly have on the system when no reflected energy is allowed to reach the source?
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73, Cecil, W5DXP
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RE: A Feedline Energy Analysis
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by W9PMZ on December 14, 2005
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I don't know that answer.
But here is my take,
Assume that we have a virtual short or a virtual open and then end of our transmission line.
If we assume that we can provide a match to the virtual open and the virtual short, with the requisite transmission line (between the ATU and the load); then there should be no reflection.
Since we assume that we are using a real world transmission line with an ATU that cannot in reality provide an impedance transformation to either a virtual short or a virtual open. Then the lumped circuit, looking from either the short or the open will not be matched. With these conditions, shorts or opens, an impedance transformation probably unobtainium. S22 (return loss of the lumped transmission line and ATU)) is likly to be almost 0 and almost all the power will be re-reflected. I don't believe it will be 100% (but pretty close) because we are using real world components.
73,
Carl - W9PMZ
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RE: A Feedline Energy Analysis
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by W5DXP on December 14, 2005
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>W9PMZ wrote: If we assume that we can provide a match to the virtual open and the virtual short, with the requisite transmission line (between the ATU and the load); then there should be no reflection.<
Part II of my article reveals the physics behind that "virtual short". Amazingly enough, in addition to s22, it involves s11 and s12 along with wave cancellation. We might revisit this subject after Part II has been posted. In the meanwhile, consider what conclusions can be drawn from
b1 = s11*a1 + s12*a2 = 0 and what happens to the energy in those cancelled waves.
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73, Cecil, W5DXP
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RE: A Feedline Energy Analysis
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by VE3WGO on December 14, 2005
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so many questions all at once.
1. Cecil - Electromagnetics vs Mechanoacoustic waves.
My take on it is that they do behave identically, and here's why I think so...they will both propagate in an appropriate medium with a definable velocity and transmission loss until they encounter an interface that cannot support their further propagation. At that point these waves will then reflect from or refract around that medium interface. Why can't we hear the sun? For the identical reason that we can't see it in a closed room: the propagating wave has encountered a medium that cannot support further propagation of the energy we are trying to observe. If there was a fluid or gas instead of a vacuum all the way between the Sun and Earth, we'd hear the Sun, but there isn't so we can't. Similarly, if there was just a fluid or gas instead of rock between the Sun and me right now, I could see it, but it's night time here and the solid earth is in the way so I can't. In both cases, propagtion has been stopped by a medium that can't support the sound or light I'm looking for. It's interesting that a vacuum allows light/electromagnetic waves but not acoustic waves to propagate, and conversely rock allows mechanoacoustic waves but not electromagnetic ones to propagate (the attenuation value must be immense). Air allows almost all energy wave forms to propagate. Quiz - is there any form of energy that CAN'T propagate through air? I failed, I can't think of one offhand.
You challenged me so I answered. I'm probably being way too diabolical about it all, but I hope it's as interesting to you as it is to me.
2. In a perfectly matched system....
"System", unless otherwise indicated, means an end to end interconnection. A perfectly matched system means that a source, transmission medium (line), and load form this end to end system and are all perfectly matched to each other at all points without exception. The obvious way that this could exist is with identical impedances everywhere. Being pedantic? Perhaps, but in a perfectly matched system, there are no impedance mismatches, so there are no reflections, and all power traveles from the source toward the load. Some of it may be dissipated along the way in the line as loss, but that is all that will be observed.
3. Carl - Measuring Amplifier S22 (source impedance)
Yes we can and do measure S22 that way, and if an amp has high return loss it is well matched. A wave reflected by a load mismatch will be dissipated by the amp. The amp may have an output VSWR monitor (measures reflected waves) built-in to back off the output power and to save the finals from self-destruction from high load mismatches. A Class C amp should still fit the S22 model since its output matching network would be designed as a compromise between the various operating points on the output transistor(s) or tube(s), so that the output impedance of the amp (ie the "source" impedance) is as close to the transission line value as possible. All output devices are non-linear regardless of operating Class, but Class C is the most since the output device is actually off for most of the RF cycle. That's where the high efficicny comes from. The output matching/filtering network's job is to remove the harmonic (and out of passband intermod) distortion products and transform the output stage's impedance to the desired value, usually 50 or 75 ohms these days. In a really well designed amp, S22 will not vary with the operating point (power level), but there is a point of diminishing "returns"....
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RE: A Feedline Energy Analysis
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by VE3WGO on December 14, 2005
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so many questions all at once.
1. Cecil - Electromagnetics vs Mechanoacoustic waves.
My take on it is that they do behave identically, and here's why I think so...they will both propagate in an appropriate medium with a definable velocity and transmission loss until they encounter an interface that cannot support their further propagation. At that point these waves will then reflect from or refract around that medium interface. Why can't we hear the sun? For the identical reason that we can't see it in a closed room: the propagating wave has encountered a medium that cannot support further propagation of the energy we are trying to observe. If there was a fluid or gas instead of a vacuum all the way between the Sun and Earth, we'd hear the Sun, but there isn't so we can't. Similarly, if there was just a fluid or gas instead of rock between the Sun and me right now, I could see it, but it's night time here and the solid earth is in the way so I can't. In both cases, propagtion has been stopped by a medium that can't support the sound or light I'm looking for. It's interesting that a vacuum allows light/electromagnetic waves but not acoustic waves to propagate, and conversely rock allows mechanoacoustic waves but not electromagnetic ones to propagate (the attenuation value must be immense). Air allows almost all energy wave forms to propagate. Quiz - is there any form of energy that CAN'T propagate through air? I failed, I can't think of one offhand. Note however, that many forms of energy are attenuated strongly by air, which thankfully is what keeps us alive on Earth.
You challenged me so I answered. I'm probably being way too diabolical about it all, but I hope it's as interesting to you as it is to me.
2. In a perfectly matched system....
"System", unless otherwise indicated, means an end to end interconnection. A perfectly matched system means that a source, transmission medium (line), and load form this end to end system and are all perfectly matched to each other at all points without exception. The obvious way that this could exist is with identical impedances everywhere. Being pedantic? Perhaps, but in a perfectly matched system, there are no impedance mismatches, so there are no reflections, and all power traveles from the source toward the load. Some of it may be dissipated along the way in the line as loss, but that is all that will be observed.
3. Carl - Measuring Amplifier S22 (source impedance)
Yes we can and do measure S22 that way, and if an amp has high return loss it is well matched. A wave reflected by a load mismatch will be dissipated by the amp. The amp may have an output VSWR monitor (measures reflected waves) built-in to back off the output power and to save the finals from self-destruction from high load mismatches. A Class C amp should still fit the S22 model since its output matching network would be designed as a compromise between the various operating points on the output transistor(s) or tube(s), so that the output impedance of the amp (ie the "source" impedance) is as close to the transission line value as possible. All output devices are non-linear regardless of operating Class, but Class C is the most since the output device is actually off for most of the RF cycle. That's where the high efficicny comes from. The output matching/filtering network's job is to remove the harmonic (and out of passband intermod) distortion products and transform the output stage's impedance to the desired value, usually 50 or 75 ohms these days. In a really well designed amp, S22 will not vary with the operating point (power level), but there is a point of diminishing "returns"....
73, Ed VE3WGO
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RE: A Feedline Energy Analysis
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by W5DXP on December 14, 2005
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>VE3WGO wrote: 1. Cecil - Electromagnetics vs Mechanoacoustic waves. My take on it is that they do behave identically, ...<
Any good physics book will explain why TEM waves in empty space are different from sound waves in air. If there is any difference whatsoever, they do not "behave identically", by definition.
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73, Cecil, W5DXP
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A Feedline Energy Analysis
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by W4EF on December 14, 2005
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W5DXP wrote:
<<<An RF engineer will usually say there are three things that can cause 100% reflection. Those are a short-circuit, an open-circuit, or a purely reactive impedance. This is true at a load. But at an impedance discontinuity with waves incident from both directions, to that list of three, we can add a fourth, namely wave cancellation due to total destructive interference. In general:
The destructive interference energy resulting from wave cancellation at an impedance discontinuity becomes an equal magnitude of constructive interference in the opposite direction. Since there are only two directions in a transmission line, wave cancellation is the equivalent of an energy reflection. 100% wave cancellation means 100% energy reflection. [9] >>>>>>
I have been sort of half-following this debate between the VE9SRB camp and the W2DU camp on and off for the past several years. I say "half-following" becuase I couldn't bring myself to work through all of the math. Prior to this evening, I was leaning towards VE9SRB, mostly because I couldn't wrap my head around the idea of the W2DU's virtual short-circuit and the accompanying concept of 100% re-reflection. These concepts struck me as cheap tricks that gave the right answer, but didn't reflect (no pun intended) the physical reality of what was going on along the transmission line. Then I had a eureka moment tonight. A simple thought experiment illustrates the concepts of the virtual short quite easily and clearly (at least for me). Imagine two phase-locked RF signal generators each driving opposite ends of a 50 ohm transmission line. Each generator has a circulator on its output so that the energy arriving from the opposite generate won't effect it. Both generators are set at equal amplitudes (say 50 volts x 1 amp = 50 Watts), but with opposite phase (the phase of generator A as measured at its output is 180 degrees measured relative to the phase of generator B at its respective output). Now imagine that both generators having been off, are turned on at time, t = to. The RF energy from each generator races towards the other generator like two trains racing towards each other on a common railroad track. At some time, t = t1 (equal to half the total transit time of the transmission line) the two RF waves coming from opposite directions meet in the center of the interconnecting transmission line. At this point the electric fields from the two waveforms have equal amplitudes and opposite phase. The net electric field amplitude at this meeting point will go to zero. The magnetic fields from the two waveforms will have equal amplitudes and equal phase, thus the net magnetic field at this meeting point will be double the magnetic field due to either incident wave. Power flow will be 50 watts in the leftward direction and 50 watts in the rightward direction. From the point of view of either generator, this situation will be indistinguishable from a short circuit placed at the center of the transmission line. Looking at either side of the centerline of the transmission line, the standing wave pattern will look identical to that which would occur if a short circuit were place at the centerpoint. There is no short circuit, however, just a 50 ohms transmission line with a 50 ohm termination at either end. The rightward traveling wave meeting the rearward traveling wave with equal amplitude and opposite phase are what creates the "virtual" short circuit at the center. When viewed in this way, I find the idea of the virtual short circuit and 100% re-reflection quite plausible.
73, Mike W4EF.................................
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RE: A Feedline Energy Analysis
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by W9PMZ on December 15, 2005
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This is a good illustration Mike.
But just one question:
Your example assumes that the reflection is 180 degrees out of phase. Is this always the case for the general argument that the re-reflection is 100% (that the re-reflection is 180 degrees out of phase) reflected under the conditions of a virtual short or open?
Perhaps if the transmission line is exactly 1 wavelength long. However, if the transmission line were a random length, to be reflected 180 degrees out of phase there would be an abrupt discontinuity. Is this possible?
73,
Carl - W9PMZ
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RE: A Feedline Energy Analysis
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by K5DVW on December 15, 2005
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Cecil, Gig 'em, nice write up and interesting comments following.
K5DVW -- Texas A&M Alum '90 also operated W5AC back when.
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A Feedline Energy Analysis
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by NT4XT on December 15, 2005
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Great article/thread.
The article takes a complex analysis and describes it in a way that an average radio hobbiest can understand.
I think I have a rudimentary grasp of the concept of Conservation of Energy, as in, when complete wave cancellations occur in a transmission line the power had to go somewhere.
In my simple mind, I would apply that concept, erroneously or otherwise- to how perhaps, a resonant vertical with 2 resonant ground plane radials dissipates/radiates signal. When current flows with equal magnitude but opposite directions in like time, such as on 1/4 wave radials that are aligned 180 degrees, it is often said, those two fields, or waves, cancel.
So I'm thinking but what happens to the Watts of those waves? RF AC Current flows, so power fields have to be present, in each leg of a resonant two legged ground plane vertical system.
I imagine, the Watts of the cancelled fields reflect back toward the feed-point, just in time to travel up the resonant vertical, and are dissipated by the vertical "radiator."
But since we're talking about the feedline, I'll remain all eyes and ears for parts II and III.
I have been wondering about qualitatively quantifying- or estimating/ball-park-figuring, Power lost or possibly even gained, via feed-line radiation for quite some time now.
In this vein, according to some notions I've read, an Isotron, a no radial ground mounted vertical, an OCF dipole with coaxial vertical radiator, an antenna that appears to be a living tree, all come to mind. As well, claims that the vertical radiation from the OCF with specially designed vertical feedline radiator, is insignificant, while others say in practice there is a worthwhile difference.
Since this discussion is concerning transmission line power, I hope these questions will be somewhat clarified in parts 2 and 3.
Actually, my layman's understanding of the ground plane vertical might even be flawed, if so- someone please set it straight.
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RE: A Feedline Energy Analysis
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by W5DXP on December 15, 2005
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> W4EF wrote: From the point of view of either generator, this situation will be indistinguishable from a short circuit placed at the center of the transmission line.<
There are a couple of incorrect concepts embedded in your example. For simplicity, let's assume a lossless system. The transmission line is Z0=50 ohms being driven by 50 ohm, 50 watt, (50v,1a) generators equipped with 50 ohm circulator load resistors. THERE ARE NO IMPEDANCE DISCONTINUITIES ANYWHERE IN THE SYSTEM!
Two waves traveling in opposite directions in a transmission line have NO EFFECT on each other until they encounter an impedance discontinuity. Since no impedance discontinuity exists in your example, the two waves have NO EFFECT on each other. The short circuit effect that you perceive is in your mind and doesn't exist in reality. Nothing happens at that "short circuit" point except superposition (phasor addition) of two waves that are completely unaware of each other's existence. That's how standing waves are formed but the forward wave is completely unaffected by the reverse wave and vice versa. There is a net voltage and a net current that varies according to the SWR pattern but the 50 volts and 1 amp of forward wave flows right through the 50 volts and 1 amp of reverse wave.
Waves flowing in OPPOSITE directions in a transmission line superpose but they do not cancel. In the above example, there are points where all the net wave energy is contained in the electric field (voltage maximum point) and 1/4WL away, all the net wave energy is contained in the magnetic field (current maximum point). But the wave energy components (forward and reverse) always maintain a constant ratio of 50 ohms between the E-field (voltage) and the H-field (current). An SWR pattern is akin to the light ring/dark ring interference pattern that we have all seen with light rays.
The forward wave is a constant 50 volts at 1 amp. The reverse wave is a constant 50 volts at 1 amp. At an SWR voltage maximum, those two voltages add in phase and the currents cancel. At the SWR voltage maximum point, the net voltage is 100 volts and the net current is zero. But the 1 amp of forward current and 1 amp of reverse current continue to flow at that point. The net current is zero simply because they are flowing in opposite directions but even at the net current equal zero point, the forward current has no effect on the reflected current and vice versa. It is the human mind that makes a big deal out of the net current being zero. The forward current and reverse current is completely unaffected by the net current being zero.
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73, Cecil, W5DXP
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RE: A Feedline Energy Analysis
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by W5DXP on December 15, 2005
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>nt4xt wrote: Since this discussion is concerning transmission line power, I hope these questions will be somewhat clarified in parts 2 and 3.<
My article is very narrow and confined to an impedance discontinuity *POINT* in a transmission line. It's best to understand what happens at that *POINT* before proceeding. Note that my article says nothing about what happens at the source or at the antenna. My basic thrust is a divide-and-conquer approach. I certainly have other bombs to drop in the future but let's limit the discussion to a single impedance discontinuity in a transmission line for now.
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73, Cecil, W5DXP
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RE: A Feedline Energy Analysis
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by W4EF on December 15, 2005
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W5DXP wrote:
>>>The forward wave is a constant 50 volts at 1 amp. The reverse wave is a constant 50 volts at 1 amp. At an SWR voltage maximum, those two voltages add in phase and the currents cancel. At the SWR voltage maximum point, the net voltage is 100 volts and the net current is zero. But the 1 amp of forward current and 1 amp of reverse current continue to flow at that point. The net current is zero simply because they are flowing in opposite directions but even at the net current equal zero point, the forward current has no effect on the reflected current and vice versa. It is the human mind that makes a big deal out of the net current being zero. The forward current and reverse current is completely unaffected by the net current being zero. <<<<<<
That was my point exactly, Cecil. I referred to the condition at the center of the line as a "virtual short circuit" because it mimics a real short-circuit in terms of the standing wave pattern. The standing wave pattern (and the real physical voltages and currents you would measure with real instruments) on the line remains unchanged if I replace the fictitious virtual short circuit with a real physical short circuit. An observer can't tell the difference unless they turn-off one of the sources.
In any case, if my example doesn't illustrate the idea of a virtual short circuit or 100% re-reflection (I am not even sure these terms are supposed mean the same thing in the context of this discussion), please provide an example (preferably on a transmission line) that does illustrate these concepts.
Thanks,
Mike, W4EF.............................
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RE: A Feedline Energy Analysis
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by W5DXP on December 15, 2005
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>w4ef wrote: That was my point exactly, Cecil. I referred to the condition at the center of the line as a "virtual short circuit" because it mimics a real short-circuit in terms of the standing wave pattern.<
I apologize for being blunt, but a "virtual short circuit" existing in an environment with no impedance discontinuities doesn't mimic a real short circuit at all. A real short circuit reflects everything. A "virtual short circuit" existing in the center of a line without an impedance discontinuity reflects nothing. Walter Maxwell's "virtual shorts" always exist at a Z0-match impedance discontinuity. (Please forgive me if I misunderstood what you are trying to say.)
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73, Cecil, W5DXP
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RE: A Feedline Energy Analysis
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by W4EF on December 15, 2005
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W5DXP
>>>>I apologize for being blunt, but a "virtual short circuit" existing in an environment with no impedance discontinuities doesn't mimic a real short circuit at all. A real short circuit reflects everything. A "virtual short circuit" existing in the center of a line without an impedance discontinuity reflects nothing. Walter Maxwell's "virtual shorts" always exist at a Z0-match impedance discontinuity. (Please forgive me if I misunderstood what you are trying to say.) <<<<<<<
Okay, now that we have established that the "virtual short circuit" in my example doesn't really reflect anything (and I agree with that), what is happening at Walt's "virtual short circuit" at the match point? What makes it different than my example? Is the rearward traveling wave totally re-reflected? If so, why? Unless the rearward travelling wave encounters a reflection coefficient of 1.0 at the match point, how is the rearward travelling wave totally re-reflected? And if it is, how do you get a 1.0 reflection coefficient in the rearward direction? That would imply that all the incident power would have been reflected initially at that interface which in turn implies that none would ever reach the load. IOW, in a real system the real reflection coefficient at an impedance discontinuity with a finite transmission coefficient will have to be less than 1.0 (by conservation of energy). So how do you get from the real <1.0 reflection coefficient, to the virtual short circuit (reflection coefficient = 1.0)?
Thanks,
Mike, W4EF.................
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RE: A Feedline Energy Analysis
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by W5DXP on December 16, 2005
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>w4ef wrote: what is happening at Walt's "virtual short circuit" at the match point?<
As my article says, Walt's "virtual short" is a two-step process that results in 100% re-reflection of reflected energy in a Z0-matched system.
The first step is the reflection at the physical impedance discontinuity. Assume Pref2 is the reflected power from the load. This first step will involve Pref2*rho^2 where rho^2 is the same as |s22|^2.
The second step involves wave cancellation of two waves. Assume Pfor1 is the forward power from the source. One of the waves is associated with Pfor1*rho^2 where rho^2 is the same as |s11|^2. The other wave is what is left over from step one above and is associated with Pref2(1-rho^2). When these two waves are equal in magnitude and 180 degrees out of phase, they cancel which eliminates reflections traveling toward the source. Since the energy components in those two waves cannot be canceled, the total energy involved must necessarily join the forward wave toward the load as constructive interference. Walt said it years ago in "Reflections"
Maxwell, Walter, "Reflections II", (c) 2001 Worldradio Books, ISBN 0-9705206-0-3
page 4-3, "The destructive wave interference between these two complementary waves ... causes a complete cancellation of energy flow in the direction toward the generator. Conversely, the constructive wave interference produces an energy maximum in the direction toward the load, ..."
page 23-9, "Consequently, all corresponding voltage and current phasors are 180 degrees out of phase at the matching point. ... With equal magnitudes and opposite phase at the same point (point A, the matching point), the sum of the two (reflected) waves is zero."
To summarize:
Step 1: Pref2*rho^2 is re-reflected normally. (The associated s-parameter voltage component is s22*a2)
Step 2: Pfor1*rho^2 + Pref2(1-rho^2) is re-reflected by wave cancellation due to destructive interference toward the source. (The associated s-parameter voltage components are s11*a1 + s12*a2 = b1 = zero.)
In his QEX article, Dr. Best embraced step 1 and denied step 2. In denying step 2, he disagreed with the technical descriptions of non-reflective glass involving 1/4WL thin films and HP's Ap Note 95-1 on S-Parameters.
Stand by for Part II which explains the above equations.
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73, Cecil, W5DXP
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RE: A Feedline Energy Analysis
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by W4EF on December 16, 2005
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W5DXP wrote:
<<<<<<<<
To summarize:
Step 1: Pref2*rho^2 is re-reflected normally. (The associated s-parameter voltage component is s22*a2)
Step 2: Pfor1*rho^2 + Pref2(1-rho^2) is re-reflected by wave cancellation due to destructive interference toward the source. (The associated s-parameter voltage components are s11*a1 + s12*a2 = b1 = zero.)
In his QEX article, Dr. Best embraced step 1 and denied step 2. In denying step 2, he disagreed with the technical descriptions of non-reflective glass involving 1/4WL thin films and HP's Ap Note 95-1 on S-Parameters.
Stand by for Part II which explains the above equations.
>>>>>>>>>>>>>>>>>>>
Yes, step 1 is pretty obvious and goes along with the standard view of things. I dug out Walt's book last night and read up on his virtual open/short concept. It sounds plausible (something has to happen to cancel the reflections at the match point), but I didn't have the eureka moment yet (guess I need to read it again). I may dig out VE9SRB's articles from QEX as well to see if I can understand what he is offering as an aternative explanation.
73, Mike W4EF.....................
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RE: A Feedline Energy Analysis
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by W5DXP on December 16, 2005
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> w4ef wrote: I didn't have the eureka moment yet (guess I need to read it again). I may dig out VE9SRB's articles from QEX as well to see if I can understand what he is offering as an aternative explanation.<
Consider how reflections are canceled at the 1/4WL thin-film surface coating on reflection-cancelling glass. The same thing happens at a match point in a transmission line.
Dr. Best's explaination is more of a half explanation than an alternative explanation. He correctly tells us what happens to the voltage in the direction of the load. He essentially ignores what happens to the energy in the direction of the source. He doesn't even mention interference and once claimed over on rec.radio.amateur.antenna that interference was unnecessary and irrelevant while asserting that
75w + 8.33w = 133.33w. (He left out the 50w interference term.)
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73, Cecil, W5DXP
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RE: A Feedline Energy Analysis
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by W5DXP on December 19, 2005
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Part II of my energy analysis article has been submitted to eHam.net and is also available on my web page: http://www.qsl.net/w5dxp/energy.htm
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73, Cecil, W5DXP
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