Where does reflected power go?
Scott M. McClements
where does reflected power (SWR) actually go? Is it eventually radiated
from the antenna, dissipated by PA transistors or is it more complex
than that?
Next, if you use an antenna tuner, this will improve how well your
antenna "couples" with the feedline, correct? It can actually improve
how much power is put into the feedline? However, the mismatch still
occurs at the antenna feedpoint and the percentage of reflected power is
still constant? With the antenna tuner in place, does this effect how or
where a reflected wave acts?
I know this sounds troll'ish..but I am serious about what I am asking.
Thanks,
Scott McClements.
Adam
You are having some interesting questions there. Maybe I can help you to
answer them. First of all the reflected power will be changed into heat by
the PA transistor or tube. As long as the PA can dissipate the heat there is
not too much danger (within limits).
The antenna tuner can be seen as a sort of impedance-transformer that
transforms the 50 Ohm of your rig into the impedance of your whole antenna
(including the feedline etc.). When the rig can see 50 Ohms it can give it's
full power and no power is coming back into the rig. So the PA of your
transceiver will be very happy. However there could be still a 'standing
wave' at your antenna which would mean that the antenna is not dipped for
the required frequency. It will mean that it doesn't have the efficiency of
a antenna that is dipped for the right frequency.
So you see, the antenna tuner doesn't do anything about standing waves. It
just corrects the impedance mismatch between the transceiver and the antenna
system. If your antenna is dipped for the right frequency, for instance 1/2
wave end fed, then the antenna is okay but has not the 50 Ohm impedance that
your feedline and rig like to see. Normally one would use a balun
transformer (1:4 for instance) at the feedingpoint of the antenna to
transform the antenna impedance to 50 Ohms unballenced. If you don't have
the right connection between feedline and antenna the feedline will become
part of the antenna and starts to radiate energy. Your neighbors might not
like it very much because TVI can be the result.
Where does the reflected power go? Well, I guess that the answer is blowing
in the wind. If you could feel the heat of the coil in the tuner when you
tune a bad SWR you will know where the power goes (don't do it if you don't
want blisters on your hands or worse).
Well, there are many nice books about antenna's and feedlines. There is much
more to it than I can write here. And I'm sure that there are many radio
amateurs who have lots of experience with homemade antenna's and can tell
you more.
'73 From the Netherlands
Adam
Scott M. McClements <yet...@grove.iup.edu> schreef in berichtnieuws
3A1433B7...@grove.iup.edu...
Adam
Richard Harrison
"where does reflected SWR actually go?"
Ideally, to a ratio of one.
SWR is an interference pattern created by forward and reflected
traveling waves on a transmission line. Voltage and current create their
own similar patterns, but it is easy to measure and plot the VSWR using
a slotted (trough) line.
In a lossless system, all reflected power is finally radiated. None is
absorbed back in the transmitter or spent in transmission line or
matching networks. SWR indicates temporary storage of circulating energy
in the transmission line. Some energy is wasted in all the actual
equipment used.
Typically, losses are small and often less than many people think when
they see the SWR number.
Next, an antenna tuner should let you transform whatever load impedance
appears to your desired load impedance. This is often the impedance of
the coax you are using. The impedance desired is frequently 50 ohms
which is the impedance many transmitters are specified for as a load at
which the transmitter will perform as expected. Other impedances are
sometimes specified.
The tuner eliminates a point of reflection between itself and the coax
by providing a match. The coax then has an SWR of one.
Best regards, Richard Harrison, KB5WZI
Cecil
> With the antenna tuner in place, does this effect how or
> where a reflected wave acts?
An antenna tuner cannot change the SWR at the antenna but a tuner
can achieve a Z0-match so no reflected power reaches the transmitter.
From a magnitude standpoint: Generated Power plus Reflected Power
equals Forward Power. Because of transmission line losses, Reflected
Power will be maximum and Forward Power will be minimum at the
antenna. Thus the SWR is highest at the antenna and lowest at the
antenna tuner (assuming the tuner is located at the transmitter).
--
http://www.mindspring.com/~w6rca
R.P.Haviland
Adam wrote in message <8v183g$150n$1...@news.kabelfoon.nl>...
. First of all the reflected power will be changed into heat by
>the PA transistor or tube. As long as the PA can dissipate the heat there
is
>not too much danger (within limits).
Not necessarily. If the source generator does not habe the same impedance as
the transmision line, part of the reflected power will be re-reflected,
adding to the power from the generator. This is always true in 60 cycle
power systems, where the source impedance is a small fraction of the load
impedance. It is also true in most RF amplifiers, those where the efficiency
is greater than 50%.
A typical situation for an amplifier is:
power to line= power from amplifier + re-reflected power: say 130 watts=100
watts + 30 watts
Power to load = power to line - reflected power: 130 watts-30 watts=100
watts for this case: the transmitter output is delivered to the
oad( neglecting line loss, which causes the SWR at the ampllifier to be less
than that at the load)
w4mb.
VE9SRB
"R.P.Haviland" <bo...@iag.net> wrote in message
news:cxWQ5.46133$Ds3.3...@e420r-sjo3.usenetserver.com...
>
> <snip>
>
>
> A typical situation for an amplifier is:
> power to line= power from amplifier + re-reflected power: say 130
watts=100
> watts + 30 watts
> Power to load = power to line - reflected power: 130 watts-30 watts=100
> watts for this case: the transmitter output is delivered to the
> oad( neglecting line loss, which causes the SWR at the ampllifier to be
less
> than that at the load)
> w4mb.
>
Hi Bob:
> Power to load = power to line - reflected power: 130 watts-30 watts=100
This statement is correct in all TL systems, I do however, respectfully
disagree with the following statement:
> power to line= power from amplifier + re-reflected power: say 130
watts=100
> watts + 30 watts
power to line= power from amplifier + re-reflected power is true but these
powers add as vectors since the V and I associated with the powers are
complex quantities. 130 watts = 100 watts + 30 watts is only true when the
100 watts and 30 watts are 90 degrees out of phase.
Steve, VE9SRB
Cecil
> 130 watts = 100 watts + 30 watts is only true when the
> 100 watts and 30 watts are 90 degrees out of phase.
Power is a scalar and has no phase. How about talking about
voltage phases and current phases instead?
--
http://www.mindspring.com/~w6rca
VE9SRB
"Cecil" <Cecil....@IEEE.org> wrote in message
news:3A14447B...@IEEE.org...
Hi Cecil,
You are correct, power is a scalar. I should have said: 130 watts = 100
watts + 30 watts is only true when the voltages and currents associated with
the 100 watts and 30 watts are 90 degrees out of phase.
In a 50 ohm transmission line, 100 watts of power has an associated voltage
magnitude of 70.711 V and a current magnitude of 1.414 A. 30 watts of power
has an associated voltage magnitude of 38.73 V and a current magnitude of
0.775 A. 130 watts of power has an associated voltage magnitude of 80.623 V
and a current magnitude of 0.62 A. This all based on the relationship that
P = |V| |I| cos(theta), where theta is the phase angle difference between V
and I. V and I both being in a 50 ohm system.
In the transmission line, the two forward complex voltages and currents add
to become the total forward voltage and current. Recognizing that both
magnitude and phase are factors in complex addition, the only condition
under which a voltage with a magnitude of 70.711 V adds to a voltage with a
magnitude of 38.73 V to result in a total voltage having a magnitude of
80.623 V is when the two voltages are 90 degrees out of phase.
Steve
Richard Harrison
"Power is a scalar and has no phase."
Verily, and Old Pythagoras and Old Watt must both be gyrating in their
tombs. 100 + 30 = 130 only when 100 and 30 are at quadrature?
I have no right to poke fun with as many goofs as I`ve made in recent
days. I welcome the company.
Reg Edwards
Reflected power doesn't go anywhere. It is just a figment of
human imagination put there by unscrupulous SWR meter
salesmen and other old wives. It wasn't generated in the
first place.
In a sensibly lossless transmission system, which most
systems are, all the RF power outputted by the transmitter
is absorbed by the intended load, ie., the antenna. There's
simply no-where else for it to go.
--
***********************************
Regards from Reg, G4FGQ
For free radio modelling software go to:
http://www.btinternet.com/~g4fgq.regp
***********************************
Scott M. McClements wrote ..
> I have a few questions about reflected and antenna tuners.
First off,
> where does reflected power (SWR) actually go? Is it
eventually radiated
> from the antenna, dissipated by PA transistors or is it
more complex
> than that?
>
> Next, if you use an antenna tuner, this will improve how
well your
> antenna "couples" with the feedline, correct? It can
actually improve
> how much power is put into the feedline? However, the
mismatch still
> occurs at the antenna feedpoint and the percentage of
reflected power is
> still constant? With the antenna tuner in place, does this
effect how or
> where a reflected wave acts?
>
Richard Harrison
"Generated Power plus Reflected Power equals Forward Power."
True if all reflected power is bounced by the generator.
I`m only sure that the reflected power subtracts from the forward power
when using the Bird wattmeter to give an indication of generated power
and that is the same power which is delivered to the load. This has to
be because there is no place in the transmission system for long term
storage of r-f energy.
Cecil`s logic holds because when you subtract reflected power from both
sides of Cecil`s equation, it`s the same as mine.
Richard Clark
<yet...@grove.iup.edu> wrote:
>I have a few questions about reflected and antenna tuners. First off,
>where does reflected power (SWR) actually go? Is it eventually radiated
>from the antenna, dissipated by PA transistors or is it more complex
>than that?
>
>Next, if you use an antenna tuner, this will improve how well your
>antenna "couples" with the feedline, correct? It can actually improve
>how much power is put into the feedline? However, the mismatch still
>occurs at the antenna feedpoint and the percentage of reflected power is
>still constant? With the antenna tuner in place, does this effect how or
>where a reflected wave acts?
>
>I know this sounds troll'ish..but I am serious about what I am asking.
>
>Thanks,
>
>Scott McClements.
>
>
Hi Scott,
Nothing trollish about a topic that incites so many to spend so much
time responding to. Hmm, that does sound trollish doesn't it?
But, to answer your question:
Reflected power going back to the transmitter is destructive because
the transmitter will attempt to absorb it as would any load. Now the
level of destruction may simply be heat, but consider that heat can be
measured in comparison to body temperature or relative to a flame
temperature.
The principle failure mode for solid state sources strained with
mismatched loads is found in the excess current passing through the
device where its ability to support local current density is exceeded.
At this level we are talking about melting of the crystal lattice and
catastrophic destruction.
If you insert a tuner between the mismatch and the transmitter, you
are providing another mismatch (to the reflections on their way to the
transmitter) that is the reciprocal of the load mismatch. This
creates a situation wherein the power is reflected, re-reflected, and
re-re-.... you get the point. However, with each successive
reflection, some energy escapes the antenna in spite of the original
mismatch (it always does) and the significance of that power sloshing
back and forth diminishes with each pass back and forth (not to speak
of that power lost to the transmission line supporting all this
sloshing - in fact that is the major source of real loss, or in the
tuner itself).
Now keep in mind the discussion above presumes a single wave to
support this sloshing analogy, and if you keep in mind that every
single cycle undergoes this process , independent of all cycles before
it, then things will work fine.
Keep your eyes open for an announcement from Walt Maxwell, W2DU, as to
when his new edition of "Reflections" becomes available, and buy a
copy quick.
73's
Richard Clark, KB7QHC
Fred Hambrecht
reflections" Sub Chapter "Too low a SWR can kill you" said "judging from
what we hear on the air, nearly everyone is looking for a VSWR of 1:1
Question why and the answer may be, I'm not getting out on this frequency
because my SWR 2.5:1... there's too much power coming back and not enough
getting to the antenna...or I feel a line having that much SWR, the
reflected power flowing back in to the amplifier will burn it up..." He then
goes on to Poo Poo any of these ideas. If interested I will be glad to post
the reasons.
"Richard Clark" <rwc...@seanet.com> >
William E. Sabin
VE9SRB wrote:
> I should have said: 130 watts = 100
> watts + 30 watts is only true when the voltages and currents associated with
> the 100 watts and 30 watts are 90 degrees out of phase.
>
> In a 50 ohm transmission line, 100 watts of power has an associated voltage
> magnitude of 70.71 V and a current magnitude of 1.414 A. 30 watts of power
> has an associated voltage magnitude of 38.73 V and a current magnitude of
> 0.775 A. 130 watts of power has an associated voltage magnitude of 80.623 V
> and a current magnitude of 1.62 A. This all based on the relationship that
> P = |V| |I| cos(theta), where theta is the phase angle difference between V
> and I. V and I both being in a 50 ohm system.
>
> In the transmission line, the two forward complex voltages and currents add
> to become the total forward voltage and current. Recognizing that both
> magnitude and phase are factors in complex addition, the only condition
> under which a voltage with a magnitude of 70.711 V adds to a voltage with a
> magnitude of 38.73 V to result in a total voltage having a magnitude of
> 80.623 V is when the two voltages are 90 degrees out of phase.
This is incorrect for the following reasons:
1) The current corresponding to 130 W is 1.613 A, not 0.62 A. But we cannot look
at this problem in this manner.
2) The forward power is 130 W. In terms of the forward wave this is
(V+)^2 / 50, so (V+) = 80.62 V RMS
3) The reflected power is 30 W. In terms of the reflected wave this is
(V-)^2 / 50, so (V-) = 38.73 V RMS.
4) The difference between the forward and reflected power waves is
(V+)^2 / 50 - (V-)^2 /50 = 130 - 30 = 100 W.
These two values are what a thermoelectric directional wattmeter measures. The
phase between (V+) and (V-) is *not* specified. The phase difference between
(V+) and (V-) is controlled by the reflection coefficient at the load, which can
be anywhere on a certain circle on the Smith chart.
For example, at a resistive load, (V+) and (V-) can be completely in-phase (or
completely out-of-phase). But as we move along the line the phase between (V+)
and (V-) rotates. But the forward (130), reflected (30) and net (100) power
values are constant everywhere on the line.
Bill W0IYH
Ed Hare, W1RFI
news:8v1us5$g8o$1...@iac5.navix.net...
> Might I presume to disagree Sir? Herr Maxwell in "Another look at
> reflections" Sub Chapter "Too low a SWR can kill you" said "judging from
> what we hear on the air, nearly everyone is looking for a VSWR of 1:1
Maxwell also wrote a series of QST articles. They are all available among
the articles posted at:
http://www.arrl.org/tis/info/reflections.html
The Reflections series is found at
http://www.arrl.org/tis/info/pdf/Reflect.pdf. It is just over a megabyte,
but about 58 pages.
While I do not agree with the presentation of every single technical detail
in the Reflections articles, they are great reading for those who would like
a better understanding of the concepts of SWR. Walt and I have corresponded
a bit and he told me that the presentation of some of the more controversial
concepts in the series and his Reflections book will be improved for the
upcoming new version.
73,
Ed Hare, W1RFI
William E. Sabin
R.P.Haviland wrote:
> Adam wrote in message <8v183g$150n$1...@news.kabelfoon.nl>...
> . First of all the reflected power will be changed into heat by
> >the PA transistor or tube. As long as the PA can dissipate the heat there
> is
> >not too much danger (within limits).
>
> Not necessarily. If the source generator does not have the same impedance as
> the transmision line, part of the reflected power will be re-reflected,
> adding to the power from the generator.
This is an oversimplification. The result of load reflection is that the
impedance presented to the amplifier may not be a) not resistive and b) not of
the right magnitude.
In vacuum tube amplifiers the PI network can accommodate a range of impedances
and transform them to the correct plate load resistance. No problem.
In solid state broadband amplifier the load presented to the transistors can be
too low, too high and/or reactive, all of which are troublesome in one way or
another. The transistors can overheat because the efficiency is too low, or the
transistors can zap because of overvoltage on the drain/collector. Reactive load
lines are bad news for transistors.
The question "Where does the reflected power go?" is a phoney baloney question
anyway. Reflected power is the ratio (V-)^2 /50, where (V-) is the reflected
voltage wave. At the amplifier output this reflected wave (V-) adds vectorially
with the forward wave (V+) that is leaving the amplifier. These two waves
establish a value of complex impedance at the amplifier output. This is the
impedance that the amplifier delivers its *net* 100 watts (or whatever) to the
load. The task then is to make the amplifer deliver this 100 watts into the
resistive component of this coax input impedance.
Bill W0IYH
Cecil
> For example, at a resistive load, (V+) and (V-) can be completely in-phase (or
> completely out-of-phase). But as we move along the line the phase between (V+)
> and (V-) rotates. But the forward (130), reflected (30) and net (100) power
> values are constant everywhere on the line.
Thanks Bill, I was trying to figure out how to put that into words.
--
http://www.mindspring.com/~w6rca
Cecil
>
> Cecil W6RCA wrote:
> "Generated Power plus Reflected Power equals Forward Power."
>
> True if all reflected power is bounced by the generator.
That was assuming a Z0-match caused by an antenna tuner where zero
reflected power reaches the transmitter (generator).
--
http://www.mindspring.com/~w6rca
Cecil
>
> Might I presume to disagree Sir? Herr Maxwell in "Another look at
> reflections" Sub Chapter "Too low a SWR can kill you" said "judging from
> what we hear on the air, nearly everyone is looking for a VSWR of 1:1
>
> because my SWR 2.5:1... there's too much power coming back and not enough
> getting to the antenna...or I feel a line having that much SWR, the
> reflected power flowing back in to the amplifier will burn it up..." He then
> goes on to Poo Poo any of these ideas. If interested I will be glad to post
> the reasons.
But Walter Maxwell has expanded on that very subject. He was talking, in
context, ONLY about tube transmitters with built-in pi net tuner circuits.
That's about all there was when he wrote that article. He has said that he
was NOT talking about solid-state amplifiers with fixed 50 ohm outputs. He
has said that solid-state amplifiers can indeed eat enough reflected energy
to overheat or blow themselves up. "Reflections II" will probably correct
that misunderstanding.
--
http://www.mindspring.com/~w6rca
Cecil
> In a sensibly lossless transmission system, which most
> systems are, all the RF power outputted by the transmitter
> is absorbed by the intended load, ie., the antenna. There's
> simply no-where else for it to go.
Here we go again. :-) Assume a 100 joules/second transmitter with
a one second long lossless transmission line and a load which causes
reflected power to equal half of the forward power. Key the transmitter.
After one second, no power has reached the load and there is 100 joules
in the transmission line.
After two seconds, 50 joules has reached the load, 50 joules has been
reflected, and 100 joules is in the forward wave. NOTE THAT 150 JOULES
EXISTS IN THE TRANSMISSION LINE, MORE THAN THE GENERATED ENERGY.
Assuming a Z0-match which blocks reflected power from reaching the
transmitter, after three seconds forward energy equals 150 joules and
reflected energy equals 50 joules.
After four seconds, 75 joules has reached the load, 75 joules has been
reflected, etc. etc. etc.
I think you get the idea. After steady-state has been reached there is
200 joules of forward energy and 100 joules of reflected energy. The
transmitter has generated 300 more joules than has been absorbed by
the load. There exists 200 watts of forward power and 100 watts of
reflected power. It really and truly exists because it hasn't made
it to the load yet.
There **IS** somewhere else for it to go, Reg. It is *stored* in the
transmission line until the transmitter is turned off. Without the
reflected energy stored in the transmission line, all the generated
power would not make it to the load.
--
http://www.mindspring.com/~w6rca
Cecil
> I'll be dipped.
I thought that was what I was smelling. :-)
> Not a word from the experts about low loss feedline
> preserving more of the power for radiation rather than dissipation,
> whether or not we're talking of the forward incident wave or a reflected
> wave.
Before we talk about lossy lines, we need to talk about lossless lines
in order to answer his question, where does the reflected power go. See
my followup posting to Reg. Even Reg forgets that the transmission line
must fill up with energy before the load can accept all the generated
power.
--
http://www.mindspring.com/~w6rca
Fred Hambrecht
line so the impedance is matched. I think you would agree that the SWR on
the line has not changed one iota, (disregarding lengths of wire in the
tuner) now we are back to Maxwell's original premise.
"Cecil" <Cecil....@IEEE.org> wrote in message
news:3A1498DC...@IEEE.org...
Korinna & Peter
so - with all the experts here:
What is the big difference between:
A - A built in Tuner like in a FT1000 etc. that 'matches' up to a SWR of
1:3
B - A external autotuner like the Kenwood AT250, the Yaesu FC-1000 etc.
C - The all famous SGC - Antenna coupler.
Many thanks in advance,
73,
Peter
--
Korinna Imle & Peter Hovorka
ICQ #74200239
Peter: DG1GPQ
http://WWW.CYBERCATS.DE
--
Richard Clark
wrote:
>Might I presume to disagree Sir? Herr Maxwell in "Another look at
>reflections" Sub Chapter "Too low a SWR can kill you" said "judging from
>what we hear on the air, nearly everyone is looking for a VSWR of 1:1
>
>Question why and the answer may be, I'm not getting out on this frequency
>because my SWR 2.5:1... there's too much power coming back and not enough
>getting to the antenna...or I feel a line having that much SWR, the
>reflected power flowing back in to the amplifier will burn it up..." He then
>goes on to Poo Poo any of these ideas. If interested I will be glad to post
>the reasons.
>
Thanks Fred,
Walt and I have corresponded to this matter considerably (both online
and off). I enjoy his style and the points made as his work is very
accessible and should be required reading.
Differences you may note certainly does not inhibit either of us to
express those same differences and hold to them, but I would reckon
that even with such divergence, he and I are much closer together than
we are far apart.
Dick Carroll
>
> I have a few questions about reflected and antenna tuners. First off,
> where does reflected power (SWR) actually go? Is it eventually radiated
> from the antenna, dissipated by PA transistors or is it more complex
> than that?
>
> Next, if you use an antenna tuner, this will improve how well your
> antenna "couples" with the feedline, correct? It can actually improve
> how much power is put into the feedline? However, the mismatch still
> occurs at the antenna feedpoint and the percentage of reflected power is
> still constant? With the antenna tuner in place, does this effect how or
> where a reflected wave acts?
>
> I know this sounds troll'ish..but I am serious about what I am asking.
>
> Thanks,
>
> Scott McClements.
I'll be dipped. Not a word from the experts about low loss feedline
George, W5YR
voltages and currents, as well as handling higher SWR situations than
interna; tuners.
Internal tuners are noted for being relatively "tender" and easily
damaged by voltage/current extremes.
72/73, George W5YR - the Yellow Rose of Texas NETXQRP 6
Fairview, TX 30 mi NE Dallas in Collin county QRP-L 1373
Amateur Radio W5YR, in the 55th year and it just keeps getting better!
Icom IC-756 PRO #02121 (9/00) Kachina #91900556 (12/99) IC-765 (6/90)
Dick Carroll
>
> I have a few questions about reflected and antenna tuners. First off,
> where does reflected power (SWR) actually go? Is it eventually radiated
> from the antenna, dissipated by PA transistors or is it more complex
> than that?
>
> Next, if you use an antenna tuner, this will improve how well your
> antenna "couples" with the feedline, correct? It can actually improve
> how much power is put into the feedline? However, the mismatch still
> occurs at the antenna feedpoint and the percentage of reflected power is
> still constant? With the antenna tuner in place, does this effect how or
> where a reflected wave acts?
>
I don't know what happens elsewhere, but here's how it goes at my
station.
When I transmit through my tuner then through the coax feeding my
antenna, if I'm on a frequency where the antenna is perfectly matched to
the feedline, and the tuner is then matched to both the feedline at its
output, and to the transmitter at its input, there is no reflected wave
present. SWR can be said to be 1:1 both between the transmitter and the
tuner, between the tuner and the feedline and the antenna.
When I move off the resonant frequency of my antenna, there then exists
a reflected wave component that returns from the antenna through the
coaxial feedline to the tuner. After a slight readjustment of the tuner,
there then exists no reflected wave between the transmitter and the
tuner, but there still exists a reflected wave component travelling back
and forth between the tuner and the antenna on the coaxial feedline.
Where does reflected "power" go? That part that successfully travels
back and forth between the tuner and the antenna and reaches the antenna
again (the entire reflected component less that amount that is
dissipated as heat due to line loss) is radiated as RF energy into space
- except that amount that is again re-reflected because of the mismatch
existing at the coax/antenna junction. It makes the trip again, and once
again suffers the lossikntroduced by the feedline loss characteristic.
So it goes, back and forth, with most of the original RF that enters
the feedline being radiated by the antenna and some small amount being
reflected, re-reflected and partly radiated, and so on for as long as
the transmitter is keyed, assuming some acceptable level of SWR that
doesn't create too much loss or cause voltages too high for the coax to
tolerate.
The bottom line here is that you can construct a dipole antenna for the
center of any HF band, then using a tuner located at the tranmitter,
operate with perfectly acceptable reflected wave level and no detectable
degernation of service on either end of the RF circuit. I use
parallelled dipoles for multiple band operation.
By now it should be obvious where some of the pitfalls of this system
lie. Since most coaxial line exhibits loss at high SWR, and high
voltages can exist at those high reflected levels, it behoove one to use
a good quality coax selected for good high voltage rating and low loss.
Since coax loss increases with frequency, and this factor becomes much
worse above HF, one has less latitude with such installations at VHF and
above than at HF. Many hams prefer to use balanced open wire or ladder
line feedlines which have much lower loss than coax, and which then can
be used at much higher SWR than coax. There are some advantages to this
approach but it is more restrictive from a hardware standpoint than
simple coaxial line intallations.
Dick W0EX
William E. Sabin
Cecil wrote:
> He
> has said that solid-state amplifiers can indeed eat enough reflected energy
> to overheat or blow themselves up. "Reflections II" will probably correct
> that misunderstanding.
This is not a good way to look at it. What happens is that the standing wave on
the line causes the load impedance for the transistor to be the wrong value. This
means that the collector/drain load impedance is not what is needed for proper
efficiency. Suppose we want 100 W *net* power output. If this impedance is too
low the transistor gets too hot. If the impedance is too high the transistor may
not get too hot but the RF voltage on the collector/drain becomes too large and
may destroy the transistor.
From a circuit standpoint this is what happens. The *circuit* behavior is
affected by the *wave* behavior on the line.
A good technique, when thinking about RF systems, is to use the left side of the
head to think "waves" and the right side of the head to think "circuit" and the
middle of the head to see how the two sides are related and work together.
Basic principles.
Bill W0IYH
Jeff Anderson
wrote:
> Assume a 100 joules/second transmitter with
>a one second long lossless transmission line and a load which causes
>reflected power to equal half of the forward power. Key the transmitter.
>
>After one second, no power has reached the load and there is 100 joules
>in the transmission line.
>
>After two seconds, 50 joules has reached the load, 50 joules has been
>reflected, and 100 joules is in the forward wave. NOTE THAT 150 JOULES
>EXISTS IN THE TRANSMISSION LINE, MORE THAN THE GENERATED ENERGY.
>
>Assuming a Z0-match which blocks reflected power from reaching the
>transmitter, after three seconds forward energy equals 150 joules and
>reflected energy equals 50 joules.
>
Given the 1-second long transmission line, how does the Z0-match block
reflected power at the 2nd second?
- Jeff
Reg Edwards
joules to be transported from the hydro-electric generators
at Boulder to your shack ? ;o)
The question is "Where" does the reflected power go, not
"When".
So we are not concerned with transients but with steady
state.
So I repeat - all the RF power outputted by the transmitter
is absorbed (for the want of a better word) by the antenna.
THERE IS SIMPLY NOWHERE ELSE FOR IT TO GO.
Why unnecessarily complicate matters by dragging red
herrings all over the place ? ;o)
--
***********************************
Regards from Reg, G4FGQ
For free radio modelling software go to:
http://www.btinternet.com/~g4fgq.regp
***********************************
Cecil <Cecil....@IEEE.org> wrote in message
news:3A149C3C...@IEEE.org...
> Reg Edwards wrote:
> > In a sensibly lossless transmission system, which most
> > systems are, all the RF power outputted by the
transmitter
> > is absorbed by the intended load, ie., the antenna.
There's
> > simply no-where else for it to go.
>
> Here we go again. :-) Assume a 100 joules/second
transmitter with
> a one second long lossless transmission line and a load
which causes
> reflected power to equal half of the forward power. Key
the transmitter.
>
> After one second, no power has reached the load and there
is 100 joules
> in the transmission line.
>
> After two seconds, 50 joules has reached the load, 50
joules has been
> reflected, and 100 joules is in the forward wave. NOTE
THAT 150 JOULES
> EXISTS IN THE TRANSMISSION LINE, MORE THAN THE GENERATED
ENERGY.
>
> Assuming a Z0-match which blocks reflected power from
reaching the
> transmitter, after three seconds forward energy equals 150
joules and
> reflected energy equals 50 joules.
>
VE9SRB
"William E. Sabin" <sab...@mwci.net> wrote in message
news:3A1486DF...@mwci.net...
>
>
> VE9SRB wrote:
>
> > I should have said: 130 watts = 100
> > watts + 30 watts is only true when the voltages and currents associated
with
> > the 100 watts and 30 watts are 90 degrees out of phase.
> >
> > In a 50 ohm transmission line, 100 watts of power has an associated
voltage
> > magnitude of 70.71 V and a current magnitude of 1.414 A. 30 watts of
power
> > has an associated voltage magnitude of 38.73 V and a current magnitude
of
> > 0.775 A. 130 watts of power has an associated voltage magnitude of
80.623 V
> > and a current magnitude of 1.62 A. This all based on the relationship
that
> > P = |V| |I| cos(theta), where theta is the phase angle difference
between V
> > and I. V and I both being in a 50 ohm system.
> >
>
>
> 1) The current corresponding to 130 W is 1.613 A, not 0.62 A. But we
cannot look
> at this problem in this manner.
Hi Bill:
If you check my post again you will see that I made the following statement
:
"130 watts of power has an associated voltage magnitude of 80.623 V and a
current magnitude of 1.62 A."
So, I am not sure where the 0.62 A figure you quote comes from.
>
> 2) The forward power is 130 W. In terms of the forward wave this is
> (V+)^2 / 50, so (V+) = 80.62 V RMS
>
This is exactly as I have stated above.
> 3) The reflected power is 30 W. In terms of the reflected wave this is
> (V-)^2 / 50, so (V-) = 38.73 V RMS.
This is exactly as I have stated above.
>
> 4) The difference between the forward and reflected power waves is
> (V+)^2 / 50 - (V-)^2 /50 = 130 - 30 = 100 W.
This is true in all transmission line systems and is not a point of
disagreement from my perspective. Regardless of the level of re-reflection
at the transmitter, the net power delivered to the transmission line is
always equal to the steady state forward power minus the steady state
reflected power. It is important to understand that the 100 W of delivered
power is a net value and not the forward source power of the transmitter.
The point of issue is what power "components" combine to become the total
steady state forward power of 130 watts. It is not a source power of 100 W
"adding" to a re-reflected power of 30 W to become 130 W. The 100 watts is
the effective net power delivered to the transmission line. The actual
forward source power delivered to the transmission line is not 100 watts.
>
> These two values are what a thermoelectric directional wattmeter measures.
The
> phase between (V+) and (V-) is *not* specified. The phase difference
between
> (V+) and (V-) is controlled by the reflection coefficient at the load,
which can
> be anywhere on a certain circle on the Smith chart.
>
> For example, at a resistive load, (V+) and (V-) can be completely in-phase
(or
> completely out-of-phase). But as we move along the line the phase between
(V+)
> and (V-) rotates. But the forward (130), reflected (30) and net (100)
power
> values are constant everywhere on the line.
>
I agree with these statements. With the directional wattmeter, we are
measuring steady state forward and reflected power in the transmission line.
We are not measuring the true forward source power nor are we measuring the
re-reflected power. All we can "see" with the wattmeter is the net forward
power of 130 watts which is the result of the re-reflected power combining
with the true forward source power. If these two powers are "in-phase", the
130 watts of total forward power will not be the result of 100 watts of true
forward source power adding to 30 watts of re-reflected power.
Steve
> Bill W0IYH
>
Dick Carroll
> all the RF power outputted by the transmitter
> is absorbed (for the want of a better word) by the antenna.
> THERE IS SIMPLY NOWHERE ELSE FOR IT TO GO.
Hving caused noticible temperature rise in a few coaxial feedlines over
the years I can't entirely agree, Reg.
Dick
VE9SRB
"Reg Edwards" <g4fgq...@btinternet.com> wrote in message
news:8v3a5r$h1k$1...@plutonium.btinternet.com...
>
> So I repeat - all the RF power outputted by the transmitter
> is absorbed (for the want of a better word) by the antenna.
> THERE IS SIMPLY NOWHERE ELSE FOR IT TO GO.
>
IMHO, here's where the total power outputted by the transmitter goes (no
tuner between TX and antenna):
1) A portion of it is delivered to the antenna.
2) A portion of it is lost in the feedline
3) A portion of it is returned to the transmitter as a function of the
transmitter output impedance.
Steve
Cecil
> They eat themselves up, do to the mismatch of impedance. Lets put a tuner in
> line so the impedance is matched. I think you would agree that the SWR on
> the line has not changed one iota, (disregarding lengths of wire in the
> tuner) now we are back to Maxwell's original premise.
I agree. The misunderstanding arises because Walter Maxwell assumed a
tuner function with its accompanying Z0-match but he did not state his
assumption in words. Because he didn't state his assumption, some people
have inferred that he was talking about systems without a tuner. He wasn't.
--
http://www.mindspring.com/~w6rca
Cecil
>
> Cecil wrote:
> > He
> > has said that solid-state amplifiers can indeed eat enough reflected energy
> > to overheat or blow themselves up. "Reflections II" will probably correct
> > that misunderstanding.
>
> low the transistor gets too hot. If the impedance is too high the transistor may
> not get too hot but the RF voltage on the collector/drain becomes too large and
> may destroy the transistor.
That's what I said. The impedance seen depends on the phase of the reflections
that make it back to the transmitter.
--
http://www.mindspring.com/~w6rca
Cecil
> So we are not concerned with transients but with steady
> state.
Whadda mean we, white man? :-)
> So I repeat - all the RF power outputted by the transmitter
> is absorbed (for the want of a better word) by the antenna.
> THERE IS SIMPLY NOWHERE ELSE FOR IT TO GO.
But there is somewhere else for it to go, Reg. It is stored
in the transmission line until the transmitter is turned off.
This is easy to prove by observing TV signals with reflections
and the resultant ghosts. It is easy to prove that there is more
energy in the transmission line with reflections than without.
In my example where 300 joules is stored in the one second long
lossless transmission line, the transmission line continues to
deliver power to the load for many seconds after the transmitter
has been disconnected. 2/3 of that 300 joules is real live reflected
energy. It is really there. It is not imaginary.
--
http://www.mindspring.com/~w6rca
Cecil
>
> "Reg Edwards" <g4fgq...@btinternet.com> wrote in message
> news:8v3a5r$h1k$1...@plutonium.btinternet.com...
> >
> > is absorbed (for the want of a better word) by the antenna.
> > THERE IS SIMPLY NOWHERE ELSE FOR IT TO GO.
>
> tuner between TX and antenna):
>
> 1) A portion of it is delivered to the antenna.
> 2) A portion of it is lost in the feedline
> 3) A portion of it is returned to the transmitter as a function of the
> transmitter output impedance.
4) A portion of it is stored in the transmission line until the
transmitter is disconnected.
We're talking about lossless systems, so number 2 doesn't apply.
We're talking about Z0-matched systems, so number 3 doesn't apply.
--
http://www.mindspring.com/~w6rca
Cecil
>
> On Thu, 16 Nov 2000 20:47:24 -0600, Cecil <Cecil....@IEEE.org>
> wrote:
> >a one second long lossless transmission line and a load which causes
> >reflected power to equal half of the forward power. Key the transmitter.
> >
> >After one second, no power has reached the load and there is 100 joules
> >in the transmission line.
> >
> >After two seconds, 50 joules has reached the load, 50 joules has been
> >reflected, and 100 joules is in the forward wave. NOTE THAT 150 JOULES
> >EXISTS IN THE TRANSMISSION LINE, MORE THAN THE GENERATED ENERGY.
> >
> >Assuming a Z0-match which blocks reflected power from reaching the
> >transmitter, after three seconds forward energy equals 150 joules and
> >reflected energy equals 50 joules.
>
> reflected power at the 2nd second?
A lossless antenna tuner provides a Z0-match which doesn't allow any
reflected energy to reach the transmitter. The "SWR" between the transmitter
and the antenna tuner is 1:1. There is simply nowhere else for the reflected
energy to go except back towards the antenna. We know "what" happens from
a scalar standpoint. "How" is really not important except to academics.
--
http://www.mindspring.com/~w6rca
Richard Harrison
"The 100 watts is the effective net power delivered to the transmission
line. The actual forward source power delivered to the transmission line
is not 100 watts."
These sentences seem contradictory to me.
Cecil gave the formula for the inline wattmeter determined power early
in this thread. It is:
forward power minus reflected power is the actual power at the
measurement point. On a lossless line, power is the same at any point
along the line, regardless of SWR. If the line is lossy, power measured
at the load is less than that measured at the transmitter.
Re-reflected power and fresh from the transmitter power are
indistinguishable to the wattmeter. So, the forward power indication can
contain old power and fresh power.
If there is no significant loss, between the transmitter and its load,
the power output of the transmitter and the power delivered to the load
are the same quantity. Forward and reflected indicated powers should be
the same at both ends of the transmission line. With loss in the line,
the difference in net power indications at opposite ends of the
transmission line is the power lost in the line.
Best regards, Richard Harrison, KB5WZI
VE9SRB
"Cecil" <Cecil....@IEEE.org> wrote in message
Sorry.
eh...@bellatlantic.net
The amount of power that is lost (dissipated as heat, rather than
radiated) depends upon the loss figure of the transmission line
at the frequency of operation and the degree of mismatch between
the antenna and the transmission line. That is the simple answer.
An overview of the theory is below.
The power travels from the rig through the tuner trough the transmission
line to the antenna. Some of the power is dissipated (lost) as heat along
the way. When the power reaches the antenna, which has an impeadance
that does not match the transmission line impedance, some of it is radiated
by the antenna, and some of it is reflected back toward the tuner. On its
way back to the tuner, some of the reflected power is dissipated in the
transmission line. When it arrives at the tuner, it is re-reflected back
toward
the antenna. Again, part of that power is dissipated in the transmission
line
as it travels toward the antenna.
The amount (percentage) of arriving power that is reflected depends upon
the amount of mismatch; the amount of loss in the transmission line depends
on the type of line and the frequency.
"Scott M. McClements" wrote:
> I have a few questions about reflected and antenna tuners. First off,
> where does reflected power (SWR) actually go? Is it eventually radiated
> from the antenna, dissipated by PA transistors or is it more complex
> than that?
>
> Next, if you use an antenna tuner, this will improve how well your
> antenna "couples" with the feedline, correct? It can actually improve
> how much power is put into the feedline? However, the mismatch still
> occurs at the antenna feedpoint and the percentage of reflected power is
> still constant? With the antenna tuner in place, does this effect how or
eh...@bellatlantic.net
Dick Carroll wrote:
Actually, Mr. Harrison alluded to it. His post was very good, in
my opinion. I added what I hope is a simple overview explanation of
what goes on. I hope it did not muddy the waters - and I would have
preferred that one of the experts had done it. I'm sure they can do a far
better job than I.
I really felt that for this particular question the overall basic description
was in question, not the more in depth stuff in the ensuing discussion. And
like you, I would have like to have seen a stronger mention of feedline loss.
Not that the discussion was out of place - it's always interesting what
these guys have to say! Just that the basics needed to be outlined first.
Fred Hambrecht
allows us to understand why the darn SWR bridge works.
"Richard Clark" <rwc...@seanet.com> wrote in message
news:3a5e77db....@news.seanet.com...
VE9SRB
"Richard Harrison" <richard...@webtv.net> wrote in message
news:12462-3A...@storefull-117.iap.bryant.webtv.net...
> Steve best wrote:
> "The 100 watts is the effective net power delivered to the transmission
> line. The actual forward source power delivered to the transmission line
> is not 100 watts."
>
> These sentences seem contradictory to me.
>
Hi Richard,
I'm not sure why. I think we all agree that there may be some level of
re-reflection at the transmitter output and that this re-reflection adds to
the transmitter's forward source power to become the total forward traveling
power that we measure with a wattmeter. The effective net power delivered
to the transmission line is the difference between the total forward and
reflected powers. The transmitter's forward source power and the effective
net power it delivers to the system are generally two different things.
Let's assume that we connect a transmitter to a TL connected to a matched
load and we measure a forward traveling power of 100 watts and a reflected
power of 0 watts. In this case, the net power delivered to the TL is 100
watts AND the transmitter's forward source power is also 100 watts.
Now let's change the load to 100 ohms. Numerous things could happen in this
case to affect the forward and reflected powers in the TL.
One possibility is the following. We again measure a forward power of 100
watts and a reflected power of 11.11 watts. In this case, the transmitter's
output impedance is ZO and again, no re-reflection occurs at the transmitter
output. The 11.11 watts of reflected power is returned to the transmitter.
The transmitter's forward source power is still 100 watts but the net power
delivered to the TL system is only 88.89 watts.
Now, let's complicate the problem by creating the situation where
re-reflections do occur at the transmitter output. Let's still assume that
the transmitter delivers 100 watts of power to a matched load but that the
transmitter's output impedance is no longer matched to ZO. Let's again
assume that we change the load to 100 ohms. As one possibility, let's
assume that we measure a forward power of 126.56 watts and a reflected power
of 14.06 watts. The effective net power delivered to the TL is now 112.5
watts. However, the transmitter's forward source power still remains at 100
watts. Most people would immediately assume that the 126.56 watts of total
forward power is the result of 14.06 watts of reflected power being totally
re-reflected and then adding in-phase to the 112.5 watts of delivered power
such that 126.56 W = 112.5 W + 14.06 W. However, when we look at voltages
and currents we see that in-phase power addition cannot occur in this
manner. What really happens? In this case, my numbers were derived by
assuming that the transmitter's output impedance was 100 ohms. Therefore,
the actual re-reflected power is only 1.56 watts. This 1.56 watts of
forward power "adds" to the transmitter's 100 watts of forward source power
to become the total forward power of 126.56 watts. At this point, most
people would say that 100 W + 1.56 Watts can't possibly equal 126.56 watts.
So, let's check the "math" with voltage addition. 1.56 watts has a voltage
magnitude of 8.83 V. 100 watts has a voltage magnitude of 70.71 V. These
two voltages add to a total voltage magnitude of 79.54 V (they add in-phase)
resulting in a total forward power of 126.56 W. In phase powers always add
as Ptotal = P1 + P2 + 2 sqrt(P1) sqrt(P2).
Although it is always correct to determine effective net power delivery as
follows: Net Power Delivery = Pforward - Preflected, it is incorrect to take
this relationship and assume that Pforward is a result of Preflected being
totally re-reflected and adding to Net Power Delivery such that Pforward =
Net Power Delivery + Pre-reflected. On the surface it is not immediately
obvious but these concepts are totally unrelated.
> Cecil gave the formula for the inline wattmeter determined power early
> in this thread. It is:
> forward power minus reflected power is the actual power at the
> measurement point. On a lossless line, power is the same at any point
> along the line, regardless of SWR. If the line is lossy, power measured
> at the load is less than that measured at the transmitter.
>
Correct, with the following understanding: From the TL's perspective, there
are only two steady state powers. The forward traveling power and the
reflected traveling power. The difference between the two is the "net
power" delivered to the point where the measurements are made.
> Re-reflected power and fresh from the transmitter power are
> indistinguishable to the wattmeter. So, the forward power indication can
> contain old power and fresh power.
>
Correct.
> If there is no significant loss, between the transmitter and its load,
> the power output of the transmitter and the power delivered to the load
> are the same quantity.
This depends again on how we define the power output of the transmitter.
Based on your statement, the "power output" of the transmitter would be
defined as the "net power" delivered to the TL which is the difference
between the forward and reflected power at the transmitter output. I have
no problem with this definition except in the case where it is used to
describe the relationship between total forward power and the re-reflected
power at the transmitter output. The net power delivered to the TL is the
end result of the wave reflection and addition process, it is not the
beginning of it. The re-reflected power "adds" to the transmitter's actual
forward source power not the net power the transmitter "delivers". This is
where the concepts start to mixed and used incorrectly.
> Forward and reflected indicated powers should be
> the same at both ends of the transmission line. With loss in the line,
> the difference in net power indications at opposite ends of the
> transmission line is the power lost in the line.
>
> Best regards, Richard Harrison, KB5WZI
>
Steve, VE9SRB
VE9SRB
"Scott M. McClements" <yet...@grove.iup.edu> wrote in message
news:3A1433B7...@grove.iup.edu...
> I have a few questions about reflected and antenna tuners. First off,
> where does reflected power (SWR) actually go? Is it eventually radiated
> from the antenna, dissipated by PA transistors or is it more complex
> than that?
>
> Next, if you use an antenna tuner, this will improve how well your
> antenna "couples" with the feedline, correct? It can actually improve
> how much power is put into the feedline? However, the mismatch still
> occurs at the antenna feedpoint and the percentage of reflected power is
> still constant? With the antenna tuner in place, does this effect how or
> where a reflected wave acts?
>
> I know this sounds troll'ish..but I am serious about what I am asking.
>
> Thanks,
>
> Scott McClements.
>
With the antenna tuner in-line between the antenna and the PA, none of the
reflected power will arrive back at the PA, so this is not an issue.
Without getting into any physics, the SWR of the antenna does not change and
as a result, there will still be reflected power on the line between the
antenna and the tuner. With the tuner matched, there will be no reflected
power between the tuner and the PA.
The reflected power from the antenna arrives at the tuner output where a
portion of it is re-reflected back towards the antenna. As a result, there
is more forward power in the line between the tuner and the antenna than in
the line between the tuner and the PA. Therefore, more total power is
delivered to the antenna for radiation than in the case where no tuner is
used. The total power delivered to the antenna can never exceed the power
delivered to the tuner. At the same time, since there is more forward and
reflected power in the line than in the case where no tuner is used, the
power dissipation in the transmission line increases. There will also be a
small loss of power in the tuner.
The transmitter does not put more power into the feedline. The power that
is reflected from the antenna is just now distributed differently in the
system and more power is radiated by the antenna.
Ultimately, the reflected power from the antenna goes to three places:
1) A portion of it is re-delivered to the antenna.
2) A portion of it is lost in the transmission line.
3) A portion of it is returned to the tuner (this aspect of the discussion
is complex and is beyond the scope of this response).
Steve
Richard Clark
wrote:
>I'm not sure why. I think we all agree that there may be some level of
>re-reflection at the transmitter output and that this re-reflection adds to
>the transmitter's forward source power to become the total forward traveling
>power that we measure with a wattmeter.
Hi Steve, All,
When the condition described above exists it is an acknowledgement
that there is a measurement plane between two planes of reflection.
Under such circumstances the Mismatch Uncertainty prevails and barring
knowledge of the phase of the components of the waves, any claims to
even poor accuracy of power determination are at best a dream.
A 2:1 mismatched source facing a 2:1 mismatched load will present an
uncertainty in power determination of 30% minimum - IN ADDITION TO ALL
OTHER INACCURACIES.
As such, further discussion of voltages and currents listed out to 2
decimal places of resolution are interesting, but hardly real.
VE9SRB
>
> As such, further discussion of voltages and currents listed out to 2
> decimal places of resolution are interesting, but hardly real.
>
> 73's
> Richard Clark, KB7QHC
Hi Richard:
The discussion were only meant to illustrate the concepts, which is where
the real issues appear to be. As for the 2 decimal places, I only present
them to avoid rounding errors if anyone wanted to verify the calculations.
Whether the conditons described are real or not is secondary to the concepts
discussed. Sorry for the trouble.
Steve
William E. Sabin
VE9SRB wrote:
> If you check my post again you will see that I made the following statement
> :
>
> "130 watts of power has an associated voltage magnitude of 80.623 V and a
> current magnitude of 1.62 A."
I checked your original post again and it said 0.62 A. I corrected it in my
reply to that post. Much ado about nothing, eh?
My recent replies to Cecil Moore and Bob Haviland have further opinions on this
subject that I believe are OK.
Let's forget about waves and transmission lines for a moment, and connect a
resistor directly to the output of the directional wattmeter. We will look at
the *circuit*.
The directional wattmeter is an impedance bridge that is calibrated in watts for
a Z0 ohm system. If the load is Z0 ohms the reflected port reads 0.0 W and the
forward scale reads 100 watts. The bridge is *balanced*. The delivered power is
100 watts.
Suppose the reflected port reads 130watts and the forward port reads 30 watts.
This means that there is an impedance difference between the load and the Z0
ohms that the bridge was designed for. In other words the readings of the
meters are a somewhat indirect indication of the degree of the Z0 mismatch.
Consider a solid state broadband amplifier. If we want to, we might be able to
deliver 100 watts to this resistor anyway. Crank up the amplifier until the
difference between forward reading and reflected reading is 100 watts. Keep in
mind that we are not talking about waves, but a *lumped* circuit.
The wattmeter makes two voltage measurements, one proportional to load current
and one proportional to load voltage. These voltages are added at the forward
port and subtracted at the reflected port. If the bridge is unbalanced then the
load is not Z0, but we don't whether the load is less than Z0 or greater than
Z0.
Now consider the amplifier. If the load is less than Z0 and we try to force 100
watts into it the amplifier may overheat. If the load is greater than Z0 and we
try to force 100 watts, the transistors may be zapped by excessive
collector/drain voltage. This has zero to do with waves, it has to do with the
load impedance that the amplifier sees.
What I am suggesting is that the concepts of forward power and reflected power
are really nothing more than a way of describing the degree of Z0 mismatch. The
actual power delivered is the product of the voltage across, and the current
through, the load resistor. But this crazy directional wattmeter is calibrated
in terms of a so-called "forward" power and a "reflected" power. The whole
thing is just a game with words, *if* the load is a lumped impedance.
If we replace the lumped load with a Z0 coax and a distant load of some kind,
the readings do tell us two values that we call forward and reflected power
waves, from which we can find SWR. But the bottom line is that the bridge is in
reality responding to the actual value of the impedance that the line presents
to the wattmeter. This impedance is the result of the vector sum of forward and
reflected voltage and current waves. The power numbers are "derived" from the
voltage and current measurements.
We should keep in mind that voltage and current waves are the real entities that
are being measured and that power is a "derived" unit, and not a basic entity
that flows back and forth on the line. We have established this repeatedly in
the past. The wattmeter responds to voltage and current, and not "directly" to
watts. And that is why the question "where does the reflected power go?" is
moot. The power doesn't go anywhere, it is voltage and current that travel, and
power is merely a "value" that goes along for the ride.
And the crucial parameter for the broadband solid state amplifier is its complex
load impedance, which can cause mayhem.
Bill W0IYH
Richard Clark
wrote:
Hi Steve,
My point was not so much the reported values, but that those values
were meaningless even conceptually.
When the measurement plane falls between two planes of reflection,
even for modest mismatches (2:1) there is so much error injected into
the debate as to render it useless.
Also, this is not a problem of Trouble, per sé, but rather a learning
opportunity as to the nature of Power determination and sources of
error. Conceptually every discussion point attempted here would be
well served if all such reportings of current and voltage were in
phasor form.
There is no mandate that this argument be devoid of the data necessary
to conduct the debate. Relying on simple SWR reporting fails in this
regard, but the scenario may be recovered by simply (and arbitrarily)
introducing the phase back into the discussion.
William E. Sabin
William E. Sabin wrote:
> VE9SRB wrote:
>
> > If you check my post again you will see that I made the following statement
> > :
> >
> > "130 watts of power has an associated voltage magnitude of 80.623 V and a
> > current magnitude of 1.62 A."
>
> I checked your original post again and it said 0.62 A. I corrected it in my
> reply to that post. Much ado about nothing, eh?
>
> My recent replies to Cecil Moore and Bob Haviland have further opinions on this
> subject that I believe are OK.
>
> Let's forget about waves and transmission lines for a moment, and connect a
> resistor directly to the output of the directional wattmeter. We will look at
> the *circuit*.
>
> The directional wattmeter is an impedance bridge that is calibrated in watts for
> a Z0 ohm system. If the load is Z0 ohms the reflected port reads 0.0 W and the
> forward scale reads 100 watts. The bridge is *balanced*. The delivered power is
> 100 watts.
>
> Suppose the reflected port reads 130watts and the forward port reads 30 watts.
I mean forward 130 and reflected 30.
VE9SRB
>
>
> VE9SRB wrote:
>
> > If you check my post again you will see that I made the following
statement
> > :
> >
> > "130 watts of power has an associated voltage magnitude of 80.623 V and
a
> > current magnitude of 1.62 A."
>
> I checked your original post again and it said 0.62 A. I corrected it in
my
> reply to that post. Much ado about nothing, eh?
>
As usual!! Bill, you should have known what I meant and just ignored what I
wrote.
> My recent replies to Cecil Moore and Bob Haviland have further opinions on
this
> subject that I believe are OK.
>
I looked at them and they seem fine to me.
> Let's forget about waves and transmission lines for a moment, and connect
a
> resistor directly to the output of the directional wattmeter. We will look
at
> the *circuit*.
>
> The directional wattmeter is an impedance bridge that is calibrated in
watts for
> a Z0 ohm system. If the load is Z0 ohms the reflected port reads 0.0 W
and the
> forward scale reads 100 watts. The bridge is *balanced*. The delivered
power is
> 100 watts.
>
Agreed.
> Suppose the reflected port reads 30watts and the forward port reads 130
watts.
I would agree that the real parameters to consider in this discussion are
the traveling voltage and current. They are the quantities that must be
considered to understand wave behavior and the relationships between the
forward and reflected waves in the TL. In any measurement or analysis,
power is always the derived quantity. I would like to continue the
discussion as it relates to the complex "load impedance" and your comments
above:
"If the load is less than Z0 and we try to force 100 watts into it the
amplifier may overheat. If the load is greater than Z0 and we try to force
100 watts, the transistors may be zapped by excessive collector/drain
voltage. This has zero to do with waves, it has to do with the load
impedance that the amplifier sees."
In the transmission line problem, the load impedance the amplifier "sees"
has everything to do with the waves traveling in the TL. The only impedance
that the amplifier ever "sees" is the transmission line characteristic
impedance ZO. It never really ever "sees" any other physical impedance.
From the perspective of your discussion above, just like power, the steady
state load or line impedance at the amplifier output is a derived quantity
and it too is a function of the traveling voltages and currents. Consider
first that the amplifier drives the TL with a forward source voltage and
current that is entirely a function of ZO. At the amplifier output, there
will also be a rearward traveling reflected voltage and current and, if the
amplifier output impedance is not equal to ZO, there will also be a forward
traveling re-reflected voltage and current. The total voltage developed at
the amplifier output is equal to Vsource + Vreflected + Vre-reflected. The
total current developed at the amplifier output is equal to Isource -
Ireflected + Ire-reflected (the + and - sign with current is a function of
the math convention selected for the analysis). The effective load or line
impedance is simply derived from Vtotal / Itotal. The steady state power
delivered to this effective impedance is in fact the steady state power
delivered to the TL and is found from Pdel = |Vtotal| |Itotal| cos (theta),
where theta is the phase angle between Vtotal and Itotal. We also know that
in all cases, Vforward total is equal to Vsource + Vre-reflected.
From the perspective of the internal components in the amplifier, there are
two "sources" in the system. One is the internal amplifier forward driving
source and the other is the rearward driving source created from the
reflections that arrive at the amplifier output. In all cases, except where
the amplifier has a purely reactive output impedance, some level of voltage
and current travel rearward into the amplifier. The level of voltage
delivered rearward into the amplifier is equal to Vreflected +
Vre-reflected. The total steady state voltages developed within the
amplifier are the combination of the forward voltage developed within the
amplifier and the rearward voltage developed within the amplifier. Since
the forward and rearward voltages are traveling in opposite directions, the
total voltage developed in any one series component is equal to Vforward -
Vrearward. The total voltage developed in anyone parallel component is
equal to Vforward + Vrearward.
The total power dissipation occurring within the amplifier is a function of
both the amplifier's forward source and the wave reflections within the
system. If the wave reflections and re-reflections are such that the
effective load impedance is less than ZO, Vforward and Vrearward typically
combine to increase the total voltages within the amplifier such that more
power is dissipated in the amplifier relative to the situation where the
amplifier operates into a matched load.
The load impedance "seen" by the amplifier is a direct function of the
forward and reflected waves within the transmission line. Given that the
internal power dissipation within the amplifier is a function of this load
impedance, it must follow that the internal power dissipation is a function
of the reflected waves traveling within the system. For this reason,
reflected waves are quite real and are significant as to how the voltages
and currents develop within any system component. They cannot be ignored as
if they don't really exist.
Steve VE9SRB
William E. Sabin
VE9SRB wrote:
> The only impedance
> that the amplifier ever "sees" is the transmission line characteristic
> impedance ZO. It never really ever "sees" any other physical impedance.
This is not true. The transmission line equations definitely predict very
accurately the input admittance (conductance and susceptance) of a transmission
line that is connected to some arbitrary load impedance. The real power that is
delivered to that admittance from the amplifier is the voltage squared times the
input conductance. [This power, minus line loss, is the load power. Extremely
easy to calculate.] This complex admittance is what the amplifier sees. The
value of this admittance is determined by the composite voltage (forward and
reflected) and by the composite current (forward and reflected).
The output admittance of the amplifier, which is essentially a lossless dynamic
resistance, should be considered to be part of the amplifier, and not of the
coax input admittance. Also, re-reflections do not have any influence on the
input admittance of the coax.
The output connector of the amplifier is a "reference plane", just like in a
network analyzer. To the left of this plane is the generator, a current source
in parallel with a dynamic resistance (this is slightly idealized, of course).
To the right of the plane is a coax input admittance that is due entirely to the
vector sum of forward and reflected voltages and forward and reflected currents.
This plane is a point of precise equilibrium between the voltage and current
arriving from the left and the voltage and current leaving to the right. At this
point of equilibrium, all so-called re-reflections are reconciled, and as a
result they do not play any further role. For example, re-reflections do not
influence coax input admittance.
Re-reflections influence the voltage and current levels into the coax (but not
the admittance), but only if there is a reflection. No reflection, no
re-reflection. From a circuit standpoint, the bottom line on re-reflection is
that the output admittance of the amplifier is in parallel with the actual load
admittance. If the load is not Z0 then the dynamic output admittance produces
an effect that we have been describing by the so-called "re-reflection". This
is an example of how a very simple circuit idea gets screwed up with a lot of
wave mechanical claptrap.
I guess I must be pretty dense, but this wave-chasing drives me nuts. The
"circuit" approach in the amplifier and the transmission line equations on the
coax (derived directly from Maxwell's equations) are so elegant and
straightforward and correct, that I really do get bored with the confusions of
chasing waves back and forth. Once we have mastered the basic ideas of how wave
mechanics works, its time to move on.
Back in the old days when complex computations by hand were difficult, we needed
these mental intuition-aids. Nowadays, with personal computers, the Smith chart
and the highly perfected mathematical equations and simulations, I keep
wondering why we don't move into the new millenium.
Bill W0IYH
Reg Edwards
> I guess I must be pretty dense, but this wave-chasing
drives me nuts.
Bill, you are wrong. Your IQ is at least average. Probably
appreciably higher. It is you who have retained sanity. It
is only nutters who chase imaginary power waves and
reflections.
For them, with a vested interest in waves and reflections,
to abolish waves means there is nothing left to waffle and
bafflegab about.
Listen to the silence . . . . . .
----
Reg.
Jeff Anderson
wrote:
Simply illustrating a small nit...
Every two seconds (when the next "wave" of reflections arrives from
the antenna end of the xmission line), you'll have to readjust your
tuner for a Z0 match until, many seconds later, you finally reach
steady state.
Richard Harrison
"And like you, I would have liked to seen a stronger mention of feedline
loss."
A perfect r-f transmission line has uniformly distributed series
inductance and shunt capacitance. These reactive characteristics store
energy per unit length. When r-f sets out on its journey through the
line it is slightly retarded in velocity by the time needed to get
current flowing through the line`s series inductance and to get voltage
across the line`s shunt capacitance charged up to the applied voltage.
It takes more time for r-f to flow through a transmission line than
through free space.
There is energy stored in the line. When a transmitter stops, r-f
continues flowing from the transmission line for the short while
required to empty the line.
A perfect transmission line has no resistance in its conductors and no
leakage conductance. Paradoxically, the ideal line having no series
resistance and no shunt conductance has a current flow which is exactly
in phase with the voltage across the transmission line.
Any loss in the line produces some phase shift from the pure resistive
in-phase relation of an ideal line. Nothing is perfect, but most
transmission lines are good enough for practical purposes. The
characteristic impedance (surge impedance) , or Zo is a resistance, but
not a resistor which consumes power.
Zo is called a surge impedance because any burst of energy applied to
the line is forced by the line`s construction to conform to the voltage
to current ratio dicted by that construstion. If we put 100 volts across
a Zo of 50 ohms, 2 amps flows, unless and until an energy reflection
comes back to alter the apparent impedance of the line. Obviously, if a
line were infinitely long, no reflection could return.
If a real resistor equal to the Zo is placed across the far end of a
transmission line, all energy is accepted from the line. So, none
returns as a reflection to produce an impedance on the line other than
Zo. If a reactance, or resistance other than Zo, terminates the
transmission line it can`t absorb all the energy the transmission line
is able to povide, so some or all of the traveling energy is reflected.
The cyclical energy traveling toward the load, and the part reflected,
produce an SWR pattern on the transmission line, given the mismatch and
enough line to show a pattern. The varying phase in a long enough line
produces a repeating pattern of a wide range of complex impedances as
indicated by SWR.
High impedance is shown by voltage maxima. Low impedance is shown by
current maxima. These are 90 degrees apart in a line.
A circle diagram or Smith chart can be used to determine the impedance
at any point on a mismatched line. All the values fall on a VSWR circle
on the chart if line loss is negligible. We need to know the load
impedance and the characteristic impedance of the line to use a chart.
The rub is that the charts used to show impedance at any point are far
easier to use by assuming lossless transmission line, so that will be
the usual starting place.
klo...@toad.net
Reg,
Speaking only for myself, I'm one of those "nutters" that is simply
trying to understand how things work from the point of view of waves
(voltage and current waves, not power). This may not be necessary
when designing and building things, but it is educational, and that's
the point. Sorry you don't want to participate.
Bill N3WK
Sent via Deja.com http://www.deja.com/
Before you buy.
Dan Yemiola
It's over there on the floor right next to the bit-bucket for the computers
kill files.
: ) ; 0 ; )
This message uses 100% recycled electrons.
NO elementary particles were harmed in the production of this missive.
LOL! its good for you!
Any serious content in this message is purely co-incidental.
klo...@toad.net
Cecil <Cecil....@IEEE.org> wrote:
> A lossless antenna tuner provides a Z0-match which doesn't allow any
> reflected energy to reach the transmitter. The "SWR" between the
> transmitter and the antenna tuner is 1:1. There is simply nowhere else
> for the reflected > energy to go except back towards the antenna.
Let's see now, you're saying that the antenna tuner passes ALL power
going from the transmitter toward the antenna. It would have to since
you've already said that the SWR is 1:1. You're also saying that the
antenna tuner passes NO power going from the antenna back toward the
transmitter. In your description, the tuner has to be a non-reciprocal
device. It just doesn't make sense for a tuner to treat incident waves
differently only because they're going different directions.
> We know "what" happens from a scalar standpoint. "How" is really not
> important except to academics.
Not true. "How" is really important if you should try to explain what
happens to a novice who is trying to reconcile your description with
standard theory concepts like reciprocity.
Richard Harrison
"It just doesn`t make sense for a tuner to treat incident waves
differently only because they`re going different directions."
You`re on to something. The tuner is behaving as a bidirectional
transformer.
The tuner may transform whatever load it feeds, say 20 ohms or 200 ohms,
to 50 ohms so that standing waves (sorry Reg) are minimized on the coax
connecting the tuner and the radio. The radio should then behave as it
was designed.
Who knows what the radio output impedance is or needs to know what the
radio output impedance is? Or, do you think we have a conjugate match or
need a conjugate match with the radio? This might affect re-reflection.
Cecil
> My personal opinion is reflected power dos not exist! It is a figment that
> allows us to understand why the darn SWR bridge works.
Will you let me pulse a long unterminated transmission line with a
megawatt pulse and grab the wires to prove your personal opinion?
Of course reflected power exists. If you don't want to blow your
hands off, you can at least boil water with it.
--
http://www.mindspring.com/~w6rca
Cecil
> Simply illustrating a small nit...
> Every two seconds (when the next "wave" of reflections arrives from
> the antenna end of the xmission line), you'll have to readjust your
> tuner for a Z0 match until, many seconds later, you finally reach
> steady state.
Naaaahhhhhh Jeff, it's an autotuner. :-)
--
http://www.mindspring.com/~w6rca
Cecil
> Let's see now, you're saying that the antenna tuner passes ALL power
> going from the transmitter toward the antenna. It would have to since
> you've already said that the SWR is 1:1. You're also saying that the
> antenna tuner passes NO power going from the antenna back toward the
> transmitter. In your description, the tuner has to be a non-reciprocal
> differently only because they're going different directions.
The generated energy and the reflected energy indeed do see different
conditions because part of what the generated energy sees is the reflected
energy and part of what the reflected energy sees is the generated energy.
A transformer treats reflected energy differently from generated energy.
It steps one up and the other down. Why can't an antenna tuner treat
the two signals differently?
There is only generated power into the tuner input. There is forward power
out of and reflected power into the tuner output. Forward power out of the
tuner equals generated power into the tuner plus reflected power into the
tuner. Conservation of energy indeed does make sense. With a Z0-match in a
lossless system, there is simply no place else for the reflected energy to
go except back towards the antenna.
Dennis Ferguson
>In the transmission line problem, the load impedance the amplifier "sees"
>has everything to do with the waves traveling in the TL. The only impedance
>that the amplifier ever "sees" is the transmission line characteristic
>impedance ZO. It never really ever "sees" any other physical impedance.
>From the perspective of your discussion above, just like power, the steady
>state load or line impedance at the amplifier output is a derived quantity
>and it too is a function of the traveling voltages and currents.
Steve,
After Bill Sabin's response I am so keen to see the punch line for
this that I want to ask with an example I can understand.
Suppose I plug a transmission line terminated by something I'm unaware
of into my amplifier and turn it on. At the amplifier output terminal I
measure a voltage of 80V, a current of 1.6A and a phase angle of 36.9
degrees between them. I can immediately tell that the amplifier is
delivering about 102W to its load. If I know Z0 of the transmission line
is 50 ohms I might also be able to deduce that the VSWR is about 2:1, the
forward power is about 115W and the reflected power is about 13W, but
there doesn't seem to be any way for the amplifier to know this.
Suppose I now turn off the amplifier, unplug the transmission line and
replace it with lumped components with an impedance of 40+j30 ohms tied
directly to the amplifer output terminal. I turn the amplifier back on
and at the output terminal measure a voltage of 80V, a current of 1.6A
and a phase angle of 36.9 degrees between them. I can tell that the
amplifer is again delivering about 102W to the load.
Is there any possible way, from inside the amplifier, to tell any
difference at all between these two cases in the steady state? If the
amplifier "sees" Z0 in the first case isn't it seeing something that
doesn't exist in the second case? If I didn't tell you the transmission
line Z0 was 50 ohms in the first case, how could you have even determined
that this was what the amplifier was "seeing" from the only conditions
visible inside the amplifier (i.e. the voltage and current at its output
terminals)?
It seems to me that Bill Sabin must be precisely correct. The amplifier
can't "know" whether this is a transmission line problem or not from
anything that is measureable at its output. It can't determine Z0 from
anything it can see at the terminals in steady state, and if Z0 is
irrelevant to the operation of the amplifier then how could voltage
and current "waves" matter to it?
>From the perspective of the internal components in the amplifier, there are
>two "sources" in the system. One is the internal amplifier forward driving
>source and the other is the rearward driving source created from the
>reflections that arrive at the amplifier output. In all cases, except where
>the amplifier has a purely reactive output impedance, some level of voltage
>and current travel rearward into the amplifier.
The amplifier sees 80V and 1.6A out of phase by 36.9 degrees at its
output terminals. These are scalar (or phasor) voltage and current
values, they aren't "travelling" anywhere. It can't "see" Z0 in steady
state as far as I can tell, or distinguish a transmission line problem
from a circuit problem, so how could the distinction possibly matter to
its operation? Where's this other "source" when it is lumped components
instead of the transmission line?
I understand that amplifiers will run at higher output currents and/or
voltages higher when delivering power to loads other than their nominal
design output impedance, and that this may cause them to run less
efficiently or even suffer damage. This is what Bill Sabin said. I'm
still willing to believe there is something more to know about this
but I'm really not getting what it is.
Thanks,
Dennis Ferguson
VE9SRB
>
>
> VE9SRB wrote:
>
> > The only impedance
> > that the amplifier ever "sees" is the transmission line characteristic
> > impedance ZO. It never really ever "sees" any other physical impedance.
>
Hi Bill:
Thank you for the comments.
I would like to respond to this from two perspectives. One is understanding
what happens at an antenna and the other is from the derivation of the
general transmission line equation, re: Johnson, since I know you have the
book to look at as I discuss it.
First as a general comment, in the steady state, the total voltage and
current developed across a physical load (ZL) connected directly to an
amplifier is exactly the same as that developed at the input to a
transmission line having an input impedance equal to ZL. However, in one
case we have a circuit problem and in the other we have a combined
circuit-transmission line problem. The issue is relating the
circuit-transmission line problem to the purely circuit problem.
Let's assume that we have an antenna connected at the end of a
one-wavelength lossless transmission line. Let's assume that there is a
steady state forward traveling voltage arriving at this antenna. What is
the total steady state voltage developed across the antenna? It is equal to
the vector sum of the steady state forward voltage and the steady state
reflected voltage. This is demanded by general transmission line theory.
At this point, we see that the total voltage developed at the end of the TL
is a direct function of the reflected voltage developed at the end of the
TL. Why would TL theory work any differently at the input end of the line?
In fact, it does not. At the input end of the transmission line, if the
amplifier output impedance is not equal to ZO, there will be some level of
voltage reflection (or re-reflection) which must be a component of the total
voltage developed at the line input as TL theory demands.
The problem that most people have is that they consider only the steady
state conditions and only understand that there is a forward and a reflected
voltage at the line input. The real issue is understanding how these
forward and reflected voltages develop.
In the example above, at the input to the line, the total voltage will be
the same as the total voltage at the antenna since the TL is lossless and
one-wavelength long. At the input to the transmission line, what voltage
components comprise this total voltage? Obviously, they are the steady
state forward and reflected voltages, which are exactly the same as those
which exist at the antenna. So, at least at this point, we should agree
that the total voltage at any point on the transmission line is equal to the
forward voltage + the reflected or rearward voltage. The input impedance at
the input to the transmission line is equal to the ratio of total steady
state voltage / total steady state current.
So now, the issue is what voltage components comprise the total steady state
voltage at the TL input? And, does the re-reflected voltage have any effect
on the TL input impedance? In order for the re-reflected voltage to have an
effect on the TL input impedance, it must be a component of the total
voltage developed at the input to the TL. My post of Friday, indicated that
the total voltage at the input to the TL is comprised of three components:
1) The forward source voltage which is a function of ZO, 2) the reflected
voltage and 3) the re-reflected voltage.
Let's look at the general transmission line equation as presented by
Johnson:
The equation is (4.23) on page 100: The total voltage at any point on the
line is given as follows:
E = A B
where A = Eg Zo / [ (Zo + Zg) (1 - rhoA rhoG e^-(2 gamma L) ]
and
where B = [ e^-(gamma x) + rhoA e^-(2 gamma L) e^+(gamma x) ]
Eg is the source voltage. rhoA and rhoG are the antenna and source
reflection coefficients, respectively.
The equation for total voltage at any point on the line is explicitly a
function of rhoG, which is the reflection coefficient of the source
indicating the level of re-reflected voltage is in fact a component of the
total voltage. The input impedance of the line is therefore a function of
the re-reflected voltage.
Consider the derivation of the equation further beginning on page 98 and the
three voltage components comprising the total are clearly evident. One is
the forward source voltage, a function of Zo. Vsource = Eg Zo / (Zo +
Zg). The other two components are the reflected and re-reflected voltages
as discussed on page 99.
To see all of this we need to look beyond our view of only the steady state.
The amplifer really only sees the Zo of the TL and this is the basis for the
true forward source voltage. The other voltages in the TL only develop as a
function of reflections and re-reflections on the line. No physical
impedances in the system change, only the total voltage and total current.
Steve VE9SRB
William E. Sabin
VE9SRB wrote:
> The problem that most people have is that they consider only the steady
> state conditions and only understand that there is a forward and a reflected
> voltage at the line input. The real issue is understanding how these
> forward and reflected voltages develop.
I know exactly how the steady state is developed, and I don't need (or want) any
"lectures" on this subject.
In the steady state the power delivered to a 10 ohm resistor at the output of
the amplifier, and the power delivered to the same 10 ohm resistor at the end of
a 1 wavelength coax are exactly the same. The amplifier looks out and sees
exactly the same impedance in both cases. That is the essence of Johnson's
equations.
And as for Johnson's Equation 4.23, I understand all of that discussion and
derivation as least as well, and (I'm beginning to think) maybe even better than
you do. I know that it is a steady state solution and that the final result is
exactly and precisely what I have been posting.
Believe it or not, I am an expert on Equation 4.23.
This "thing" is obviously turning into some kind of very unpleasant macho
pissing contest and dominance game, and I really don't want anything more to do
with it.
I stand by what I have been saying, and for me personally that is the *end* of
it.
Bill W0IYH
Fred Hambrecht
If it exists, where does it go?
"Cecil" <Cecil....@IEEE.org> was heard muttering
Roger Halstead
N833R World's Oldest Debonair? s# CD-2
"Fred Hambrecht" <Fr...@w4jle.com> wrote in message
news:8v924s$mb6$1...@iac5.navix.net...
>
> If it exists, where does it go?
I came in the middle of this and did not see the original, nor do I find it
on my server.
In plain terms, the power just keeps going back and forth with bits getting
radiated at the antenna each time it gets there and bits getting turned into
heat by resistive losses.
If you had a theoretical lossless line, all of the power would get
radiated... eventually, although some would be more than a bit late.
Starting with a lossless line and a conjugetae match *at* the transmitter,
lets say that 50% of the power is reflected at the antenna. 50 watts would
be radiated and 50 watts would be reflected back tot he transmitter, where
due to the cojnugetate match it would head back out again, in phase with,
and added to, the output power. We'd now have 150 watts on the line andwhen
getting to the antenna we'd find 75 watts radiated and 75 watts reflected
back.
If you do the math you will find 100 watts being pumped into the line and
100 watts radiated from the antenna. The feed line would have 200 watts and
they are real.
Unfortunately we don't have lossless lines, some and the mismatch at the
antenna would result in some of the reflected power probably ending up on
the outside of the shield which would radiate. Most of the power is lost in
the form of heat as it runs back anf forth.
But if the run is relativley short and of good quality and low loss, most of
the reflected power (assuming a match at the transmitter end) will end up
being radiated from the antenna.
However that's not all good as there will be many cycles (depending on
frequency)which are well delayed from the initial radiated power.
In CW and SSB these are not usually a problem, but with data and television
they can be real problems.
Now, if you were talking about reflected power and no tuning network (as I
understand it) which eliminates tube type equipment, then it depends on the
sign of the returning power. Then you can end up with either excessive
voltage, or current in the final network and transistors. The power is
again reflected, or most of it is, but the excess voltage, or current can
create a bit of mischief .
Roger (K8RI)
VE9SRB
I apolgize if you felt lectured and offended.
Steve
"William E. Sabin" <sab...@mwci.net> wrote in message
news:3A180BE3...@mwci.net...
>
>
> VE9SRB wrote:
>
> > The problem that most people have is that they consider only the steady
> > state conditions and only understand that there is a forward and a
reflected
> > voltage at the line input. The real issue is understanding how these
> > forward and reflected voltages develop.
>
> I know exactly how the steady state is developed, and I don't need (or
want) any
> "lectures" on this subject.
>
> In the steady state the power delivered to a 10 ohm resistor at the output
of
> the amplifier, and the power delivered to the same 10 ohm resistor at the
end of
Reg Edwards
> I stand by what I have been saying, and for me personally
that is the *end* of
> it.
Good for you Bill. Well, at least for the time being,
anyway.
;o) Reg.
William E. Sabin
VE9SRB wrote:
> That's fine Bill. I simply thought we were having a technical discussion.
> I apolgize if you felt lectured and offended.
>
> Steve
Thanks, Steve. I just feel that I have nothing more to talk about on this
particular topic. I am *really* weary of it and I am not an enthusiastic
journeyman-in-training wave mechanic.
Bill W0IYH
Cecil
> If it exists, where does it go?
I don't understand the question. The reflected energy stored in the
transmission line is dissipated/radiated *AFTER* the transmitter is
disconnected. I've already shown that 300 joules is stored in a
one second long transmission line when the SWR is 5.83:1 and the
transmitter is putting out 100 joules/second. There is a transient
condition when the transmitter is turned off. The reflected energy
is dissipated and radiated during that transient time.
--
http://www.mindspring.com/~w6rca
David J. Windisch
totally-reflecting sphere. What, when, and how many times do you hear ..
it, or them .. ?
K3BHJ
Richard Harrison
"Clap your hands once when you are in the center of a 10,000 ft radius
totally reflecting sphere."
You will hear echos repeat until the clapper soaks up so much sound that
it`s no longer perceptible. Sound velocity depends on several variables
and we don`t know how absorbant the clapper is.
Cecil
> All a wave component sees are the components of the tuner. What you see
> when you perform a measurement is the sum of the generated and the
> reflected components. Are you saying an incident wave component behaves
> differently based on the conditions at the incidence point?
If a Z0-match exists, the reflections at the Z0-point will be exactly
canceled by complimentary reflections arriving from the load. There
will be no reflections on the coax to the transmitter. The reflected
energy has nowhere to go except back toward the antenna.
--
http://www.mindspring.com/~w6rca
klo...@toad.net
Cecil <Cecil....@IEEE.org> wrote:
> The generated energy and the reflected energy indeed do see different
reflected
> energy and part of what the reflected energy sees is the generated
energy.
All a wave component sees are the components of the tuner. What you see
when you perform a measurement is the sum of the generated and the
reflected components. Are you saying an incident wave component behaves
differently based on the conditions at the incidence point?
> A transformer treats reflected energy differently from generated
energy.
> It steps one up and the other down. Why can't an antenna tuner treat
> the two signals differently?
How does the tuner or the transformer "know" that a wave incident on it
is either coming from the generator or coming from the load? How does it
"know" to completely pass one and completely block the other?
BTW, the transformer is a reciprocal device. If Z12 = V1/I2 and Z21 =
V2/I1, then for a step-up/step-down, Z12 = Z21, which makes it
reciprocal.
Roger Halstead
N833R World's Oldest Debonair? s# CD-2
"Cecil" <Cecil....@IEEE.org> wrote in message
news:3A18AECC...@IEEE.org...
> klo...@toad.net wrote:
> > All a wave component sees are the components of the tuner. What you see
> > when you perform a measurement is the sum of the generated and the
> > reflected components. Are you saying an incident wave component behaves
> > differently based on the conditions at the incidence point?
Ahhh...Who are you (Klocko) answering?
Roger (K8RI)
VE9SRB
"Dennis Ferguson" <den...@wawa.juniper.net> wrote in message
news:8v7iq3$27v5$1...@heap.juniper.net...
> VE9SRB <srb...@worldnet.att.net> wrote:
> >In the transmission line problem, the load impedance the amplifier "sees"
> >has everything to do with the waves traveling in the TL. The only
impedance
> >that the amplifier ever "sees" is the transmission line characteristic
> >impedance ZO. It never really ever "sees" any other physical impedance.
> >From the perspective of your discussion above, just like power, the
steady
> >state load or line impedance at the amplifier output is a derived
quantity
> >and it too is a function of the traveling voltages and currents.
>
> Steve,
>
> After Bill Sabin's response I am so keen to see the punch line for
> this that I want to ask with an example I can understand.
>
Dennis:
I sent a response to this example via e-mail and it bounced. If you don't
mind, could you forward me a valid e-mail address to srb...@att.net and I
will resend my response.
Thanks,
Steve
Fred Hambrecht
asked where did the reflected power go. In the future, I will preface all
such messages with large letters stating "JOKE ALERT"
"Cecil" <Cecil....@IEEE.org> wrote in message
news:3A184A61...@IEEE.org...
Richard Harrison
"The generated energy and the reflected energy indeed do see different
For example:
1/4-wave line section (50 ohms =Zo).
Load = 25 ohms.
The line input will be = 100 ohms.
This is standard impedance inversion.
None of the reflected energy is returned to the source because the
source has higher voltage at every point in the cycle when the volts are
not zero.
We assume our source has an impedance of 100 ohms and is matched to the
load we provide with our 1/4-wave matching section.
The source`s energy is not reflected at its terminals because the
transformed load`s 100 ohms matches its input impedance.
It seems to me that the incident wave is delayed as it travels to the
load by 90 degrees. At the load the voltage experiences another 180
degree phase shift, but no phase shift occurs on reflection in the
current through the low load impedance. The reflected wave is delayed by
another 90 degrees as it travels back to the source.
The returning voltage has gone through 360 degrees and has been reduced
by the energy absorbed in the load. The reflected voltage fraction is:
(Rload - Zo) / (Rload + Zo).
The line is lossless.
At the source`s terminals, as Cecil implied, there is a different
condition from that at the load. It`s not just the impedance that`s
different. There is also an energy source at the source.
We have a source voltage on one side of an internal impedance and a
smaller reflected in-phase voltage at the output terminals. The current
which flows out the internal impedance is motivated by the difference
between the in-phase voltages on opposite sides of the internal
resistance of the source. This causes us to see 100 ohms as a load, and
not the Zo of the cable.
Cecil
> Why don't yawl put some numbers in the classical
> transmission line formulae and discover, before its too
> late, what really happens.
What really happens in a lossless steady-state purely sinusoidal
condition with no modulation? :-) Reg, we know reflections exist
because we can see them as ghosts on a TV screen.
--
http://www.mindspring.com/~w6rca
Bill Aycock
so he gets to be the 'most recent reference'- ie, the one I answer.
Funny thing happened when I was looking for a reference to an Amp I saw-
I picked up the January 1995 issue of CQ, which is the 'Collectors
Edition' covering 50 years of Amateur Radio. Gues what the lead article was?
"Reflected Power Stays in the Coax", by Warren B. Bruene, W5OLY.
This article has several VERY well worked out diagrams, showing which
way the stuff goes, and where the waves are-
Very helpful- I need to study it, in detail, and compare with the LARGE
stack of print out from the recent "debate" (?)
Bill- W4BSG
Cecil
> "Reflected Power Stays in the Coax", by Warren B. Bruene, W5OLY.
The scalar magnitude of reflected power is constant given a continuous
wave signal but the reflected power does not stay in the coax. Hang
a stub on a TV receiver and observe the ghosting which proves that
the reflected energy is continuously being refreshed. Reflected power
is dynamic and cannot be stored forever.
--
http://www.mindspring.com/~w6rca
Reg Edwards
Why don't yawl put some numbers in the classical
transmission line formulae and discover, before its too
late, what really happens.
Or, if you have no faith in formulae, just set up a signal
generator + line + attenuator + load + voltmeter and
discover the same things.
It's really very simple. Please don't drive yourselves into
nut-houses by haggling and quibbling about imaginary
quantities such as reflected watts and the time taken to
transfer from A to B. I cannot promise to pay visits to
your solitary confinement cells because of travelling
expenses. But I wish you all well.
--
***********************************
Regards from Reg, G4FGQ
For free radio modelling software go to:
http://www.btinternet.com/~g4fgq.regp
***********************************
Richard Harrison <richard...@webtv.net> wrote in
message
news:12569-3A...@storefull-111.iap.bryant.webtv.net...
Richard Harrison
"Reflected Power Stays in the Coax", by Warren B. Bruene, W5OLY."
Warren is about as infallible as I am. Reflected power loses the same
percentage to the load on each successive pass as on the first. So most
of the power is radiated. It simply can`t stay in the coax because there
is no place for it there.
EMF does no work until current flows. When current flows, volts times
amps is power in watts. Watts measures the rate of doing work.
In a transmission line, power is delivered to a load. The load for a
radio transmitter is its antenna`s radiation resistance.
Work is done by the energy supplied which is power times time.
One measure of energy is the watt second. Most coax accepts input at a
speed of about 120,000 miles per second, so a piece of coax about
120,000 miles long is needed to contain one second of energy at however
many watts are being supplied. The energy is multiplied by the number of
watts. Shorter lines contain proportionately less energy. But, the coax
can hold energy traveling in the reverse direction, in addition to
energy traveling in the forward direction. Total energy contained in the
line is the sum of forward and reverse energies.
So, where does the power go? It enters the coax and travels to the load
and is radiated except for some reflected power which circulates before
being radiated. Losses in the system take a small percentage toll on
each trip through the coax.
Richard Harrison, KB5WZI
scottmc...@my-deja.com
Man, I would like to reply to everybody with this response.
After a bunch of reading here and the "Reflections" series, I
feel a little bit better. Still, lots of uncertainty. After reading
through the reflection series and consulting the ARRL Antenna book,
I couldn't find "conjugate match" in the index of the book. I don't
understand how those articles can talk about it like its so common,
but I can't find it in the index.
First off, It obvious its hard to explain things when there are so
many variables. So, people must be clear about what the exact context
is when they're explaining things. And before you jump on somebody,
ask them to clarify.
Now, from what I understand....
Taking the solid-state rig, into 50 Ohm coax into a simple dipole
(no common mode currents ;) and no tuner example. The transmitter
transmits (on 80 meters), the signal goes out to the antenna. Part
is reflected back towards the transmitter.
Now, since there isn't a Z0 Match, does this mean *some* of the
reflected
power is absorbed by the final PA and some is re-reflected back towards
the load?
Its my understanding if there is a Z0 Match, the reflected wave sees the
reciprocal mismatch and is *completely* reflected back towards the
load.
First uncertainty, incident wave makes it back to the load...isn't
part of the signal old, and possibly out of phase (the wavelength of
the line) with the originally (already) transmitted signal? Wouldn't
this cause an echo or some cancellation/re-enforcement of the signal
overall?
The "Reflection" series seems to stress that if there is a Z0 match and
line loss is low, power eventually gets radiated so there isn't a
good reason to get SWR super low (from what I understood), especially if
the SWR is normally not supposed to be 1.1 (ground mounted 1/4 wave
vertical).
Next question, "conjugate match"..is this the same as "Z0" match?
What are some ways to "conjugate match"? (1/4 wave transformers?)
Next scenario, what if we take the above scenario and add a solid-state
amplifier in-line...still no tuner. Should there be a "Z0 match"
between
the radio and amp under all circumstances, or does a mismatch at the
load
in turn effect the Z0 match between the radio and amp?
I won't get email sent to the deja.com email address.
Thanks,
Scott McClements
In article <19645-3A...@storefull-111.iap.bryant.webtv.net>,
richard...@webtv.net (Richard Harrison) wrote:
> Scott McClements wrote:
> "where does reflected SWR actually go?"
>
> Ideally, to a ratio of one.
>
> SWR is an interference pattern created by forward and reflected
> traveling waves on a transmission line. Voltage and current create
their
> own similar patterns, but it is easy to measure and plot the VSWR
using
> a slotted (trough) line.
>
> In a lossless system, all reflected power is finally radiated. None is
> absorbed back in the transmitter or spent in transmission line or
> matching networks. SWR indicates temporary storage of circulating
energy
> in the transmission line. Some energy is wasted in all the actual
> equipment used.
>
> Typically, losses are small and often less than many people think when
> they see the SWR number.
>
> Next, an antenna tuner should let you transform whatever load
impedance
> appears to your desired load impedance. This is often the impedance of
> the coax you are using. The impedance desired is frequently 50 ohms
> which is the impedance many transmitters are specified for as a load
at
> which the transmitter will perform as expected. Other impedances are
> sometimes specified.
>
> The tuner eliminates a point of reflection between itself and the coax
> by providing a match. The coax then has an SWR of one.
>
> Best regards, Richard Harrison, KB5WZI
>
>
Cecil
> After reading
> through the reflection series and consulting the ARRL Antenna book,
> I couldn't find "conjugate match" in the index of the book. I don't
> understand how those articles can talk about it like its so common,
> but I can't find it in the index.
There's a little history involved there. Go back and read an ARRL
Antenna Book from the mid-1980s and you will find 'conjugate match'
mentioned quite often. A few years ago, the ARRL purged 'conjugate
match' from their publications allegedly because they believe a
conjugate match rarely exists in reality.
> Now, since there isn't a Z0 Match, does this mean *some* of the reflected
> power is absorbed by the final PA and some is re-reflected back towards
> the load?
That's probably what happens. Depending on the phase of the reflections,
over-current or over-voltage may result. In any case, the final PA is not
seeing its designed-for impedance. It's possible for the reflected energy
to make it all the way back to the DC power supply or battery.
However, conventional "wisdom" will tell you that reflected power absorbed
by the transmitter wasn't ever generated in the first place.
> Its my understanding if there is a Z0 Match, the reflected wave sees the
> reciprocal mismatch and is *completely* reflected back towards the
> load.
Reflected waves are technically not re-reflected at a Z0-match. What
actually happens is two complimentary sets of reflections perfectly
cancel each other through an interference pattern. The "re-reflection"
is a *scalar* process involving the conservation of energy. The canceled
reflected energy simply has no other place to go except back towards the
antenna (assuming a Z0-match).
> First uncertainty, incident wave makes it back to the load...isn't
> part of the signal old, and possibly out of phase (the wavelength of
> the line) with the originally (already) transmitted signal? Wouldn't
> this cause an echo or some cancellation/re-enforcement of the signal
> overall?
At a perfect Z0-match, everything is perfectly in phase or perfectly
180 degrees out of phase. And yes, an echo (ghosting) can and does
occur where modulation is involved.
> The "Reflection" series seems to stress that if there is a Z0 match and
> line loss is low, power eventually gets radiated so there isn't a
> good reason to get SWR super low (from what I understood), especially if
> the SWR is normally not supposed to be 1.1 (ground mounted 1/4 wave
> vertical).
The reason for getting the SWR low is feedline loss. Ordinary coax is
much more lossy than open-wire line.
> Next question, "conjugate match"..is this the same as "Z0" match?
Nope, each is possible without the other. IMO, a Z0-match is all the
average ham needs to worry about since he rarely knows the internal
impedance of his transmitter.
--
http://www.mindspring.com/~w6rca
Bill Aycock
These help, especially put in a one-to-one context, as you did.
Comment about the indexing system used in ARRL publications- "they do
not understand the concept"- for example- try to find the word "Coax" as
an entry in any ARRL handbook before about four years ago- and "Coax" is
MUCH more commonly used than "Conjugate match".
Bill
Richard Clark
>Hello,
>
>Man, I would like to reply to everybody with this response.
>After a bunch of reading here and the "Reflections" series, I
>feel a little bit better. Still, lots of uncertainty. After reading
> through the reflection series and consulting the ARRL Antenna book,
>I couldn't find "conjugate match" in the index of the book. I don't
>understand how those articles can talk about it like its so common,
>but I can't find it in the index.
>
See my re-post below for complete and authoritative definitions.
>
>First off, It obvious its hard to explain things when there are so
>many variables. So, people must be clear about what the exact context
>is when they're explaining things. And before you jump on somebody,
>ask them to clarify.
>
>Now, from what I understand....
>
>Taking the solid-state rig, into 50 Ohm coax into a simple dipole
>(no common mode currents ;) and no tuner example. The transmitter
>transmits (on 80 meters), the signal goes out to the antenna. Part
>is reflected back towards the transmitter.
>
>Now, since there isn't a Z0 Match, does this mean *some* of the
>reflected
>power is absorbed by the final PA and some is re-reflected back towards
>the load?
>
By and large, yes, unless you are running at the power specified by
the manufacturer where the output Z is specified (50 Ohms). If you
have any doubt as of this specification, simple refer to the
characteristics listed in your tech manual for your rig.
Even here, there can be some variation from 50 Ohms. My own rigs are
specified as 50 Ohms, but under less than full power they exhibit Z's
that are not 50 Ohms and yet do not fall below 35 Ohms nor rise above
70 Ohms.
You probably can assume fairly similar characteristics given that
finals design for transistor rigs has not changed much in the last 20
years.
>
>Its my understanding if there is a Z0 Match, the reflected wave sees the
>reciprocal mismatch and is *completely* reflected back towards the
>load.
>
>First uncertainty, incident wave makes it back to the load...isn't
>part of the signal old, and possibly out of phase (the wavelength of
> the line) with the originally (already) transmitted signal? Wouldn't
>this cause an echo or some cancellation/re-enforcement of the signal
>overall?
>
Yes, but do not dwell on this greatly as SWR does not contain phase
information and trying to pursue this thought without all the
information will only frustrate.
>
>The "Reflection" series seems to stress that if there is a Z0 match and
>line loss is low, power eventually gets radiated so there isn't a
>good reason to get SWR super low (from what I understood), especially if
>the SWR is normally not supposed to be 1.1 (ground mounted 1/4 wave
>vertical).
>
You learned that lesson well.
>
>Next question, "conjugate match"..is this the same as "Z0" match?
>What are some ways to "conjugate match"? (1/4 wave transformers?)
>
Simply put NO, but with the proviso - sometimes. Both have specific
meanings - see my repost of this topic that follows.
>
>Next scenario, what if we take the above scenario and add a solid-state
>amplifier in-line...still no tuner. Should there be a "Z0 match"
>between
>the radio and amp under all circumstances, or does a mismatch at the
>load
>in turn effect the Z0 match between the radio and amp?
>
>I won't get email sent to the deja.com email address.
>
>Thanks,
>
>Scott McClements
>
Hi All,
During my tenure in Metrology, the standard text on the subjects
Conjugate Match and Z0 Match were formalized. Insofar as your access
to the NBS standard methods, these definitions may be found in
"Microwave Theory and Applications," Stephen F. Adam; or they
may be found in Hewlett-Packard's Application Note 56, or they may be
found in the original works by R. W. Beatty published in IEEE
Transactions in the early 60's.
The following terms are those understood by Metrologists at NBS and
Primary/Secondary Standards Labs:
Substitution Loss:
A general term referring to the change in power absorbed by a load
when first one waveguide junction (these words are used in the
broadest sense) and then another (such as an attenuator) is used to
connect a source to the load.
Transducer Loss:
What the substitution loss becomes when the first waveguide junction
is a perfect transducer and initially the maximum available power is
delivered to the load.
Insertion Loss:
What the substitution loss becomes when the first waveguide junction
is a perfect (lossless and phase-shift-less) connector, so that in
general there is some initial mismatch loss.
Attenuation:
What any of the above losses become when the source and load both have
Z0 impedance. Under these conditions, what is measured on an
attenuator is a property of the attenuator alone, so that this is the
ideal system in which to make measurements.
Residual Attenuation:
The minimum attenuations of a variable attenuator. What is measured
when a variable attenuator is set to it minimum position and inserted
into an ideal system.
Incremental Attenuation:
The change in attenuation between minimum setting and any other
setting on a variable attenuator. Residual Attenuation and Incremental
Attenuation together make up Attenuation.
Conjugate Match:
The condition for maximum power absorption by a load, in which the
impedance seen looking toward the load at a point in a transmission
line is the complex conjugate of that seen looking toward the source.
Conjugate Mismatch:
The condition in the situation above in which the load impedance is
not the conjugate of the source impedance.
Conjugate Mismatch Loss:
The loss resulting from conjugate mismatch.
Z0 Match:
The condition in which the impedance seen looking into a transmission
line is equal to the characteristic impedance of the line.
Z0 Mismatch:
The condition in which the impedance seen looking into a transmission
line is not equal to the transmission line characteristic impedance
Z0. In general, conjugate match is a case of Z0 mismatch.
Z0 Mismatch Loss:
The loss resulting from a Z0 mismatch.
Conjugate Available Power:
Maximum available power.
Z0 Available Power:
The power a source will deliver to a Z0 load.
73's
Richard Clark, KB7QHC
eh...@bellatlantic.net
Bill Aycock wrote:
> Funny thing happened when I was looking for a reference to an Amp I saw-
> I picked up the January 1995 issue of CQ, which is the 'Collectors
> Edition' covering 50 years of Amateur Radio. Gues what the lead article was?
>
> "Reflected Power Stays in the Coax", by Warren B. Bruene, W5OLY.
So THAT'S why my coax keeps getting heavier and heavier!
Dick Carroll
>
> Bill Aycock wrote:
> "Reflected Power Stays in the Coax", by Warren B. Bruene, W5OLY."
>
> percentage to the load on each successive pass as on the first. So most
> of the power is radiated. It simply can`t stay in the coax because there
> is no place for it there.
I can't imagine *anyone* believing that. The powre is going to get out
of that coax one way or the other, if it's by only heat loss resulting
from the natural losses present. If it's to high a level for that, it'll
excape as light-when the coax burns. If the coax is lossless, it has no
REASON to stay within. Nuts.
Cecil
>
> Richard Harrison wrote:
> >
> > Bill Aycock wrote:
> > "Reflected Power Stays in the Coax", by Warren B. Bruene, W5OLY."
> >
> > Warren is about as infallible as I am. Reflected power loses the same
> > percentage to the load on each successive pass as on the first. So most
> > of the power is radiated. It simply can`t stay in the coax because there
> > is no place for it there.
>
> I can't imagine *anyone* believing that.
Well, to be fair, we should say that during steady-state conditions
with continuous wave signals, the magnitude of the reflected power
in the coax is constant. Perhaps that is what Bruene meant. For instance,
200 watts stays in the coax until the transmitter is shut off. It's not
the same energy but is the same *amount* of power.
--
http://www.mindspring.com/~w6rca
Bill Aycock
just the title. In the article, his point is that the reflected power
stays in the coax on the way back toward the source, but when
re-reflected, is the same as applied power, and is dissipated over a few
cycles, in the load. the energy stored in these 'bounces' in the coax,
is small at any time.
His illustrations are in terms of voltage and current, rather than
power, and he makes the point that the re-reflection is caused by the
input voltage, not the hardware. This causes the phase of the outward
reflection to be controlled by the source.
I wish a person with more qualifications that I have would read the
article, and comment.
Bill-W4BSG
Bill Nelson
VE9SRB <srb...@worldnet.att.net> wrote:
: In the transmission line problem, the load impedance the amplifier "sees"
: has everything to do with the waves traveling in the TL. The only
: impedance
: impedance ZO. It never really ever "sees" any other physical impedance.
I would disagree with this, or at least with your terminology.
The transmission line impedance, either actual or what the amplifier sees,
has nothing to do with the waves that travel in the transmission line. What
it will effect, is how much power is loaded from the amplifier into that
line.
By the same token, the impedance that the amplifer sees will only be Z0
if there is a perfect impedance match between the line and the load.
You may have meant/said that above - but it is unclear to me based on what
you wrote.
--
Bill Nelson (bi...@peak.org)
Keith Dysart
that there is no such THING as reflected power.
It seemed to me that any acceptable explanation of where 'reflected
power' goes should work:
a. at a frequency of 0 hertz.
b. with the transmission line open-circuited.
c. with the transmission line short-circuited.
d. with any generator. In particular, it should work if the
generator is replaced with:
d1. its Thevenin equivalent circuit.
d2. its Norton equivalent circuit.
Since these boundary conditions lead to fairly simple arithmetic,
it can all be done in your head.
Let us start with a transmission line (TL) of impedance 50 Ohms,
length t (the time for one way propagation along the line) and a
generator with an output impedance of 50 Ohms.
Leave the right end of the TL open-circuit.
Set the generator to 50 V (into 50 Ohms) and attach it to the left
end of the TL.
One Amp begins to flow into the TL.
This charge propagates down the TL creating a voltage step of 50 V.
To the right of the voltage step, no current is flowing and the
voltage is zero; to the left of the voltage step, 1 A is flowing
and the voltage is 50 V.
At time t, the voltage step reaches the end of the TL and the current
can flow no further since the TL is open-circuit.
But charge is still arriving so a voltage step from 50 to 100 V occurs
which propagates back along the line.
To the right of this step, the voltage is 100 V and no current is flowing;
while to the left of the step, the voltage is still 50 V and 1 A is still
flowing.
At time 2*t, this voltage step has propagated all the way back to the start
of the TL.
While the line was charging, the generator was providing 50V * 1A -> 50 Watts
into the line. This continued for 2 * t seconds.
Now the TL is fully charged to 100 V and no current is flowing anywhere in
the line.
The generator is no longer adding energy to the TL.
The energy delivered to charge the TL is stored in the capacitance of the TL.
Since the current is zero everywhere in the TL, there cannot be energy flow
anywhere in the TL so the power must be zero.
And yet believers in reflected power would like me to accept that there is
50 W of forward power and 50 W of reflected power for a net 0.
But I can easily see that there is no power flowing in this TL, anywhere.
It is only a charged capacitor.
As for the 50 forward and reverse, one could just as easily argue that it
is a megawatt forward and reverse. It yields the same result.
So I conclude there is no such thing as 'reflected power' because it does
not explain the 0 frequency situation and therefore the question 'Where
does the reflected power go?' is meaningless.
To carry the study a bit further we can ask what is happening in the
generator after time 2 * t.
If we assume a Thevenin circuit for the generator it is dissipating 0 W.
If we assume Norton, it is dissipating 200 W.
Dissipation in generators depends more on generator design than 'reflected
power'. Any argument that 'reflected power' is dissipated in the generator
can easily be shown to be false by substituting Norton for Thevenin or vice
versa.
For a short-circuited TL after 2 * t:
- the voltage will be 0 V everywhere
- the current will be 2 A
- this still leads to 0 W
- the energy is stored in the inductance of the TL
- Thevenin will dissipate 200 W
- Norton will dissipate 0 W
The so called 'reflected power' is the same but the dissipations are all
different.
Some might argue that it is the 0 frequency case that is special and that
'reflected power' makes sense for non-zero frequency.
Let us consider an open-circuited TL of several wavelength driven by a
matched generator.
After time 2 * t there is a standing wave on the line.
At the end of the line the current must be zero since it is open-circuit.
Every 0.5 wavelength back from the end the current is 0.
0.25 wavelengths from the end the voltage is 0.
Every 0.5 wavelength back from there, the voltage is 0.
We know that power is voltage times current.
Since every quarter wavelength back from the end, either the voltage or
the current is always 0, the power must be 0, so there is no energy
flowing across these points.
If there is no energy flowing, there can not be any 'reflected power'
flowing unless 'reflected power' is not actually a flow of energy and
therefore not a power.
And this is the answer. 'Reflected power' results from a computation
that yields a dimension of power but does not result in anything that
is real. This mathematical convenience does have value though, because
when it is combined with another computation yielding a non-existent
'forward power', one can compute SWR and net delivered power.
This is an example of analysis where the intermediate results have no
physical meaning but the final result is useful. Remembering the type
of circuit analysis (I can not remember the name) where all voltage
sources are set to zero and then one-by-one they are set to their
desired value and the resulting currents are computed. Later all the
computed currents are summed to find the real currents. While the
intermediate computations yield values with dimensions of current,
these currents do not physically exist. A simple example will suffice.
Consider two 10 VDC sources whose negatives are grounded and whose
positives are connected with a 10 Ohm resistor. Analysis yields 1 Amp
flowing clockwise and 1 Amp flowing counter-clockwise for a net of
0 A in the resistor. The final answer is correct; the intermediate
answers do not represent currents which are flowing.
To make this very clear, replace the 10 Ohm resistor with a zero Ohm
resistor. Result: infinite current clockwise, infinite counter-clockwise,
but the net is still correct at 0.
Never assume that intermediate results, no matter how useful to solving
the problem, represent real things.
...Keith
William E. Sabin
Keith Dysart wrote:
> Let us start with a transmission line (TL) of impedance 50 Ohms,
> length t (the time for one way propagation along the line) and a
> generator with an output impedance of 50 Ohms.
> Leave the right end of the TL open-circuit.
> Set the generator to 50 V (into 50 Ohms) and attach it to the left
> end of the TL.
The open-circuit generator volts is 100 V.
> At time t, the voltage step reaches the end of the TL and the current
> can flow no further since the TL is open-circuit.
> But charge is still arriving so a voltage step from 50 to 100 V occurs
> which propagates back along the line.
The return current is -1.0 A (note the change of polarity). The net current to
the right of the returning wave is 1.0 - 1.0 = 0.0 A. To the right of the return
wave the power is 50 W forward and 50 W reflected. This is what a directional
wattmeter would read. To the left of the return wave it reads 50 W forward and
0.0 W reflected.
> Now the TL is fully charged to 100 V and no current is flowing anywhere in
> the line.
> The generator is no longer adding energy to the TL.
> The energy delivered to charge the TL is stored in the capacitance of the TL.
> Since the current is zero everywhere in the TL, there cannot be energy flow
> anywhere in the TL so the power must be zero.
OK. The wattmeter reads 50 W forward and 50 W reflected, for a net of 50 - 50 =
0.0 W
> And yet believers in reflected power would like me to accept that > there is
> 50 W of forward power and 50 W of reflected power for a net 0.
> But I can easily see that there is no power flowing in this TL, anywhere.
> It is only a charged capacitor.
> As for the 50 forward and reverse, one could just as easily argue that it
> is a megawatt forward and reverse. It yields the same result.
The wattmeter in your example reads 50 W forward and 50 W refelected.
> So I conclude there is no such thing as 'reflected power'
It has been said several times (suggest you review) in this forum that because
of the way it is designed, the wattmeter (an impedance bridge) is in actuality
responding to the voltage *across* its output terminals and the current *into*
its output terminals. It does *not* respond *directly* to a power value. The
thing is that the meter scale is marked in terms of watts. The concept of
forward and reflected *power* is convenient. One reason is that these two values
can easily be used to calculate SWR. And when the return power reading is zero,
the true load power is what is the forward meter reads. And the actual load power
is always forward minus reflected. When we talk about forward and reflected power
we are actually thinking in terms of what a directional wattmeter scale reads.
But the real action is in the forward and reflected voltage and current waves.
> 'Where
> does the reflected power go?' is meaningless.
OK. That has been said several times also.
> To carry the study a bit further we can ask what is happening in the
> generator after time 2 * t.
> If we assume a Thevenin circuit for the generator it is dissipating 0 W.
> If we assume Norton, it is dissipating 200 W.
Not OK. The conversion from Thevenin to Norton is not that simple. In the steady
state the Thevenin current is 0.0 because the Thevenin impedance is infinite
(generator plus line). The Norton current must therefore also be 0.0 in the
steady state. The condition "no current flows" must always be satisfied when the
load is an open-circuit. Stand at the end of the line and look back. We must see
0.0 A in both cases. In other words, Norton and Thevenin *must* produce
identical results or they are not equivalent.
Bill W0IYH
Cecil
> So I conclude there is no such thing as 'reflected power' because it does
> not explain the 0 frequency situation and therefore the question 'Where
> does the reflected power go?' is meaningless.
Does a pendulum clock keep time when the pendulum is swinging? How about
when the pendulum is not swinging? An AC system is simply different, in
many ways, from a DC system.
Reflected power is a *dynamic* exchange of energy. It simply doesn't
exist under steady-state DC conditions. Note that reactance is either
zero (short) or infinite (open) at 0 frequency. Does that mean there
is no such thing as reactance?
We know that reflected power exists because it causes additional losses
in the transmission line over and above the matched line loss.
--
http://www.mindspring.com/~w6rca
William E. Sabin
William E. Sabin wrote:
> Keith Dysart wrote:
>
> > Let us start with a transmission line (TL) of impedance 50 Ohms,
> > length t (the time for one way propagation along the line) and a
> > generator with an output impedance of 50 Ohms.
> > Leave the right end of the TL open-circuit.
> > Set the generator to 50 V (into 50 Ohms) and attach it to the left
> > end of the TL.
>
> The open-circuit generator volts is 100 V.
>
> > At time t, the voltage step reaches the end of the TL and the current
> > can flow no further since the TL is open-circuit.
> > But charge is still arriving so a voltage step from 50 to 100 V occurs
> > which propagates back along the line.
>
> The return current is -1.0 A (note the change of polarity). The net current to
> the right of the returning wave is 1.0 - 1.0 = 0.0 A. To the right of the return
> wave the power is 50 W forward and 50 W reflected. This is what a directional
> wattmeter would read. To the left of the return wave it reads 50 W forward and
> 0.0 W reflected.
>
> > Now the TL is fully charged to 100 V and no current is flowing anywhere in
> > the line.
> > The generator is no longer adding energy to the TL.
> > The energy delivered to charge the TL is stored in the capacitance of the TL.
> > Since the current is zero everywhere in the TL, there cannot be energy flow
> > anywhere in the TL so the power must be zero.
>
> OK. The wattmeter reads 50 W forward and 50 W reflected, for a net of 50 - 50 =
> 0.0 W
>
> > And yet believers in reflected power would like me to accept that > there is
>
> > 50 W of forward power and 50 W of reflected power for a net 0.
> > But I can easily see that there is no power flowing in this TL, anywhere.
> > It is only a charged capacitor.
> > As for the 50 forward and reverse, one could just as easily argue that it
> > is a megawatt forward and reverse. It yields the same result.
>
> The wattmeter in your example reads 50 W forward and 50 W refelected.
>
> > So I conclude there is no such thing as 'reflected power'
>
> It has been said several times (suggest you review) in this forum that because
> of the way it is designed, the wattmeter (an impedance bridge) is in actuality
> responding to the voltage *across* its output terminals and the current *into*
> its output terminals. It does *not* respond *directly* to a power value. The
> thing is that the meter scale is marked in terms of watts. The concept of
> forward and reflected *power* is convenient. One reason is that these two values
> can easily be used to calculate SWR. And when the return power reading is zero,
> the true load power is what is the forward meter reads. And the actual load power
> is always forward minus reflected. When we talk about forward and reflected power
> we are actually thinking in terms of what a directional wattmeter scale reads.
> But the real action is in the forward and reflected voltage and current waves.
>
> > 'Where
> > does the reflected power go?' is meaningless.
>
> OK. That has been said several times also.
>
> > To carry the study a bit further we can ask what is happening in the
> > generator after time 2 * t.
> > If we assume a Thevenin circuit for the generator it is dissipating 0 W.
> > If we assume Norton, it is dissipating 200 W.
>
> Not OK. The conversion from Thevenin to Norton is not that simple. In the steady
> state the Thevenin current is 0.0 because the Thevenin impedance is infinite
> (generator plus line). The Norton current must therefore also be 0.0 in the
> steady state. The condition "no current flows" must always be satisfied when the
> load is an open-circuit. Stand at the end of the line and look back. We must see
> 0.0 A in both cases. In other words, Norton and Thevenin *must* produce
> identical results or they are not equivalent.
>
> Bill W0IYH
There is something else that I overlooked. In an RF power amplifier the output
series resistance (Thevenin) or the shunt conductance (Norton) do not dissipate
power. They are almost entirely lossless, *dynamic resistances* that are due to a
*negative feedback* effect within the amplifier tubes/transistors. This idea has been
very well established, but is still having some acceptance problems for some folks.
The result is that in your example, the power dissipated in both the Thevenin and
Norton circuits are 0.0 watts. The current generator is 2.0 A and develops 100 V
across a 50 ohm *dynamic resistance*. In the Thevenin the 50 ohm generator *dynamic
resistance* dissipates 0.0 W and the output voltage is 100 V. Both circuits are
equivalent *inside* the circuit..
In ordinary situations of real resistors your comments about Thevenin and Norton do
illustrate a problem. In general, the equivalence can only be measured by the
results that they produce *beyond* the equivalent circuit. The insides of the
equivalent circuit may not produce the same actions, just as you said.
The rule is that we are not supposed to look inside the equivalent circuit, but only
look beyond the equivalent circuit, to see how the rest of the circuit is behaving.
Bill W0IYH
eh...@bellatlantic.net
is no such thing as reflected power.
Anyone who has ever done any TDR work knows that
to be false.
Peter Bertini
"Keith Dysart" <dys...@nortel.ca> wrote in message
news:8vis99$4f4$1...@bcarh8ab.ca.nortel.com...
> Allow me to provide the simple analysis that caused me to conclude
> that there is no such THING as reflected power.
Why do they keep putting reject loads on ferrite isolators??
Pete