Parallel coax
 

On 10/5/2015 4:17 AM, Ian Jackson wrote:
In message , rickman writes



You keep saying that the 1:1 match between the TX and the ATU prevents
any power from being sent to the TX which is not true. You are
confusing the power from the TX which is not reflected and the power
reflected from the antenna which passes through the ATU to the TX.

If the SWR meter between the TX output indicates a 1:1 SWR, then there
can be NO power travelling between the ATU input and the TX output - ie
there IS no reflected power. QED, surely?

If you ignore the losses in the ATU, all the power that the mismatched
antenna reflects, and that makes it back to the ATU output, MUST be
re-reflected by the ATU output impedance, and head off back towards the
antenna. This is because the reflected signal cannot heat up a lossless
ATU, and the SWR meter says it isn't coming back through the ATU. It
simply has nowhere to go except back down the coax.

You saying something is true or imagining a SWR reading is not the same
as understanding what is going on. What SWR reading are you imagining?
Can you explain this in terms of the circuit analysis? The ATU
consists of what circuit? The TX has some source impedance, what would
that be? I don't think you can design an ATU circuit that will isolate
the real source impedance of the TX from the reflected wave from the
antenna.

--

Rick

Parallel coax
 

rickman wrote:

On 10/4/2015 6:01 PM, Ian Jackson wrote:
In message , rickman writes
On 10/4/2015 3:00 PM, Ian Jackson wrote:
In message , rickman
writes
On 10/4/2015 5:51 AM, Jeff wrote:

Is that good? If the cable length is 1/4 wavelength (as it is
in the
info provided for this case) the reflected power is nearly 180
degrees
out of phase with the initial power at the antenna. I think
reducing
this through cable losses would not be so bad, or better to dump
it in
the ATU?

It is good in as much as some of the re-reflected power is radiated
(and
some re-re-reflected) since the ATU causes the phase of the
re-reflected
wave to be 'in-phase' at the antenna.

It is bad in as much as the reflected power suffers 2 times the
cable
loss, and dissipates that in heat, on each return trip, up and down
the
coax.

With a high VSWR at the antenna there will be many return trips
before
the re-reflected power drops to a negligible level.

No power is 'dumped' in the ATU; although there will be losses, but
that
is another story.

If not the ATU, then the transmitter. I'm sure not all of the
power is
reflected back from the ATU. Exactly what is the phase of the
reflected
power from the ATU? I haven't seen an actual circuit for the ATU in
question. For that matter, what is the phase of the power reflected
from the antenna? I'm pretty confident we are not looking at the
return
of the reflected wave in phase with the incident wave.


If the ATU is adjusted so that the Tx sees a 1:1 match then no
power is
reflected to the transmitter. A 1:1 match means that there is no power
reflected to the Tx by definition (and can be proved by measurement).

Fine, but not related to my question if that is what you were
responding to.


By conservation of energy then all of the power must be radiated or
lost
as heat, (mostly in the coax).

Uh, I'm not sure of that since there are other paths and I'm pretty
sure power can be sent back into the driver of the transmitter.


When the ATU gives a conjugate match, (ie the Tx sees a 1:1 vswr) the
phase of the re-reflected wave at the antenna is 0, ie it is in-phase
with the original forward wave and so any power that is not
re-re-reflected again adds to the power supplied to the antenna.

You seem to be mixing two things here. We have been talking about an
ATU at the transmitter. The feed line cable is between the ATU and
the antenna. You then talk about reflections both at the TX to ATU
interface and the cable to antenna interface. You also make
assumptions about the reflected power phase when it returns to the
antenna. If you want to talk about the phase at that point you have
to take into account the electrical length of the cable and *three*
reflections.


If the coax were lossless, and there were no losses in-the ATU,
then ALL
of the power that was supplied by the tx would be radiated
regardless of
the mismatch at coax to antenna interface.
However, in reality even small coax losses add up to a significant
loss
when the mismatch at the coax to antenna interface is high due to the
number of times that the power bounces up & down the coax suffering
loss
on each trip..

One poster has tried to say the reflections are all such that the
phase of the reflected power ends up in phase with the original signal
delivered to the antenna. But no one has done a power phase analysis.
You seem to be saying the same thing.

Power is reflected from the complex impedance of the antenna. Is that
power not out of phase from the incident signal by some amount
determined by the complex impedance? On traveling down the cable
there is a phase shift by an amount related to the ratio of the
electrical length and the wavelength. The ATU reflects some of the
signal back up the cable and some is passed directly to the TX (since
there is no or a very short cable between them) and is partially
reflected by the TX, again with some phase shift determined by the two
impedances (TX output and the ATU input). Another phase shift by the
cable and the reflected wave arrives at the antenna.

Is anyone going to tell me that all this has to add up to a reflected
signal arriving *in phase* with the incident signal?

I don't think you need to think about phase. The forward-going signal
has no need to know about the reflected signal, so just think of them as
being independent. One simply passes through the other.

When they are reaching the antenna you damn well *do* need to know
about phase. If the incident wave and the reflected wave returning to
the antenna are 180 degrees out of phase the power subtracts!


I understand Jeff correctly, if the ATU input presents a perfect 1-to-1
SWR to the transmitter output, it follows that none of power in the
multiple reflections in the coax is getting back into the TX.

I think you do *not* understand Jeff. The TX to ATU match has nothing
to do with the energy reaching the TX, just the opposite, it means all
power from the antenna passing through the ATU is absorbed by the TX.
But I have not seen anyone say the ATU matches the TX impedance, only
that it is a conjugate to the antenna.

Should be not be a conjugate match to the impedance as seen looking into
the coax at the TX end?

There are three interfaces, TX/ATU, ATU/cable, cable/antenna. Each one
can have reflections and each of the four parts of the system can
absorb (or radiate in the case of the antenna) power. This is not so
easy to analyze fully. At least until some of the fundamentals are
revealed that we have not talked about.


Therefore,
if we ignore the losses in the ATU, all the TX output power must
eventually, after countless ever-diminishing reflections between the
antenna and the ATU output, be either reaching the antenna, or being
lost in heating up the coax.

A given reflection may "reach" the antenna, but will it contribute
positively to the radiated field or counter it?

I think we are looking at the scenario in two different ways. You are
trying to see a 'snapshot' of the vectorial additions of all the forward
and reverse signals, and I am simply looking at them as being
independent entities, rattling up and down the coax, and progressively
either escaping via the antenna feedpoint and being radiated, or heating
up the coax.

You seem to want to ignore the possibility that the reflected waves can
interfere with your incident wave. This is entirely possible. Even if
reflected energy is radiated it can cancel the incident wave and reduce
your signal strength.

Conservation of energy suggests that it can only reduce it in one
direction by increasing it in another. If feeder reflections can have
this effect it should be possible to demonstrate it in small model
aerials at high frequencies.


--
Roger Hayter

Parallel coax
 

rickman wrote:

On 10/5/2015 4:17 AM, Ian Jackson wrote:
In message , rickman writes



You keep saying that the 1:1 match between the TX and the ATU prevents
any power from being sent to the TX which is not true. You are
confusing the power from the TX which is not reflected and the power
reflected from the antenna which passes through the ATU to the TX.

If the SWR meter between the TX output indicates a 1:1 SWR, then there
can be NO power travelling between the ATU input and the TX output - ie
there IS no reflected power. QED, surely?

If you ignore the losses in the ATU, all the power that the mismatched
antenna reflects, and that makes it back to the ATU output, MUST be
re-reflected by the ATU output impedance, and head off back towards the
antenna. This is because the reflected signal cannot heat up a lossless
ATU, and the SWR meter says it isn't coming back through the ATU. It
simply has nowhere to go except back down the coax.

You saying something is true or imagining a SWR reading is not the same
as understanding what is going on. What SWR reading are you imagining?
Can you explain this in terms of the circuit analysis? The ATU
consists of what circuit? The TX has some source impedance, what would
that be? I don't think you can design an ATU circuit that will isolate
the real source impedance of the TX from the reflected wave from the
antenna.

So, when you tune your transmitter-end ATU for minimum SWR *between the
Tx and the ATU*, what exactly are you doing? You are making the
feeder/aerilal/ATU combination look like a resistive load from the
transmitter side, what are you making the ATU/transmitter combination
look like from the aerial feeder side? It must be something that
reflects back incident waves, otherwise your SWR meter wouldn't be
reading 1.0. That seems to be the argument, and it sounds moderately
convincing to me.


--
Roger Hayter

Parallel coax
 

In message , rickman
writes
On 10/5/2015 4:17 AM, Ian Jackson wrote:
In message , rickman writes



You keep saying that the 1:1 match between the TX and the ATU prevents
any power from being sent to the TX which is not true. You are
confusing the power from the TX which is not reflected and the power
reflected from the antenna which passes through the ATU to the TX.

If the SWR meter between the TX output indicates a 1:1 SWR, then there
can be NO power travelling between the ATU input and the TX output - ie
there IS no reflected power. QED, surely?

If you ignore the losses in the ATU, all the power that the mismatched
antenna reflects, and that makes it back to the ATU output, MUST be
re-reflected by the ATU output impedance, and head off back towards the
antenna. This is because the reflected signal cannot heat up a lossless
ATU, and the SWR meter says it isn't coming back through the ATU. It
simply has nowhere to go except back down the coax.

You saying something is true or imagining a SWR reading is not the same
as understanding what is going on. What SWR reading are you imagining?

A typical HF station setup is TX - SWR meter - ATU - feeder - antenna.
[Please no one start saying they don't use this configuration.]

Can you explain this in terms of the circuit analysis?

The only analysis required is to ask yourself why, after much careful
twiddling with the knobs, you eventually get the SWR meter to show 1:1.
Does this not indicate that no power is coming back to the TX from the
ATU? [If not, what does it indicate?]

The ATU consists of what circuit?

I haven't a clue. It's a large shiny black box with three silver knobs
and two RF connectors. All the spec says is "Insertion loss 0dB". [Mind
you, it was very, very expensive.]

The TX has some source impedance, what would that be?

Somewhere between not-a-lot and probably not-too-high (because, in
operation, the TX doesn't get unduly hot). It really doesn't matter what
it is. The SWR meter shows that zero power is coming back out the ATU
input, so there is nothing to get re-reflected from the TX output.

I don't think you can design an ATU circuit that will isolate the
real source impedance of the TX from the reflected wave from the
antenna.

The purpose of what many of us (often slightly incorrectly) call an ATU
is to present the TX output with its design load impedance. In most
cases, this is 50 ohms resistive.

The TX output impedance is really of no concern to the designers of the
ATU. As long as they can produce an ATU that will present the TX with a
load of 50 ohms resistive, then it will do its job perfectly - and this
will be indicated by the 50 ohm SWR meter between the TX and the ATU
showing 1:1. And when an SWR meter shows 1:1, it means that there is no
power coming back towards the TX output.

--
Ian

Parallel coax
 

In message , Jeff writes

NO, as above, if the ATU is adjusted for a 1:1 match then there is no
power reflected back to the TX.

You are saying the ATU doesn't reflect power back from the TX. If the
match is 1:1 that is true. But you aren't considering the power
reflected from the antenna. The antenna reflects power back to the ATU
and there is nothing in the ATU to prevent that power from being handed
to the transmitter.

The whole function of the ATU is to provide a 1:1 match for the Tx,
when that is the case NO POWER GOES BACK TO THE TX.

Yes the antenna reflects power back towards the ATU, BUT the ATU then
reflects ALL of that power back again towards the antenna and none into
the TX (assuming that the ATU achieved a conjugate match.)


When adjusted for a 1:1 match there are no reflections between the Tx
and ATU, but there are multiple reflections between the ATU and teh
antenna. That is where a lot of power is dissipated in the antenna is a
poor match to the feeder impedance.

No one is talking about the power from the TX being reflected back to
the TX by the ATU.



The reflected wave will start off at the antenna out of phase with the
forward wave, (the actual phase depends on the complex impedance of the
mismatch), what the conjugate match that the ATU provides, when adjusted
so that there is a 1:1 match and no power reflected power sent to the
Tx,

This doesn't cover the reflected power from the antenna, just the power
from the TX.

NO. It covers the reflected power from the antenna. There is no
reflected power between the ATU and Tx if the match is 1:1.



is a phase shift such that the re-reflected wave from the ATU
towards the antenna is in phase with the original forward wave, so when
it reaches the antenna the portion of that re-reflected wave that is not
bounced back again down the coax by the mismatch is delivered to the
antenna.

How does that work? There is phase shift in the reflections which may
be compensated for by the ATU, but there is also phase shift in the cable.

The ATU applies a conjugate match at the end of the cable, the
impedance that it sees, and applies the conjugate of, is the impedance
of the antenna modified by the length of cable.

It just so happens that the conditions for a 1:1 vswr, and that of a
conjugate match, are that match causes the phase at the antenna end of
the cable to be the same phase as the original forward wave. That is
the physics of a conjugate match.


NO, no power gets back to the tx if the ATU is adjusted for a 1:1 match.
after all that is the definition of a 1:1 match; no reflected power.

You keep saying that the 1:1 match between the TX and the ATU prevents
any power from being sent to the TX which is not true. You are
confusing the power from the TX which is not reflected and the power
reflected from the antenna which passes through the ATU to the TX.


YES IT IS TRUE. I am not confusing anything.


Is anyone going to tell me that all this has to add up to a reflected
signal arriving *in phase* with the incident signal?


YES.

I may have to look at the math for this. How does the ATU reflect all
the power from the antenna back to the antenna? I thought only an open
or a short can reflect all the power.


Please look it up, do some experiments yourself. Try some simulations
in Spice or similar. You will find that I am correct.

While I've being saying "Let's ignore the losses in the ATU", presumably
you can assign a loss to it, and the reflected signal will suffer this
loss each time it bounces off the ATU output. If so, the loss can be
treated in the same way as the loss in the coax. In fact as 15' of
decent coax will only have (say) 0.5dB matched loss on 14MHz, the
summation of the 'return and go' ATU losses (say 2dB each time?) could
be more significant than the 1dB each time the signal traverses the
coax.
--
Ian

Parallel coax
 

On 09/27/2015 11:22 AM, John S wrote:
On 9/27/2015 9:41 AM, kg7fu wrote:
Doubling the number of feedlines would double the losses. Not only that
but each connector in the system inserts losses so that number
would be 4x.

How can that be? Each line carries half the power.

Connector loss at 14MHz is insignificant.

Connector losses are static up to UHF at around .5db per connector for a
typical PL-259/SO-239 pair. Insertion losses for professional grade
connectors on high quality rigid and semi-rigid coax such as Andrew
Heliax are around .05db.


Matching the antenna won't make the Return Loss go away but it will make
the transmitter happy.


I should have said matching the line.

Maybe you are working from the assumption that a mismatch between the
source and line causes return loss. It does not. Only the mismatch
between the load and line causes return loss.

KG7FU

Parallel coax
 

On 10/5/2015 5:29 AM, Roger Hayter wrote:
rickman wrote:

On 10/4/2015 6:01 PM, Ian Jackson wrote:
In message , rickman writes
On 10/4/2015 3:00 PM, Ian Jackson wrote:
In message , rickman
writes
On 10/4/2015 5:51 AM, Jeff wrote:

Is that good? If the cable length is 1/4 wavelength (as it is
in the
info provided for this case) the reflected power is nearly 180
degrees
out of phase with the initial power at the antenna. I think
reducing
this through cable losses would not be so bad, or better to dump
it in
the ATU?

It is good in as much as some of the re-reflected power is radiated
(and
some re-re-reflected) since the ATU causes the phase of the
re-reflected
wave to be 'in-phase' at the antenna.

It is bad in as much as the reflected power suffers 2 times the
cable
loss, and dissipates that in heat, on each return trip, up and down
the
coax.

With a high VSWR at the antenna there will be many return trips
before
the re-reflected power drops to a negligible level.

No power is 'dumped' in the ATU; although there will be losses, but
that
is another story.

If not the ATU, then the transmitter. I'm sure not all of the
power is
reflected back from the ATU. Exactly what is the phase of the
reflected
power from the ATU? I haven't seen an actual circuit for the ATU in
question. For that matter, what is the phase of the power reflected
from the antenna? I'm pretty confident we are not looking at the
return
of the reflected wave in phase with the incident wave.


If the ATU is adjusted so that the Tx sees a 1:1 match then no
power is
reflected to the transmitter. A 1:1 match means that there is no power
reflected to the Tx by definition (and can be proved by measurement).

Fine, but not related to my question if that is what you were
responding to.


By conservation of energy then all of the power must be radiated or
lost
as heat, (mostly in the coax).

Uh, I'm not sure of that since there are other paths and I'm pretty
sure power can be sent back into the driver of the transmitter.


When the ATU gives a conjugate match, (ie the Tx sees a 1:1 vswr) the
phase of the re-reflected wave at the antenna is 0, ie it is in-phase
with the original forward wave and so any power that is not
re-re-reflected again adds to the power supplied to the antenna.

You seem to be mixing two things here. We have been talking about an
ATU at the transmitter. The feed line cable is between the ATU and
the antenna. You then talk about reflections both at the TX to ATU
interface and the cable to antenna interface. You also make
assumptions about the reflected power phase when it returns to the
antenna. If you want to talk about the phase at that point you have
to take into account the electrical length of the cable and *three*
reflections.


If the coax were lossless, and there were no losses in-the ATU,
then ALL
of the power that was supplied by the tx would be radiated
regardless of
the mismatch at coax to antenna interface.
However, in reality even small coax losses add up to a significant
loss
when the mismatch at the coax to antenna interface is high due to the
number of times that the power bounces up & down the coax suffering
loss
on each trip..

One poster has tried to say the reflections are all such that the
phase of the reflected power ends up in phase with the original signal
delivered to the antenna. But no one has done a power phase analysis.
You seem to be saying the same thing.

Power is reflected from the complex impedance of the antenna. Is that
power not out of phase from the incident signal by some amount
determined by the complex impedance? On traveling down the cable
there is a phase shift by an amount related to the ratio of the
electrical length and the wavelength. The ATU reflects some of the
signal back up the cable and some is passed directly to the TX (since
there is no or a very short cable between them) and is partially
reflected by the TX, again with some phase shift determined by the two
impedances (TX output and the ATU input). Another phase shift by the
cable and the reflected wave arrives at the antenna.

Is anyone going to tell me that all this has to add up to a reflected
signal arriving *in phase* with the incident signal?

I don't think you need to think about phase. The forward-going signal
has no need to know about the reflected signal, so just think of them as
being independent. One simply passes through the other.

When they are reaching the antenna you damn well *do* need to know
about phase. If the incident wave and the reflected wave returning to
the antenna are 180 degrees out of phase the power subtracts!


I understand Jeff correctly, if the ATU input presents a perfect 1-to-1
SWR to the transmitter output, it follows that none of power in the
multiple reflections in the coax is getting back into the TX.

I think you do *not* understand Jeff. The TX to ATU match has nothing
to do with the energy reaching the TX, just the opposite, it means all
power from the antenna passing through the ATU is absorbed by the TX.
But I have not seen anyone say the ATU matches the TX impedance, only
that it is a conjugate to the antenna.

Should be not be a conjugate match to the impedance as seen looking into
the coax at the TX end?

There are three interfaces, TX/ATU, ATU/cable, cable/antenna. Each one
can have reflections and each of the four parts of the system can
absorb (or radiate in the case of the antenna) power. This is not so
easy to analyze fully. At least until some of the fundamentals are
revealed that we have not talked about.


Therefore,
if we ignore the losses in the ATU, all the TX output power must
eventually, after countless ever-diminishing reflections between the
antenna and the ATU output, be either reaching the antenna, or being
lost in heating up the coax.

A given reflection may "reach" the antenna, but will it contribute
positively to the radiated field or counter it?

I think we are looking at the scenario in two different ways. You are
trying to see a 'snapshot' of the vectorial additions of all the forward
and reverse signals, and I am simply looking at them as being
independent entities, rattling up and down the coax, and progressively
either escaping via the antenna feedpoint and being radiated, or heating
up the coax.

You seem to want to ignore the possibility that the reflected waves can
interfere with your incident wave. This is entirely possible. Even if
reflected energy is radiated it can cancel the incident wave and reduce
your signal strength.

Conservation of energy suggests that it can only reduce it in one
direction by increasing it in another. If feeder reflections can have
this effect it should be possible to demonstrate it in small model
aerials at high frequencies.

Conservation of energy? Just show the mechanics. The energy will take
care of itself.

--

Rick

Parallel coax
 

On 10/5/2015 5:43 AM, Roger Hayter wrote:
rickman wrote:

On 10/5/2015 4:17 AM, Ian Jackson wrote:
In message , rickman writes



You keep saying that the 1:1 match between the TX and the ATU prevents
any power from being sent to the TX which is not true. You are
confusing the power from the TX which is not reflected and the power
reflected from the antenna which passes through the ATU to the TX.

If the SWR meter between the TX output indicates a 1:1 SWR, then there
can be NO power travelling between the ATU input and the TX output - ie
there IS no reflected power. QED, surely?

If you ignore the losses in the ATU, all the power that the mismatched
antenna reflects, and that makes it back to the ATU output, MUST be
re-reflected by the ATU output impedance, and head off back towards the
antenna. This is because the reflected signal cannot heat up a lossless
ATU, and the SWR meter says it isn't coming back through the ATU. It
simply has nowhere to go except back down the coax.

You saying something is true or imagining a SWR reading is not the same
as understanding what is going on. What SWR reading are you imagining?
Can you explain this in terms of the circuit analysis? The ATU
consists of what circuit? The TX has some source impedance, what would
that be? I don't think you can design an ATU circuit that will isolate
the real source impedance of the TX from the reflected wave from the
antenna.

So, when you tune your transmitter-end ATU for minimum SWR *between the
Tx and the ATU*, what exactly are you doing? You are making the
feeder/aerilal/ATU combination look like a resistive load from the
transmitter side, what are you making the ATU/transmitter combination
look like from the aerial feeder side? It must be something that
reflects back incident waves, otherwise your SWR meter wouldn't be
reading 1.0. That seems to be the argument, and it sounds moderately
convincing to me.

I don't know what you are doing. I am talking about analysis of the
circuits involved. If analyzing the circuits is not appropriate we can
just drop this discussion.

Any chance of coming up with a specific circuit? Forget tuning, let's
set parameters for the TX, cable, antenna, calculate the circuit for the
ATU and then we can see what happens at the various interfaces.

--

Rick

Parallel coax
 

On 10/5/2015 5:58 AM, Ian Jackson wrote:
In message , rickman writes
On 10/5/2015 4:17 AM, Ian Jackson wrote:
In message , rickman
writes



You keep saying that the 1:1 match between the TX and the ATU prevents
any power from being sent to the TX which is not true. You are
confusing the power from the TX which is not reflected and the power
reflected from the antenna which passes through the ATU to the TX.

If the SWR meter between the TX output indicates a 1:1 SWR, then there
can be NO power travelling between the ATU input and the TX output - ie
there IS no reflected power. QED, surely?

If you ignore the losses in the ATU, all the power that the mismatched
antenna reflects, and that makes it back to the ATU output, MUST be
re-reflected by the ATU output impedance, and head off back towards the
antenna. This is because the reflected signal cannot heat up a lossless
ATU, and the SWR meter says it isn't coming back through the ATU. It
simply has nowhere to go except back down the coax.

You saying something is true or imagining a SWR reading is not the
same as understanding what is going on. What SWR reading are you
imagining?

A typical HF station setup is TX - SWR meter - ATU - feeder - antenna.
[Please no one start saying they don't use this configuration.]

Can you explain this in terms of the circuit analysis?

The only analysis required is to ask yourself why, after much careful
twiddling with the knobs, you eventually get the SWR meter to show 1:1.
Does this not indicate that no power is coming back to the TX from the
ATU? [If not, what does it indicate?]

Not sending power to the TX is not the same as reflecting all power back
to the antenna.


The ATU consists of what circuit?

I haven't a clue. It's a large shiny black box with three silver knobs
and two RF connectors. All the spec says is "Insertion loss 0dB". [Mind
you, it was very, very expensive.]

The TX has some source impedance, what would that be?

Somewhere between not-a-lot and probably not-too-high (because, in
operation, the TX doesn't get unduly hot). It really doesn't matter what
it is. The SWR meter shows that zero power is coming back out the ATU
input, so there is nothing to get re-reflected from the TX output.

I don't think you can design an ATU circuit that will isolate the
real source impedance of the TX from the reflected wave from the antenna.

The purpose of what many of us (often slightly incorrectly) call an ATU
is to present the TX output with its design load impedance. In most
cases, this is 50 ohms resistive.

The TX output impedance is really of no concern to the designers of the
ATU. As long as they can produce an ATU that will present the TX with a
load of 50 ohms resistive, then it will do its job perfectly - and this
will be indicated by the 50 ohm SWR meter between the TX and the ATU
showing 1:1. And when an SWR meter shows 1:1, it means that there is no
power coming back towards the TX output.



--

Rick

Parallel coax
 

On 10/5/2015 7:29 AM, Jeff wrote:

NO, as above, if the ATU is adjusted for a 1:1 match then there is no
power reflected back to the TX.

You are saying the ATU doesn't reflect power back from the TX. If the
match is 1:1 that is true. But you aren't considering the power
reflected from the antenna. The antenna reflects power back to the ATU
and there is nothing in the ATU to prevent that power from being handed
to the transmitter.

The whole function of the ATU is to provide a 1:1 match for the Tx, when
that is the case NO POWER GOES BACK TO THE TX.

That is not technically correct. A 1:1 match prevents TX power from
being reflected, it does not prevent other power from being sent to the TX.


Yes the antenna reflects power back towards the ATU, BUT the ATU then
reflects ALL of that power back again towards the antenna and none into
the TX (assuming that the ATU achieved a conjugate match.)

That is the part I am trying to understand. No one seems to be able to
explain how the ATU does all this, present a 50 ohm resistive load to
the TX, reflect all power back toward the antenna *and* align the phase
of that reflected signal so it is phase aligned to the incident
radiation. I am told it is a "black box" with a knob.

Ok, I'll accept that this is possible, but I'd like to see the math that
shows it. So far no one seems to understand the math. If that is true,
then I'll stop talking about this. I'm not trying to bug people.


When adjusted for a 1:1 match there are no reflections between the Tx
and ATU, but there are multiple reflections between the ATU and teh
antenna. That is where a lot of power is dissipated in the antenna is a
poor match to the feeder impedance.

No one is talking about the power from the TX being reflected back to
the TX by the ATU.



The reflected wave will start off at the antenna out of phase with the
forward wave, (the actual phase depends on the complex impedance of the
mismatch), what the conjugate match that the ATU provides, when adjusted
so that there is a 1:1 match and no power reflected power sent to the
Tx,

This doesn't cover the reflected power from the antenna, just the power
from the TX.

NO. It covers the reflected power from the antenna. There is no
reflected power between the ATU and Tx if the match is 1:1.

You keep saying that, but it isn't true. Power can come through the ATU
circuit into the TX even when there is a 1:1 impedance match. In a
silly case lightning can strike the antenna. I don't think the 1:1
impedance match will prevent that power from reaching the TX.


is a phase shift such that the re-reflected wave from the ATU
towards the antenna is in phase with the original forward wave, so when
it reaches the antenna the portion of that re-reflected wave that is not
bounced back again down the coax by the mismatch is delivered to the
antenna.

How does that work? There is phase shift in the reflections which may
be compensated for by the ATU, but there is also phase shift in the
cable.

The ATU applies a conjugate match at the end of the cable, the impedance
that it sees, and applies the conjugate of, is the impedance of the
antenna modified by the length of cable.

It just so happens that the conditions for a 1:1 vswr, and that of a
conjugate match, are that match causes the phase at the antenna end of
the cable to be the same phase as the original forward wave. That is the
physics of a conjugate match.

This is what I'd like to see. Any math that you can share?


NO, no power gets back to the tx if the ATU is adjusted for a 1:1 match.
after all that is the definition of a 1:1 match; no reflected power.

You keep saying that the 1:1 match between the TX and the ATU prevents
any power from being sent to the TX which is not true. You are
confusing the power from the TX which is not reflected and the power
reflected from the antenna which passes through the ATU to the TX.


YES IT IS TRUE. I am not confusing anything.

Yes, what you say it true, no reflected power... "read my lips, no new
taxes". But the 1:1 match does not imply anything other than the
reflected power from the TX.


Is anyone going to tell me that all this has to add up to a reflected
signal arriving *in phase* with the incident signal?


YES.

I may have to look at the math for this. How does the ATU reflect all
the power from the antenna back to the antenna? I thought only an open
or a short can reflect all the power.


Please look it up, do some experiments yourself. Try some simulations in
Spice or similar. You will find that I am correct.

Jeff


--

Rick