Parallel coax
 

Roger Hayter wrote:

rickman wrote:

On 10/4/2015 4:48 AM, Jeff wrote:

What is the reflection coefficient at the ATU/feedline interface?


Assuming a 'perfect' ATU and a prefect conjugate match, giving 1:1 at
the TX, then it is 1. ie all of the reflected power that reaches the ATU
is re-reflected back up towards the antenna.

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.

Having specified the transmitter power output, by definiton, no power
can be "lost" in the amplifier, because transmitter power output is
*defined* as the net power it actually manages to get out of its output
socket net of any reflections. The mismatch may make it harder for the
transmitter to achieve that; but, by definition, whatever it does
achieve is its power output. You can't draw a valid defiinition of a
difference between power it never produced at all and power sent back to
it.

Sorry, typo. I mean you can't draw a valid distinction between power
never produced and power reflected. Whatever power the transmitter will
produce in a given load *is* its power output.

--
Roger Hayter

Parallel coax
 

In message , Jeff writes

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).

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

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.

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..


A concise explanation.

What is the easiest way of calculating the power loss (say, assuming
it's all in coax, and none in the ATU)? Is it simply a case of adding up
a large number of diminishing losses as the signal repeatedly rattles up
and down the coax (until it becomes so small that the losses can be
ignored), or are there some more-elegant (and accurate) methods?
--
Ian

Parallel coax
 

In message , Jeff writes

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).

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

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.

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..


A concise explanation.

What is the easiest way of calculating the power loss (say, assuming
it's all in coax, and none in the ATU)? Is it simply a case of adding up
a large number of diminishing losses as the signal repeatedly rattles up
and down the coax (until it becomes so small that the losses can be
ignored), or are there some more-elegant (and accurate) methods?

The easiest way is to make an Excel spreadsheet.
Convert the VSWR to return loss and convert to a ratio.
Take the Tx power subtract the line loss and multiply the RL ratio to
get the reflected power. Subtract that power from the incident power to
find the power transmitted.
Take the reflected power and subtract twice the line loss, then
multiply that incident power by the RL ratio and continue as above,
adding the proportion that goes to that antenna to the transmitted
power. Also add the 2x line loss the the original line loss to get the
power dissipated in the coax.
Repeat above until the reflected power becomes insignificant; probably
at least six ox seven time if the vswr is very high.

Ah yes! Thanks. It's a while since I honed my skills in driving an Excel
spreadsheet! I'll certainly give it a go.


--
Ian

Parallel coax
 

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?

--

Rick

Parallel coax
 

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.

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. 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.
--
Ian

Parallel coax
 

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.

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?

--

Rick

Parallel coax
 

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.
--
Ian

Parallel coax
 

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.

--

Rick

Parallel coax
 

On 10/5/2015 2:48 AM, Jeff wrote:

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.


Of course it is. If the ATU is adjusted for a 1:1 match then NO power is
sent back to the Tx.




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.

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.


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.

No, I and talking about a Tx then an ATU then a feeder to a mismatched
antenna.

I believe that is what we are all talking about.


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.


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.

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.


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.


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.

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.


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.

--

Rick

Parallel coax
 

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.



--
Ian