Parallel coax
 

On 10/5/2015 9:36 AM, Ian Jackson wrote:
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.

I don't mind treating the cable and ATU as "ideal". I'd like to
understand how you can turn one knob and get a match to the transmitter
*and* a perfect reflection *with* phase matching to the cable. It seems
to me there are three variables you need to tune for and if such a
circuit is possible will require more than one knob.

--

Rick

Parallel coax
 

"kg7fu" wrote in message
...


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.


Where did you get the .5 db loss per connector ?

If you would think for about 1 minuit,you would see that has to be wrong.

To make things simple and using round numbers. Take 2 PL-259 connectors and
a barrel connector and put them together. Not counting the barrel that
would be 1 db of loss. That is about 25 %. Using a 100 watt transmitter,
that is 25 watts of heat in a couple of inches. It would get warm. Bumping
it up to the 1 KW level that lots of hams use, that is 250 watts. It would
melt down in a short order.

The loss is around .01 db per connector around 100 MHz, about half that at
10 MHz.

Parallel coax
 

In message , rickman
writes
On 10/5/2015 7:29 AM, Jeff wrote:





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.

Am I mistaken, or is the discussion starting to get a bit silly? The ATU
is certainly NOT a one-way device!

Of course power can flow back through the ATU. If it didn't, when the
rig is receiving (assuming it's a transceiver), you would never hear any
of the signals that the antenna picks up. This includes a lightning
strike! For the same reasons, if you have any strong radio transmissions
in the neighbourhood, these too pass through the ATU to the TX output,
and you may not be possible to get a 1:1 reading on the SWR meter!

However, apart from these sorts of situations, if the SWR meter is
responding to only to the signal transmitted by your TX, and it is
showing a 1:1 SWR reading, none of the reflections of TX signal are
coming back through the ATU and into the TX output. The portion of the
TX signal that is not making it into the antenna can only be accounted
for by it heating up the coax, and the ATU.


--
Ian

Parallel coax
 

On 10/5/2015 4:02 AM, 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.


Okay, Rick, here ya go...

Using the antenna info given by Wayne (20-j130) and his transmission
line, I find the following:

Using characteristics of the line I found on the 'Net, I see that the
velocity factor is .66 and the loss is about 0.7dB/100 feet
(insignificant in this case, but I feel I must mention it).

This should transform the antenna to about 6.453+j52.544 ohms at the far
end (don't forget the velocity factor). I used a Smith chart but if you
wish to verify using the Telegraphers Equations, be my guest.

What does it take to make this impedance look like 50+j0 ohms?

One solution is a shunt capacitance at the far end of the cable of
140.5pF and then a series capacitance of 80.9pF to the transmitter. That
would be a shunt Xc of about 80.9 ohms and a series Xc of about -138.64
ohms respectively.

Please verify that this will present a 50 ohm load to the transmitter.
If so, the transmitter will see a 1:1 SWR and no power will be returned
from the matching network.

Parallel coax
 

On 10/5/2015 9:53 PM, John S wrote:
On 10/5/2015 4:02 AM, 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.


Okay, Rick, here ya go...

Using the antenna info given by Wayne (20-j130) and his transmission
line, I find the following:

Using characteristics of the line I found on the 'Net, I see that the
velocity factor is .66 and the loss is about 0.7dB/100 feet
(insignificant in this case, but I feel I must mention it).

This should transform the antenna to about 6.453+j52.544 ohms at the far
end (don't forget the velocity factor). I used a Smith chart but if you
wish to verify using the Telegraphers Equations, be my guest.

What does it take to make this impedance look like 50+j0 ohms?

One solution is a shunt capacitance at the far end of the cable of
140.5pF and then a series capacitance of 80.9pF to the transmitter. That
would be a shunt Xc of about 80.9 ohms and a series Xc of about -138.64
ohms respectively.

Please verify that this will present a 50 ohm load to the transmitter.
If so, the transmitter will see a 1:1 SWR and no power will be returned
from the matching network.

Thanks for this. I've got some work for the next day or two, but I'll
dig into this and see what comes out. What is the length of the cable
again? I believe we are talking about 14 MHz, right?

To figure what the reflected signal sees when reaching the ATU, I would
need to know the characteristic impedance of the TX. Should I assume 50
ohms? I've read this is often not the case. If not, the 50 ohm ATU
impedance assumption is not valid. When you tune for VSWR of 1:1, the
impedance of the TX side of the ATU will match the TX. You can't say it
is 50 ohms unless the TX is 50 ohms, no?

--

Rick

Parallel coax
 

On 10/5/2015 10:23 PM, rickman wrote:
On 10/5/2015 9:53 PM, John S wrote:
On 10/5/2015 4:02 AM, 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.


Okay, Rick, here ya go...

Using the antenna info given by Wayne (20-j130) and his transmission
line, I find the following:

Using characteristics of the line I found on the 'Net, I see that the
velocity factor is .66 and the loss is about 0.7dB/100 feet
(insignificant in this case, but I feel I must mention it).

This should transform the antenna to about 6.453+j52.544 ohms at the far
end (don't forget the velocity factor). I used a Smith chart but if you
wish to verify using the Telegraphers Equations, be my guest.

What does it take to make this impedance look like 50+j0 ohms?

One solution is a shunt capacitance at the far end of the cable of
140.5pF and then a series capacitance of 80.9pF to the transmitter. That
would be a shunt Xc of about 80.9 ohms and a series Xc of about -138.64
ohms respectively.

Please verify that this will present a 50 ohm load to the transmitter.
If so, the transmitter will see a 1:1 SWR and no power will be returned
from the matching network.

Thanks for this. I've got some work for the next day or two, but I'll
dig into this and see what comes out. What is the length of the cable
again? I believe we are talking about 14 MHz, right?

Yes, 14Mhz. The line is 15ft of RG-213.

To figure what the reflected signal sees when reaching the ATU, I would
need to know the characteristic impedance of the TX.

Work the problem. You will find that the capacitor network (ATU) will
change the antenna impedance so that the transmitter sees 50 + j0 ohms.


Should I assume 50
ohms? I've read this is often not the case. If not, the 50 ohm ATU
impedance assumption is not valid. When you tune for VSWR of 1:1, the
impedance of the TX side of the ATU will match the TX. You can't say it
is 50 ohms unless the TX is 50 ohms, no?

Yes, I can. Work the problem. In fact, if you have trouble with the
transmission line situation, you can forget the line and just use the
6.453+j52.544 that my Smith chart shows and the two capacitor values. It
really doesn't matter what impedance you put on the ATU (as long as it
is designed to handle it).

It is obvious that you think a transmitter is designed with some output
impedance. This is not so. It is designed to supply a maximum current at
some maximum voltage just as any source is designed to do. If you
yourself design an amplifier of any sort, do you worry about its output
impedance, or do you worry about the load it drives? I know, sometimes
your circuit impedance is high enough to cause problems. That's why RF
amplifiers are designed with only the voltage and current constraints in
mind. Besides, how would you propose that the source impedance be
adjusted to be 50 ohms without using a power resistor in the final?

Parallel coax
 

In message , rickman
writes



When you tune for VSWR of 1:1, the impedance of the TX side of the ATU
will match the TX. You can't say it is 50 ohms unless the TX is 50
ohms, no?

No, no, no.

As I keep saying, the reading on the SWR meter has nothing to do with
the output impedance of the TX feeding RF into it. It is determined by
the reference resistors in the meter's directional coupler circuits and
the impedance of the load attached to its output. If the load is 50
ohms, a 50 ohm SWR meter will read 1:1, regardless of the TX output
impedance.

The purpose of the ATU is not to match the TX output impedance to the
outside world. It is to convert the impedance of the outside world to 50
ohms - which is the impedance the TX is designed to work into.
--
Ian

Parallel coax
 

In message , Ian Jackson
writes
In message , rickman
writes



When you tune for VSWR of 1:1, the impedance of the TX side of the ATU
will match the TX. You can't say it is 50 ohms unless the TX is 50
ohms, no?

No, no, no.

As I keep saying, the reading on the SWR meter has nothing to do with
the output impedance of the TX feeding RF into it. It is determined by
the reference resistors in the meter's directional coupler circuits and
the impedance of the load attached to its output. If the load is 50
ohms, a 50 ohm SWR meter will read 1:1, regardless of the TX output
impedance.

The purpose of the ATU is not to match the TX output impedance to the
outside world. It is to convert the impedance of the outside world to
50 ohms - which is the impedance the TX is designed to work into.

And just to illustrate the point, consider the following:
You have a TX with a 50 ohm output impedance feeding through a 50 ohm
SWR meter into 50 feet of 50 ohm coax terminated with a 50 ohm resistor.
What does the SWR meter read, and why?

You then replace the TX with one with an output impedance of 25 ohms.
What does the SWR meter read, and why?

Finally, you replace the TX with one with an output impedance of 100
ohms. What does the SWR meter read, and why?
--
Ian

Parallel coax
 

In message , Jeff writes



Anyway this discussion has now got to the silly stage so I am off.

Great minds seem to be thinking alike! However, I'll keep a watching
brief. I've started that Excel sheet.



--
Ian

Parallel coax
 

On 10/6/2015 12:45 AM, John S wrote:
On 10/5/2015 10:23 PM, rickman wrote:
On 10/5/2015 9:53 PM, John S wrote:
On 10/5/2015 4:02 AM, 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.


Okay, Rick, here ya go...

Using the antenna info given by Wayne (20-j130) and his transmission
line, I find the following:

Using characteristics of the line I found on the 'Net, I see that the
velocity factor is .66 and the loss is about 0.7dB/100 feet
(insignificant in this case, but I feel I must mention it).

This should transform the antenna to about 6.453+j52.544 ohms at the far
end (don't forget the velocity factor). I used a Smith chart but if you
wish to verify using the Telegraphers Equations, be my guest.

What does it take to make this impedance look like 50+j0 ohms?

One solution is a shunt capacitance at the far end of the cable of
140.5pF and then a series capacitance of 80.9pF to the transmitter. That
would be a shunt Xc of about 80.9 ohms and a series Xc of about -138.64
ohms respectively.

Please verify that this will present a 50 ohm load to the transmitter.
If so, the transmitter will see a 1:1 SWR and no power will be returned
from the matching network.

Thanks for this. I've got some work for the next day or two, but I'll
dig into this and see what comes out. What is the length of the cable
again? I believe we are talking about 14 MHz, right?

Yes, 14Mhz. The line is 15ft of RG-213.

To figure what the reflected signal sees when reaching the ATU, I would
need to know the characteristic impedance of the TX.

Work the problem. You will find that the capacitor network (ATU) will
change the antenna impedance so that the transmitter sees 50 + j0 ohms.

That's not what I am asking. I'm asking the impedance of the TX to see
what is seen from the other side of the ATU looking in. This determines
the reflection of the signal coming from the antenna by the ATU.
Remember there are three interfaces which can reflect the signal, two on
the ATU and one at the antenna. Four if you count the TX and any signal
passing through the ATU.


Should I assume 50
ohms? I've read this is often not the case. If not, the 50 ohm ATU
impedance assumption is not valid. When you tune for VSWR of 1:1, the
impedance of the TX side of the ATU will match the TX. You can't say it
is 50 ohms unless the TX is 50 ohms, no?

Yes, I can. Work the problem. In fact, if you have trouble with the
transmission line situation, you can forget the line and just use the
6.453+j52.544 that my Smith chart shows and the two capacitor values. It
really doesn't matter what impedance you put on the ATU (as long as it
is designed to handle it).

The rest of the system does not determine the output impedance of the
TX. Others have said here that transmitters are usually *not* 50 ohms
output impedance. Rather they are designed to drive a 50 ohm load,
which is not the same thing at all.


It is obvious that you think a transmitter is designed with some output
impedance. This is not so. It is designed to supply a maximum current at
some maximum voltage just as any source is designed to do. If you
yourself design an amplifier of any sort, do you worry about its output
impedance, or do you worry about the load it drives?

Yes, I have designed outputs to drive a known impedance matched. So the
source impedance was designed to appear to be 50 ohms.

No, it is not required that that output impedance of an amplifier match
that of the load, but if I want to deal with power returning to the
output I can't determine the reflection unless I know the impedance of
that output. It *will* have an output impedance. That is a given. It
may be a very low impedance, but it will exist.


I know, sometimes
your circuit impedance is high enough to cause problems. That's why RF
amplifiers are designed with only the voltage and current constraints in
mind. Besides, how would you propose that the source impedance be
adjusted to be 50 ohms without using a power resistor in the final?

I'm not saying the output impedance is 50 ohms. I'm asking what the
output impedance is. I will say it would be impossible to get full
power out of the TX into an ATU presenting a 50 ohm load, without
reflection if the TX did not have a 50 ohm output impedance, no?

Thinking the output impedance has to dissipate power is not correct. A
circuit I designed provides 50 ohm output impedance using a 12.5 ohm
resistor. This was does expressly to get an 8 Vpp signal using a 12
volt power supply which would not be possible using a 50 output
resistor. If I had wanted to, I could have designed the circuit to use
an even smaller resistor and less dissipation.

--

Rick