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

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

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

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

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

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.

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

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?

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

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