On 8/31/2010 3:09 PM, Richard Fry wrote:
On Aug 30, 5:52 pm, Roy wrote:

In my example, the antenna is not matched to the transmission line. Nor,
for that matter, is the transmitter matched to the transmission line. My
point is that power transfer doesn't depend on either of these points
being matched.

Roy:

My post showing very high input SWR at the base of an unloaded, base-
driven, 10 foot vertical on 3.8 MHz described an UNMATCHED system
resulting from its connection to transmission lines of typical
impedance values. It did not include matching networks, whether
located at the base of the vertical radiator, the output connector of
the transmitter, or wherever.

Then you posted, "Using your antenna as an example, suppose that a
transmitter with output Z of 50 ohms is connected to a tuner that
transforms its output impedance to 0.6 + j1250 ohms. ... The
transmitter will see 50 + j0 ohms, the antenna will see an impedance
of 0.6 + j1250 ohms, and full power will be transferred."

That configuration you posted is a MATCHED system, and its performance
does not disprove the accuracy of my post.

RF

So you're saying that the mismatch between the impedance of an antenna
and the transmission line connected to it doesn't inhibit power flow
when there's a matching network anywhere in the system. But it does
interfere with power flow when there's no matching network?

What if the antenna is 50 ohms and the transmission line is a half
wavelength of 600 ohms, for a 12:1 mismatch? There's no matching
network. The transmitter sees 50 ohms. The antenna sees 50 ohms. What
will interfere with the power flow?

Roy Lewallen, W7EL

On Aug 31, 5:25*pm, Roy Lewallen wrote:

What if the antenna is 50 ohms and the transmission line is a half
wavelength of 600 ohms, for a 12:1 mismatch? There's no matching
network. The transmitter sees 50 ohms. The antenna sees 50 ohms.
What will interfere with the power flow ?

This, and your other what ifs describe a system *different* than the
one in my post.

The performance of the systems you described does not apply to the
system I described, and vise-versa..

RF

In message , Roy Lewallen
writes
On 8/31/2010 3:09 PM, Richard Fry wrote:
On Aug 30, 5:52 pm, Roy wrote:

In my example, the antenna is not matched to the transmission line. Nor,
for that matter, is the transmitter matched to the transmission line. My
point is that power transfer doesn't depend on either of these points
being matched.

Roy:

My post showing very high input SWR at the base of an unloaded, base-
driven, 10 foot vertical on 3.8 MHz described an UNMATCHED system
resulting from its connection to transmission lines of typical
impedance values. It did not include matching networks, whether
located at the base of the vertical radiator, the output connector of
the transmitter, or wherever.

Then you posted, "Using your antenna as an example, suppose that a
transmitter with output Z of 50 ohms is connected to a tuner that
transforms its output impedance to 0.6 + j1250 ohms. ... The
transmitter will see 50 + j0 ohms, the antenna will see an impedance
of 0.6 + j1250 ohms, and full power will be transferred."

That configuration you posted is a MATCHED system, and its performance
does not disprove the accuracy of my post.

RF

So you're saying that the mismatch between the impedance of an antenna
and the transmission line connected to it doesn't inhibit power flow
when there's a matching network anywhere in the system. But it does
interfere with power flow when there's no matching network?

What if the antenna is 50 ohms and the transmission line is a half
wavelength of 600 ohms, for a 12:1 mismatch? There's no matching
network. The transmitter sees 50 ohms. The antenna sees 50 ohms. What
will interfere with the power flow?


In my simplistic world, this is how I understand things:

Provided the TX is followed by a tuner/matcher, which matches whatever
is attached to the tuner output to 50 ohms at the tuner input, the TX
will be happy.

The power loss in the feeder is essentially a function of its inherent
loss (when matched) per unit length, and the SWR on it.

The SWR is a function of the mismatch between the load on the antenna
end of the feeder, and the feeder characteristic impedance. The greater
the SWR and the longer the feeder, the higher will be the loss on the
feeder. 'Lossless' feeder have no loss, regardless of length and SWR.
However, with real-world feeders, the losses rise increasingly rapidly
with increasing SWR.

The impedance looking into the tuner end of the feeder is the impedance
of the load, transformed by length of the feeder, and is also modified
by the loss of the feeder.

The higher the feeder loss, the closer the impedance seen at the tuner
end will approach the characteristic impedance of the feeder. [Long
lengths of lossy feeder - maybe with a nominal termination on the far
end - can make good dummy loads at VHF and UHF.]
--
Ian

Ian Jackson wrote in
:

....
In my simplistic world, this is how I understand things:

Provided the TX is followed by a tuner/matcher, which matches whatever
is attached to the tuner output to 50 ohms at the tuner input, the TX
will be happy.

"Happey" eh!

The power loss in the feeder is essentially a function of its inherent
loss (when matched) per unit length, and the SWR on it.

Wrong.


The SWR is a function of the mismatch between the load on the antenna
end of the feeder, and the feeder characteristic impedance. The

Well, the SWR at the load end is a function of Zl and Zo (both complex
quantities)... but the 'notional' SWR varies along the line as accounted
for by the complex propagation coefficient, in most practical cases at
HF, SWR decreases smoothly from load to source.

greater the SWR and the longer the feeder, the higher will be the loss
on the feeder. 'Lossless' feeder have no loss, regardless of length
and SWR. However, with real-world feeders, the losses rise
increasingly rapidly with increasing SWR.

See your earlier misconception regarding loss being a simple function of
SWR.


The impedance looking into the tuner end of the feeder is the
impedance of the load, transformed by length of the feeder, and is
also modified by the loss of the feeder.

You got that right.


The higher the feeder loss, the closer the impedance seen at the tuner
end will approach the characteristic impedance of the feeder. [Long
lengths of lossy feeder - maybe with a nominal termination on the far
end - can make good dummy loads at VHF and UHF.]

Yes, but is it of practical application in a transmit scenario? If the
input impedance due to a severly mismatched load is at all close to Zo,
then you have lost most of the transmitter power in the line.

The "make a good dummy load" recipe doesn't address the power rating,
especially where most of the power is dissipated in a very short length
of cable.

Owen

Owen Duffy wrote in news:Xns9DE6CD1DCDA37nonenowhere@
61.9.134.55:

"Happey" eh!

How did that e get in there...

"Happy", a new dumbed down attribute of a transmitter. Everyone loves a
happy transmittter.

What happens if they are not happy? Well, everyone knows you will let the
smoke out of them (AKA reflected power is dissipated in the PA and kills
transmitters).

Yep, a simple world... even if wrong.

Owen

On Sep 1, 4:11*am, Ian Jackson
wrote:
The impedance looking into the tuner end of the feeder is the impedance
of the load, transformed by length of the feeder, and is also modified
by the loss of the feeder.

That's the result of the 50 ohm Z0-match achieved at the input of the
tuner which is responsible for the transmitter's "happiness". A Z0-
match causes the incident reflected energy to be redistributed back
toward the antenna through a combination of reflection and destructive
interference.
--
73, Cecil, w5dxp.com

In message , Owen Duffy
writes
Ian Jackson wrote in
:

...
In my simplistic world, this is how I understand things:

Provided the TX is followed by a tuner/matcher, which matches whatever
is attached to the tuner output to 50 ohms at the tuner input, the TX
will be happy.

"Happey" eh!

Your later correction is noted!

The power loss in the feeder is essentially a function of its inherent
loss (when matched) per unit length, and the SWR on it.

Wrong.

OK, the loss may not be uniform if the SWR is greater at the load end.
But when is the power loss* NOT a function of its matched loss per unit
length, and the SWR on it. [*Should I have said 'in dB'? ]

The SWR is a function of the mismatch between the load on the antenna
end of the feeder, and the feeder characteristic impedance. The

Well, the SWR at the load end is a function of Zl and Zo (both complex
quantities)... but the 'notional' SWR varies along the line as accounted
for by the complex propagation coefficient, in most practical cases at
HF, SWR decreases smoothly from load to source.

greater the SWR and the longer the feeder, the higher will be the loss
on the feeder. 'Lossless' feeder have no loss, regardless of length
and SWR. However, with real-world feeders, the losses rise
increasingly rapidly with increasing SWR.

See your earlier misconception regarding loss being a simple function of
SWR.

I didn't say 'simple'. I said 'essentially'.

The impedance looking into the tuner end of the feeder is the
impedance of the load, transformed by length of the feeder, and is
also modified by the loss of the feeder.

You got that right.

Yes, I know I did.

The higher the feeder loss, the closer the impedance seen at the tuner
end will approach the characteristic impedance of the feeder. [Long
lengths of lossy feeder - maybe with a nominal termination on the far
end - can make good dummy loads at VHF and UHF.]

Yes, but is it of practical application in a transmit scenario?

As a termination, I've used a fairly long piece of coax plus, for good
luck, a low-power termination hung on the end.

If the
input impedance due to a severly mismatched load is at all close to Zo,
then you have lost most of the transmitter power in the line.

Quite.

The "make a good dummy load" recipe doesn't address the power rating,
especially where most of the power is dissipated in a very short length
of cable.

Oh, indeed. But I have to confess that I've realised that I'm suddenly
unsure of where, on a long, lossy, feeder with a mismatched load, most
of the absolute power is actually lost. I'm assuming that it is at the
TX end, where the power is greatest - even though that's where the SWR
is best. Please advise!
--
Ian

On Sep 1, 10:29 am, Ian Jackson
wrote:
Oh, indeed. But I have to confess that I've realised that I'm suddenly
unsure of where, on a long, lossy, feeder with a mismatched load, most
of the absolute power is actually lost. I'm assuming that it is at the
TX end, where the power is greatest - even though that's where the SWR
is best. Please advise!

It seems logical that the highest I^2*R losses would be where the
standing-wave current is maximum and the highest dielectric losses
wold be where the the standing-wave voltage is maximum. As Owen is
fond of pointing out, the locations of those points are very
important. If a current maximum point exists at the load and a voltage
maximum point exists at the source, the losses at the load are
probably higher than the losses at the source on HF. If a voltage
maximum point exists at the load and a current maximum point exists at
the source, the losses at the load are probably lower than the losses
at the source on HF (assuming that losses due to SWR are mostly I^2*R
losses on HF).
--
73, Cecil, w5dxp.com

On 9/1/2010 9:20 AM, Cecil Moore wrote:

...
It seems logical that the highest I^2*R losses would be where the
standing-wave current is maximum and the highest dielectric losses
wold be where the the standing-wave voltage is maximum. As Owen is
fond of pointing out, the locations of those points are very
important. If a current maximum point exists at the load and a voltage
maximum point exists at the source, the losses at the load are
probably higher than the losses at the source on HF. If a voltage
maximum point exists at the load and a current maximum point exists at
the source, the losses at the load are probably lower than the losses
at the source on HF (assuming that losses due to SWR are mostly I^2*R
losses on HF).
--
73, Cecil, w5dxp.com

Uhhh, sorry to pose this question to you here but, doesn't a "tuner"
really just "shorten" and "lengthen" the feed line? I mean, not
physically, of course. But, it would seem to me that, this is exactly
what my xmitter and ant are "seeing."

Regards,
JS

On Sep 1, 11:43*am, John Smith wrote:
Uhhh, sorry to pose this question to you here but, doesn't a "tuner"
really just "shorten" and "lengthen" the feed line? *I mean, not
physically, of course. *But, it would seem to me that, this is exactly
what my xmitter and ant are "seeing."

If all the tuner did was shorten and lengthen the feedline, there
would be only one purely resistive low resistance available on the SWR
circle on the Smith Chart and it wouldn't be exactly 50 ohms. So a
tuner does one other thing - it transforms the non-50 ohm resistive
impedance to a 50 ohm resistive impedance. It is not only the
equivalent of shortening and lengthening the transmission line but
also performs an N:1 transformer function. Of course, it does that
seamlessly, i.e. it is not actually a two-step process.

When I vary the length of my ladder-line to obtain system resonance, I
have to be satisfied with purely resistive impedances between 35 ohms
and 85 ohms. I cannot achieve a perfect 50 ohms on all HF bands with
my matching method. That would require the addition of an actual N:1
transformer which is certainly possible but probably not worth the
effort. Since 35-85 ohms is perfectly acceptable to my SC-500 amp, I
don't need a high-power tuner. And since the SC-500 is spec'ed to
handle an SWR of 6:1, I doubt that an 35-85 ohm load makes it
"unhappy". If my SC-500 has ever been "unhappy", I failed to observe
that "unhappiness" but maybe I am just oblivious to such?
--
73, Cecil, w5dxp.com