"rickman" wrote in message ...

On 7/2/2015 3:52 PM, Wayne wrote:


"rickman" wrote in message ...

On 7/2/2015 12:18 PM, Wayne wrote:


"John S" wrote in message ...

On 7/1/2015 10:56 AM, Ian Jackson wrote:
In message , John S
writes
On 6/29/2015 3:47 PM, Wayne wrote:

snipped to shorten

Ok. Well, 43ft is a half wavelength at about 12MHz. The vertical will
be very high impedance at that frequency and a 1:4 unun will
theoretically bring that impedance down closer to the feed line
impedance.

Does this help?

It was been pointed out to me that the figures for feeder loss with an
imperfect SWR are only correct when the length is fairly long (at least
an electrical wavelength?). How much loss does 25' of RG-8 really have
at 12MHz, when there's a halfwave hanging on the far end?

# A *resonant* half wave at 12MHz is about 36.7 feet long and it presents
# an impedance of about 1063 + j0 ohms to the RG-8 at the antenna end.
The
# current at the antenna end is 0.0245A while one watt is applied at the
# source end. This means that the power applied to the antenna is about
# 0.687W. So, about 68% of the applied power reaches the antenna.

# So, about 32% of the power is lost in the RG-8 for this example.

I'm just trying to understand this, so let me ask a question about your
example.

Isn't the 32% lost a function of not having a conjugate match maximum
power transfer?
If the transmitter had a Z of 1063 -j0, and a lossless RG8 feedline,
wouldn't maximum power be transferred?
(Even with a SWR of about 21:1)

# Transferred where? The match at the transmitter output only matches the
# output to the line. There are still reflections from the mismatch at
# the antenna. These reflections result in extra losses in the line as
# well as power delivered back into the transmitter output stage
# (especially with a perfect impedance match).

Well, I put a few (unrealistic) qualifiers into my question: a
transmitter with a a 1063 ohm output (not 50), and a lossless RG-8.

Thus, the back and forth reflections would not have attenuation.
And the transmitter and load are conjugately matched for maximum power
transfer.

# Your quoting style is very confusing. If you use with a space at the
# front of lines you are quoting it will show up the same as everyone
# else's quotes.

It's a problem with my newsreader not doing the proper job.



#Why will the reflections not have losses?

Because the assumption I posed was for a lossless line. In that case, with
a conjugate match on both ends, wouldn't there be maximum power transmission
regardless of the SWR?

......just a question I'm posing to the group.

With no line losses, and a conjugate match, is the SWR of any consequence?






A matched impedance does not mean no losses. It means the maximum
transfer of power. These are not at all the same thing.


# But I don't see anyone taking wavelength vs. feed line length into
# account. If the wavelength is long compared to the feed line I believe
# a lot of the "bad" stuff goes away. But then I am used to the digital
# transmission line where we aren't really concerned with delivering
# power, rather keeping a clean waveform of our (relatively) square waves.
# So I guess a short feed line doesn't solve the SWR problems... or does
# it?

The attenuation at a given high SWR depends upon the the matched
feedline loss, as reflections encounter that loss with every forward or
backward trip.
Thus feedline length/attenuation should be considered.

As a young man I was given a problem of solving poor antenna performance
on an aircraft band fixed station antenna. The SWR at the transmitter
was close to 1:1, but the antenna didn't work well.
I climbed up on the tower and found that the coax had never been
connected to the antenna. That was with about 400 feet of coax at 120
MHz.



# So how was the SWR 1:1?

It was a long run of coax at 120 MHz. The reflected wave was was attenuated
considerably by the time it returned to the source.

On 7/2/2015 8:53 PM, Wayne wrote:


"rickman" wrote in message ...

On 7/2/2015 3:52 PM, Wayne wrote:


"rickman" wrote in message ...

On 7/2/2015 12:18 PM, Wayne wrote:


"John S" wrote in message ...

On 7/1/2015 10:56 AM, Ian Jackson wrote:
In message , John S
writes
On 6/29/2015 3:47 PM, Wayne wrote:

snipped to shorten

Ok. Well, 43ft is a half wavelength at about 12MHz. The vertical will
be very high impedance at that frequency and a 1:4 unun will
theoretically bring that impedance down closer to the feed line
impedance.

Does this help?

It was been pointed out to me that the figures for feeder loss with an
imperfect SWR are only correct when the length is fairly long (at least
an electrical wavelength?). How much loss does 25' of RG-8 really have
at 12MHz, when there's a halfwave hanging on the far end?

# A *resonant* half wave at 12MHz is about 36.7 feet long and it
presents
# an impedance of about 1063 + j0 ohms to the RG-8 at the antenna end.
The
# current at the antenna end is 0.0245A while one watt is applied at the
# source end. This means that the power applied to the antenna is about
# 0.687W. So, about 68% of the applied power reaches the antenna.

# So, about 32% of the power is lost in the RG-8 for this example.

I'm just trying to understand this, so let me ask a question about your
example.

Isn't the 32% lost a function of not having a conjugate match maximum
power transfer?
If the transmitter had a Z of 1063 -j0, and a lossless RG8 feedline,
wouldn't maximum power be transferred?
(Even with a SWR of about 21:1)

# Transferred where? The match at the transmitter output only matches
the
# output to the line. There are still reflections from the mismatch at
# the antenna. These reflections result in extra losses in the line as
# well as power delivered back into the transmitter output stage
# (especially with a perfect impedance match).

Well, I put a few (unrealistic) qualifiers into my question: a
transmitter with a a 1063 ohm output (not 50), and a lossless RG-8.

Thus, the back and forth reflections would not have attenuation.
And the transmitter and load are conjugately matched for maximum power
transfer.

# Your quoting style is very confusing. If you use with a space at the
# front of lines you are quoting it will show up the same as everyone
# else's quotes.

It's a problem with my newsreader not doing the proper job.


#Why will the reflections not have losses?

Because the assumption I posed was for a lossless line. In that case,
with a conjugate match on both ends, wouldn't there be maximum power
transmission regardless of the SWR?

You aren't grasping the issue. Losses are *not* only in the
transmission line. When a reflected wave returns to the transmitter
output, it is not reflected 100%. If the output and transmission line
are matched exactly, 50% of the reflected wave reaching the output will
be reflected and 50% will be dissipated in the output stage.

Are you suggesting that the conjugate match will reflect back to the
antenna 100% of the original reflected wave from the antenna?

--

Rick

"rickman" wrote in message ...

On 7/2/2015 8:53 PM, Wayne wrote:


"rickman" wrote in message ...

On 7/2/2015 3:52 PM, Wayne wrote:

Why will the reflections not have losses?

Because the assumption I posed was for a lossless line. In that case,
with a conjugate match on both ends, wouldn't there be maximum power
transmission regardless of the SWR?

You aren't grasping the issue. Losses are *not* only in the transmission
line. When a reflected wave returns to the transmitter output, it is not
reflected 100%. If the output and transmission line are matched exactly,
50% of the reflected wave reaching the output will be reflected and 50%
will be dissipated in the output stage.

I don't think I've ever heard that anywhere before. Could you elaborate?


Are you suggesting that the conjugate match will reflect back to the
antenna 100% of the original reflected wave from the antenna?

Well, yes. Minus losses in matching networks and transmission lines.

In examples with lossless lines and lossless matching networks, wouldn't it
be 100%.

On 7/3/2015 3:27 PM, Wayne wrote:


"rickman" wrote in message ...

On 7/2/2015 8:53 PM, Wayne wrote:


"rickman" wrote in message ...

On 7/2/2015 3:52 PM, Wayne wrote:

Why will the reflections not have losses?

Because the assumption I posed was for a lossless line. In that case,
with a conjugate match on both ends, wouldn't there be maximum power
transmission regardless of the SWR?

You aren't grasping the issue. Losses are *not* only in the
transmission line. When a reflected wave returns to the transmitter
output, it is not reflected 100%. If the output and transmission line
are matched exactly, 50% of the reflected wave reaching the output
will be reflected and 50% will be dissipated in the output stage.

I don't think I've ever heard that anywhere before. Could you elaborate?

I'm not so sure now. I think I mentioned before that I learned about
transmission lines in the digital context where source and loads are
largely resistive. Resistance dissipates power. So when matched the
source dissipates as much power as delivered to the load (or
transmission line). Likewise, matched impedance will not reflect power,
but rather it is all absorbed. That is what happens at the antenna for
sure. But I'm not clear about what this conjugate network is really.
If it is purely reactive, then it will not have losses other than the
parasitics.

I have to admit I am not fluent in the complex math of networks. So off
hand an impedance of 1063 -j0 says to me resistive. The imaginary part
implies phase shifting, no? With that term being 0 doesn't that say the
capacitive and inductive parts cancel out leaving only resistance? If
you can, please explain how I am wrong.


Are you suggesting that the conjugate match will reflect back to the
antenna 100% of the original reflected wave from the antenna?

Well, yes. Minus losses in matching networks and transmission lines.

In examples with lossless lines and lossless matching networks, wouldn't
it be 100%.

I don't get how the matching network will reflect the wave from the
antenna 100%. Is that something you can explain?

--

Rick

On 7/4/2015 4:07 PM, rickman wrote:
On 7/3/2015 3:27 PM, Wayne wrote:


"rickman" wrote in message ...

On 7/2/2015 8:53 PM, Wayne wrote:


"rickman" wrote in message ...

On 7/2/2015 3:52 PM, Wayne wrote:

Why will the reflections not have losses?

Because the assumption I posed was for a lossless line. In that case,
with a conjugate match on both ends, wouldn't there be maximum power
transmission regardless of the SWR?

You aren't grasping the issue. Losses are *not* only in the
transmission line. When a reflected wave returns to the transmitter
output, it is not reflected 100%. If the output and transmission line
are matched exactly, 50% of the reflected wave reaching the output
will be reflected and 50% will be dissipated in the output stage.

I don't think I've ever heard that anywhere before. Could you elaborate?

I'm not so sure now. I think I mentioned before that I learned about
transmission lines in the digital context where source and loads are
largely resistive. Resistance dissipates power. So when matched the
source dissipates as much power as delivered to the load (or
transmission line). Likewise, matched impedance will not reflect power,
but rather it is all absorbed. That is what happens at the antenna for
sure. But I'm not clear about what this conjugate network is really. If
it is purely reactive, then it will not have losses other than the
parasitics.

I have to admit I am not fluent in the complex math of networks. So off
hand an impedance of 1063 -j0 says to me resistive. The imaginary part
implies phase shifting, no? With that term being 0 doesn't that say the
capacitive and inductive parts cancel out leaving only resistance? If
you can, please explain how I am wrong.


Are you suggesting that the conjugate match will reflect back to the
antenna 100% of the original reflected wave from the antenna?

Well, yes. Minus losses in matching networks and transmission lines.

In examples with lossless lines and lossless matching networks, wouldn't
it be 100%.

I don't get how the matching network will reflect the wave from the
antenna 100%. Is that something you can explain?


Yes, he is correct.

Think of it this way, without the math. On the transmitter side of the
network, the match is 1:1, with nothing reflected back to the transmitter.

So you have a signal coming back from the antenna. You have a perfect
matching network, which means nothing is lost in the network. The
feedline is perfect, so there is no loss in it. The only place for the
signal to go is back to the antenna.

In a perfect system, all power is transferred to the antenna, even with
a large mismatch between the feedline and the antenna. However, that's
still not the same as having a match at the antenna, because reflected
signals most likely will arrive out of phase with the original signal.


-- ==================
Remove the "x" from my email address
Jerry, AI0K

==================

On Sat, 04 Jul 2015 19:04:01 -0400, Jerry Stuckle
wrote:
Think of it this way, without the math. On the transmitter side of the
network, the match is 1:1, with nothing reflected back to the transmitter.

So you have a signal coming back from the antenna. You have a perfect
matching network, which means nothing is lost in the network. The
feedline is perfect, so there is no loss in it. The only place for the
signal to go is back to the antenna.

Wikipedia says that if the source is matched to the line, any
reflections that come back are absorbed, not reflected back to the
antenna:

https://en.wikipedia.org/wiki/Impedance_matching
"If the source impedance matches the line, reflections
from the load end will be absorbed at the source end.
If the transmission line is not matched at both ends
reflections from the load will be re-reflected at the
source and re-re-reflected at the load end ad infinitum,
losing energy on each transit of the transmission line."

--
Jeff Liebermann
150 Felker St #D http://www.LearnByDestroying.com
Santa Cruz CA 95060 http://802.11junk.com
Skype: JeffLiebermann AE6KS 831-336-2558

On 7/4/2015 7:22 PM, Jeff Liebermann wrote:
On Sat, 04 Jul 2015 19:04:01 -0400, Jerry Stuckle
wrote:
Think of it this way, without the math. On the transmitter side of the
network, the match is 1:1, with nothing reflected back to the transmitter.

So you have a signal coming back from the antenna. You have a perfect
matching network, which means nothing is lost in the network. The
feedline is perfect, so there is no loss in it. The only place for the
signal to go is back to the antenna.

Wikipedia says that if the source is matched to the line, any
reflections that come back are absorbed, not reflected back to the
antenna:

https://en.wikipedia.org/wiki/Impedance_matching
"If the source impedance matches the line, reflections
from the load end will be absorbed at the source end.
If the transmission line is not matched at both ends
reflections from the load will be re-reflected at the
source and re-re-reflected at the load end ad infinitum,
losing energy on each transit of the transmission line."


And you believe everything Wikipedia says? ROFLMAO.

But that also explains your ignorance.


--
==================
Remove the "x" from my email address
Jerry Stuckle

==================

"Jeff Liebermann" wrote in message
...

On Sat, 04 Jul 2015 19:04:01 -0400, Jerry Stuckle
wrote:
Think of it this way, without the math. On the transmitter side of the
network, the match is 1:1, with nothing reflected back to the transmitter.

So you have a signal coming back from the antenna. You have a perfect
matching network, which means nothing is lost in the network. The
feedline is perfect, so there is no loss in it. The only place for the
signal to go is back to the antenna.

Wikipedia says that if the source is matched to the line, any
reflections that come back are absorbed, not reflected back to the
antenna:

https://en.wikipedia.org/wiki/Impedance_matching
"If the source impedance matches the line, reflections
from the load end will be absorbed at the source end.
If the transmission line is not matched at both ends
reflections from the load will be re-reflected at the
source and re-re-reflected at the load end ad infinitum,
losing energy on each transit of the transmission line."

Well, I looked at that section of the writeup.
And, I have no idea what the hell they are talking about.
Looks like a good section for a knowledgeable person to edit.

On 7/3/2015 2:50 AM, Jeff wrote:

Are you suggesting that the conjugate match will reflect back to the
antenna 100% of the original reflected wave from the antenna?


Yes, it must.

For example with an external ATU that provides a conjugate match it is
clearly the case that if a 1:1 VSWR is achieved then no reflected power
reaches the TX. (as shown on an SWR meter between the Tx and ATU.)

I am very certain that this assumption is not correct. I wish I had the
math to back me up. The only total reflection I am aware of is an open
circuit which of course absorbs no power at all.

Here is a point. The VSWR only shows no power being sent back to the
txmt output. That does not mean no power is absorbed from the reflected
wave by the matching circuit. I believe the example you gave was Z of
1063 -j0. Isn't that a real impedance of 1063 ohms which is equivalent
to a resistor? Resistors dissipate power don't they?

--

Rick

In message , rickman
writes
On 7/3/2015 2:50 AM, Jeff wrote:

Are you suggesting that the conjugate match will reflect back to the
antenna 100% of the original reflected wave from the antenna?


Yes, it must.

For example with an external ATU that provides a conjugate match it is
clearly the case that if a 1:1 VSWR is achieved then no reflected power
reaches the TX. (as shown on an SWR meter between the Tx and ATU.)

I am very certain that this assumption is not correct. I wish I had
the math to back me up. The only total reflection I am aware of is an
open circuit which of course absorbs no power at all.

Here is a point. The VSWR only shows no power being sent back to the
txmt output. That does not mean no power is absorbed from the
reflected wave by the matching circuit. I believe the example you gave
was Z of 1063 -j0. Isn't that a real impedance of 1063 ohms which is
equivalent to a resistor? Resistors dissipate power don't they?

I have to admit that I am, to some extent, confused.

Maybe it helps to look at the situation from the point of view that the
matching circuit doesn't 'know' that there is a reflected wave. All it
sees is the impedance looking into the sending end of the coax - and
this is whatever is on the antenna end, transformed by the length of
coax. The load the matching unit sees could be replaced with the same
physical values of L, C and R, so there IS nowhere for a reflected wave
to exist.

Provided the TX sees a 50 ohm load when looking into the input of the
matcher, there will be no theoretical losses. However, a real-life
matcher WILL have loss, and so will the coax. Also, the coax will have a
loss greater than when it is matched, mainly because of the 'I -squared
R' (literal) hot-spots.
--
Ian