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

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

On Sat, 04 Jul 2015 19:33:30 -0400, Jerry Stuckle
wrote:

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.

Let's see if I understand you correctly. You claim that with a power
amplifier (source) output impedance that is perfectly matched to the
coax cable, but not necessarily the load (antenna), any reflected
power from the load (antenna) is bounced back to the load (antenna) by
the perfectly matched source (power amp). Is that what you're saying?

Yet, when I have a perfectly matched load (antenna), all the power it
is fed is radiated and nothing is reflected. You can't have it both
ways because the reflected power from the load (antenna), becomes the
incident power going towards the source (power amp). Matched and
mismatched loads do NOT act differently depending on the direction of
travel. If you claim were true, then transmitting into a matched
antenna or dummy load would reflect all the power back towards the
transmitter.



--
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 9:43 PM, Jeff Liebermann wrote:
On Sat, 04 Jul 2015 19:33:30 -0400, Jerry Stuckle
wrote:

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.

Let's see if I understand you correctly. You claim that with a power
amplifier (source) output impedance that is perfectly matched to the
coax cable, but not necessarily the load (antenna), any reflected
power from the load (antenna) is bounced back to the load (antenna) by
the perfectly matched source (power amp). Is that what you're saying?


With a perfect matching network and a perfect feedline (which is what we
are discussing), that is true. But I also know that is far beyond your
limited intelligence.

Yet, when I have a perfectly matched load (antenna), all the power it
is fed is radiated and nothing is reflected. You can't have it both
ways because the reflected power from the load (antenna), becomes the
incident power going towards the source (power amp). Matched and
mismatched loads do NOT act differently depending on the direction of
travel. If you claim were true, then transmitting into a matched
antenna or dummy load would reflect all the power back towards the
transmitter.


Nothing wrong with it at all - except your limited intelligence can't
understand simple physics. But then that's nothing new.

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

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

On 7/4/2015 10:37 PM, Jerry Stuckle wrote:

With a perfect matching network and a perfect feedline (which is what we
are discussing), that is true. But I also know that is far beyond your
limited intelligence.

Nothing wrong with it at all - except your limited intelligence can't
understand simple physics. But then that's nothing new.

Jeff, this is the sort of reply that you can expect from Jerry so that I
just don't even respond to him anymore unless... well, at the moment I
can't think of a reason to respond to what he writes.

--

Rick

On Sat, 04 Jul 2015 22:37:41 -0400, Jerry Stuckle
wrote:

On 7/4/2015 9:43 PM, Jeff Liebermann wrote:
Let's see if I understand you correctly. You claim that with a power
amplifier (source) output impedance that is perfectly matched to the
coax cable, but not necessarily the load (antenna), any reflected
power from the load (antenna) is bounced back to the load (antenna) by
the perfectly matched source (power amp). Is that what you're saying?

With a perfect matching network and a perfect feedline (which is what we
are discussing), that is true.

Ok, so we agree on that part.

Yet, when I have a perfectly matched load (antenna), all the power it
is fed is radiated and nothing is reflected. You can't have it both
ways because the reflected power from the load (antenna), becomes the
incident power going towards the source (power amp). Matched and
mismatched loads do NOT act differently depending on the direction of
travel. If you claim were true, then transmitting into a matched
antenna or dummy load would reflect all the power back towards the
transmitter.

Nothing wrong with it at all

Ah, but there's plenty wrong with your view. At the load (antenna)
end of the coax, we both agree that with a perfect match, perfect
coax, and perfect load, there's is no reflection. Yet when you look
at the other end of the same coax, the same perfectly matched coax
(S21 back into the PA) suddenly decides to reflect any RF that might
be returned from a load (antenna) mismatch. It seems rather odd that
RF would act differently at opposing ends of the coax cable. In the
forward direction, a matched load either absorbs or radiate. In the
reverse direction, a matched load changes its mind and decides to
reflect? I don't think so.

There's another problem with your view of how VSWR works. If I
transmitted into an open or short circuit load (antenna), all the
forward RF would be reflected back to the source (PA). That would
mean that the PA will need to protect itself from over voltage or over
current using the traditional VSWR protection circuit. Yet, if the PA
were perfectly matched into a perfect coax cable, all that reflected
RF would bounce off the PA and back to the load (antenna). The PA
would not see any of that RF, and there would not be any need for a
VSWR protection circuit. I don't think so.

Personal experience with blowing up finals has demonstrated to my
satisfaction that a perfectly matched PA is quite capable of being
blown up by transmitting into an open or short with no VSWR
protection. Presumably, the damage was done by the reflected RF
(causing over voltage or over current in the PA) which would not be
present in your scheme of things, with a perfectly matched PA and
coax.

Another problem is IMD (intermodulation) products produced in the
power amplifier (PA). This is not a major problem with HF radios, but
is a serious problem with mountain top repeater sites. The antennas
on the towers tend to be rather close together. RF from a nearby
transmitter can couple into adjacent antennas, travel down the coax to
the PA, mix with the transmit signal in the PA, get amplified by the
PA, and get re-radiated by the antenna. The effect is typically
blocked by cavity filters and one-way isolators or circulators. The
problem here is that if the perfectly matched PA really did reflect
anything coming down from the antenna back to the antenna, there would
be no need for such IMD protection. The RF from the adjacent antenna
would simply bounce back towards the antenna and be re-radiated
without any mixing taking place. It would be a wonderful world if it
worked that way, but it obviously does not.

There are other problems, but they require math to explain, which
requires more time than I have available right now.



--
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 9:43 PM, Jeff Liebermann wrote:
On Sat, 04 Jul 2015 19:33:30 -0400, Jerry Stuckle
wrote:

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.

Let's see if I understand you correctly. You claim that with a power
amplifier (source) output impedance that is perfectly matched to the
coax cable, but not necessarily the load (antenna), any reflected
power from the load (antenna) is bounced back to the load (antenna) by
the perfectly matched source (power amp). Is that what you're saying?

Yet, when I have a perfectly matched load (antenna), all the power it
is fed is radiated and nothing is reflected. You can't have it both
ways because the reflected power from the load (antenna), becomes the
incident power going towards the source (power amp). Matched and
mismatched loads do NOT act differently depending on the direction of
travel. If you claim were true, then transmitting into a matched
antenna or dummy load would reflect all the power back towards the
transmitter.

I think this is one of those situations where a casual explanation won't
work. You can use a "casual" explanation when the various
qualifications for a simplification apply. But to do that, the
qualifiers have to be fully understood and no one here is showing what
the qualifiers are much less that they are met. So until we get a real
explanation I will stick with what I recall. In the end, to settle this
we may have to use the math.

I'm sure someone in s.e.d could explain this properly. Some of them may
be purely argumentative, but some really know their stuff. I believe
the description of a conjugate match is the mathematical inverse of the
complex impedance of the antenna "viewed" through the feed line, but I
have to admit I don't really know what that implies or if it is even an
accurate description.

--

Rick

On 5.7.15 18:56, rickman wrote:
On 7/4/2015 9:43 PM, Jeff Liebermann wrote:
On Sat, 04 Jul 2015 19:33:30 -0400, Jerry Stuckle
wrote:

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.

Let's see if I understand you correctly. You claim that with a power
amplifier (source) output impedance that is perfectly matched to the
coax cable, but not necessarily the load (antenna), any reflected
power from the load (antenna) is bounced back to the load (antenna) by
the perfectly matched source (power amp). Is that what you're saying?

Yet, when I have a perfectly matched load (antenna), all the power it
is fed is radiated and nothing is reflected. You can't have it both
ways because the reflected power from the load (antenna), becomes the
incident power going towards the source (power amp). Matched and
mismatched loads do NOT act differently depending on the direction of
travel. If you claim were true, then transmitting into a matched
antenna or dummy load would reflect all the power back towards the
transmitter.

I think this is one of those situations where a casual explanation won't
work. You can use a "casual" explanation when the various
qualifications for a simplification apply. But to do that, the
qualifiers have to be fully understood and no one here is showing what
the qualifiers are much less that they are met. So until we get a real
explanation I will stick with what I recall. In the end, to settle this
we may have to use the math.

I'm sure someone in s.e.d could explain this properly. Some of them may
be purely argumentative, but some really know their stuff. I believe
the description of a conjugate match is the mathematical inverse of the
complex impedance of the antenna "viewed" through the feed line, but I
have to admit I don't really know what that implies or if it is even an
accurate description.


You're right. To get a 1:1 match for the piece of feedline between
the transmitter and the antenna tuning unit, the tuning unit has
to present a conjugate match to the feedline from the tuning unit
to the antenna. If there is a reflected wave from the antenna, it
will be re-reflected back toward the antenna from the tuning unit.
The ping-ponging signal will die out by antenna radiation or feedline
losses. The situation is quite OK with slow modulations (like voice),
but the ping-ponging is unacceptable for fast signals (like analog TV).

--

-TV