On 8/31/2013 5:41 AM, Jeff wrote:


Note that the 0.35 dB loss is not converted to heat or dissipated. The
antenna (or coax) does not get warmer because of mismatch loss. All
that happens is that some of the power gets reflected around and does
not get radiated out the antenna.

So the reflected wave is somehow mysteriously exempt from the loss/m of
the coax then!!

Jeff

He is correct. That 0.35 db loss exists even if you have zero feet of
coax. It is a "point loss", unrelated to coax length.

The loss in the coax is separate.

--
==================
Remove the "x" from my email address
Jerry Stuckle
JDS Computer Training Corp.

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On Sat, 31 Aug 2013 15:44:50 +0100, Jeff wrote:

On 31/08/2013 15:15, Jerry Stuckle wrote:
On 8/31/2013 5:41 AM, Jeff wrote:


Note that the 0.35 dB loss is not converted to heat or dissipated. The
antenna (or coax) does not get warmer because of mismatch loss. All
that happens is that some of the power gets reflected around and does
not get radiated out the antenna.

So the reflected wave is somehow mysteriously exempt from the loss/m of
the coax then!!

Jeff

He is correct. That 0.35 db loss exists even if you have zero feet of
coax. It is a "point loss", unrelated to coax length.

The loss in the coax is separate.

The loss may be 'separate' but that coax does *get warmer* as the
reflected power also experiences loss in the cable, so he is not correct.
Jeff

Ok, let's try a different approach. Assumptions:
1. Only resistive losses generate heat. Reactive loads and
transmission lines do not generate any heat.
2. Below about 1GHz, the dominant loss mechanism in coax cable
is I^2*R heating losses in the copper conductors.
3. The coax is assumed to be non-radiating.
4. Coax looks resistive because the distributed capacitance
and inductive reactances cancel, leaving only the I^2*R losses.

Therefore, if I replace a length of 50 ohm coax, with a physically
similar length of 75 ohm coax, the I^2*R losses do not change. What
does change are the standing waves along the coax, which will cause
mismatch losses. However, the basic coax loss, as controlled by the
I^2*R losses, remains unchanged. Therefore, since the mismatch losses
are all inspired by changes in reactance, there is no additional
heating losses produced by the mismatch losses, since reactive loads
and transmission lines do not generate any heat.

Anyway, please note my use of the forms at:
http://vk1od.net/calc/tl/tllc.php
to calculate the mismatch loss for various cable lengths. I
previously demonstrated that the mismatch loss is constant, no matter
how long or short the transmission line. I'm fairly sure the
calculations are correct. I'm not so certain of my explanation.

--
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 Saturday, August 31, 2013 11:37:20 AM UTC-5, Jeff Liebermann wrote:
Therefore, if I replace a length of 50 ohm coax, with a physically
similar length of 75 ohm coax, the I^2*R losses do not change.

What you may be missing is that the RMS value of the current is higher when reflections are present than when they are not present. Therefore, the I^2*R losses in the transmission line are higher when reflections are present. Part of the reflected energy from the load (used to calculate mismatch loss) is dissipated as heat in the I^2*R of the copper transmission line as illustrated by the following example.

Consider 200 ft. of RG-58 used on 440 MHz driving a 291.5 ohm load. The mismatch loss at the load is 3dB but the loss in the coax is 29.4 dB and the impedance looking into the coax at the source is 50.12-j0.19 ohms, almost a perfect match. Would you still argue that none of the power involved in the mismatch loss is dissipated in the coax?
--
73, Cecil, w5dxp.com

"W5DXP" wrote in message
...

On Saturday, August 31, 2013 11:37:20 AM UTC-5, Jeff Liebermann wrote:
Therefore, if I replace a length of 50 ohm coax, with a physically
similar length of 75 ohm coax, the I^2*R losses do not change.

# What you may be missing is that the RMS value of the current is higher
when reflections are present than when
# they are not present. Therefore, the I^2*R losses in the transmission line
are higher when reflections are
#present. Part of the reflected energy from the load (used to calculate
mismatch loss) is dissipated
# as heat in the I^2*R of the copper transmission line as illustrated by
the following example.

# Consider 200 ft. of RG-58 used on 440 MHz driving a 291.5 ohm load. The
mismatch loss at the load
# is 3dB but the loss in the coax is 29.4 dB and the impedance looking into
the coax at the
# source is 50.12-j0.19 ohms, almost a perfect match. Would you still argue
that none of the power
# involved in the mismatch loss is dissipated in the coax?
--
# 73, Cecil, w5dxp.com

This chart has been around a long time and indicates what is going on.
See figure 1 at:

http://www.arrl.org/files/file/Techn...f/q1106037.pdf

As a practical example, my elevated vertical (on a metal patio cover) is fed
with about 20 feet of RG-8. Matching is via a tuner right at the rig, and
the vertical element connects to the coax with no other matching.

RG-8 has a loss of about 0.55 db per 100 feet.

Assume that my 20 foot feedline has a full 0.55 dB of loss when matched. On
bands where the VSWR is 20:1,
according to the chart, the system will have additional loss of less than 3
dB.

And it works fine.

On Sunday, September 1, 2013 12:40:51 PM UTC-5, Wayne wrote:
... the system will have additional loss of less than 3 dB.
And it works fine.

So the question is: Is any part of the reflected power in the
mismatch loss calculation included in that 3 dB of additional
loss? The answer is 'yes' and whether it works fine or not is
irrelevant to the argument.
--
73, Cecil, w5dxp.com

"W5DXP" wrote in message
...

On Sunday, September 1, 2013 12:40:51 PM UTC-5, Wayne wrote:
... the system will have additional loss of less than 3 dB.
And it works fine.

So the question is: Is any part of the reflected power in the
mismatch loss calculation included in that 3 dB of additional
loss? The answer is 'yes' and whether it works fine or not is
irrelevant to the argument.
--
73, Cecil, w5dxp.com

Indeed. If there is loss going one direction on a line, reflected power
going the other way suffers loss also.

The part about "works fine" was thrown in, not for you, but for those who
might be horrified by a high vswr.

On Sunday, September 1, 2013 6:09:09 PM UTC-5, Wayne wrote:
The part about "works fine" was thrown in, not for you, but for those who
might be horrified by a high vswr.

Sorry, I misunderstood what you were trying to say.
--
73, Cecil, w5dxp.com