"Jeff Liebermann" wrote in message
...

Drivel: All of my 50 ohm antennas on my roof are connected to their
respective radios with 75 ohm RG-6/u coax, F connectors, and various
adapters. Also some RG-6/u with BNC compression connectors. No
problems and very little additional mismatch loss.


Often simple dipoles are closer to 70 ohms than 50. Not enough to make any
differance in most ham instalations.

I saw on youtube where you could take the rg-6 and after you strip it back ,
wrap about 6 turns of duck tape around it just where the outer jacket stops.
You only need a strip about 1/2 or 1/4 of an inch wide. Then you can fold
the braid back and use crimp connectors designed for rg-8 size. That helps
solve the aluminum jacket problem with the PL259s.

On Thu, 29 Aug 2013 17:13:39 -0400, "Ralph Mowery"
wrote:


"Jeff Liebermann" wrote in message
.. .

Drivel: All of my 50 ohm antennas on my roof are connected to their
respective radios with 75 ohm RG-6/u coax, F connectors, and various
adapters. Also some RG-6/u with BNC compression connectors. No
problems and very little additional mismatch loss.


Often simple dipoles are closer to 70 ohms than 50. Not enough to make any
differance in most ham instalations.

Actually, the mismatched RG-6/u can be better than the properly
matched RG-58c/u. For a given diameter, 75 ohm coax has less loss
than 50 ohm coax. 50 ohms has the advantage of being able to handle
more power, but at the expense of some additional loss.
http://www.microwaves101.com/encyclopedia/why50ohms.cfm

However, what was referring to was mismatch loss.
Voltage_reflection off of a 75 ohm input:
Vr = (50-75)/(50+75) = 0.2
Voltage transmission into 75 ohm input:
T = 1 - (0.2^2) = 0.96
Converting to decibels, the loss will be:
Mismatch_Loss = 20 Log 0.96 = -0.35 dB
mismatch loss. Note that the mismatch loss is independent of the
length of coax cable.

If you compare various common cables with RG-6/u, the benefits of the
better RG-6/u coax are obvious. 0.35dB of mismatch loss isn't going
to make much difference when there's 2 to 5 dB/100m difference in
attenuation.
http://vk1od.net/calc/tl/tllc.php
RG-8x = -12.6 dB/100m at 150 MHz.
LMR-240 = -9.89 dB/100m at 150 MHz.
RG-6/u = -7.78 dB/100m at 150 Mhz.

I saw on youtube where you could take the rg-6 and after you strip it back ,
wrap about 6 turns of duck tape around it just where the outer jacket stops.
You only need a strip about 1/2 or 1/4 of an inch wide. Then you can fold
the braid back and use crimp connectors designed for rg-8 size. That helps
solve the aluminum jacket problem with the PL259s.

Retch. There's no way to tightly crimp a few layers of tape.
Compressing the tape will cause it to cold flow at the glue junctions,
eventually causing the tape to slither out of the connector. Add a
little hot weather and the connector falls apart as the duct tape wrap
unravels. I know because I've done tricks like that eventually
failed. However, several layers of shrink tube might work because
shrink tube doesn't slide.


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

"Jeff Liebermann" wrote in message
...
However, what was referring to was mismatch loss.
Voltage_reflection off of a 75 ohm input:
Vr = (50-75)/(50+75) = 0.2
Voltage transmission into 75 ohm input:
T = 1 - (0.2^2) = 0.96
Converting to decibels, the loss will be:
Mismatch_Loss = 20 Log 0.96 = -0.35 dB

mismatch loss. Note that the mismatch loss is independent of the
length of coax cable.

I agree with what you have been saying.

I may not be reading the part above the way you wrote it, but think I am.

I am thinking that that .35 db loss due to the mismatch is in adition to
the loss per unit length. That is if you have 100 feet of coax with a loss
of say 3 db when matched, you will have a loss of 3.35 per 100 feet, and if
you go to 200 feet you will have a loss of 6.70 db and not 6.35.

On Fri, 30 Aug 2013 11:03:22 -0400, "Ralph Mowery"
wrote:


"Jeff Liebermann" wrote in message
.. .
However, what was referring to was mismatch loss.
Voltage_reflection off of a 75 ohm input:
Vr = (50-75)/(50+75) = 0.2
Voltage transmission into 75 ohm input:
T = 1 - (0.2^2) = 0.96
Converting to decibels, the loss will be:
Mismatch_Loss = 20 Log 0.96 = -0.35 dB

mismatch loss. Note that the mismatch loss is independent of the
length of coax cable.

I am thinking that that .35 db loss due to the mismatch is in adition to
the loss per unit length. That is if you have 100 feet of coax with a loss
of say 3 db when matched, you will have a loss of 3.35 per 100 feet, and if
you go to 200 feet you will have a loss of 6.70 db and not 6.35.

http://en.wikipedia.org/wiki/Mismatch_loss

I've always assumed that it was independent of length because the
mismatch loss can only occur at two points (source and load) and is
not a "bulk" phenomenon. Note that this is a 75 ohm system, not a 50
ohm system, where both the 50 ohm source and load are mismatched to
the 75 ohm transmission media. That makes things a bit easier to
visualize. My

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.

Sanity check:
http://vk1od.net/calc/tl/tllc.php
Plug in:
Belden 1530A (RG-6/u)
100 meters
150 MHz
Zload = 50
which results in:
Line Loss (matched) 7.924 dB
Line Loss 8.097 dB
VSWR(50)in 1.59
Mismatch loss = 8.097 - 7.924 = 0.1730 dB

Now, changing on the 100 meters to 500 meters should produce 5 times
the mismatch loss if your method is correct. It doesn't:
Line Loss (matched) 39.622 dB
Line Loss 39.799 dB
VSWR(50)in 1.50
Mismatch loss = 39.799 - 39.622 = 0.01770 dB
which is almost identical to the 100 meter caculation.

Note that this is for the load end of the coax only. A mismatch at
the source would produce an additional 0.1730 dB loss or:
2 * 0.1730 = 0.3460 dB
total mismatch loss, which corresponds nicely to my original 0.35 dB
loss calculation.

--
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 8/31/2013 4:41 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 buys his coax from "The Lossless Coax Store".

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.

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

On Friday, August 30, 2013 11:26:03 AM UTC-5, Jeff Liebermann wrote:
The antenna (or coax) does not get warmer because of mismatch loss.

Sounds like a confusing play on words to me.

IF (the greater the mismatch loss) THEN (the higher the SWR) is TRUE
AND
IF (the higher the SWR) THEN (the greater the heat loss in the transmission line) is TRUE
THEN (the greater the mismatch loss, the greater the heat loss in the transmission line). LOGIC 101
--
73, Cecil, w5dxp.com

On Sat, 31 Aug 2013 10:41:24 +0100, 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

Coax attenuation is part of a separate loss calculation that does not
involve matching. Mismatch loss is in addition to the coax losses.
Actually, that's not quite right. Mismatch loss is not a real loss,
where RF is converted to heat. It's simply the amount of additional
power that could have been delivered to the load had the system been
properly matched.

Let's try the boundary conditions and see what breaks. The coax
attenuation (in both directions) changes the measured VSWR and
therefore the mismatch loss. For example, if you had a ridiculously
long length of coax, with plenty of attenuation, the reflected RF at
the source is sufficiently attenuated so that the VSWR looks very
close to 1:1. Therefore, there's no mismatch, and therefore no
mismatch loss.[1]

At the other extreme, very short lengths of coax cable, have almost no
effect on the end point VSWR's. For this example, we have a 50 ohm
source, 75 ohm coax, and 50 ohm load. Reduce the 75 ohm coax cable to
near zero length, and there's no coax attenuation. Since the source
and load are matched, there's no mismatch, and therefore no mismatch
loss.

So, by your interpretation, there's no mismatch loss at the boundary
conditions (short coax and very long coax), while there's allegedly
mismatch loss for coax cable lengths in between? I don't think so.
More likely that the mismatch loss is unchanged, no matter how long or
short the cable.


[1] In the distant past, I wired 10base2 ethernet (cheapernet) at
several customers using existing 75 ohm CATV coax cables. 50 ohm
transceivers, 75 ohm coax, and 50 ohm resistive terminators. No
problems (other than crappy crimps). I also played with two 1000ft
rolls of RG-6/u and RG-58a/u. The 75 ohm RG-6/u was better because of
much lower losses (6dB versus 14dB at 10 Mhz).



--
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 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 11:15:50 AM UTC-5, Jeff Liebermann wrote:
Mismatch loss is not a real loss ... It's simply the amount of
additional power that could have been delivered to the load had
the system been properly matched.

Not quite correct yet. Consider the following system:

100w source---1/2WL 291.5 ohm twinlead---50 ohm load

The mismatch loss at the load is 3dB but the source is already delivering its maximum available power so there is ZERO additional power available.

(Forward Power) minus (Maximum Source Power) is NOT available for delivery to the load no matter what the mismatch loss happens to be.
--
73, Cecil, w5dxp.com