Coaxial Antenna question
 

I'm trying to build a decent performing 2M coaxial skirted antenna and
have a question about its design for maximum efficiency.

I based the "hub" on an SO-239 connector. I soldered my RG8X cable
center conductor to the solder pin center conductor and brought the
shield braid out in two places.

I cut a 19" piece of half inch copper pipe, cut 4 half inch slots on
one end, fanned the slotted end out slightly to fit nicely against the
SO-239, and slid this pipe over my coax and up to the SO-239. I brought
the two braid lengths previously prepared out through two of the rather
fat slots, soldered the pipe to the SO-239 and the braids were soldered
to the pipe where they came protruded out the slots.

I soldered an 18 1/2" brass welding rod to a PL-259 center conductor
and screwed that the the SO-239 for my radiator. The copper pipe assy
and coax slide nicely down into a length of 3/4" PVC . This makes a very
nice break-down package for transportation and portable work.

Back to my question: I am measuring about 125 watts forward and 4-5
watts reflected at my desired frequency ( 146.000 ). While this may be
acceptable to some, I would like to get the match down to 1:1 SWR.

Are there any design deficiencies in my proto-type? Can anyone
suggest something I might try to improve the match? Is there a "rule of
thumb" regarding the construction of such antennas?... length of coaxial
skirt vs. antenna element?

I appreciate any feedback on this. Thanks.

I suppose I could provide a picture if anyone requires it....

Ed K7AAT

Coaxial Antenna question
 

"Ed" wrote in message
. 192.196...

I'm trying to build a decent performing 2M coaxial skirted antenna and
have a question about its design for maximum efficiency.

I based the "hub" on an SO-239 connector. I soldered my RG8X cable
center conductor to the solder pin center conductor and brought the
shield braid out in two places.

I cut a 19" piece of half inch copper pipe, cut 4 half inch slots on
one end, fanned the slotted end out slightly to fit nicely against the
SO-239, and slid this pipe over my coax and up to the SO-239. I brought
the two braid lengths previously prepared out through two of the rather
fat slots, soldered the pipe to the SO-239 and the braids were soldered
to the pipe where they came protruded out the slots.

I soldered an 18 1/2" brass welding rod to a PL-259 center conductor
and screwed that the the SO-239 for my radiator. The copper pipe assy
and coax slide nicely down into a length of 3/4" PVC . This makes a very
nice break-down package for transportation and portable work.

Back to my question: I am measuring about 125 watts forward and 4-5
watts reflected at my desired frequency ( 146.000 ). While this may be
acceptable to some, I would like to get the match down to 1:1 SWR.

Are there any design deficiencies in my proto-type? Can anyone
suggest something I might try to improve the match? Is there a "rule of
thumb" regarding the construction of such antennas?... length of coaxial
skirt vs. antenna element?

I appreciate any feedback on this. Thanks.

I suppose I could provide a picture if anyone requires it....

Ed K7AAT

remember, too low an swr can kill you!

To get exactly 50 ohms you may need a fatter copper pipe, or some other
change to the geometry. if ti works well as it is i would leave it alone.

Coaxial Antenna question
 

On 02 Nov 2008 00:12:29 GMT, Ed
wrote:

I'm trying to build a decent performing 2M coaxial skirted antenna and
have a question about its design for maximum efficiency.
(...)
Back to my question: I am measuring about 125 watts forward and 4-5
watts reflected at my desired frequency ( 146.000 ). While this may be
acceptable to some, I would like to get the match down to 1:1 SWR.

125 watts forward with 5 watts back is a VSWR = 1.5:1

The characteristic impedance of a coaxial antenna is about 75 ohms.
The best you can do with your present arrangement is therefore about
1.5:1 which is what you're getting.

To do any better, you'll need some way to match the 75 ohm antenna to
your 50 ohm system. That's usually an odd multiple of 1/4 wave coax
section, with an impedance of 61 ohms. Something like:
http://www.repeater-builder.com/rbtip/matchingstubs.html

I suppose I could provide a picture if anyone requires it....

A JPG is worth 1000 guesses.

Also, you might want to do your testing with something less than 125
watts, as accidents and miscalculations might become expensive.

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

Coaxial Antenna question
 

Ed wrote:
. . .
Back to my question: I am measuring about 125 watts forward and 4-5
watts reflected at my desired frequency ( 146.000 ). While this may be
acceptable to some, I would like to get the match down to 1:1 SWR.
. . .

Actually, this is acceptable to just about everyone. And those for whom
it's not don't usually have a rational reason for it.

Here are some ways to do it:

1. Lower the impedance at resonance by making the top of the antenna
into an inverted cone shape, by using fanned out multiple conductors.
You'll have to adjust the length of the conductors and possibly the
sleeve to maintain the same resonant frequency.
2. Build and use a simple matching network, preferably placed as close
to the antenna as possible. Then you can adjust it for a perfect match.
3. Use some other form of impedance transformation such as a stub
matching network or transmission line transformer.

If you do things right, it'll work the same when you finish as it did
when you start. But you'll feel better seeing zero "reflected" power on
your meter, and the placebo effect is not to be sneezed at.

Roy Lewallen, W7EL

Coaxial Antenna question
 

Ed,

You might have expected the feedpoint impedance to be around 70 ohms. It
will depend on the feedline configuration, because you haven't taken much
is the way of measures to decouple the feedline.

Your measured fwd and ref indicates VSWR~=1.5 which is consistent with 70
ohms, but you haven't measured 70 ohms.

Assuming though that such an antenna should be close to 70+j0 at
resonance...

If you did want to incorporate an impedance matching system that doesn't
compromise the portability you have described, you could try a twelfth wave
transformer with 29.3° of 50 ohm coax from the feedpoint, then 29.3° of 75
ohm coax then any length of 50 ohm coax to the transmitter.

For example, for 146MHz, that could be 137mm of Belden 9258 (RG8/X) then
139mm of Belden 1189A (RG6/U) then any length of 50 ohm coax to the tx.

Owen

Coaxial Antenna question
 

Dave, Jeff, Roy, ...

Thank you for the excellent responses to my questions. I am
considering any and all of them.

One note: I erred on my previously stated power levels.... it was
1.25 Watts forward and about .05 watts reflected. Same ratio, but less
energy.

Ed

Coaxial Antenna question
 

"Ed" wrote in message
. 192.196...

I'm trying to build a decent performing 2M coaxial skirted antenna and
have a question about its design for maximum efficiency.

I based the "hub" on an SO-239 connector. I soldered my RG8X cable
center conductor to the solder pin center conductor and brought the
shield braid out in two places.

I cut a 19" piece of half inch copper pipe, cut 4 half inch slots on
one end, fanned the slotted end out slightly to fit nicely against the
SO-239, and slid this pipe over my coax and up to the SO-239. I brought
the two braid lengths previously prepared out through two of the rather
fat slots, soldered the pipe to the SO-239 and the braids were soldered
to the pipe where they came protruded out the slots.

I soldered an 18 1/2" brass welding rod to a PL-259 center conductor
and screwed that the the SO-239 for my radiator. The copper pipe assy
and coax slide nicely down into a length of 3/4" PVC . This makes a very
nice break-down package for transportation and portable work.

Back to my question: I am measuring about 125 watts forward and 4-5
watts reflected at my desired frequency ( 146.000 ). While this may be
acceptable to some, I would like to get the match down to 1:1 SWR.

Are there any design deficiencies in my proto-type? Can anyone
suggest something I might try to improve the match? Is there a "rule of
thumb" regarding the construction of such antennas?... length of coaxial
skirt vs. antenna element?

I appreciate any feedback on this. Thanks.

I suppose I could provide a picture if anyone requires it....

Ed K7AAT
Hi Ed

If you are in search of precission, I'd suggest that you consider using 70
ohm coax to feed the 70 ohm antenna. There are some very good 70 ohm coax
available at low cost. Their loss will be a fraction of a dB less than
your RG-8X. Their OD may be smaller (that helps a little to decouple the
stub from the coax). TheVSWR on the 70 ohm line will be lower than on the
50 ohm line. You will then be able to construct a tuning network at the
junction from the transmitter to the 70 ohm line. There will be
essentially no meaureable VSWR anywhere in the system.

Jerry KD6JDJ

Coaxial Antenna question
 

Owen Duffy wrote in
:


Ed,

You might have expected the feedpoint impedance to be around 70 ohms.
It will depend on the feedline configuration, because you haven't
taken much is the way of measures to decouple the feedline.

Your measured fwd and ref indicates VSWR~=1.5 which is consistent with
70 ohms, but you haven't measured 70 ohms.

Assuming though that such an antenna should be close to 70+j0 at
resonance...

If you did want to incorporate an impedance matching system that
doesn't compromise the portability you have described, you could try a
twelfth wave transformer with 29.3° of 50 ohm coax from the feedpoint,
then 29.3° of 75 ohm coax then any length of 50 ohm coax to the
transmitter.

For example, for 146MHz, that could be 137mm of Belden 9258 (RG8/X)
then 139mm of Belden 1189A (RG6/U) then any length of 50 ohm coax to
the tx.

Owen


Very nice, Owen. Saved me a lot of difficult math.... since I have
those materials on hand I may see what I can throw together tomorrow.

Ed

Coaxial Antenna question
 

On Sun, 02 Nov 2008 03:22:09 GMT, Owen Duffy wrote:

If you did want to incorporate an impedance matching system that doesn't
compromise the portability you have described, you could try a twelfth wave
transformer with 29.3° of 50 ohm coax from the feedpoint, then 29.3° of 75
ohm coax then any length of 50 ohm coax to the transmitter.

For example, for 146MHz, that could be 137mm of Belden 9258 (RG8/X) then
139mm of Belden 1189A (RG6/U) then any length of 50 ohm coax to the tx.

Owen

Very nice. Here's a bit more on how it works:
http://ourworld.compuserve.com/homepages/demerson/twelfth.htm
http://ourworld.compuserve.com/homepages/demerson/12thfdbk.htm

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

Coaxial Antenna question
 

Ed wrote in
36.82:

Owen Duffy wrote in
:


Ed,

You might have expected the feedpoint impedance to be around 70 ohms.
It will depend on the feedline configuration, because you haven't
taken much is the way of measures to decouple the feedline.

Your measured fwd and ref indicates VSWR~=1.5 which is consistent
with 70 ohms, but you haven't measured 70 ohms.

Assuming though that such an antenna should be close to 70+j0 at
resonance...

If you did want to incorporate an impedance matching system that
doesn't compromise the portability you have described, you could try
a twelfth wave transformer with 29.3° of 50 ohm coax from the
feedpoint, then 29.3° of 75 ohm coax then any length of 50 ohm coax
to the transmitter.

For example, for 146MHz, that could be 137mm of Belden 9258 (RG8/X)
then 139mm of Belden 1189A (RG6/U) then any length of 50 ohm coax to
the tx.

Owen


Very nice, Owen. Saved me a lot of difficult math.... since I
have
those materials on hand I may see what I can throw together tomorrow.

I didn't do the math, I punched the numbers into TLLC
(http://www.vk1od.net/tl/tllc.php).

Of course, the reason I was so specific is that translation from the
29.3° depends on the velocity factor... so use the velocity factor for
the cables you have at hand. (For example, if you use RG59, it has a
very different velocity factor to te 1189A, and you need to adjust
accordingly.)

There is a little on the twelfth wave transformer, including a graph of
the lengths for different transformation ratios at
http://www.vk1od.net/RG6/index.htm .

Have fun.

Owen

Coaxial Antenna question
 

Owen Duffy wrote in news:Xns9B4AAC084EA4nonenowhere@
61.9.191.5:

....
There is a little on the twelfth wave transformer, including a graph of
the lengths for different transformation ratios at
http://www.vk1od.net/RG6/index.htm .

If you read the article, you might try evaluating the systemusing TLLC with
say 10m of feedline configured with:
- the twelfth wave transformer near the antenna and 10m of RG8/X, and
- the twelfth wave transformer near the tx and 10m of RG6.

Intesting, the cheapest option (mostly RG6) is the one with the least loss!

Owen

Coaxial Antenna question
 

"Roy Lewallen" wrote:

But you'll feel better seeing zero "reflected" power on your meter, ...
_____________

The meter indication will be essentially zero only if the directivity of the
device measuring reflected energy is extremely good (as in 60+ dB).

The leakage of forward energy into the reflected sample that is present in
typical commercial products will lead to (mis)adjusting the load Z to
whatever produces zero on the reflected meter. However some amount of
reflected energy will be needed to cancel the leakage of the reflected
coupler, in order to obtain that "zero" reading.

And then the reverse energy is not really zero, and the true load SWR is not
really 1:1.

RF

Coaxial Antenna question
 

"Owen Duffy" wrote in message
...
Ed wrote in
36.82:

Owen Duffy wrote in
:


Ed,

You might have expected the feedpoint impedance to be around 70 ohms.
It will depend on the feedline configuration, because you haven't
taken much is the way of measures to decouple the feedline.

Your measured fwd and ref indicates VSWR~=1.5 which is consistent
with 70 ohms, but you haven't measured 70 ohms.

Assuming though that such an antenna should be close to 70+j0 at
resonance...

If you did want to incorporate an impedance matching system that
doesn't compromise the portability you have described, you could try
a twelfth wave transformer with 29.3° of 50 ohm coax from the
feedpoint, then 29.3° of 75 ohm coax then any length of 50 ohm coax
to the transmitter.

For example, for 146MHz, that could be 137mm of Belden 9258 (RG8/X)
then 139mm of Belden 1189A (RG6/U) then any length of 50 ohm coax to
the tx.

Owen


Very nice, Owen. Saved me a lot of difficult math.... since I
have
those materials on hand I may see what I can throw together tomorrow.

I didn't do the math, I punched the numbers into TLLC
(http://www.vk1od.net/tl/tllc.php).

Of course, the reason I was so specific is that translation from the
29.3° depends on the velocity factor... so use the velocity factor for
the cables you have at hand. (For example, if you use RG59, it has a
very different velocity factor to te 1189A, and you need to adjust
accordingly.)

There is a little on the twelfth wave transformer, including a graph of
the lengths for different transformation ratios at
http://www.vk1od.net/RG6/index.htm .

Have fun.

Owen


Hi Owen

Another way of avoiding the math is to use both a Smith Chart and an
overlay of a Z Theta Chart. The problem of choosing line lengths and their
Zo them becomes intuitive. But any "perfect match" does depend heavily on
knowing impedance rather than VSWR, as you know.
The load impedance ploted on the Smith Chart can be assummed to translate
to any impedance on the circle of constant VSWR for any load impedance.
The impedance moves along the line of constant "Theta" on the Z Theta Chart
for a change of Chart Z.
With the overlay of the two charts, it is fairly easy to see what lengths
and Zo will produce the best match.

Jerry KD6JDJ

Coaxial Antenna question
 

"Richard Clark" wrote in message
...
On Sun, 02 Nov 2008 14:00:40 GMT, "Jerry"
wrote:

Hi Owen

Another way of avoiding the math is to use both a Smith Chart and an
overlay of a Z Theta Chart. The problem of choosing line lengths and
their
Zo them becomes intuitive. But any "perfect match" does depend heavily
on
knowing impedance rather than VSWR, as you know.
The load impedance ploted on the Smith Chart can be assummed to
translate
to any impedance on the circle of constant VSWR for any load impedance.
The impedance moves along the line of constant "Theta" on the Z Theta
Chart
for a change of Chart Z.
With the overlay of the two charts, it is fairly easy to see what
lengths
and Zo will produce the best match.

Jerry KD6JDJ

Hi Jerry,

Your solution is rather exotic for this group, but I have encountered
it in my Metrology days as part of the HP legacy. The method you
described is missing from this article, but it gives the group a
picture of the chart, none-the-less:
http://www.hpl.hp.com/hpjournal/pdfs...Fs/1950-04.pdf

73's
Richard Clark, KB7QHC

Hi Richard

As you know, I have been far away from the antenna design community for
such a long time that I feel like a beginner today. So, I wouldnt have
considered anything i know to be "exotic".
For those who havent used the Z Theta Chart, it is identical to the Smith
Chart, but expressed in polar coordinates ( Impedance magnitude with its
angle).
With both a Smith Chart and a Z Theta Chart of equal radius and overlayed,
a pin hole thru both will identify any impedance with real Resistance. The
value identified on the Smith Chart is given as R+/- jX.. The value
identified on the Z Theta Chart is given as Z Angle Theta.

The use of both charts together is a great tool for impedance matching
with transmission line sections. It is quick and simple yet it is quite
accurate. An impedance, plotted on the Z Theta Chart moves along the lines
of constant angle when the chart Impedance is changed. That lets the
designer plot any impedance on the Z Theta Chart with, for instance 50 ohms
as its center, then immediately see where the impedance will move if the
Chart impedance is changed to the value of the new transmission line
impedance chosen for the transformer, for instance 70 ohms.

Jerry

Coaxial Antenna question
 

On Sun, 02 Nov 2008 14:00:40 GMT, "Jerry"
wrote:

Hi Owen

Another way of avoiding the math is to use both a Smith Chart and an
overlay of a Z Theta Chart. The problem of choosing line lengths and their
Zo them becomes intuitive. But any "perfect match" does depend heavily on
knowing impedance rather than VSWR, as you know.
The load impedance ploted on the Smith Chart can be assummed to translate
to any impedance on the circle of constant VSWR for any load impedance.
The impedance moves along the line of constant "Theta" on the Z Theta Chart
for a change of Chart Z.
With the overlay of the two charts, it is fairly easy to see what lengths
and Zo will produce the best match.

Jerry KD6JDJ

Hi Jerry,

Your solution is rather exotic for this group, but I have encountered
it in my Metrology days as part of the HP legacy. The method you
described is missing from this article, but it gives the group a
picture of the chart, none-the-less:
http://www.hpl.hp.com/hpjournal/pdfs...Fs/1950-04.pdf

73's
Richard Clark, KB7QHC

Coaxial Antenna question
 

Ed wrote:

...
Back to my question: I am measuring about 125 watts forward and 4-5
watts reflected at my desired frequency ( 146.000 ). While this may be
acceptable to some, I would like to get the match down to 1:1 SWR.
...

Ed K7AAT

You could always construct one of the "fancy" and esoteric 50:75 ohm
ununs to accomplish, however, you would probably not gain much if
anything, as there will be some loss--even if you construct the unun
from optimum material.

However, if ONLY the mismatch bothers you (and not the insignificant 4-5
watts) and not the actual loss of watts--it is an option. But any
"gain" you would get from this option would be virtually
"insignificant", or so in my humble opinion.

Regards,
JS

Coaxial Antenna question
 

Hi Jerry,

The Z-Theta Chart has actually been around a long time, well known as the
'Carter' Chart, developed by Philip Carter, an early RCA transmission-line and
antenna inventor, around the same time as Philip Smith developed the Smith
Chart. In one of HP's Application Notes it is incorrectly called the 'Charter
Chart'. But you are correct, overlaying those two charts is often very helpful.

Walt, W2DU

Coaxial Antenna question
 

Gee it's too bad he didn't have a bunch of CATV hardline and a Motrac. All
of this would be real simple.

I recall that there was a commercial AS "fire engine" antenna that never
bothered with the matching at all because adding all the extra hardware for
matching, wouldn't have justified the potential losses that might be
introduced. Of course the main advantage of the antenna was that it could
be elevated without need for reflecting plane or radials and thus wouldn't
poke eyes out or get tangled. Otherwise a regular mobile mount or base
radial kit would be advantageous.

Coaxial Antenna question
 

On Nov 1, 7:12*pm, Ed wrote:
* I'm trying to build a decent performing 2M coaxial skirted antenna and
have a question about its design for maximum efficiency.

* I based the "hub" on an SO-239 connector. *I soldered my RG8X cable
center conductor to the solder pin center conductor and brought the
shield braid out in two places.

* *I cut a 19" piece of half inch copper pipe, *cut 4 half inch slots on
one end, *fanned the slotted end out slightly to fit nicely against the
SO-239, *and slid this pipe over my coax and up to the SO-239. *I brought
the two braid lengths previously prepared out through two of the rather
fat slots, * soldered the pipe to the SO-239 and the braids were soldered
to the pipe where they came protruded out the slots.

* *I soldered an 18 1/2" brass welding rod to a PL-259 center conductor
and screwed that the the SO-239 for my radiator. * The copper pipe assy
and coax slide nicely down into a length of 3/4" PVC . *This makes a very
nice break-down package for transportation and portable work.

* * Back to my question: *I am measuring about 125 watts forward and 4-5
watts reflected at my desired frequency *( 146.000 ). *While this may be
acceptable to some, *I would like to get the match down to 1:1 SWR.

* * Are there any design deficiencies in my proto-type? * Can anyone
suggest something I might try to improve the match? * Is there a "rule of
thumb" regarding the construction of such antennas?... length of coaxial
skirt vs. antenna element?

* * I appreciate any feedback on this. *Thanks.

* *I suppose I could provide a picture if anyone requires it....

* *Ed * K7AAT

Hi Ed, I think your SWR is about as good as you are going to get it.
What you have created is a center feed dipole with a feedpoint
impedance of about 70 ohms or so. One way to get a better match to 50
ohms is to flair out your coaxial skirt. Skirt at 90 degrees to the
radiator the impedance will be 36 ohms, 70 with the skirt at 180
degree. You will find 50 ohms somewhere in between, about 45 degrees I
think. This may be done at the expense of your radiation pattern. I
built something like what you are building a few years ago using sheet
metal rolled in a trumpet shape for the skirt. I think my SWR was
about 1.3:1. My metal mast and skirt connected together at the
feedpoint. Whether or not the mast was insulated from the skirt of not
didnt make any difference. If itis not clear what I am talking about
think AEA isopole. While theirs was a 5/8ths we are talking about a
1/2 wl antenna.


Jimmie

Coaxial Antenna question
 

Kreco Antennas in Cresco, PA makes a line of
coaxial dipole basestation antennas that exhibit
a 50-ohm feedpoint impedance. Here's the
website for their high-band basic model:

http://www.krecoantennas.com/hbcaxial.htm

They pull off this trick by, *I THINK*, shortening
the top element slightly and lengthening the skirt in
*just the right way* to achieve a match at a spot
frequency.

An interesting variant on the basic antenna is their
"shunt-fed" coaxial dipole that places the entire
antenna at DC ground for lightning protection. Here's
the webpage for it:

http://www.krecoantennas.com/shuntfed.htm

I've used their antennas in the past with excellent
results, but they are a bit pricey.

Jim, K7JEB

Coaxial Antenna question
 

On Nov 2, 7:30*pm, "Jim, K7JEB" wrote:
Kreco Antennas in Cresco, PA makes a line of
coaxial dipole basestation antennas that exhibit
a 50-ohm feedpoint impedance. *Here's the
website for their high-band basic model:

* *http://www.krecoantennas.com/hbcaxial.htm

They pull off this trick by, *I THINK*, shortening
the top element slightly and lengthening the skirt in
*just the right way* to achieve a match at a spot
frequency.

An interesting variant on the basic antenna is their
"shunt-fed" coaxial dipole that places the entire
antenna at DC ground for lightning protection. *Here's
the webpage for it:

*http://www.krecoantennas.com/shuntfed.htm

I've used their antennas in the past with excellent
results, but they are a bit pricey.

Jim, K7JEB

We have used these at work. The 50 ohm value is very "nominal".


http://www.krecoantennas.com/hbcaxial.htm

Jimmie

Coaxial Antenna question
 

On Sun, 02 Nov 2008 23:23:26 GMT, "JB" wrote:

Gee it's too bad he didn't have a bunch of CATV hardline and a Motrac. All
of this would be real simple.

I recall that there was a commercial AS "fire engine" antenna that never
bothered with the matching at all because adding all the extra hardware for
matching, wouldn't have justified the potential losses that might be
introduced. Of course the main advantage of the antenna was that it could
be elevated without need for reflecting plane or radials and thus wouldn't
poke eyes out or get tangled. Otherwise a regular mobile mount or base
radial kit would be advantageous.

Yep. However, they recommended using 75 ohm coax cable. The loss of
equal lengths of similar size 75 ohm coax is less than 50 ohm. For
example:
RG-58c/u 0.20dB/meter at 150 Mhz (cheap 50 ohms coax)
LMR-240 0.09dB/meter at 150 Mhz (much better 50 ohm coax)
RG-6/u 0.07dB/meter at 150 Mhz (75 ohm CATV coax)

However, if you wanna run 50 ohm coax, the mismatch loss at the 75 ohm
antenna is about:
reflection_coef = (75-50)/(50+75)= 0.20
voltage = 1 - (0.2^2) = 0.96
20 * log(0.96) = 0.35 dB mismatch loss.
No big deal.

Hmmmm...
0.35 dB / 0.02dB/meter = 17.5 meters
At 17.5 meters, the losses of the 50 and 75 coax systems are
identical. Beyond 17.5 meters of coax, the 75 ohm coax delivers more
power.

I've been using RG-6/u for 2.4GHz wireless for quite a while. The
main incentive is that I can get the 75 ohm coax quite cheaply. For a
while, Hyperlink (http://www.hyperlinktech.com) had a rooftop 2.4Ghz
amplifier that was fed with 75 ohm coax. Alvarion/Breezecom also used
75 ohm coax in some of their BreezeAccess LB radios.

Someone eventually asks why 50 or 75 ohms. See:
http://www.microwaves101.com/encyclopedia/why50ohms.cfm

Motrac? Those are 30-40 years ancient. I used them for boat anchors.
Back then, I preferred GE radios:
http://802.11junk.com/jeffl/pics/Old%20Repeaters/index.html



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

Coaxial Antenna question
 

Jeff Liebermann wrote in
:

....
However, if you wanna run 50 ohm coax, the mismatch loss at the 75 ohm
antenna is about:
reflection_coef = (75-50)/(50+75)= 0.20
voltage = 1 - (0.2^2) = 0.96
20 * log(0.96) = 0.35 dB mismatch loss.

The analysis you give assumes that the notional 'reflected power' is lost
from the system as heat. The old 'reflected power is dissipated as
increased heat in the PA' line.

In the real world, the power that a transmitter delivers to a non ideal
load is not so simply predicted, and it is entirely possible that it
delivers more power to the mismatched load.

Owen

Coaxial Antenna question
 

Owen Duffy wrote:

...
In the real world, the power that a transmitter delivers to a non ideal
load is not so simply predicted, and it is entirely possible that it
delivers more power to the mismatched load.

Owen

That may well be; I am no expert on this; and, the point has missed my
detailed investigation.

But, if memory serves me correct, when I fed a 50 ohm antenna with an
old 75 ohm PA, equipped with plate voltage and current meters, I would
have expected a dip (although it might appear slight) in voltage and a
rise in plate current--indicating, that indeed, I was feeding "more
power" to the load ... although not desirable ... extrapolating from
this, feeding a 75 ohm antenna with a 50 ohm rig, I would expect the
opposite.

Regards,
JS

Coaxial Antenna question
 

On Mon, 03 Nov 2008 03:08:55 GMT, Owen Duffy wrote:

Jeff Liebermann wrote in
:

...
However, if you wanna run 50 ohm coax, the mismatch loss at the 75 ohm
antenna is about:
reflection_coef = (75-50)/(50+75)= 0.20
voltage = 1 - (0.2^2) = 0.96
20 * log(0.96) = 0.35 dB mismatch loss.

The analysis you give assumes that the notional 'reflected power' is lost
from the system as heat. The old 'reflected power is dissipated as
increased heat in the PA' line.

In the real world, the power that a transmitter delivers to a non ideal
load is not so simply predicted, and it is entirely possible that it
delivers more power to the mismatched load.

Owen

I beg to differ somewhat. In order for the reflected power to
contribute to the incident power, the reflected power would first be
attenuated by the coax loss. It would then require a substantial
mismatch at the transmitter, which is unlikely. However, assuming
there is a mismatch at the source, some of the reflected power will be
sent back to the load (antenna), after getting attenuated by the coax
for a 2nd time. There may be some contribution, but it will very very
very very small.

Let's try some more or less real numbers. I kinda prefer doing
everything in dBm but hams have this thing about using watts...

Start with a 50 watt xmitter and 20 meters of LMR-240 coax at
0.09dB/meter for an attenuation of 1.8dB.

The power delivered to the antenna is:
50 watts / ((1.8/10)^10) = 50 / 1.5 = 33 watts
The 1.5:1 VSWR reflects 4% of 33 watts for 1.3 watts reflected.

The 1.3 watts is again attenuated by the 1.8dB coax loss resulting in:
1.3 watts / (1.8/10)^10) = 1.3 / 1.5 = 0.87 watts

Now, lets assume that the xmitter has a broadband output stage,
optimized for 50 ohms and lacks the ability to properly match 75 ohms.
Once again, 4% of the power if reflected, resulting in:
0.87 watts * 4% = 0.035 watts reflected

Once again, the power reflected from the source end (xmitter end) is
attenuated by the 1.8dB coax loss for:
0.035 watts / ((1.8/10^10) = 0.035 / 1.5 = 0.023 watts.

Therefore, you're correct. It's possible that some of the reflected
power adds to the incident power. However, it's a really small
amount. In this case, it's only 23 milliwatts added to 33 watts
delivered to the antenna.


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

Coaxial Antenna question
 

Jeff Liebermann wrote in
:

....
In the real world, the power that a transmitter delivers to a non
ideal load is not so simply predicted, and it is entirely possible
that it delivers more power to the mismatched load.

Owen

I beg to differ somewhat. In order for the reflected power to
contribute to the incident power, the reflected power would first be
attenuated by the coax loss. It would then require a substantial

Are you proposing vector addition of power?

mismatch at the transmitter, which is unlikely. However, assuming
there is a mismatch at the source, some of the reflected power will be
sent back to the load (antenna), after getting attenuated by the coax
for a 2nd time. There may be some contribution, but it will very very
very very small.

Let's try some more or less real numbers. I kinda prefer doing
everything in dBm but hams have this thing about using watts...

Of course it doesn't matter, which unit system you use, but if you start
adding 'forward' and 'reflected' power in dBm because it is real
convenient, you have peformed a vector addition of power. Is that valid?


Start with a 50 watt xmitter and 20 meters of LMR-240 coax at
0.09dB/meter for an attenuation of 1.8dB.

The power delivered to the antenna is:
50 watts / ((1.8/10)^10) = 50 / 1.5 = 33 watts
The 1.5:1 VSWR reflects 4% of 33 watts for 1.3 watts reflected.

The 1.3 watts is again attenuated by the 1.8dB coax loss resulting in:
1.3 watts / (1.8/10)^10) = 1.3 / 1.5 = 0.87 watts

You start with a limited view of the mismatch, VSWR conveys only one
dimension of a two dimensional mismatch.

Your treatment of the forward wave and reflected waves as independently
attenuated is an approximation that will lead to significant errors in
some cases.

For example, what percentage of the power at the source end of the line
is lost as heat in 1m of LMR400 at 1MHz with a) a 5+j0 ohm load, and b) a
500+j0 ohm load. The VSWR is approximatly the same in both cases but the
answers are very different, one is almost 100 times the other.

Doesn't it stand to reason that as the length of the transmission line
approaches zero, that the power lost transmission in this type of line in
the high voltage low current load scenario is lower than the low voltage
high current load scenario.

Another issue is that the V/I characteristics of a transmitter output
stage is not necessarily (or usually for most ham transmitters) a
straight line, in other words it does not exibit a constant Thevenin
equivalent source impedance with varying loads and the application of
some linear circuit analysis techniques to the output stage are
inappropriate.

Owen

Coaxial Antenna question
 

Owen Duffy wrote in
:

....
Of course it doesn't matter, which unit system you use, but if you
start adding 'forward' and 'reflected' power in dBm because it is real
convenient, you have peformed a vector addition of power. Is that
valid?

I should have said "...you have peformed a flawed vector addition of
power..."

Owen

Coaxial Antenna question
 

On Mon, 03 Nov 2008 04:46:57 GMT, Owen Duffy wrote:

Jeff Liebermann wrote in
:

...
In the real world, the power that a transmitter delivers to a non
ideal load is not so simply predicted, and it is entirely possible
that it delivers more power to the mismatched load.

Owen

I beg to differ somewhat. In order for the reflected power to
contribute to the incident power, the reflected power would first be
attenuated by the coax loss. It would then require a substantial

Are you proposing vector addition of power?

Nope. No need to complexicate things. I can safely assume a real 75
ohm load. I can cheat a bit and assume some multiple of 1/2 wave coax
cables, thus eliminating any imaginary contributions from the coax.

Please note that my purpose was to demonstrate that 75 ohm antennas
and 75 ohm coax will work adequately in a 50 ohm system. I think I've
done most of that. Including complex impedances to the calculations
will yield a more accurate result, but the resultant reflected power
that will be added to the forward delivered power, will be LESS than
the results produced by my calculations using only the real part of
the impedances.

mismatch at the transmitter, which is unlikely. However, assuming
there is a mismatch at the source, some of the reflected power will be
sent back to the load (antenna), after getting attenuated by the coax
for a 2nd time. There may be some contribution, but it will very very
very very small.

Let's try some more or less real numbers. I kinda prefer doing
everything in dBm but hams have this thing about using watts...

Of course it doesn't matter, which unit system you use, but if you start
adding 'forward' and 'reflected' power in dBm because it is real
convenient, you have peformed a flawed vector addition of power.
Is that valid?

It's only valid for the level of accuracy with which you are working.
For convenience, perhaps we can just assume that the re-reflected
contribution to the forward power is in phase, thus yielding the
maximum delivered power. Any phase shifts between the two signals
will result in LESS delivered power than the in phase simplistic
calculation. I'm sure the accuracy might be useful for academic
purposes, but my example demonstrated that only 35 mw was added to 33
watts, an error of 0.1%. Of course, that's ridiculous because the
initial measurement of the originating 50 watts is probably only
accurate to 2 significant figures.

Start with a 50 watt xmitter and 20 meters of LMR-240 coax at
0.09dB/meter for an attenuation of 1.8dB.

The power delivered to the antenna is:
50 watts / ((1.8/10)^10) = 50 / 1.5 = 33 watts
The 1.5:1 VSWR reflects 4% of 33 watts for 1.3 watts reflected.

The 1.3 watts is again attenuated by the 1.8dB coax loss resulting in:
1.3 watts / (1.8/10)^10) = 1.3 / 1.5 = 0.87 watts

You start with a limited view of the mismatch, VSWR conveys only one
dimension of a two dimensional mismatch.

Sure. It's good enough for a back of the envelope estimate of how
much power the re-reflected signal can possibly add to the forward
power.

Your treatment of the forward wave and reflected waves as independently
attenuated is an approximation that will lead to significant errors in
some cases.

True. However, as long as I assume a 1.8dB coax cable loss, the
reflected and re-reflected powers will be sufficiently low to be
considered negligible. Including the necessary vector arithmetic to
include the possibility of random coax cable lengths will improve
accuracy, but not affect the result very much.

For example, what percentage of the power at the source end of the line
is lost as heat in 1m of LMR400 at 1MHz with a) a 5+j0 ohm load, and b) a
500+j0 ohm load. The VSWR is approximatly the same in both cases but the
answers are very different, one is almost 100 times the other.

I'll work out the exact numbers tomorrow, but I see your point.
However, please note that I made an effort to use a REALISTIC example
of a typical 2m radio, coax, and coaxial antenna arrangement. Of
course, you can conjure a set of numbers that will result in a
substantially increased calculation error. I can do the same thing if
I take my example and simply reduce the coax cable length to the point
where coax attenuation is dramatically smaller. A 100:1 load
impedance change is not the same as a 1.5:1 impedance change (from 50
to 75 ohms)

Doesn't it stand to reason that as the length of the transmission line
approaches zero, that the power lost transmission in this type of line in
the high voltage low current load scenario is lower than the low voltage
high current load scenario.

Ummmm... you lost me there. I've got a headache tonite. I'll see it
makes more sense tomorrow morning.

Another issue is that the V/I characteristics of a transmitter output
stage is not necessarily (or usually for most ham transmitters) a
straight line, in other words it does not exibit a constant Thevenin
equivalent source impedance with varying loads and the application of
some linear circuit analysis techniques to the output stage are
inappropriate.

I really don't know if that's true for a 2m FM transmitter. I'm not
sure it even matters. The V/I characteristic (slope) is just the
source impedance of the output stage. Whether it's 50 or 75 ohms is
close enough for my simplistic calculation to be accurate without
throwing in non-linearities. The source impedance may be different
for a 50 watt radio, running at 1 watt, but not enough to make a big
difference. If the source impedance were magically 10 times as high,
the 35 milliwatts of re-reflected RF would become 350 mw and still be
a fairly negligible contribution to the delivered 33 watts.

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

Coaxial Antenna question
 

Jeff, your re-reflection concept is complicating things.

Just consider a typical ham FM voice transmitter after steady state has
been substantially established.

If the load was for example 70+j0 and the transmitter was connected by 2
wavelengths of Belden 9258, accounting for line loss we would expect the
transmitter to see a load of about 68+j0 (VSWR(50)=1.4), ie the ratio of
v/i at the output terminals of the transmitter would be about 68+j0.

Consider also the case where the transmitter was connected by 2.25
wavelengths of Belden 9258, accounting for line loss we would expect the
transmitter to see a load of about 37+j0 (VSWR(50)=1.4), ie the ratio of
v/i at the output terminals of the transmitter would be about 37+j0.

If you were to measure the output power of a range of such transmitters,
it is unlikely that they will produce substantially identical power
output under both conditions though the VSWR is similar, that the power
is changed from that with a 50+j0 load by the same amount in all cases,
or that the change is reliably predicted by your analysis technique
(based on VSWR).

Your treatment of the 'forward power' as a constant with different loads,
and the approximation of transmisssion line behaviour contribute error.

Additionally, the formulae you use do not account for non linear
behaviour of typical output stages, gain variation at different output
level, whether they reach voltage or current saturation with a given
load, the effects of PA protection schemes that might limit current,
'relected power', 'power output' etc.

I would agree that a 70 ohm antenna at the end of 4m or so of RG8/X will
*probably* not result in a large loss of output power, but I wouldn't
agree with your results or method because it is not sound. Nevertheless,
I understand why Ed might want to transform the load to 50 ohms, and he
shouldn't be discouraged by flawed estimates.

Owen

PS:
You might dismiss my example of the 5 and 500 ohm loads on a short line
as unrealistic, but it exposes a common misunderstanding that the loss
per unit length when VSWR1 is uniform along the line. If you want to
explore the idea further, I have written some notes at
http://www.vk1od.net/VSWR/displacement.htm .

Coaxial Antenna question
 

Jeff Liebermann wrote:
I beg to differ somewhat. In order for the reflected power to
contribute to the incident power, the reflected power would first be
attenuated by the coax loss. It would then require a substantial
mismatch at the transmitter, which is unlikely. However, assuming
there is a mismatch at the source, some of the reflected power will be
sent back to the load (antenna), after getting attenuated by the coax
for a 2nd time. There may be some contribution, but it will very very
very very small.

Reflection of a single reflected wave is not the only mechanism
that can redistribute energy back toward the load. Superposition
of two waves at the source impedance (or at an impedance
discontinuity in a feedline) accompanied by destructive
interference can accomplish a similar feat. Non-reflective glass
is a 1/4WL matching section of thin-film that accomplishes the
same thing as a Z0-match.

To the best of my knowledge, nobody is taking wave cancellation
at the source impedance into account although it may be the
major source of the redistribution of reflected energy back
toward the antenna. It's explained on the following web pages:

http://www.mellesgriot.com/products/optics/oc_2_1.htm

"Clearly, if the wavelength of the incident light and
the thickness of the film are such that a phase difference
exists between reflections of p, then *reflected wavefronts*
*interfere destructively*, and overall reflected intensity is
a minimum. If the two reflections are of equal amplitude,
then this amplitude (and hence intensity) minimum will be
zero." (Referring to 1/4 wavelength thin films.)

"In the absence of absorption or scatter, the principle of
conservation of energy indicates all 'lost' reflected intensity
will appear as *enhanced intensity in the transmitted beam*.
The sum of the reflected and transmitted beam intensities is
always equal to the incident intensity. This important fact
has been confirmed experimentally."

http://micro.magnet.fsu.edu/primer/j...ons/index.html

"... when two waves of equal amplitude and wavelength that are
180-degrees ... out of phase with each other meet, they are not
actually annihilated, ... All of the photon energy present in
these waves must somehow be recovered or *redistributed* in a new
direction, according to the law of energy conservation ... Instead,
upon meeting, the photons are *redistributed* to regions that permit
*constructive interference*, so the effect should be considered as
a *redistribution* of light waves and photon energy rather than
the spontaneous construction or destruction of light."

Why does almost everyone seem to consider reflection the only way
to redistribute reflected energy back toward the antenna?
--
73, Cecil http://www.w5dxp.com

Coaxial Antenna question
 

Owen Duffy wrote:
I should have said "...you have peformed a flawed vector addition of
power..."

The correct method of adding power comes to us from the
field of optics in the form of the irradiance equation.
If we multiply irradiance by the cross-sectional area of
coax, we get power.

Ptot = P1 + P2 + 2*SQRT(P1*P2)cos(A)

where 'A' is the angle between the electric fields
of the two phasors.
--
73, Cecil http://www.w5dxp.com

Coaxial Antenna question
 

Motrac? Those are 30-40 years ancient. I used them for boat anchors.
Back then, I preferred GE radios:

Back in the day we were radio techs and weight lifters!

I used both. I had Pre-Progs and Progs. Both were good for keeping warm
and digging out of the snow, but the MHT motracs were extremely easy to
duplex, bullet proof and had excellent audio. With 110 watts VHF and the
optional preamp for .12uv for 20dbq, that radio talked like no other I have
owned before or since, and I could do it parked at a jam packed radio site.
I ran a UHF 35 watt radio, duplexed, to access a high level remote base
system with an autopatch. We had wide area flat rate dialing and I had a
switch on my WE TT pad to key the radio continuously. A resistor across the
relay and a small cavity for the RX and it sounded better than the IMTS
system. I never graduated to Micor for a mobile because it could cost you
$500 just to fill up a 12 channel mobile. I was using a duplexed 73MHT up
to 5 years ago because I always wanted to monitor that frequency and had
little trouble as NCS.

IMNSHO the Micor and MastrII have yet to be improved upon for performance
and practicality for a repeater.

OH - the point was the coupling in the PA tube would allow you to match
directly to 75 ohm.