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.

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

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

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

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

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

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

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

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

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