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

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

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 .

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