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Old June 30th 05, 07:45 PM
Cecil Moore
 
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wrote:

When you superpose two 100w coherent laser beams, the resultant
power is indeed 400w


This is correct if the two beams are coherent and have the
same polarization. Very hard to do at optical frequencies, much
easier at lower frequencies.


As in coherent RF waves confined to a transmission line, eh?
--
73, Cecil,
http://www.qsl.net/w5dxp

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Old June 30th 05, 08:57 PM
 
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When you superpose two 100w coherent laser beams, the resultant
power is indeed 400w


This is correct if the two beams are coherent and have the
same polarization. Very hard to do at optical frequencies, much
easier at lower frequencies.


Hi Tor,

We've seen the math pencil-whipped both ways now to cover all the
available answers. The devil's in the details that are not found in:
Itot = I1 + I2 + 2*Sqrt(I1*I2)cos(theta)

not to mention the glaring mistakes of the first posting of this
formula.


So? Superposition works. With a yagi antenna, through superposition you
get an EM wave which has larger intensity in certain directions than
for a single dipole with the same power. Someone far away can't tell
the difference between switching to a yagi and turning on a linear.

What the formula doesn't say is that in any real system, the wave
must have a finite extent (not be a infinite plane wave). Then there
must be destructive interference in some directions. So there isn't
a problem with conservation of energy.

Tor
N4OGW
  #104   Report Post  
Old June 30th 05, 09:01 PM
 
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When you superpose two 100w coherent laser beams, the resultant
power is indeed 400w

This is correct if the two beams are coherent and have the
same polarization. Very hard to do at optical frequencies, much
easier at lower frequencies.


As in coherent RF waves confined to a transmission line, eh?


That's nothing new. Why do they call them "waveguides" ? Jackson
has a chapter with all the hairy details.

Tor
N4OGW
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Old June 30th 05, 09:36 PM
Owen
 
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On Thu, 30 Jun 2005 06:47:07 GMT, Owen wrote:


The only other inference you can make from one instrument with regard
to the other will be if one of the instruments shows zero reflected
power, then you know the VSWR that the other instrument will indicate.


As Cecil has hinted ), you can
also infer that the impedance at the point where you have two sets of
Pf,Pr readings is that the impedance (R, X) is one of two values
(which can be visualised as the intersection of two circles on a Smith
Chart), one only in the cases VSWR1*VSWR2=Zo1/Zo2 (or the inverse)
(which can be visualised as the kissing of two circles on a Smith
Chart). The latter includes the case above where one or other
instruments indicates VSWR=1, but is more exact because one of the
circles is infinitely small.

These cases are not reliably of practical value, the conversion of
error in the measurements made with each instrument, into error in the
estimated R and X at the point could be huge.

So, I "correct" my statement to "The only other inference that you can
reasonably reliably make from one instrument with regard to the other
will be if one of the instruments shows zero reflected power, then you
know R and X, and the VSWR that the other instrument should indicate.

Thanks Cecil.

Owen
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Old June 30th 05, 10:39 PM
Frank Gilliland
 
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On Thu, 30 Jun 2005 07:36:21 -0500, Cecil Moore
wrote in :

Frank Gilliland wrote:

Cecil Moore wrote:
But I specifically stated above the Z0 environment was different
from 50 ohms. The same type of error happens when one uses a
50 ohm SWR meter in a 75 ohm coaxial line.


If that were true then the mere existence of standing waves could
render any measurements worthless. Regardless, I did the experiment a
long time ago -- take a 50 ohm SWR meter and plug it into a 75 ohm
line -- it gives you almost the same measurement (in fact, I didn't
see -any- difference at all).


Please run it again in the following configuration:

Xmtr--1/4WL 75 ohm line--SWR meter--1/4WL 75 ohm line--50 ohm load

The SWR meter will read 2.25:1 when the actual SWR is 1.5:1

Xmtr--1/2WL 75 ohm line--SWR meter--1/2WL 75 ohm line--50 ohm load

The SWR meter will read 1:1 when the actual SWR is 1.5:1



I'm not going to argue this -- either you can play with theory and
speculate about the results, or you can do the experiment yourself,
observe the empirical evidence, and -then- use theory to explain the
results. When you get around to doing the latter give me a holler in
rrcb since I'm done cross posting on this topic.

And BTW, the best location for the directional coupler is at the
feedpoint of the antenna. Barring that, the next best place is at the
transmitter. Regardless of it's location, you should -never- leave the
coupler floating with the coax or you will end up with results like
what you describe above.





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Old June 30th 05, 10:55 PM
Cecil Moore
 
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Frank Gilliland wrote:
Regardless of it's location, you should -never- leave the
coupler floating with the coax or you will end up with results like
what you describe above.


The results above obey the laws of physics. What laws do your
results obey?
--
73, Cecil http://www.qsl.net/w5dxp

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Old June 30th 05, 10:57 PM
Steveo
 
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Cecil Moore wrote:
Frank Gilliland wrote:
Regardless of it's location, you should -never- leave the
coupler floating with the coax or you will end up with results like
what you describe above.


The results above obey the laws of physics. What laws do your
results obey?

You guys are just itchin' for a visit from the coax length police.
  #110   Report Post  
Old July 1st 05, 12:36 AM
Walter Maxwell
 
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"Tom Ring" wrote in message
. ..
Walter Maxwell wrote:
Reading the mail appearing in this thread is more fun than watching Saturday
Night Live!

Walt, W2DU


Some of the people involved appear to be listening from inside Faraday Cages!

tom
K0TAR


Faraday used the cages after the monkeys were through with them. The monkeys
left them so fouled up that EM waves couldn't penetrate the walls.

Walt, W2DU


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