K7ITM wrote:
Field strength alone is not acceptable to me as a means to adjust an
antenna load to a transmitter, ...

Doesn't being located in the near field introduce
a measurement error?
--
73, Cecil http://www.qsl.net/w5dxp

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Tom Donaly wrote:
Cecil was talking about current, not power. You can't add
power the way you can voltage and current.

That's absolutely correct. When one adds powers, one must
include the interference term which takes care of conservation
of energy. The equation is:

Ptot = P1 + P2 + SQRT(P1*P2)cos(theta)

where theta is the phase angle between V1 (associated with
P1) and V2 (associated with P2). The last term is labeled
the "interference term" and is absolutely necessary when
adding powers. If the interference term is positive, the
interference is constructive. If the interference term is
negative, the interference is destructive.

The best reference on interference and the adding of powers
that I have found is Chapter 9 in "Optics", by Hecht.
--
73, Cecil http://www.qsl.net/w5dxp

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Frank Gilliland wrote:
There lies our misperceptions; I was not referring to using an HF SWR
meter designed for coax and plugging it into 450 ohm ladder line.

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.
--
73, Cecil http://www.qsl.net/w5dxp

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james wrote:
The problem is that current is not reflected back from the load, power
is. Thus the you can add magnitudes of power.

If power (ExH) is reflected then, of course current is reflected.
Powers can be added but you *must* include the interference term.
--
73, Cecil http://www.qsl.net/w5dxp

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james wrote:
Agreed that a rig cannot detect the difference between forward and
reflected power. If the reflection coeffiecient of the source is zero
then final stage of a transmiter will look purely resistive to any
power reflected by the load. Thereby that refelcted power is
dissapated as heat. Other reflection coefficients at the source will
yield lesser amounts of reflected power from the load as heat.

This is true for a source equipped with a circulator and load but
most ham transmitters are not equipped with a circulator and load.
You must take the phase of the voltages or currents into account
in order to calculate the interference power term. Only then will
you be able to tell where the power goes.
--
73, Cecil http://www.qsl.net/w5dxp

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Roy Lewallen wrote:
But I do want to caution
readers that this view of "reflected power" is demonstrably incorrect.

So is your concept of "sloshing" energy. Reflected energy
waves are demonstrably real. One can find out exactly where
the reflected power goes by taking the interference power
terms into account. Optics engineers figured it out a long
time ago.
--
73, Cecil http://www.qsl.net/w5dxp

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Frank Gilliland wrote:

Wes Stewart wrote:
In a word, baloney. The error is independent of length. A zero length
bridge calibrated at 75 ohm is in error when measuring in a 50 ohm
system. Period.

Prove it.

A 75 ohm bridge is expecting the ratio of voltage to current
to be 75 for a matched system. In a 50 ohm matched system, the
ratio of voltage to current will be 50. Therefore, the 75 ohm
bridge won't be balanced. A 50 ohm bridge would be balanced.
--
73, Cecil http://www.qsl.net/w5dxp

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james wrote:

snip

*****

Agreed that a rig cannot detect the difference between forward and
reflected power. If the reflection coeffiecient of the source is zero
then final stage of a transmiter will look purely resistive to any
power reflected by the load. Thereby that refelcted power is
dissapated as heat. Other reflection coefficients at the source will
yield lesser amounts of reflected power from the load as heat.

james

So if I have a perfect voltage source in series with a 50 ohm resistor,
and I set the voltage source for 100Vrms (50W to the load, 50W to the
source resistor) and I leave the output terminals open (100% reflected
power) then the resistor will dissipate 100W? With no current flowing
through it?

Wow. I gotta review my basic electronics.

-------------------------------------------
Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

On Wed, 29 Jun 2005 23:00:17 -0500, Cecil Moore
wrote:
Optics engineers figured it out a long time ago.
And you have consistently failed in its demonstration - so what?

Depends on what you mean by error. Is it a linear system? If the near
field strength increases with no change in the radiating structure
itself (and propagation is stable, etc.), does the far field not
increase by the same ratio? But of course, with a repositionable
(rotatable) directional antenna, it's pretty hard to calibrate the FSM
in a meaningful way since the antenna system changes (quite a lot, with
respect to the FSM) as you rotate it, so you don't know from one time
to the next that you have the RIGHT field strength. I'd (ideally) like
to know that the transmitter is properly adjusted to output a clean
signal, and that the antenna system presents the proper load to the
transmitter, AND that the antenna system is radiating like I'd like it
to. The "SWR meter" is one component that helps me, but with only one
of those tasks. (And yes, it's fine with me if you care also about the
SWR on your 450 ohm balanced line...there may also be good reason for
wanting to know that.)

Cheers,
Tom

PS--Frank, if you look back in the archives from this group, you'll
find directional couplers (of the sort that measure the line at a
single point) explained in great detail with four-part harmony and the
whole nine yards. Go study them and you may understand why calibration
is important.