On Wed, 29 Jun 2005 19:06:14 -0700, Wes Stewart
wrote in :

On Wed, 29 Jun 2005 14:07:26 -0700, Frank Gilliland
wrote:

On Tue, 28 Jun 2005 23:17:15 -0500, Cecil Moore
wrote in :

Frank Gilliland wrote:
Impedance matching of an SWR meter is generally unimportant since most
SWR meters used for HF have a directional coupler that is much shorter
than the operating wavelength.

Point is that they are usually calibrated for Z0=50 ohms
and are in error when used in Z0 environments differing
from Z0=50 ohms, e.g. Z0=75 ohms.


The point is that the error is insignificant when the directional
coupler is much shorter than the wavelength.

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.


The error is even more
insignificant when there are a host of variables and confounds between
the SWR meter and the transmitted field that can (and frequently do)
affect the objective -- field strength.

Often, field strength is of zero importance. What do you do when the
device under test isn't supposed to radiate?


That device probably wouldn't make a very good radio, would it?


The simplest example of
this would be a CATV system, yet VSWR is *extremely* important in
cascaded networks.


Thank you for making my point.


It's much simpler (and just
plain logical) to measure the field strength directly instead of
measuring an abstract value halfway towards the objective and relying
on nothing more than speculation that the rest is working according as
expected.

More baloney and it isn't even sliced.


The word is "blarney". And although the syntax of my statement was
somewhat 'convoluted', the logic is sound -- you can dyno your engine
all day, but the only way to know for sure how fast you can get down
the quarter mile is to run the race.





----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==----
http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups
----= East and West-Coast Server Farms - Total Privacy via Encryption =----

james wrote:
Power is power. Phase is not a problem. Take the mafnitude of the
transmitted power and teh magnitude of the reflected power. The
results are phaseless. The magnitudes add linearly.

Strangely enough, the results are not phaseless. The equation
for reflected power at an impedance discontinuity is:

Pref = P3 + P4 - SQRT(P3*P4)cos(theta)

Where theta is the phase angle between V3 and V4, the
associated reflected interferring voltages.

Reference "Optics", by Hecht, Chapter 9 - Interference

The last term in the equation above is known as the "interference"
term. Unless you take the interference term into account, you
don't have a ghost of a chance of ascertaining where the power
goes.
--
73, Cecil http://www.qsl.net/w5dxp

----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==----
http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups
----= East and West-Coast Server Farms - Total Privacy via Encryption =----

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

----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==----
http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups
----= East and West-Coast Server Farms - Total Privacy via Encryption =----

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

----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==----
http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups
----= East and West-Coast Server Farms - Total Privacy via Encryption =----

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

----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==----
http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups
----= East and West-Coast Server Farms - Total Privacy via Encryption =----

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

----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==----
http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups
----= East and West-Coast Server Farms - Total Privacy via Encryption =----

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

----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==----
http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups
----= East and West-Coast Server Farms - Total Privacy via Encryption =----

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

----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==----
http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups
----= East and West-Coast Server Farms - Total Privacy via Encryption =----

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

----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==----
http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups
----= East and West-Coast Server Farms - Total Privacy via Encryption =----

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