On Thu, 30 Jun 2005 00:19:22 GMT, "Tom Donaly"
wrote:

You need to read _Reflections II, Transmission Lines and Antennas_
by M. Walter Maxwell, W2DU. Even better, get a book on electromagnetics.
You might be able to puzzle some of it out although much of
the math might be too esoteric for you.
73,
Tom Donaly, KA6RUH
*****

Whats wrong with stating that power is reflected by the load? Isn't
power delivered to the load from the source? Elementary electronics
states that power is voltage time current.

Currents in a transmission line are induced currents. They are induced
from the E and H fields of the TEM wave. I hope that you don't think
that current races up and down the coax millions a times per second?

james

On Thu, 30 Jun 2005 00:22:48 GMT, james wrote
in :

On Wed, 29 Jun 2005 22:45:03 GMT, "Tom Donaly"
wrote:

james wrote:
On Wed, 29 Jun 2005 11:07:17 -0400, "Fred W4JLE"
wrote:


What is the reason a 2:1 SWR can cause such havoc?

How can I avoid this catastrophic condition?

I feed my dipoles with 450 Ohm ladder line, but the last 20 feet or so is 50
Ohm coax, I guess that makes it work ok. I haven't blown up my finals yet.

Lions and tigers and bears Oh my...

*****

Actually can happen if you push the finals to where there is
insufficeint margin to the maximum heat dissapation. Tubes are a bit
more forgiving. Transistor inadequately heatsinked and overdriven,
typical CB usage, often have little of no margin for heat dissapation.

If the transmitter has a refelction coefficient of zero and the load
say .3, then that reflected power from the load is dissapated as heat
in the output circuits and any final transistors or tubes. Now if the
radio has a reflection coefficient other than zero that will lessen
the heat dissapation on the transimiiter. Now you get load and source
reflections convoluting within the transmission line.

You ought to model a 400 Mhz square wave with source and load
refelctions coefficients other than zero. It can get ugly


james



Consider the MRF 140, a 150 Watt 2.0 - 150.0 Mhz N-Channel
linear RF power fet. From the technical data sheet: "100% Tested
For Load Mismatch At All Phase Angles With 30:1 VSWR." You'd have
a tough time damaging this device with a mere 2:1 VSWR.
How do load and source reflections convolute within the
transmission line? That's a new one on me. My old dictionary
defines 'convolute' as "Rolled or folded together with one part
over another; twisted; coiled." The rest of the post is pretty
fanciful, too. A trip to the library would do wonders.
73,
Tom Donaly, KA6RUH
******

In electrical engineering it is the instantaneous power density of two
signals passing at the same spot from two directions. That is called
Convolution.


I'm an EE and I've NEVER heard the term used in reference to
electronics, let alone used to describe standing waves.


It also is a nice mathematical means of modeling SWR at
any point on a transmisison line at a particular time.


I know how to model standing waves, but how do you model a standing
wave ratio?


Well if you knew CBers, they are not satidied getting 150 watts from
a transistor rated for 150 watts. But in my first paragraph I thought
I made it clear but evidently I did not. I guess I must strive to
better explain myslef.


Agreed.







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On Thu, 30 Jun 2005 00:25:23 GMT, james wrote
in :

On Thu, 30 Jun 2005 00:12:31 GMT, "Tom Donaly"
wrote:

Nope. You need a course in electromagnetics. Who put all these
ideas into your head, anyway?
73,
Tom
Electromagnetics


Did "Electromagnetics" teach you that power = voltage * current?






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On Wed, 29 Jun 2005 17:31:36 -0700, Richard Clark
wrote:

On Thu, 30 Jun 2005 00:22:48 GMT, james wrote:

In electrical engineering it is the instantaneous power density of two
signals passing at the same spot from two directions. That is called
Convolution.

Hi James,

No, it is called Superposition, and that is done only with voltage or
current. What you are describing may be associated with the Fourier
convolution of power series - an entirely different field (and not
even additive).

73's
Richard Clark, KB7QHC
*****

Okay maybe I am not expressing my self correctly and right now I don't
realy have the time or patience to look back through my old text
books. It has been several years since I have done a lot of RF work
and some things are not as fresh in my mind. It does seem like the
less you use the more you forget or have trouble explaining what you
think.

Most of my work lately has been away from RF.

james

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.

james

I've posted many, many times on this topic and have shown a number of
cases where the load is perfectly matched but the power dissipated in
the source resistor is less than or greater than the "reverse power",
clearly demonstrating that this concept is incorrect. There are several
examples at Food for thought.txt available at
http://eznec.com/misc/food_for_thought/.

Because I've posted so much on the topic I won't do it all again. But I
know at least one person on this newsgroup would be glad to have an
opportunity to express his views once again. I'll leave this discussion
to those who want to revisit it; I don't. But I do want to caution
readers that this view of "reflected power" is demonstrably incorrect.

Roy Lewallen, W7EL

Richard Clark wrote:
On Thu, 30 Jun 2005 00:22:48 GMT, james wrote:


In electrical engineering it is the instantaneous power density of two
signals passing at the same spot from two directions. That is called
Convolution.


Hi James,

No, it is called Superposition, and that is done only with voltage or
current. What you are describing may be associated with the Fourier
convolution of power series - an entirely different field (and not
even additive).

73's
Richard Clark, KB7QHC

Convolution is a mathematical stunt you can perform with
two functions: f(x)* g(x) = (integral from 0 to x) f(t)g(x-t) dt.
At least that's how it's explained in Schaum's Outline book
_Differential Equations_. It's pretty tough to see how it relates
to power in a transmission line. Maybe someone has a use for it
there.
73,
Tom Donaly, KA6RUH

On Thu, 30 Jun 2005 01:02:06 GMT, james wrote:



Most of my work lately has been away from RF.

Uh huh. And do you drive a Kenworth or a Volvo?

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.


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


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.

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.





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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

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