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Old June 20th 05, 08:25 AM
Richard Harrison
 
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Cecil, W5DXP, quoting others, wrote:
"The reflected power wave doesn`t travel from the load back toward the
source and the extra 4.17 watts in the forward wave doesn`t travel from
the match point back toward the load."

Users of the Bird and Similar wattmeters know that is what they see.
Source and load power is the forward power minus the reflercted power.

Zo of the coax enforces its volt to amp ratio on both the incident and
reflected waves. Reflected power is again reflected at the match point
because the matched source sees no reflection. That`s the point of
producing a match in the load.

Best regards, Richard Harrison, KB5WZI

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Old June 20th 05, 09:30 AM
Ian White GM3SEK
 
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Richard Harrison wrote:

Users of the Bird and Similar wattmeters know that is what they see.
Source and load power is the forward power minus the reflercted power.

But that's only what is printed on the meter scale. It doesn't make the
"reflected power" real.

For the (n+1)th time: the Bird so-called "wattmeter" does NOT sense
forward and reflected power. It only senses RF voltage and current on
the line. The meter scale calibration is a mathematical operation that
depends on a lot of assumptions... most notably the assumption that
"reflected power" has some physical reality.

I am genuinely open-minded about that debate - which makes the all the
more determined to be ruthless about bogus arguments on either side.

And the most bogus argument of all is: "Users of the Bird and Similar
wattmeters know that is what they see."


--
73 from Ian GM3SEK 'In Practice' columnist for RadCom (RSGB)
http://www.ifwtech.co.uk/g3sek
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Old June 20th 05, 12:34 PM
Cecil Moore
 
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Ian White GM3SEK wrote:
But that's only what is printed on the meter scale. It doesn't make the
"reflected power" real.

I am genuinely open-minded about that debate - which makes the all the
more determined to be ruthless about bogus arguments on either side.


Ian, correct me if I'm wrong, but I infer that you are biased
toward the "no reflected energy waves" side. That bias could be
based on perceived knowledge.

Seems to me, some people are begging the question. They assume
there is no energy in reflected waves and call what the other
side says, "gobbleygook", even when presented in scientific
terms. I think we all understand the concepts behind net
energy. What is in dispute is the next lower layer of dynamic
energy movement in both directions at the same time during
steady-state.

So would you explain what happens to the reflections of a laser
beam when aimed at a perfect mirror in free space. Offset the
laser beam slightly from normal incidence to start with and
observe the reflections with your naked eye. Then bring the
beam to 90 degree incidence with the mirror. You can no
longer see the reflections but now you can detect the
superposition of the forward and reflected waves through
interference "rings" or loops with an intensity maximum
occuring every half wavelength and an intensity minimum
occuring every half wavelength in between.

Where is the EM wave energy just sloshing around and not
traveling forward and rearward at the speed of light?
--
73, Cecil http://www.qsl.net/w5dxp

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Old June 20th 05, 01:46 PM
Ian White GM3SEK
 
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Cecil Moore wrote:
But that's only what is printed on the meter scale. It doesn't make
the "reflected power" real.
I am genuinely open-minded about that debate - which makes the all
the more determined to be ruthless about bogus arguments on either
side.


Ian, correct me if I'm wrong, but I infer that you are biased
toward the "no reflected energy waves" side. That bias could be
based on perceived knowledge.

I am trying very hard not to be biased about the actual problem -
but I am very much against your methods of debate.

Why should anyone consent to follow you into the realms of optics and
laser physics, merely because that's where you want to go? Stand your
ground right here on the log, Cecil, and talk strictly and exclusively
about RF transmission lines.


--
73 from Ian GM3SEK 'In Practice' columnist for RadCom (RSGB)
http://www.ifwtech.co.uk/g3sek
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Old June 20th 05, 05:40 PM
Richard Harrison
 
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Ian, GM3SEK wrote:
"I am trying not to be biased about the actual problem - "

Open minds consider arguments.

Wave velocity = frequency x wavelength

Wave velocity in free space is 186,000 miles per second (300,000
km/sec).

Velocity of disturbances on open wire transmission lines is almost equal
to that in free space.

At high radio frequencies, a wavelength may be measured in inches and
centimeters. A transmission line containing a discontinuity produces a
reflection from the change. As a short distance can produce a phase
change of 360 degrees, the incident and reflected waves can combine to
produce voltage variations along a short line length. These can be
measured using simple instruments with confidence.

Circuit theory works on transmission lines because the proximity of
their conductors causes an effect on one conductor to be instantaneously
imposed on the other.

A transmission line cannot be analyzed as a simple series circuit
because current in the wires is not everywhere the same. Volts and amps
vary along the line depending upon construction, length. and load placed
on the line.

The line`s resistance, inductance, conductance, and capacitance are
distributed and accumulate with its length. In a wave in either
direction on the line, the volts and amps at any point on the line
conform to Zo but also depend on summation of the incident and reflected
waves at that point. Current in the linee is not independent of the
voltage.

A transmission line of any length terminated with an impedance equal to
its Zo has an input impedance of Zo.

Distribution of volts and amps on lines terminated with loads other than
Zo has been demonstrated countless times and is well understood.
Directional couplers are capable of separating forward and reverse
directions on a line, and are also well known, understood and have
proved useful. The Bird wattmeter uses a directional coupler, works as
advertised, and may be inserted anywhere in a 50-ohm coax line.

Best regards, Richard Harrison, KB5WZI



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Old June 20th 05, 10:54 PM
Ian White GM3SEK
 
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Richard Harrison wrote:
Ian, GM3SEK wrote:
"I am trying not to be biased about the actual problem - "

Open minds consider arguments.

Certainly. You begin with small steps that no-one will seriously
dispute.

Wave velocity = frequency x wavelength

Wave velocity in free space is 186,000 miles per second (300,000
km/sec).

Velocity of disturbances on open wire transmission lines is almost equal
to that in free space.

At high radio frequencies, a wavelength may be measured in inches and
centimeters. A transmission line containing a discontinuity produces a
reflection from the change. As a short distance can produce a phase
change of 360 degrees, the incident and reflected waves can combine to
produce voltage variations along a short line length. These can be
measured using simple instruments with confidence.

Circuit theory works on transmission lines because the proximity of
their conductors causes an effect on one conductor to be instantaneously
imposed on the other.

A transmission line cannot be analyzed as a simple series circuit
because current in the wires is not everywhere the same. Volts and amps
vary along the line depending upon construction, length. and load placed
on the line.

The line`s resistance, inductance, conductance, and capacitance are
distributed and accumulate with its length. In a wave in either
direction on the line, the volts and amps at any point on the line
conform to Zo but also depend on summation of the incident and reflected
waves at that point. Current in the linee is not independent of the
voltage.

A transmission line of any length terminated with an impedance equal to
its Zo has an input impedance of Zo.

But here you pick up the pace. Instead of the detailed argument above,
suddenly whole chapters flash by in a single sentence:

Distribution of volts and amps on lines terminated with loads other than
Zo has been demonstrated countless times and is well understood.


Directional couplers are capable of separating forward and reverse
directions on a line, and are also well known, understood and have
proved useful.


And it's all a calculated run-up to this huge flying leap:

The Bird wattmeter uses a directional coupler, works as
advertised, and may be inserted anywhere in a 50-ohm coax line.

Sorry, Richard, but that isn't a constructed argument any more. It's
just a declaration.


--
73 from Ian GM3SEK 'In Practice' columnist for RadCom (RSGB)
http://www.ifwtech.co.uk/g3sek
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Old June 21st 05, 12:16 AM
Cecil Moore
 
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Ian White GM3SEK wrote:
Richard Harrison wrote:
The Bird wattmeter uses a directional coupler, works as
advertised, and may be inserted anywhere in a 50-ohm coax line.

Sorry, Richard, but that isn't a constructed argument any more. It's
just a declaration.


Do we need to go into the wave reflection model in detail? The Bird
wattmeter accepts the wave reflection model as scientific fact as do
most of my reference books. Given the wave reflection model and a 50
ohm environment, the Bird wattmeter reads forward and reflected power,
i.e. the number of joules passing a point on the 50 ohm transmission
line in each direction. The Bird wattmeter assumes: Forward V is in
phase with forward I. Reflected V is in phase with reflected I.
Vfor*Ifor = Pfor Vref*Iref = Pref
Vfor/Ifor = 50 ohms, Vref/Iref = 50 ohms.
Vfor^2/50 = forward power, Vref^2/50 = reflected power
Ifor^2*50 = forward power, Iref^2*50 = reflected power
All this is in any good textbook covering the wave reflection model
and has been accepted as fact by RF engineers for the better part
of a century.
--
73, Cecil http://www.qsl.net/w5dxp


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Old June 21st 05, 02:40 PM
Richard Harrison
 
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Ian White, GM3SEK wrote:
"But here you pick up the pace. Instead of the detailed argument above,
suddenly whole chapters rush by in a single sentence."

Fair criticism. It reflects tiring of posting before its conclusion.

The Bird wattmeter`s firectional coupler distinguishes between incident
and reflected waves by their singular difference. Upon reflection of a
wave, either the voltage or the current it generates is reversed in
phase, but not both.

Bird takes equal samples of voltage and current from the wave.. When
there has been a reflection, the samples have opposite polarity and
cancel. When there has been no reflection the samples from that
direction of travel are in-phase and the sample total is double the
contribution of either sample.

To determine reverse power flow, the polarity of one of the samples is
reversed.

You don`t need to know how it works to use it and Bird never advertised
how simple it is as far as I know.

Best regards, Richard Harrison, KB5WZI

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Old June 20th 05, 04:38 PM
Cecil Moore
 
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Cecil Moore wrote:
So would you explain what happens to the reflections of a laser
beam when aimed at a perfect mirror in free space? Offset the
laser beam slightly from normal incidence to start with and
observe the reflections with your naked eye. Then bring the
beam to 90 degree incidence with the mirror. You can no
longer see the reflections but now you can detect the
superposition of the forward and reflected waves through
interference "rings" or loops with an intensity maximum
occuring every half wavelength and an intensity minimum
occuring every half wavelength in between.


Incidentally, there is a very good discussion of standing waves
in section 7.1.4 of "Optics", by Hecht, 4th Edition.
--
73, Cecil http://www.qsl.net/w5dxp

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Old June 21st 05, 01:44 AM
Jim Kelley
 
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Richard Harrison wrote:

Reflected power is again reflected at the match point
because the matched source sees no reflection.


I think you put your finger on it, Richard. That's exactly what
inspired the 'sloshing energy' comment.

73, ac6xg





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