Henry Kolesnik wrote:

I know that any power not dissipated by an antenna is reflected back to the
transmitter. Then the transmitter "reflects" this reflection back to
antenna, ad nauseum until its all gone. I also know that a short or an open
is required to reflect power and I'm searching for which it is, an open or a
short. I'm inclined to think it's a virtual open but I'm at a loss to
understand that and I wonder if someone has a good explanation or analogy
and some math wouldn't hurt.
tnx
Hank WD5JFR

Hi Hank,

Here's a link which talks about (and illustrates) the physics of waves
at a boundary.

http://www.physicsclassroom.com/Class/waves/U10L3a.html

It shows what happens at the two extremes for transverse waves along a
string - not unlike electromagnetic waves in a transmission line in some
ways. In one instance, the string is fastened directly to the boundary
- a rod in this case. This is analogous to a short across a
transmission line. It will be seen that a reflection occurs, and that
the wave becomes inverted upon reflecting.

In the other case, the string is fastened to a ring which can slide
freely up and down the rod. This case is analogous to an unterminated,
or open transmission line. It can be seen that this too causes a
reflection, only this time the wave is reflected back without a phase
reversal. The amplitude of the reflected wave in both these cases
equals the amplitude of the incident wave.

Now imagine that some friction between the sliding ring and the rod can
be added in varying amounts. This friction would be proportional to the
conductance in an electrical circuit, and would be infinite at one
extreme and zero at the other. The greater the friction, the greater
the conductance (and the lower the electrical resistance). As we begin
to increase the friction (conductance) from zero, the amplitude of the
inverted, reflected wave begins to decrease. The decrease in amplitude
continues with increasing friction until the amplitude of the reflected
wave becomes zero. It could be said that this value is equal to the Z0
of the transmission line.

As the amount of friction is increased still further, a small reflection
once again begins to appear. Only now the phase is opposite from what
it was before. Further increases in friction produce further increases
in reflection amplitude until the amplitude of the reflected wave once
again equals the amplitude of the incident wave.

What we noticed in the exercise is that there is some value of friction
(conductance or resistance) for which no reflection occurs. The exact
value depends on the medium through which the wave is propagating. All
other values produced a reflection.

There is no perfect explanation, and this certainly isn't a perfect
analogy but I hope that it will at least help give you a little more of
a feel for the idea.

73, Jim AC6XG

Jim Kelley wrote:
It shows what happens at the two extremes for transverse waves along a
string - not unlike electromagnetic waves in a transmission line in some
ways.

Where are the E and H fields in the string response?
--
73, Cecil http://www.qsl.net/w5dxp



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Thanks, I knew this but what I don't know is why a final doesn't dissipate
the reflected wave but just reflects 100% I assume. Transistor and tube
finals dissipate a bunch producing the RF but what is the mirror, check
valve or diode that keeps it reflecting

--
73
Hank WD5JFR
"Jim Kelley" wrote in message
...
Henry Kolesnik wrote:

I know that any power not dissipated by an antenna is reflected back to
the
transmitter. Then the transmitter "reflects" this reflection back to
antenna, ad nauseum until its all gone. I also know that a short or an
open
is required to reflect power and I'm searching for which it is, an open
or a
short. I'm inclined to think it's a virtual open but I'm at a loss to
understand that and I wonder if someone has a good explanation or
analogy
and some math wouldn't hurt.
tnx
Hank WD5JFR

Hi Hank,

Here's a link which talks about (and illustrates) the physics of waves
at a boundary.

http://www.physicsclassroom.com/Class/waves/U10L3a.html

It shows what happens at the two extremes for transverse waves along a
string - not unlike electromagnetic waves in a transmission line in some
ways. In one instance, the string is fastened directly to the boundary
- a rod in this case. This is analogous to a short across a
transmission line. It will be seen that a reflection occurs, and that
the wave becomes inverted upon reflecting.

In the other case, the string is fastened to a ring which can slide
freely up and down the rod. This case is analogous to an unterminated,
or open transmission line. It can be seen that this too causes a
reflection, only this time the wave is reflected back without a phase
reversal. The amplitude of the reflected wave in both these cases
equals the amplitude of the incident wave.

Now imagine that some friction between the sliding ring and the rod can
be added in varying amounts. This friction would be proportional to the
conductance in an electrical circuit, and would be infinite at one
extreme and zero at the other. The greater the friction, the greater
the conductance (and the lower the electrical resistance). As we begin
to increase the friction (conductance) from zero, the amplitude of the
inverted, reflected wave begins to decrease. The decrease in amplitude
continues with increasing friction until the amplitude of the reflected
wave becomes zero. It could be said that this value is equal to the Z0
of the transmission line.

As the amount of friction is increased still further, a small reflection
once again begins to appear. Only now the phase is opposite from what
it was before. Further increases in friction produce further increases
in reflection amplitude until the amplitude of the reflected wave once
again equals the amplitude of the incident wave.

What we noticed in the exercise is that there is some value of friction
(conductance or resistance) for which no reflection occurs. The exact
value depends on the medium through which the wave is propagating. All
other values produced a reflection.

There is no perfect explanation, and this certainly isn't a perfect
analogy but I hope that it will at least help give you a little more of
a feel for the idea.

73, Jim AC6XG

Henry Kolesnik wrote:
Thanks, I knew this but what I don't know is why a final doesn't dissipate
the reflected wave but just reflects 100% I assume. Transistor and tube
finals dissipate a bunch producing the RF but what is the mirror, check
valve or diode that keeps it reflecting

There is none! The definition is the problem. It is simply a copout.
Sources can dissipate reflected energy. The amount is unknown.
--
73, Cecil http://www.qsl.net/w5dxp



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This is absolutely the best thread I've seen in years. It's
educational, thought provoking, entertaining, and revealing about the
human psyche.

The thread reminds of some of those that I monitor from time to time
regarding aeronautics and fluid dynamics on the topic of how an
airplane wing generates lift. Believe or not, there is still no
consensus after 100+ years. It's interesting in that it parallels this
thread in many ways -- attempting to interpret various abstract
mathmatical definitions of a physical process in a way that it "makes
sense." We're fortunate in that such things don't actually have to
make sense to work.

Al

OH! NO! Vortex vs. Bernoulli
Steve N.

--
Steve N, K,9;d, c. i My email has no u's.
"alhearn" wrote in message
om...
This is absolutely the best thread I've seen in years. It's
educational, thought provoking, entertaining, and revealing about the
human psyche.

The thread reminds of some of those that I monitor from time to time
regarding aeronautics and fluid dynamics on the topic of how an
airplane wing generates lift. Believe or not, there is still no
consensus after 100+ years. It's interesting in that it parallels this
thread in many ways -- attempting to interpret various abstract
mathmatical definitions of a physical process in a way that it "makes
sense." We're fortunate in that such things don't actually have to
make sense to work.

Al

"Steve Nosko" wrote in message ...
OH! NO! Vortex vs. Bernoulli

Actually, it's Circulation vs. Newtonian vs. Bernoulli -- all three
are different mathmatical means of describing accurately and precisely
what happens when a airfoil produces lift. Actually each is simply a
different way of expressing exactly the same thing, but none of them
translates well to a real-life understanding of the concept. One of
the problems is that causes and effects get confused and
oversimplified by the math.

Much the same with reflections, transmission lines, and impedance
matching. While reflections do indeed exist on transmission lines when
mismatched to a source or load, they simply create standing waves.
Standing waves create non-optimum impedances depending on the
characteristics and length of the line. These impedances interact with
source and load impedances in very predictable and calculated ways.
Efficiency of power transfer is then determined by optimizing the
matching of these impedances. Optimimizing impedances then eliminates
reflections --- a circle of causes and effects.

Mathmatically, it's more expedient to skip much of the in-between
cause-and-effect stuff, and jump directly to describing the entire
process as a direct relationship between reflections and power
transfer -- which causes problems when attempting to visualize or
explain the process -- because that's not the way it really works.
It's not quite that simple and direct.

A standard SWR meter is a good example. It can't conveniently measure
reflections OR standing waves, so it measures mismatch. Since
everything is directly related, it could be said that it measures
reflections -- but it really doesn't. So, it doesn't really matter
unless you try to understand how the meter works in terms of how it
measures reflections or standing waves.

Al

I'm probably not the only one that is getting an adequate fill of facts,
opinions and quotes. I have only one request. Does anyone have verifiable
and repeatable evidence that a properly tuned pi network final amplifier
without a tuner does or does not dissipate power when there are reflections?
If they do can they please direct us to the source or give us an easliy
understandable write up.
tnx
--
73
Hank WD5JFR

"alhearn" wrote in message
om...
"Steve Nosko" wrote in message
...
OH! NO! Vortex vs. Bernoulli

Actually, it's Circulation vs. Newtonian vs. Bernoulli -- all three
are different mathmatical means of describing accurately and precisely
what happens when a airfoil produces lift. Actually each is simply a
different way of expressing exactly the same thing, but none of them
translates well to a real-life understanding of the concept. One of
the problems is that causes and effects get confused and
oversimplified by the math.

Much the same with reflections, transmission lines, and impedance
matching. While reflections do indeed exist on transmission lines when
mismatched to a source or load, they simply create standing waves.
Standing waves create non-optimum impedances depending on the
characteristics and length of the line. These impedances interact with
source and load impedances in very predictable and calculated ways.
Efficiency of power transfer is then determined by optimizing the
matching of these impedances. Optimimizing impedances then eliminates
reflections --- a circle of causes and effects.

Mathmatically, it's more expedient to skip much of the in-between
cause-and-effect stuff, and jump directly to describing the entire
process as a direct relationship between reflections and power
transfer -- which causes problems when attempting to visualize or
explain the process -- because that's not the way it really works.
It's not quite that simple and direct.

A standard SWR meter is a good example. It can't conveniently measure
reflections OR standing waves, so it measures mismatch. Since
everything is directly related, it could be said that it measures
reflections -- but it really doesn't. So, it doesn't really matter
unless you try to understand how the meter works in terms of how it
measures reflections or standing waves.

Al

WARNING! I use the sinusoidal steady state assumptions for this
explanation, this means that all reflection transients have died out and the
input signal is not changing. These are generally good assumptions for
general use that does not depend on signal changing on a time scale similar
to the reflection period, like radar or fast scan tv.... for cw, am, and
even ssb in amateur sized systems these are generally very good and yeild
answers that are more than adequate to answer problems like this.

given the above its very simple. look at the coax connector as if it was a
connector on a black box, there are no reflections. period. the input of
the black box looks like a simple linear impedance.

the procedure to find out how it affects your pi network is this: use your
favorite circuit modeling program and model the linear and output network to
whatever degree of detail you see fit. attach a load of 50+j0 and determine
currents and voltages in the matching network. (note that you will have to
include real world losses in the inductors and capacitors if you want to
calculate power dissipation in them). change load to however bad a
condition you want to model and compare currents and voltages to the 50 ohm
case. this will show you if more or less power is lost in the matching
network. if you have modeled a tube or fet with real world parameters it
will also tell you if it's dissipation goes up or down.


"Henry Kolesnik" wrote in message
. ..
I'm probably not the only one that is getting an adequate fill of facts,
opinions and quotes. I have only one request. Does anyone have
verifiable
and repeatable evidence that a properly tuned pi network final amplifier
without a tuner does or does not dissipate power when there are
reflections?
If they do can they please direct us to the source or give us an easliy
understandable write up.
tnx
--
73
Hank WD5JFR

"alhearn" wrote in message
om...
"Steve Nosko" wrote in message
...
OH! NO! Vortex vs. Bernoulli

Actually, it's Circulation vs. Newtonian vs. Bernoulli -- all three
are different mathmatical means of describing accurately and precisely
what happens when a airfoil produces lift. Actually each is simply a
different way of expressing exactly the same thing, but none of them
translates well to a real-life understanding of the concept. One of
the problems is that causes and effects get confused and
oversimplified by the math.

Much the same with reflections, transmission lines, and impedance
matching. While reflections do indeed exist on transmission lines when
mismatched to a source or load, they simply create standing waves.
Standing waves create non-optimum impedances depending on the
characteristics and length of the line. These impedances interact with
source and load impedances in very predictable and calculated ways.
Efficiency of power transfer is then determined by optimizing the
matching of these impedances. Optimimizing impedances then eliminates
reflections --- a circle of causes and effects.

Mathmatically, it's more expedient to skip much of the in-between
cause-and-effect stuff, and jump directly to describing the entire
process as a direct relationship between reflections and power
transfer -- which causes problems when attempting to visualize or
explain the process -- because that's not the way it really works.
It's not quite that simple and direct.

A standard SWR meter is a good example. It can't conveniently measure
reflections OR standing waves, so it measures mismatch. Since
everything is directly related, it could be said that it measures
reflections -- but it really doesn't. So, it doesn't really matter
unless you try to understand how the meter works in terms of how it
measures reflections or standing waves.

Al

Henry Kolesnik wrote:
I'm probably not the only one that is getting an adequate fill of facts,
opinions and quotes. I have only one request. Does anyone have verifiable
and repeatable evidence that a properly tuned pi network final amplifier
without a tuner does or does not dissipate power when there are reflections?
If they do can they please direct us to the source or give us an easliy
understandable write up.

Hank, when reflected current flows backwards through a pi-net loading
coil, some of the reflected power is dissipated as I^2*R losses in
the coil. Other than that, a properly tuned pi-net causes a match
point that reflects all the reflected energy back toward the load.

If a match point exists in a ham radio antenna system, no reflected
energy will reach the source. This is the great majority of amateur
radio systems and no-reflections-at-the-source is the goal of every
ham. The thing that you are worried about is the unusual case where
reflections are allowed to reach the source.
--
73, Cecil http://www.qsl.net/w5dxp



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