Cecil's Math
 

Cecil,
W5DXP wrote:
"Steven Best says that total re-reflection is a fallacy. Do you disagree
with Dr. Best?"

I`ll play along. On this issue, Dr. Best may be mistaken. Individual
testimonies may not be very reliable, but large numbers become a
preponderence of evidence. Numbers of operators claim to be able to
reach near 1:1 SWR at their transmitter outputs while having greater
SWR`s in other parts of their transmission systems. I believe them. It
agrees with my own experience. The 1:1 SWR at the transmitter means
reflections were stopped short of their return to the transmitter.
Re-reflection at the match point is therefore very near total.

I have similar practical experience that transmitter systems can be
adjusted to reject returnng reflections in many cases. Reg used to
explain such rejection by an assumed very high transmitter output
impedance, if my recollection is correct. If I misquoted Reg in any way
it is unintentional. I disagree with the common opinion that classical
plate resistance has much to do with transmitter output impedance. In my
opinion, output conductance is proportional to the switched-on time of
the final amplifier. Conduction angle controls impedance, which is much
lower than plate resistance.

So why doesn`t the transmitter take back reflected power? The impedance
bump is needed as a catalyst to cause a re-reflection. The re-reflection
is as my professors said over 50 years ago due to the fact that the
transmitter already has a surplus of energy. Energy always flows on a
line from a surplus to a dearth, and not vice versa.

A tapering charge battery charger undergoes an amperage subsidence as a
battery becomes charged. Even were the charger another battery as in the
case of a jump-start, energy doesn`t flow from a lower potential battery
to the higher potential battery.

A-C sources behave much the same. The greater-magnitude source
outsupplies the other and determines the direction of current flow when
the sources are facing-off against each other.

Best regards, Richard Harrison, KB5WZI

Richard Harrison wrote:
A-C sources behave much the same. The greater-magnitude source
outsupplies the other and determines the direction of current flow when
the sources are facing-off against each other.

Let's say we have two sources each equipped with a circulator and load
resistor. We'll call such a source an SGCR (Signal generator equipped
with a circulator and load). We set up the following experiment.

100W SGCL#1------------50 ohm lossless coax-----------50W SGCL#2

Seems to me that the 100W from SGCL#1 will flow unopposed and
dissipate in SGCL#2's circulator load resistor. Seems to me that
the 50W from SGCL#2 will flow unopposed and dissipate in SGCL#1's
circulator load resistor.

In other words, the forward wave has no effect on the reflected
wave and vice versa. There is no face off.

Now consider a normal system.

100W SGCL----------50 ohm lossless coax----------291.5 ohm load

50W will be reflected from the load and flow unopposed back to
the circulator load resistor. There is no face off. Unless there
is an impedance discontinuity, forward waves have no effect on
reflected waves.
--
73, Cecil http://www.qsl.net/w5dxp



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W5DXP wrote:

Let's say we have two sources each equipped with a circulator and load
resistor. We'll call such a source an SGCR (Signal generator equipped
with a circulator and load). We set up the following experiment.

100W SGCL#1------------50 ohm lossless coax-----------50W SGCL#2

Seems to me that the 100W from SGCL#1 will flow unopposed and
dissipate in SGCL#2's circulator load resistor. Seems to me that
the 50W from SGCL#2 will flow unopposed and dissipate in SGCL#1's
circulator load resistor.

May I modify your experiment slightly? Change SGCL#2 to produce 100W.
Restating your explanation for the new experiment:
"Seems to me that the 100W from SGCL#1 will flow unopposed and
dissipate in SGCL#2's circulator load resistor. Seems to me that
the 100W from SGCL#2 will flow unopposed and dissipate in SGCL#1's
circulator load resistor."

Let's also make the 50 ohm lossless coax long enough that we can
find a voltage maximum. At this voltage maximum the current is 0,
always.
Assuming that P = V x I, the power is 0, always.
This seems to be at odds with explanation that SGCL#1's power is
dissipated in SGCL#2's circulator load resistor since there is no
energy flowing at the voltage maximum.

The obvious (and probably controversial) answer to this
inconsistency is that the power dissipated in CL#1 originates
in SG#1. Similarly for SGCL#2.

To further the experiment slightly, whenever a conductor in a circuit
has 0 current, the conductor can be cut without changing a thing.

Let's cut the coax at the voltage maximum (where the current is 0).
Nothing changes. The voltage and currents remain the same every
where. The powers remain the same everywhere, and yet there is no
longer a path from SG#1 to CL#2. The power being dissipated in CL#1
must be originating in SG#1.

Comments and corrections invited.

....Keith

Cecil, W5DXP wrote:
"50W will be reflected from the load and flow unopposed back to the
circulator load resistor. There is no face-off. Unless there is an
impedance discontinuity, forward waves have no effect on reflected
waves."

A 291.5-ohm load on a 50-ohm Zo produces a reflection coefficient of
0.707, and the rest follows suit.

Best regards, Richard Harrison, KB5WZI

wrote:
Assuming that P = V x I, the power is 0, always.
This seems to be at odds with explanation that SGCL#1's power is
dissipated in SGCL#2's circulator load resistor since there is no
energy flowing at the voltage maximum.

Nope, a directional coupler can still separate out the forward
and reflected waves even at voltage and current nulls. Remember,
those voltage and current nulls are only a 'dl' in length as 'l'
approaches zero. Superposition of two waves traveling in opposite
directions has no effect on the individual waves. There are no
reflections except at an impedance discontinuity point which
doesn't exist in the example.

Let's cut the coax at the voltage maximum (where the current is 0).
Nothing changes.

Of course something changes. You have introduced an impedance discontinuity
where none existed before. The forward wave is 100% reflected at
the cut point and of course, the reflected power is dissipated in
the source's circulator load resistor. If you make the two sources
one Hz different in frequency, you can easily observe what happens
when you cut the line. Otherwise, it's the old steady-state Catch-22
coverup in action. :-)
--
73, Cecil http://www.qsl.net/w5dxp



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Richard Harrison wrote:

Cecil, W5DXP wrote:
"50W will be reflected from the load and flow unopposed back to the
circulator load resistor. There is no face-off. Unless there is an
impedance discontinuity, forward waves have no effect on reflected
waves."

A 291.5-ohm load on a 50-ohm Zo produces a reflection coefficient of
0.707, and the rest follows suit.

Yes, but there is no face off between forward waves and reflected
waves. Reflected waves flow unopposed back to the circulator load
resistor where they are dissipated. Forward waves can affect
reflected waves only at an impedance discontinuity where reflections
and re-reflections can occur. Reflected waves can indeed flow "back
up the fire hose" unopposed by the forward waves.
--
73, Cecil http://www.qsl.net/w5dxp



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I wrote:
"Zeros in voltage and current SWR waveforms are 180-degrees out-of-phase
with each other."

What I should have said was: Voltage is maximum when current is minimum
and vice versa.

Voltage and current peaks are just 90-degrees apart in SWR patterns.

Best regards, Richard Harrison, KB5WZI

W5DXP wrote:

wrote:
Assuming that P = V x I, the power is 0, always.
This seems to be at odds with explanation that SGCL#1's power is
dissipated in SGCL#2's circulator load resistor since there is no
energy flowing at the voltage maximum.

Nope, a directional coupler can still separate out the forward
and reflected waves even at voltage and current nulls.

It is a bit early to move to the complexity of directional couplers.
I am still stuck on how energy can flow through a point on the circuit
where the current or voltage is always 0.

Using instantaneous Power = Vinst x Iinst, the power at such a point
must always be 0, leading to the conclusion that no energy is flowing.

So how does energy flow through a point in the circuit where the voltage
or current is always 0.

Is it that Pinst != Vinst x Iinst?
Or is it that there is no point in the ideal experiment presented where
V or I is always 0?
Or have I missed something in the chain of reasoning above?

....Keith

wrote:
I am still left quite confused by this. In the experiment proposed (a
slight variation on Cecil's), at a voltage maximum the current IS a constant 0
(or similarly, at a current maximum, the voltage IS a constant 0). These 0
volts and currents can be measured.

But they are zero only because of superposition. The forward current
is NOT zero and the reflected current is NOT zero. The fact that the
*NET* current is zero just means that all the energy moved into the
electric field at that point but it still exists.

If we accept that instantaneous power is P = V x I, then at any point on
the line where the voltage or current is a constant 0, the power must be a
constant 0.

Nope, for standing waves, if the H-field is zero, all the energy is in the
E-field. If the E-field is zero, all the energy is in the H-field. The field
that is zero is undergoing destructive interference. The other field is
always undergoing an equal magnitude of constructive interference.

From the definition of power, it would seem that where ever the power is
a constant 0, the energy flowing must be a constant 0.

Nope, if the current is zero, the voltage is at its maximum value. You
cannot have a maximum voltage with zero energy.

So I do not understand how energy can get through a point in a circuit
where the voltage or current is always 0, ...

That's one of the problems with being seduced by the steady-state model. :-)
Look at the component forward and reflected waves and all will become clear.
--
73, Cecil http://www.qsl.net/w5dxp



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wrote:
It is a bit early to move to the complexity of directional couplers.
I am still stuck on how energy can flow through a point on the circuit
where the current or voltage is always 0.

There are two voltages and two currents. Sometimes they are 180 degrees
out of phase and their sum is zero. But those waves don't know each
other exists. Someone forgot to tell them that they aren't supposed
to carry any energy so they just keep on carrying energy.

Using instantaneous Power = Vinst x Iinst, the power at such a point
must always be 0, leading to the conclusion that no energy is flowing.

No *NET* energy is flowing but the forward energy and reflected energy
just keep on flowing, carrying an equal amount of energy in each direction.
In a transmission line with reflections, there are always forward and reflected
transactions. Please don't be seduced by the steady-state religion.
--
73, Cecil http://www.qsl.net/w5dxp



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