Follow up and more corrections.
Roger wrote:
Correction. Please note the change below. I apologize for the error.

Roger Sparks wrote:
"Cecil Moore" wrote in message
...
Roger wrote:
The principles of superposition are mathematically usable, not too
hard, and I think very revealing. Yes, if we use part of the model,
we must use it all the way. To do otherwise would be error, or worse.

clip some

Roger, I have explained all of this before. If you are
capable of understanding it, I take my hat off to you.
Here's what happens in that ideal voltage source using
the rules of superposition.


Well, I would like to think I could understand it, but maybe something
is wrong like Roy suggests. I think it is all falling into place, but
our readers are not all in agreement.

Could I ask a couple of questions to make sure I am understanding your
preconditions?

Is this a Thevenin source? If so, what is the internal resistor set
to in terms of Z0?

1. With the reflected voltage set to zero, the source
current is calculated which results in power in the
source resistor. Psource = Isource^2*Rsource


For condition number 2 below, is this a Thevenin equivalent resistance?

2. With the source voltage set to zero, the reflected
current is calculated which results in power in the
source resistor. Pref = Iref^2*Rsource

Now superpose the two events. The resultant power
equation is: Ptot = Ps + Pr + 2*SQRT(Ps*Pr)cos(A)
where 'A' is the angle between the source current
and the reflected current. Note this is NOT power
superposition. It is the common irradiance (power
density) equation from the field of optics which
has been in use in optical physics for a couple of
centuries so it has withstood the test of time.

OK. A few reactions.

1. To add the total power of each wave and then also add the power
reduction of the phase relationship, would be creating power unless
the cos(A) is negative. Something seems wrong here, probably my
understanding.

2. The two voltages should be equal, therfore the power delivered by
each to the source resister would be equal.

3. The power delivered to the source resistor will arrive a different
times due to the phase difference between the two waves.

4. If the reflected power returns at the same time as the delivered
power (to the source resistor), no power will flow. This because the
resistor in a Thevenin is a series resistor. Equal voltages will be
applied to each side of the resitor. The voltage difference will be
zero.

5. If the reflected power returns 180 degrees out of phase with the
applied voltage (to the source resistor), the voltage across the
resistor will double with each cycle, resulting in an ever increasing
current.(and power) into the reflected wave.

For the voltage source looking into a 1/2WL open
circuit stub, angle 'A' is 180 degrees. Therefore
the total power in the source resistor is:

Ptot = Ps + Pr + 2*SQRT(Ps*Pr)cos(180) or

Ptot = Ps + Pr - 2*SQRT(Ps*Pr) = 0

This is correct for this condition. The problem with the equation
comes when cos(90) which can easily happen.

I would suggest the equation Ptot = Ps+ Pr*(sine(A) where angle 'A' is
the angle between positive peaks of the two waves. This angle will
rotate twice as fast as the signal frequency due to the relative
velocity between the waves..
No, this angle is fixed by system design. If the system changes, this
angle will rotate twice as fast as the angle change of a single wave.
The equation should be OK.

The equation can NOT be OK for anything except a single interaction
point. It can not be used to plot the interaction between waves because
the wave location information is not present.

Ptot in the source resistor is zero watts but we
already knew that. What is important is that the
last term in that equation above is known as the
"interference term". When it is negative, it indicates
destructive interference. When Ptot = 0, we have
"total destructive interference" as defined by
Hecht in "Optics", 4th edition, page 388.

Whatever the magnitude of the destructive interference
term, an equal magnitude of constructive interference
must occur in a different direction in order to
satisfy the conservation of energy principle. Thus
we can say with certainty that the energy in the
reflected wave has been 100% re-reflected by the
source. The actual reflection coefficient of the
source in the example is |1.0| even when the source
resistor equals the Z0 of the transmission line.
I have been explaining all of this for about 5 years
now. Instead of attempting to understand these relatively
simple laws of physics from the field of optics, Roy
ploinked me. Hopefully, you will understand.

All of this is explained in my three year old Worldradio
energy article on my web page.

http://www.w5dxp.com/energy.htm
--
73, Cecil http://www.w5dxp.com

Light, especially solar energy, is a collection of waves of all
magnitudes and frequencies. We should see only about 1/2 of the power
that actually arrives due to superposition. The reflection from a
surface should create standing waves in LOCAL space that will contain
double the power of open space. I think your equation Ptot = Ps + Pr
+ 2*SQRT(Ps*Pr)cos(A) would be correct for a collection of waves, but
incorrect for the example.

I tried to read your article a couple of weeks ago, but I found myself
not understanding. I have learned a lot since then thanks to you,
Roy, and others. I will try to read it again soon.

So what do you think Cecil?

73, Roger, W7WkB


In the last 24 hours, Roy posted a revised analysis that contains
results useful here. He presented a voltage example that resulted in a
steady state with steady state voltages 4 time initial value. Under
superposition, this would equate to 4 times the initial power residing
on the transmission line under the conditions presented. This concurs
with other authors who predict power on the transmission line may exceed
the delivered power due to reflected waves.

Your equation "Ptot = Ps + Pr + 2*SQRT(Ps*Pr)cos(A)" is taken from
illumination and radiation theory to describe power existing at a point
in space near a reflecting surface. If we consider space to be a
transmission media, and the reflecting surface to be a discontinuity in
the transmission media, then we have a situation very similar to an
electrical transmission line near a line discontinuity.

It is entirely reasonable to consider that the reflection ratio between
the space transmission media and the reflective surface would result in
an storage factor equaling 4 times the peak power of the initial forward
wave. By storage factor, I simply mean the ratio of forward power to
total power on the transmission media under standing wave conditions.

Under open circuit conditions, a half wavelength transmission line will
have a storage factor of 2. Roy presented an example where the storage
factor was 4. Perhaps the space transmission media also has a storage
factor of 4 under some conditions, as described by "Ptot = Ps + Pr +
2*SQRT(Ps*Pr)cos(A)".

73, Roger, W7WKB

Roger wrote:

In the last 24 hours, Roy posted a revised analysis that contains
results useful here. He presented a voltage example that resulted in a
steady state with steady state voltages 4 time initial value. Under
superposition, this would equate to 4 times the initial power residing
on the transmission line under the conditions presented. This concurs
with other authors who predict power on the transmission line may exceed
the delivered power due to reflected waves.

Your equation "Ptot = Ps + Pr + 2*SQRT(Ps*Pr)cos(A)" is taken from
illumination and radiation theory to describe power existing at a point
in space near a reflecting surface. If we consider space to be a
transmission media, and the reflecting surface to be a discontinuity in
the transmission media, then we have a situation very similar to an
electrical transmission line near a line discontinuity.

It is entirely reasonable to consider that the reflection ratio between
the space transmission media and the reflective surface would result in
an storage factor equaling 4 times the peak power of the initial forward
wave. By storage factor, I simply mean the ratio of forward power to
total power on the transmission media under standing wave conditions.

Under open circuit conditions, a half wavelength transmission line will
have a storage factor of 2. Roy presented an example where the storage
factor was 4. Perhaps the space transmission media also has a storage
factor of 4 under some conditions, as described by "Ptot = Ps + Pr +
2*SQRT(Ps*Pr)cos(A)".

Power is neither stored nor conserved, so a power "storage factor" is
meaningless. Consider a very simple example. Let's charge a capacitor
with a constant current DC source. We'll apply 1 amp to a 1 farad
capacitor for 1 second. During that time, the power begins at zero,
since the capacitor voltage is zero, then it rises linearly to one watt
as the capacitor voltage rises to one volt at the end of the one second
period. So the average power over that period was 1/2 watt, and we put
1/2 joule of energy into the capacitor. (To confirm, the energy in a
capacitor is 1/2 * C * V^2 = 1/2 joule.) Was power "put into" or stored
in the capacitor?

Now we'll connect a 0.1 amp constant current load to the capacitor, in a
direction that discharges it. We can use an ideal current source for
this. The power measured at the capacitor or source terminals begins at
0.1 watt and drops linearly to zero as the capacitor discharges. The
average is 0.05 watt. Why are we getting less power out than we put in?

"Where did the power go?" is heard over and over, and let me assure you,
anyone taking care with his mathematics and logic is going to spend a
long time looking for it. So in this capacitor problem, where did the
power go?

It takes 10 seconds to discharge the capacitor, during which the load
receives the 1/2 joule of energy stored in the capacitor. Energy was
stored. Energy was conserved. Power was neither stored nor conserved.

Roy Lewallen, W7EL

Roy Lewallen wrote:
"Where did the power go?"

Or more correctly, where did the energy go?
Was it destroyed or created? (Rhetorical)
--
73, Cecil http://www.w5dxp.com

Roy Lewallen wrote:
Roger wrote:

By storage factor, I simply mean the ratio of
forward power to total power on the transmission media under standing
wave conditions.


Power is neither stored nor conserved, so a power "storage factor" is
meaningless. Consider a very simple example. Let's charge a capacitor
with a constant current DC source. We'll apply 1 amp to a 1 farad
capacitor for 1 second. During that time, the power begins at zero,
since the capacitor voltage is zero, then it rises linearly to one watt
as the capacitor voltage rises to one volt at the end of the one second
period. So the average power over that period was 1/2 watt, and we put
1/2 joule of energy into the capacitor. (To confirm, the energy in a
capacitor is 1/2 * C * V^2 = 1/2 joule.) Was power "put into" or stored
in the capacitor?

Now we'll connect a 0.1 amp constant current load to the capacitor, in a
direction that discharges it. We can use an ideal current source for
this. The power measured at the capacitor or source terminals begins at
0.1 watt and drops linearly to zero as the capacitor discharges. The
average is 0.05 watt. Why are we getting less power out than we put in?

"Where did the power go?" is heard over and over, and let me assure you,
anyone taking care with his mathematics and logic is going to spend a
long time looking for it. So in this capacitor problem, where did the
power go?

It takes 10 seconds to discharge the capacitor, during which the load
receives the 1/2 joule of energy stored in the capacitor. Energy was
stored. Energy was conserved. Power was neither stored nor conserved.

Roy Lewallen, W7EL

By my using the words 'power' "storage factor", you got my point, hence
the reaction.

Before dismissing the concept of "storing power", consider that when
discussing a transmission line, it could be a useful description.

As you know, power is energy delivered over a time period. It always
carries a time dimension having beginning and end. Power(watt)
=v*i/(unit time) = 1 joule/second.

In the example you give of charging a capacitor, the time dimension is
lost, so you are correct that only energy is conserved. Power is lost.

With a transmission line, we have an entirely different case. Here
power is conserved because the time information is maintained. Power is
stored on the line during the period it resides on the line. For
example, we excite the line at one end and some time period later find
that power is delivered to some destination. During the time period
that the power was on the line, the information that defines the energy
distribution over time has been preserved.

If power is stored, we implicitly store energy. Energy is v*i measured
in joules without a time factor. Obviously we store energy on a
transmission line when we store power.

So if in the future, I use the term "power storage", please take it to
mean that energy distributed over time is under consideration. I hope
the term might be useful to you as well.

73, Roger, W7WKB

Roger wrote:

By my using the words 'power' "storage factor", you got my point, hence
the reaction.

Before dismissing the concept of "storing power", consider that when
discussing a transmission line, it could be a useful description.

As you know, power is energy delivered over a time period.

No, it's the rate of energy delivery or movement, which is not quite the
same thing.

It always
carries a time dimension having beginning and end. Power(watt)
=v*i/(unit time) = 1 joule/second.

Sorry, you've got this wrong. One watt is indeed one joule/second, but
P(t) = v(t) * i(t), period. Energy is the integral of P(t) dt. Power is
the time derivative of energy, or dE(t)/dt where E is the energy.

You could as reasonably say that energy always "carries a time
dimension". After all, one joule = 1 watt-second.

In the example you give of charging a capacitor, the time dimension is
lost, so you are correct that only energy is conserved. Power is lost.

Sorry, I don't understand that.

With a transmission line, we have an entirely different case. Here
power is conserved because the time information is maintained. Power is
stored on the line during the period it resides on the line. For
example, we excite the line at one end and some time period later find
that power is delivered to some destination. During the time period
that the power was on the line, the information that defines the energy
distribution over time has been preserved.

Ok, let's test this. Please tell me exactly how many watts are stored on
the line of the second analysis (where the perfect source is in series
with a 150 ohm resistor). Next, tell me how many watts will come out of
the line if we quickly disconnect the perfect source and source
resistance and replace it with:

A: A 50 ohm resistor, or
B: A 150 ohm resistor

If power is stored, we implicitly store energy. Energy is v*i measured
in joules without a time factor.

No, Energy is not v*i. Power is v*i. Energy is the time integral of v*i.
Power is not stored; energy is.

Obviously we store energy on a
transmission line when we store power.

I guess it would be obvious if you believe you can store power. But
before going further, please demonstrate what you mean by calculating
how many watts of power are stored on the example line. I showed exactly
how many joules of energy were stored, you can show how many watts of power.

So if in the future, I use the term "power storage", please take it to
mean that energy distributed over time is under consideration.

I'm afraid I'm not very good at translating what people mean when they
say something else. Why not call energy storage "energy storage", power
"power", and energy "energy"? Then I and hopefully other readers will
know what you mean. The MKSA unit of power is the watt, and of energy,
the joule. The two are no more the same than speed and distance, or
charge and current.

I hope
the term might be useful to you as well.

No, I have enough trouble communicating when I take great care with my
terminology. The last thing I need is to be saying something which means
something else -- or means nothing at all. When I mean energy storage,
I'll say "energy storage", thank you.

Roy Lewallen, W7EL

Roger wrote:
Perhaps the space transmission media also has a storage
factor of 4 under some conditions, as described by "Ptot = Ps + Pr +
2*SQRT(Ps*Pr)cos(A)".

Visualize visible interference rings between two equal
waves each of P magnitude, where the darkest of the rings
is completely black. If it is not obvious, the power in
the brightest of the rings would have to be 4P for the
energy to average out to the 2P in the source waves. That's
all there is to "black" destructive interference vs "bright"
constructive interference. (4P+0P)/2 = 2P = average power
It's a no-brainer for most folks.

On the source side of a Z0-match, the reflections are "black".
On the load side of a Z0-match with reflections, the forward
wave is "bright". What could be simpler?
--
73, Cecil http://www.w5dxp.com