The Rest of the Story
 

On Mar 20, 1:07*pm, Roger Sparks wrote:
Keith Dysart wrote:
On Mar 16, 10:21 am, Cecil Moore wrote:
snip
Would you please explain how energy is conserved in the
following example at the zero-crossing point for Vs?

* * * * * * * *Rs * * * Vg * * * * * * * * * * Vl
* * * * *+----/\/\/-----+----------------------+
* * * * *| * *50 ohm * * * * * * * * * * * * * |
* * * * *| * * * * * * * * * * * * * * * * * * |
* * * * Vs * * * * * * * * 45 degrees * * * * *| Shorted
* * *100v RMS * * * * * * *50 ohm line * * * * | Stub
* * * * *| * * * * * * * * * * * * * * * * * * |
* * * * *| * * * * * * * * * * * * * * * * * * |
* * * * *+--------------+----------------------+
* * * * gnd

At the zero-crossing of Vs, Ps(t) = 0, i.e. the source
is supplying zero watts at that time but Prs(t) = 100w.
Where is the 100 watts coming from?

For the first 90 degrees of time, the circuit can be represented as

* * * * * * * * *Rs * * * Vg * * * * * * * * * * Vl
* * * * * *+----/\/\/-----+----------------------+
* * * * * *| * *50 ohm * * * * * * * * * * * * * |
* * * * * *| * * * * * * * * * * * * * * * * * * /
* * * * * Vs * * * * * * * * * * * * * * * * * * \ 50 ohm resistor
* * * *100v RMS * * * * * * * * * * * * * * * * */
* * * * * *| * * * * * * * * * * * * * * * * * * \
* * * * * *| * * * * * * * * * * * * * * * * * * |
* * * * * *+--------------+----------------------+
* * * * * gnd

After 90 degrees of time has passed, the circuit can be represented as

* * * * * * * * *Rs * * * Vg * * * * * * * * * * Vl
* * * * * *+----/\/\/-----+----------------------+
* * * * * *| * *50 ohm * * * * * * * * * * * * * |
* * * * * *| * * * * * * * * * * * * * * * * * * |
* * * * * Vs * * * * * * * * * * * * * * * * * *---
* * * *100v RMS * * * * * * * * * * * * * * * * --- 50 ohm inductive
* * * * * *| * * * * * * * * * * * * * * * * * * |
* * * * * *| * * * * * * * * * * * * * * * * * * |
* * * * * *+--------------+----------------------+
* * * * * gnd

The sudden switch in circuit design at time 90 degrees is not unique to
start up, but is true for any adjustment made to Vs and any returning
wave from the shorted stub,. *As a result, a true stable circuit will
never be found unless some voltage adjustment is allowed for the 90
timing shift caused by the shorted stub. *Keith (in his analysis of the
circuit) recognizes that Vs drives into a reactive circuit.

If we want to understand how constructive and destructive interference
act to cause a 50 ohm resistor to evolve into a 50 ohm capacitor, then
we need to examine how traveling waves might do this. *It would be nice
to have a formula or wave sequence that fully addressed this evolution.

*From circuit theory, we have the inductive reactance of a
short-circuited line less than 1/4 wavelength long is

* * *XL = Zo * tan (length degrees)

* * * * = 50 * tan(45)

* * * * = 50 ohms

*From traveling wave theory, we would have the applied wave from the
source arriving 90 degrees late to the stub side of resistor Rs. *This
ignores the fact that current must already be passing through resistor
Rs because voltage has been applied to Rs from the Vs side 90 degrees
earlier. *Whoa! *Things are not adding up correctly this way!

We need to treat the wave going down the 50 ohm line as a single wave
front. *The wave reverses at the short-circuit, reversing both direction
of travel and sign of voltage. *When the wave front reaches the input
end, 90 degrees after entering (for this case), the voltage/current
ratio is identical to the starting ratio (the line was 45 degrees long,
tan(45) = 1), and the returning voltage directly adds to the voltage
applied from Vs. *As a result, the current flowing through Rs will
increase, and Vs will see a changed (decreased) impedance.

After 90 degrees of signal application, we should be able to express the
voltage across Rs as

* * *Vrs(t) = Vs(t) - Vg(t) + Vref(t)

Just to ensure clarity, you are using a different convention for terms
and
signs than I do.

For me, Vg(t) is the actual voltage at point Vg. In terms of forward
and
reverse: Vg(t) = Vf.g(t) = Vr.g(t), meaning the actual voltage at
point
g is sum of the forward and reverse voltage at point g.

In my terms, this leads to
Vrs(t) = Vs(t) - Vg(t)
= Vs(t) - Vf.g(t) - Vr.g(t)

Your Vg seems to be same as my Vf.g and you seem to be using a
different
sign for Vref than I do for Vr.

You must be carrying these differences through the equations correctly
because the final results seem to agree.

Vs(t) = Vs(wt + 90), Vg(t) = Vs(wt + 90)/2, Vref(t) = Vs(wt)/2

Substitute so

Vrs(t) = Vs(wt + 90) - Vs(wt + 90)/2 + Vs(wt)/2

* * *= Vs(wt + 90)/2 + Vs(wt)/2

Allow Vs to be represented by a sine wave, we have

A small aside. It is conventional to use cos for sinusoidal waves
because
it maps more readily into phasor notation.

* * 2Vrs(t) = Vs*sin(wt + 90) + Vs*sin(wt)
* * * * * * = 2Vs*(sin(wt + 45)(cos(45))
* * Vrs(t) = Vs(sin(wt + 45)(cos(45))

Vs is defined as 100v RMS, which equals 100 * 1.414 = 141.4v Peak. *The
maximum voltage would occur when the sin term was 90 degrees and equals
1, which would occur at wt = 45 degrees. * We would have

* * *Vrs(45) = 141.42 * sin(90)(cos(45)
* * * * * * *= 141.42 * 1 * 0.7071
* * * * * * *= 100v

Now consider the current. *After the same 90 degrees of signal
application, we should be able to express the current through Rs as

* * *Irs(t) = Is(t) + Iref(t)

Is(t) = Is(wt + 90), Iref(t) = Is(wt)

The reflected current has been shifted by 90 degrees due to the
reflection so we must rewrite Iref(t) to read

* * *Iref(t) = Is(wt + 90)

Substitute,

* * * Irs(t) = Is(wt + 90) + Is(wt)

Allow Is to be represented by a sine wave, we have

* * * Irs(t) = Is*sin(wt + 90) + Is*sin(wt)

* * * * * * *= 2*Is(sin(wt + 45)(cos(45))

How do we find Is? *Is is the initial current found by dividing the
applied voltage at peak (141.42v) by the initial resistance (100 ohms).

* * *Is = 141.42/100 = 1.4142a

The maximum current would occur when the sin term was 90 degrees and
equals 1, which would occur at wt = 45 degrees. * We would have

* * * Irs(t) = 2*Is(sin(wt + 45)(cos(45))

* * * * * * *= 2 * 1.4142 * 1 * 0.7071

* * * * * * *= 2a

These results agree with the results from Keith and from circuit theory.

We have a theory and at least the peaks found from the theory agree with
the results from others. * How about Cecil's initial question which is
* At the zero-crossing of Vs, Ps(t) = 0, i.e. the source
* is supplying zero watts at that time but Prs(t) = 100w.
* Where is the 100 watts coming from?

We will use the equation

* *Vrs(t) = Vs(t) - Vg(t) + Vref(t)

Vs(t) = Vs(wt + 90), Vg(t) = Vs(wt + 90)/2, Vref(t) = Vs(wt)/2

The challenge looks at the time when Vs(t) = Vs*sin(wt + 90) = 0, which
occurs when wt = -90. *When wt = -90, all the power to the source
resistor Rs is from the reflected portion described as Vref(t) =
Vs(wt)/2. *The voltage across Rs would be

* *Vrs(-90) = Vref(-90)
* * * * * * = Vs*sin(-90)/2
* * * * * * = 141.4/2
* * * * * * = 70.7v

The power to Rs would be (Vrs^2)/50 = (70.7^2)/50 = 100w, all coming
from the reflection.

In summary, power to the resistor Rs comes via two paths, one longer
than the other by 90 degrees (in this example). *

This is not quite complete. The power to Rs does come from two places
but they are the voltage source and the line.

Ps(t) = Prs(t) + Pg(t)

where Pg(t) is the power at point g flowing out of the generator into
the line and Ps(t) is the power flowing from the source to the source
resistor and the line.

The power into the line can be seaparated into a forward component and
a reflected component:

Pg(t) = Pf(t) + Pr(t)

(Sometimes this is written as P = Pf - Pr, but that is just a question
of how the sign is used to represent the direction of energy flow.)

Substituting we have

Ps(t) = Prs(t) + Pf.g(t) + Pr.g(t)
or
Prs = Ps(t) - Pf.g(t) - Pr.g(t)

Whenever Ps(t) equal 0, then Pf.g(t) also equals 0 so at those times,
this simplifies to
Prs = - Pr.g(t)
which is the same conclusion you came to.

But it is wise not to forget the Pf.g(t) term because at times when
Ps(t) is not equal to zero, the Pf.g(t) term will be needed to
balance the energy flows.

The short path is the
series path of two resistors composed of the source resistor Rs and the
input to the 50 ohm transmission line measured by Vg. *The long path is
the series path of one resistor Rs and one capacitor composed of the
shorted transmission line. *Both paths are available at all times.
Power flows through both paths to Rs at all times, but because of the
time differential in arrival timing, at some point Rs will receive power
only from Vs, and at another point, receive power only from Vref.

And at still other times, the power in the source resistor will depend
on
all of Ps(t), Pf.g(t) and Pr.g(t), or, more simply, it will depend on
Ps(t) and Pg(t).

...Keith

The Rest of the Story
 

On Mar 20, 2:07*pm, Cecil Moore wrote:
Roger Sparks wrote:
Vrs(t) = Vs(t) - Vg(t) + Vref(t)

Good job, Roger. Let's simplify the equation through
substitution.

Let V1(t) = [Vs(t) - Vg(t)]

Let V2(t) = Vref(t)

Vrs(t) = V1(t) + V2(t)

The power to Rs would be (Vrs^2)/50 = (70.7^2)/50 = 100w, all coming
from the reflection.

Yes, because when Vs(t)=0, there is zero instantaneous
interference.

In summary, power to the resistor Rs comes via two paths, one longer
than the other by 90 degrees (in this example). *The short path is the
series path of two resistors composed of the source resistor Rs and the
input to the 50 ohm transmission line measured by Vg. *The long path is
the series path of one resistor Rs and one capacitor composed of the
shorted transmission line. *Both paths are available at all times. Power
flows through both paths to Rs at all times, but because of the time
differential in arrival timing, at some point Rs will receive power only
from Vs, and at another point, receive power only from Vref.

What Keith is missing is that:

Vrs(t)^2 = [V1(t) + V2(t)]^2 NOT= V1(t)^2 + V2(t)^2

In his math, Keith is asserting that since

Prs(t) NOT= P1(t) + P2(t), then the reflected power is not
being dissipated in the source resistor. But every sophomore
EE knows NOT to try to superpose powers like that. Roger,
I'll bet you know not to try to superpose powers?

Since Keith doesn't listen to me, would you pass that
technical fact on to him?

When Keith uses the correct equation:

Prs(t) = P1(t) + P2(t) +/- 2*SQRT[P1(t)*P2(t)]

he will see that the equation balances and therefore
100% of the reflected energy is dissipated in the
source resistor since the interference term averages
out to zero over each cycle.

It would be excellent if you would correct my exposition
using the correct formulae. We could then see if your
proposal actually provides accurate results.

...Keith

PS. It won't.

The Rest of the Story
 

On Fri, 21 Mar 2008 05:10:01 -0700 (PDT)
Keith Dysart wrote:

On Mar 20, 1:07*pm, Roger Sparks wrote:
Keith Dysart wrote:
On Mar 16, 10:21 am, Cecil Moore wrote:
snip
Would you please explain how energy is conserved in the
following example at the zero-crossing point for Vs?

* * * * * * * *Rs * * * Vg * * * * * * * * * * Vl
* * * * *+----/\/\/-----+----------------------+
* * * * *| * *50 ohm * * * * * * * * * * * * * |
* * * * *| * * * * * * * * * * * * * * * * * * |
* * * * Vs * * * * * * * * 45 degrees * * * * *| Shorted
* * *100v RMS * * * * * * *50 ohm line * * * * | Stub
* * * * *| * * * * * * * * * * * * * * * * * * |
* * * * *| * * * * * * * * * * * * * * * * * * |
* * * * *+--------------+----------------------+
* * * * gnd

At the zero-crossing of Vs, Ps(t) = 0, i.e. the source
is supplying zero watts at that time but Prs(t) = 100w.
Where is the 100 watts coming from?

For the first 90 degrees of time, the circuit can be represented as

* * * * * * * * *Rs * * * Vg * * * * * * * * * * Vl
* * * * * *+----/\/\/-----+----------------------+
* * * * * *| * *50 ohm * * * * * * * * * * * * * |
* * * * * *| * * * * * * * * * * * * * * * * * * /
* * * * * Vs * * * * * * * * * * * * * * * * * * \ 50 ohm resistor
* * * *100v RMS * * * * * * * * * * * * * * * * */
* * * * * *| * * * * * * * * * * * * * * * * * * \
* * * * * *| * * * * * * * * * * * * * * * * * * |
* * * * * *+--------------+----------------------+
* * * * * gnd

After 90 degrees of time has passed, the circuit can be represented as

* * * * * * * * *Rs * * * Vg * * * * * * * * * * Vl
* * * * * *+----/\/\/-----+----------------------+
* * * * * *| * *50 ohm * * * * * * * * * * * * * |
* * * * * *| * * * * * * * * * * * * * * * * * * |
* * * * * Vs * * * * * * * * * * * * * * * * * *---
* * * *100v RMS * * * * * * * * * * * * * * * * --- 50 ohm inductive
* * * * * *| * * * * * * * * * * * * * * * * * * |
* * * * * *| * * * * * * * * * * * * * * * * * * |
* * * * * *+--------------+----------------------+
* * * * * gnd

The sudden switch in circuit design at time 90 degrees is not unique to
start up, but is true for any adjustment made to Vs and any returning
wave from the shorted stub,. *As a result, a true stable circuit will
never be found unless some voltage adjustment is allowed for the 90
timing shift caused by the shorted stub. *Keith (in his analysis of the
circuit) recognizes that Vs drives into a reactive circuit.

If we want to understand how constructive and destructive interference
act to cause a 50 ohm resistor to evolve into a 50 ohm capacitor, then
we need to examine how traveling waves might do this. *It would be nice
to have a formula or wave sequence that fully addressed this evolution.

*From circuit theory, we have the inductive reactance of a
short-circuited line less than 1/4 wavelength long is

* * *XL = Zo * tan (length degrees)

* * * * = 50 * tan(45)

* * * * = 50 ohms

*From traveling wave theory, we would have the applied wave from the
source arriving 90 degrees late to the stub side of resistor Rs. *This
ignores the fact that current must already be passing through resistor
Rs because voltage has been applied to Rs from the Vs side 90 degrees
earlier. *Whoa! *Things are not adding up correctly this way!

We need to treat the wave going down the 50 ohm line as a single wave
front. *The wave reverses at the short-circuit, reversing both direction
of travel and sign of voltage. *When the wave front reaches the input
end, 90 degrees after entering (for this case), the voltage/current
ratio is identical to the starting ratio (the line was 45 degrees long,
tan(45) = 1), and the returning voltage directly adds to the voltage
applied from Vs. *As a result, the current flowing through Rs will
increase, and Vs will see a changed (decreased) impedance.

After 90 degrees of signal application, we should be able to express the
voltage across Rs as

* * *Vrs(t) = Vs(t) - Vg(t) + Vref(t)

Just to ensure clarity, you are using a different convention for terms
and
signs than I do.

For me, Vg(t) is the actual voltage at point Vg. In terms of forward
and
reverse: Vg(t) = Vf.g(t) = Vr.g(t), meaning the actual voltage at
point
g is sum of the forward and reverse voltage at point g.

In my terms, this leads to
Vrs(t) = Vs(t) - Vg(t)
= Vs(t) - Vf.g(t) - Vr.g(t)

Your Vg seems to be same as my Vf.g and you seem to be using a
different
sign for Vref than I do for Vr.

You must be carrying these differences through the equations correctly
because the final results seem to agree.

Thanks for looking at my posting very carefully, because I think you are drawing the same conclusions as I, at least so far.

I think you are correct in saying that
For me, Vg(t) is the actual voltage at point Vg. In terms of forward
and
reverse: Vg(t) = Vf.g(t) = Vr.g(t), meaning the actual voltage at
point
g is sum of the forward and reverse voltage at point g.

but doesn't this describe the standing wave?

At point g, the forward wave has passed through resistor Rs, loosing part of the original wave energy, and is inroute beginning the long path to the resistor Rs. The reflected wave has arrived back to one side of Rs and the voltage is already in series with the source voltage acting on Rs. I think we need to subtract the two terms so that we can separate the apples and oranges.

Vs(t) = Vs(wt + 90), Vg(t) = Vs(wt + 90)/2, Vref(t) = Vs(wt)/2

Substitute so

Vrs(t) = Vs(wt + 90) - Vs(wt + 90)/2 + Vs(wt)/2

* * *= Vs(wt + 90)/2 + Vs(wt)/2

Allow Vs to be represented by a sine wave, we have

A small aside. It is conventional to use cos for sinusoidal waves
because
it maps more readily into phasor notation.

This is an important observation. cos(0) is 1 so to me it seems like we are saying that we begin at time zero with peak voltage or peak current. I wanted to begin with zero current at time zero so I used the sine. Am I missing something important here?


* * 2Vrs(t) = Vs*sin(wt + 90) + Vs*sin(wt)
* * * * * * = 2Vs*(sin(wt + 45)(cos(45))
* * Vrs(t) = Vs(sin(wt + 45)(cos(45))

Vs is defined as 100v RMS, which equals 100 * 1.414 = 141.4v Peak. *The
maximum voltage would occur when the sin term was 90 degrees and equals
1, which would occur at wt = 45 degrees. * We would have

* * *Vrs(45) = 141.42 * sin(90)(cos(45)
* * * * * * *= 141.42 * 1 * 0.7071
* * * * * * *= 100v

Now consider the current. *After the same 90 degrees of signal
application, we should be able to express the current through Rs as

* * *Irs(t) = Is(t) + Iref(t)

Is(t) = Is(wt + 90), Iref(t) = Is(wt)

The reflected current has been shifted by 90 degrees due to the
reflection so we must rewrite Iref(t) to read

* * *Iref(t) = Is(wt + 90)

Substitute,

* * * Irs(t) = Is(wt + 90) + Is(wt)

Allow Is to be represented by a sine wave, we have

* * * Irs(t) = Is*sin(wt + 90) + Is*sin(wt)

* * * * * * *= 2*Is(sin(wt + 45)(cos(45))

How do we find Is? *Is is the initial current found by dividing the
applied voltage at peak (141.42v) by the initial resistance (100 ohms).

* * *Is = 141.42/100 = 1.4142a

The maximum current would occur when the sin term was 90 degrees and
equals 1, which would occur at wt = 45 degrees. * We would have

* * * Irs(t) = 2*Is(sin(wt + 45)(cos(45))

* * * * * * *= 2 * 1.4142 * 1 * 0.7071

* * * * * * *= 2a

These results agree with the results from Keith and from circuit theory.

We have a theory and at least the peaks found from the theory agree with
the results from others. * How about Cecil's initial question which is
* At the zero-crossing of Vs, Ps(t) = 0, i.e. the source
* is supplying zero watts at that time but Prs(t) = 100w.
* Where is the 100 watts coming from?

We will use the equation

* *Vrs(t) = Vs(t) - Vg(t) + Vref(t)

Vs(t) = Vs(wt + 90), Vg(t) = Vs(wt + 90)/2, Vref(t) = Vs(wt)/2

The challenge looks at the time when Vs(t) = Vs*sin(wt + 90) = 0, which
occurs when wt = -90. *When wt = -90, all the power to the source
resistor Rs is from the reflected portion described as Vref(t) =
Vs(wt)/2. *The voltage across Rs would be

* *Vrs(-90) = Vref(-90)
* * * * * * = Vs*sin(-90)/2
* * * * * * = 141.4/2
* * * * * * = 70.7v

The power to Rs would be (Vrs^2)/50 = (70.7^2)/50 = 100w, all coming
from the reflection.

In summary, power to the resistor Rs comes via two paths, one longer
than the other by 90 degrees (in this example). *

This is not quite complete. The power to Rs does come from two places
but they are the voltage source and the line.

Ps(t) = Prs(t) + Pg(t)

where Pg(t) is the power at point g flowing out of the generator into
the line and Ps(t) is the power flowing from the source to the source
resistor and the line.

The power into the line can be seaparated into a forward component and
a reflected component:

Pg(t) = Pf(t) + Pr(t)

(Sometimes this is written as P = Pf - Pr, but that is just a question
of how the sign is used to represent the direction of energy flow.)

Substituting we have

Ps(t) = Prs(t) + Pf.g(t) + Pr.g(t)
or
Prs = Ps(t) - Pf.g(t) - Pr.g(t)

Whenever Ps(t) equal 0, then Pf.g(t) also equals 0 so at those times,
this simplifies to
Prs = - Pr.g(t)
which is the same conclusion you came to.

No, I think my Pr.g(t) was positive, not negative. The difference is important because of the timing of the energy flows. Here is where it is important to observe that the forward wave entering the transmission line will have no further effect on resistor Rs until it has traveled 90 degrees of time. On the other hand, the reflected wave Pr.g(t) is only an instant of time away from entering the resistor Rs, but has not actually entered it because you are measuring it as being present at Vg, not as voltage Vrs.


But it is wise not to forget the Pf.g(t) term because at times when
Ps(t) is not equal to zero, the Pf.g(t) term will be needed to
balance the energy flows.

Pf.g(t) is the remaining power after the wave generated by Ps(t) has passed through Rs and left behind Prs(t). Another Prs(t + x) will be generated by the returning reflected wave.


The short path is the
series path of two resistors composed of the source resistor Rs and the
input to the 50 ohm transmission line measured by Vg. *The long path is
the series path of one resistor Rs and one capacitor composed of the
shorted transmission line. *Both paths are available at all times.
Power flows through both paths to Rs at all times, but because of the
time differential in arrival timing, at some point Rs will receive power
only from Vs, and at another point, receive power only from Vref.

And at still other times, the power in the source resistor will depend
on
all of Ps(t), Pf.g(t) and Pr.g(t), or, more simply, it will depend on
Ps(t) and Pg(t).

...Keith

We differ on how to utilize the term Pg(t). I think it describes the standing wave, not the voltage across Rs.
--
73, Roger, W7WKB

The Rest of the Story
 

Roger Sparks wrote:
So it looks to me like Keith is right in his method, at least in this case.

Roger, do you understand why EE201 professors admonish
their students not to try to superpose powers?
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Keith Dysart wrote:
But you need to clearly state your limitations and stop
flip flopping.

What you are calling "flip flopping" is me correcting
my errors. Once I correct an error, I don't flip-flop
back. My error was in assuming that the power-density
(irradiance) equation only works on average powers.
K7ITM convinced me that it works on instantaneous powers
also.

I am surprised, this being 2008, that I could actually be
offering a new way to study the question, but if you insist,
I accept the accolade.

I'm sure you are not the first, just the first to think
there is anything valid to be learned by considering
instantaneous power to be important. Everyone except
you discarded that notion a long time ago.

Analysis has shown that when examined with fine granularity,
that for the circuit of Fig 1-1, the energy in the reflected
wave is not always dissipated in the source resistor.

Yes, yes, yes, now you are starting to get it. When
interference is present, the energy in the reflected
wave is NOT dissipated in the source resistor. Those
facts will be covered in Part 2 & 3 of my web article.

Ptot = P1 + P2 + 2*SQRT(P1*P2)cos(A)

Which of the two needs the 'cos' term?

Ps(t) = Prs(t) + Pg(t)
or
Pg(t) = Pf.g(t) + Pr.g(t)

In fact neither do.

For instantaneous values of voltage, the phase angle is
either 0 or 180 degrees so the cosine term is either +1
or -1. The math is perfectly consistent. That you don't
recognize the sign of the instantaneous interference term
as the cosine of 0 or 180 is amazing.

Non-the-less do feel free to offer corrected expression that include
the 'cos(A)' term.

I did and you ignored it. There is no negative sign in the
power equation yet you come up with negative signs. That
you don't recognize your negative sign as cos(180) is
unbelievable.

The math holds as it is. But I invite you to offer an alternative
analysis that includes cos(A) terms. We can see how it holds up.

It it unfortunate that you don't comprehend that the cos(0) is
+1 and the cos(180) is -1. The sign of the instantaneous
interference term is the cosine term. The math certainly does
hold as it is - you are just ignorant of the "is" part.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Keith Dysart wrote:
In my terms, this leads to
Vrs(t) = Vs(t) - Vg(t)
= Vs(t) - Vf.g(t) - Vr.g(t)

How about expanding those equations for us?

Vs(t) = 141.4*cos(wt) ????
Vg(t) = ____*cos(wt+/-____) ????
Vf.g(t) = ____*cos(wt+/-____) ????
Vr.g(t) = ____*cos(wt+/-____) ????

If you ever did this before, I missed it.

Given the correct voltage equations, I can
prove what I am saying about destructive and
constructive interference averaging out to zero
over one cycle is a fact.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Keith Dysart wrote:
It would be excellent if you would correct my exposition
using the correct formulae. We could then see if your
proposal actually provides accurate results.

Roger didn't understand your terms and subscripts and
neither do I. I doubt that anyone understands your
formulas well enough to discuss them.

What is the equation for the forward voltage component
dropped across Rs?

What is the equation for the reflected voltage
component dropped across Rs?

Given valid equations for those two voltages, I can
prove everything I have been saying.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

On Fri, 21 Mar 2008 11:22:22 -0500
Cecil Moore wrote:

Roger Sparks wrote:
So it looks to me like Keith is right in his method, at least in this case.

Roger, do you understand why EE201 professors admonish
their students not to try to superpose powers?
--
73, Cecil http://www.w5dxp.com

No, I really don't.

I do understand that you had better not try to find the power flowing from superposed voltages because they may be the sum of apples and oranges. That is, the superposed voltages may be flowing in opposite directions so they will each be carrying power but the power will be available/reachable only for the direction toward which the wave is traveling. An SWR meter comes to mind here.

--
73, Roger, W7WKB

The Rest of the Story
 

Roger Sparks wrote:

Cecil Moore wrote:
Roger, do you understand why EE201 professors admonish
their students not to try to superpose powers?

No, I really don't.

It is because (V1 + V2)^2 usually doesn't equal V1^2+V2^2
because of interference. Keith's addition of powers
without taking interference into account is exactly
the mistake that the EE201 professors were talking about.
One cannot validly just willy-nilly add powers. It is
an ignorant/sophomoric thing to do.

If we add two one watt coherent waves, do we get a two
watt wave? Only in a very special case. For the great
majority of cases, we do *NOT* get a two watt wave. In
fact, the resultant wave can be anywhere between zero
watts and four watts. The concepts behind Keith's
calculations are invalid. If you are also trying to
willy-nilly add powers associated with coherent waves,
your calculations are also invalid.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

On Fri, 21 Mar 2008 19:43:12 GMT
Cecil Moore wrote:

Roger Sparks wrote:

Cecil Moore wrote:
Roger, do you understand why EE201 professors admonish
their students not to try to superpose powers?

No, I really don't.

It is because (V1 + V2)^2 usually doesn't equal V1^2+V2^2
because of interference. Keith's addition of powers
without taking interference into account is exactly
the mistake that the EE201 professors were talking about.
One cannot validly just willy-nilly add powers. It is
an ignorant/sophomoric thing to do.

If we add two one watt coherent waves, do we get a two
watt wave? Only in a very special case. For the great
majority of cases, we do *NOT* get a two watt wave. In
fact, the resultant wave can be anywhere between zero
watts and four watts. The concepts behind Keith's
calculations are invalid. If you are also trying to
willy-nilly add powers associated with coherent waves,
your calculations are also invalid.
--
73, Cecil http://www.w5dxp.com

OK, yes, I agree. It is OK to add powers when you are adding the power used by light bulbs. It is not OK to willy nilly multiply the voltage or current by the number of bulbs to learn the power used. You must carefully consider the circuit that connects the bulbs before selecting the proper method of calculating power, especially the possibility that the bulbs may be connected to phased power as in 3 phase or in traveling waves.

--
73, Roger, W7WKB