The Rest of the Story
 

On Mar 21, 9:12*am, Roger Sparks wrote:
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 nsign 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. *

Ooooppps. That should have been Vg(t) = Vf.g(t) + Vr.g(t)

but doesn't this describe the standing wave? *

Not by itself. It is simply the function describing the voltage at
point g.

If you were to compute Vx(g) = Vf.x(t) + Vr.x(t) for many points
x, then you might end up with something akin to a standing wave.
Note that Vg(t) = Vf.g(t) + Vr.g(t) holds true for any function
while standing waves only really appear when the excitation
function is sinusoidal.

At point g, the forward wave has passed through resistor Rs,

This is not a way that I would recommend thinking about what is
happening. The voltage source and source resistor are lumped
circuit elements and thinking that wave is passing through
them is likely to lead to confusion.

It might be valuable to study the circuit completely with a
lumped voltage source, source resistor and reactance. Once
that circuit is understood, replace the reactance with the
transmission line. When studying the circuit with the
transmission line instead of the reactance, always check
to see whether you would use the same words if the TL were
replaced with the reactance. If not, question why.

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?

It does not alter the outcome, so is not that important.

* * 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. *

I think that is just because your reference direction was different.
I use the same reference direction for both Pf and Pr. It is often
done that the reference direction is different for Pf and Pr.

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.

Again, I don't think it wise to think of the wave as passing through
Rs.

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. *

I describe Vg(t) as the actual voltage that would be measured by an
oscilloscope at the point g; i.e. the beginning of the transmission
line. It is certainly not the voltage across Rs. Perhaps it IS what
you mean by standing wave. In any case the voltage across the resistor
is
Vs(t) = Vrs(t) + Vg(t)
Vrs(t) = Vs(t) - Vg(t)

Pg(t) is the power being delivered to the transmission line at any
instance t.
Pg(t) = Vg(t) * Ig(t)

...Keith

The Rest of the Story
 

On Mar 21, 12:37*pm, Cecil Moore wrote:
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.

Actually the flip-flopping I was referring to was the
constant changes in your view of the limitations that
apply to your claim.

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.

Is this a flop or a flip? Are you now agreeing that the
energy in the reflected wave is only dissipated in the
source resistor for those instances when Vs is 0?

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.
[gratuitous insult snipped]

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

I did and you ignored it.

I could not find them in the archive. Could you kindly provide
them again, showing where the 'cos(A)' term fits in the
equations:
Ps(t) = Prs(t) + Pg(t)
and
Pg(t) = Pf.g(t) + Pr.g(t)

There is no negative sign in the
power equation yet you come up with negative signs.
[gratuitous insult snipped]

Negative signs also arise when one rearranges equations.

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.
[gratuitous insult snipped]

...Keith

The Rest of the Story
 

On Mar 21, 12:53*pm, Cecil Moore wrote:
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.

I did. And they are also conveniently in the spreadsheet at
http://keith.dysart.googlepages.com/...oad,reflection

For your convenience, in the example of Fig 1-1, 100 Vrms
sinusoidal source, 50 ohm source resistor, 45 degrees of
50 ohm line, 12.5 ohm load, after the reflection returns...

Vs(t) = 141.42135623731*cos(wt)
Vg(t) = 82.46211251*cos(wt+30.96375653)
Vf.g(t) = 70.71067812*cos(wt)
Vr.g(t) = 42.42640687*cos(wt+90)

And for completeness...

Ps(t) = 100 + 116.6190379cos(2wt-30.96375653)
Prs(t) = 68 + 68cos(2wt-61.92751306)
Pg(t) = 32 + 68cos(2wt)
Pf.g(t) = 50 + 50cos(2wt)
Pr.g(t) = -18 + 18cos(2wt)

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.

Not that anyone has disputed that. But it would
be good to see anyway.

...Keith

The Rest of the Story
 

On Mar 21, 1:00*pm, Cecil Moore wrote:
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.

Since I am not sure exactly what your are looking for,
and you could not be sure that I did them right,
it would probably be best if you were to compute these
equations which you need.

...Keith

The Rest of the Story
 

On Mar 21, 5:03*pm, Roger Sparks wrote:
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.

My analysis used voltages, currents and impedances to compute
all the voltages and currents within the circuit. Some were
derived using superposition of voltages and currents but most
were derived using basic circuit theory (E=IR, Ztot=Z1+Z2, etc.)

Having done that, the powers for the three components (the
voltage source, resistor, and entrance to the transmission line)
in the circuit were computed. These powers were not derived using
superposition but by multiplying the current through the component
by the voltage across it. This is universally accepted as a valid
operation.

Having the power functions for each of the component, we can
then turn to the conservation of energy principle: The energy
in a closed system is conserved.

This is the basis for the equation
Ps(t) = Prs(t) + Pg(t)
This equation says that for the system under consideration
(Fig 1-1), the energy delivered by the source is equal to
the energy dissipated in the resistor plus the energy
delivered to the line. This is extremely basic and satisfies
the conservation of energy principle. This is not
superposition and any inclusion of cos(theta) terms would
be incorrect.

The equation
Pg(t) = Pf.g(t) + Pr.g(t)
is more interesting. The basis for this is superposition.
The forward and reverse voltage and current are superposed
to derive the actual voltage and current. It would seem
invalid to also sum the powers. To me it was a complete
surprise that summing the voltages produces the correct
total voltages and, at the same time, summing the powers
(which are a squared function of the voltage) also
produce the correct result.

But by starting with the equations used to derive forward
and reverse voltage and current, it can be easily shown
with appropriate substitution that Ptot is always equal
to Pforward + Preverse (Or Pf - Pr if you use the other
convention for the direction of the energy flow)s. It
simply falls out from the way that Vf and Vr are derived
from Vactual and Iactual.

So
Pg(t) = Pf.g(t) + Pr.g(t)
is always true. For any arbitrary waveforms. Inclusion
of cos(theta) terms would be incorrect.

...Keith

The Rest of the Story
 

Keith Dysart wrote:
. . .
The equation
Pg(t) = Pf.g(t) + Pr.g(t)
is more interesting. The basis for this is superposition.
The forward and reverse voltage and current are superposed
to derive the actual voltage and current. It would seem
invalid to also sum the powers. To me it was a complete
surprise that summing the voltages produces the correct
total voltages and, at the same time, summing the powers
(which are a squared function of the voltage) also
produce the correct result.

But by starting with the equations used to derive forward
and reverse voltage and current, it can be easily shown
with appropriate substitution that Ptot is always equal
to Pforward + Preverse (Or Pf - Pr if you use the other
convention for the direction of the energy flow)s. It
simply falls out from the way that Vf and Vr are derived
from Vactual and Iactual.

It only holds true when Z0 is purely real. Of course, when it isn't,
time domain analysis becomes very much more cumbersome. But it's not
hard to show the problem using steady state sinusoidal analysis, and
that's where the cos term appears and is appropriate.


So
Pg(t) = Pf.g(t) + Pr.g(t)
is always true. For any arbitrary waveforms. Inclusion
of cos(theta) terms would be incorrect.

...Keith

Roy Lewallen, W7EL

The Rest of the Story
 

On Mar 21, 9:32*pm, Roy Lewallen wrote:
Keith Dysart wrote:
. . .
The equation
* Pg(t) = Pf.g(t) + Pr.g(t)
is more interesting. The basis for this is superposition.
The forward and reverse voltage and current are superposed
to derive the actual voltage and current. It would seem
invalid to also sum the powers. To me it was a complete
surprise that summing the voltages produces the correct
total voltages and, at the same time, summing the powers
(which are a squared function of the voltage) also
produce the correct result.

But by starting with the equations used to derive forward
and reverse voltage and current, it can be easily shown
with appropriate substitution that Ptot is always equal
to Pforward + Preverse (Or Pf - Pr if you use the other
convention for the direction of the energy flow)s. It
simply falls out from the way that Vf and Vr are derived
from Vactual and Iactual.

It only holds true when Z0 is purely real. Of course, when it isn't,
time domain analysis becomes very much more cumbersome. But it's not
hard to show the problem using steady state sinusoidal analysis, and
that's where the cos term appears and is appropriate.

*
* So
* * Pg(t) = Pf.g(t) + Pr.g(t)
* is always true. For any arbitrary waveforms. Inclusion
* of cos(theta) terms would be incorrect.

Thanks for providing the limitation.

But I am having difficulty articulating where the math in the
following
derivation fails.

Starting by measuring the actual voltage and current at a single
point on the line, and wishing to derive Vf and Vr we have the
following four equations:

V = Vf + Vr
I = If - Ir
Zo = Vf / If
Zo = Vr / Ir

rearranging and substituting
Vf = V - Vr
= V - Zo * Ir
= V - Zo * (If - I)
= V - Zo * (Vf/Zo - I)
= V - Vf + Zo * I
= (V + Zo * I)/2
similarly
Vr = (V - Zo * I)/2

Pf = Vf * If
= Vf**2 / Zo
= ((V + Zo * I)(V + Zo * I)/4)/Zo
= (V**2 + 2(V * Zo * I) + Zo**2 * I**2) / (4 * Zo)

Pr = Vr * Ir
= (V**2 - 2(V * Zo * I) + Zo**2 * I**2) / (4 * Zo)

So, comtemplating that
P = Pf - Pr
and substituting
P = (V**2 + 2(V * Zo * I) + Zo**2 * I**2) / (4 * Zo)
- (V**2 - 2(V * Zo * I) + Zo**2 * I**2) / (4 * Zo)
= 4(V * Zo * I) / (4 * Zo)
= V * I
as required.

So when Zo is real, i.e. can be represented by R, it is
clear that P always equals Pf - Pr. And it does not even
matter which value of R is used for R. It does not have
to be the characteristic impedance of the transmission
line, the subtraction of powers still produces the correct
answer.

But when Zo has a reactive component, it still cancels
out of the equations. So why is this not a proof that
also holds for complex Zo.

I suspect it has to do with complex Zo being a concept
that only works for single frequency sinusoids, but am
having difficulty discovering exactly where it fails.

And if it is related to Zo and single frequency sinusoids,
does that mean that P = Pf - Pr also always works for
single frequency sinusoids?

...Keith

The Rest of the Story
 

Keith Dysart wrote:
Roger Sparks wrote:

Ooooppps. That should have been Vg(t) = Vf.g(t) + Vr.g(t)

but doesn't this describe the standing wave?

Not by itself. It is simply the function describing the voltage at
point g.

The net total voltage at point g is the standing wave
voltage that is present at that point.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Keith Dysart wrote:
Actually the flip-flopping I was referring to was the
constant changes in your view of the limitations that
apply to your claim.

As I said, I corrected an error in my thinking. You
are free to consider that to be a flip-flop, I consider
it to be a step forward.

Are you now agreeing that the
energy in the reflected wave is only dissipated in the
source resistor for those instances when Vs is 0?

Yes, for instantaneous reflected energy exactly as I
previously stated, but not true for average reflected
energy. 100% of the average reflected energy is
dissipated in the source resistor when the transmission
line is 45 degrees long. That intra-cycle interference
exists, thus delaying the dissipation by 90 degrees,
is irrelevant to where the net energy winds up going.

Your conservation of power principle would have you
demanding that the power sourced by a battery charger
must be instantaneously dissipated. Everyone except
you seems to realize that is an invalid concept. The
dissipation of energy can be delayed by a battery
or a reactance. In the present example, the dissipation
of the instantaneous reflected energy is delayed by
90 degrees by the reactance.

I could not find them in the archive. Could you kindly provide
them again, showing where the 'cos(A)' term fits in the
equations:

Go back and read the part where I said cos(0)=+1 and
cos(180)=-1. There is no such thing as conservation of
power. Your equations assume a conservation of power
principle that doesn't exist in reality. The forward
power is positive power. The reflected power is positive
power. The only negative power is destructive interference
which must be offset by an equal magnitude of constructive
interference. When the instantaneous interference power is
negative, the two voltages are 180 degrees out of phase and
that is your cos(180)=-1. When the instantaneous interference
power is positive, the two voltages are in phase and that
is your cos(0)=+1. Why don't you already know all this
elementary stuff?
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

On Mar 22, 9:36*am, Cecil Moore wrote:
Keith Dysart wrote:
Roger Sparks wrote:
Ooooppps. That should have been Vg(t) = Vf.g(t) + Vr.g(t)

but doesn't this describe the standing wave? *

Not by itself. It is simply the function describing the voltage at
point g.

The net total voltage at point g is the standing wave
voltage that is present at that point.

I suppose that is a name you could use, but do you still
call it "standing wave voltage" when there is no
reflected wave, i.e. Vr is 0 for all time. There is
no standing wave when Vr is always 0.

...Keith