On Apr 4, 11:41*am, Roger Sparks wrote:
On Fri, 4 Apr 2008 06:30:08 -0700 (PDT)

Keith Dysart wrote:
On Apr 4, 1:29*am, Roger Sparks wrote:
[snip]
Another way to figure the power to the source would be by using the voltage and current through the source. *

This is how I did it.

Taking Esource.50[90..91] = 0.03046 J as an example ...

Psource.50[90] = V * I
* * * * * * * *= 0.000000 * 0.000000
* * * * * * * *= 0 W
Psource.50[91] = -2.468143 * -0.024681
* * * * * * * *= 0.060917 W

Using the trapezoid rule for numerical integration,
Esource.50[90..91] = ((Psource.50[90]+Psource.50[91])/2) * interval
* * * * * * * * * *= ((0+0.060917)/2)*1
* * * * * * * * * *= 0.030459 J

The other powers and energies in the spreadsheet are computed
similarly.

There was an error in the computation of the 'delta's; the sign was
wrong.
The spreadsheet available athttp://keith.dysart.googlepages.com/radio6
has
now been corrected. (And, for Cecil, this spreadsheet no longer has
macros
so it may be downloadable.)

To use this spreadsheet to compute my numbers above, set the formulae
for Vr.g and Ir.g in the rows for degrees 90 and 91 to zero.

Using current and voltage, the power at time 91 degrees of the reflected wave is 1.234v*1.43868a= 1.775w. *Over one second integrated, the energy should be 1.775 J.

I could not match to the above data to any rows, so I can't comment.
But
perhaps the explanation above will correct the discrepancy.

I took a look at your revised spreadsheet entitled "Reflected45degrees-1.xls". Using the numbers from row 94 (91 degrees), the voltage developed at the source would be Vs = -2.468143v. *The current folowing through the source would be found from Ig which is 1.389317a in todays version of the spreadsheet (it was 1.439a previously). *The power flowing INTO the source is 2..468143* 1.389317 = 3.429032w. *This is the power Ps found in Column 11. *

This returning power is all from the reflected wave. *

I would not say this. The power *is* from the line, but this is Pg,
and it satisfies
the equation
Ps(t) = Prs(t) + Pg(t)

The imputed power in the reflected wave is Pr.g(t) and is equal to
-99.969541 W, at
91 degrees. This can not be accounted for in any combination of Ps(91)
(-3.429023 W)
and Prs(91) (96.510050 W). And recall that expressing Cecil's claim
using instantaneous
powers requires that the imputed reflected power be accounted for in
the source
resistor, and not the source. This is column 26 and would require that
Prs(91) equal
100 W (which it does not).

The source is acting like a resistor with an impedance of 2.468143/1.389317 = 1.776 ohms. *

This is not a good way to describe the source. The ratio of the
voltage to the current
is 1.776 but this is not a resistor since if circuit conditions were
to change, the
voltage would stay the same while the current could take on any value;
this being
the definition of a voltage source. Since the voltage does not change
when the current
does, deltaV/deltaI is always 0 so the voltage source is more properly
described as
having an impedance of 0.

As a result, the returning reflection does not truely see 50 ohms but sees 50 + 1.776 = 51.776 ohms. *

The returning reflection is affectively a change in the circuit
conditions. Using
the source impedance of 0 plus the 50 ohm resistor means the
reflection sees 50
ohms, so there is no reflection.

Using your approach of computing a resistance from the instantaneous
voltage and
current yeilds a constantly changing resistance. The reflection would
alter this
computed resistance. This change in resistance would then alter the
reflection
which would change the resistance. Would the answer converge?

The only approach that works is to use the conventional approach of
considering
that a voltage source has an impedance of 0.

Of course the result is a another reflection. *Is this the idea you were trying to communicate Cecil?

To me, this is destructive interference at work, so all the power in the reflected wave does not simply disappear into the resistor Rs on the instant basis. *

I agree with latter, but not for the reason expressed. Rather, because
the imputed
power of the reflected wave is a dubious concept. This being because
it is impossible
to account for this power.

clip

When using superposition, one has to pick a reference voltage and
current
direction for each component and then add or subtract the contributing
voltage or current depending on whether it is in the reference
direction
or against the reference direction. Mistakes and confusion with
relation
to the signs are not uncommon. Careful choice of reference directions
will
sometimes help, but mostly it simply leads to some other voltage or
current that needs to be subtracted instead of added.

...Keith

Yes. *I don't want to force my conventions onto you, so I am trying to understand yours while insisting that the answers from each convention must be the same. *

I think we both agree that the reflections are carrying power now.

Not I. Not until the imputed power can be accounted for.

...Keith