On Sat, 5 Apr 2008 03:06:18 -0700 (PDT)
Keith Dysart wrote:
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).
Taking these numbers and adding the Psource.50[91] = -2.468143 * -0.024681*= 0.060917 W found previously above,
we have Total = 0.060917 + 3.429023 + 96.510050 = 99.99999. Very close to 100w, but I am not sure of exactly what I am adding here.
I think you are right, that we should always be able to add the power on the line to 100w if we truely account for the power at any instant. This results from the trig identity that sin(x) + cos(x) = 1. Sin(x+90)= cos(x) so sin(x)+sin(x+90)= 1. This explains why the sum of the forward power and reflected power should always equal 100w.
Pg(t) is the result of a standing wave, containing power from Pf(x) and
Pr(x+90). Only the power from
Pr(x+90) is available to at a later time Prs(t+90+delta). Power from Pf(x+delta) is found in the transmission line.
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.
The overriding issue is to account for all the power, which we are having a hard time doing. The current is flowing the wrong way for the source voltage at Ps(91) so the source is absorbing power. That negates the idea that the source has an impedance of zero when we also assign the source a voltage.
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.
Does it help to notice that we are applying the power through two wires/paths. We then divide the power in half and say that the power to the resistor from the close half flows seamlessly from source to resistor. On the other hand, the second half of the source power flows through the transmission line and we say that it may never get to the resistor. Something is wrong with this logic.
If we can't account for the power, it is because we are doing the accounting incorrectly.
clip
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
I am convinced that accounting for the power can be done on an instant basis, but neither of us has done it on a mutually acceptable basis yet. Maybe if we pursue the trig identity
sin(x) + sin(x+90) = 1, we can have a constant reference and avoid mixing forward and reflected powers?
--
73, Roger, W7WKB