The Rest of the Story
On Sun, 6 Apr 2008 19:21:00 -0700 (PDT)
Keith Dysart wrote:
On Apr 5, 10:06*am, Roger Sparks wrote:
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.
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