On Fri, 4 Apr 2008 06:30:08 -0700 (PDT)
Keith Dysart wrote:
On Apr 4, 1:29*am, Roger Sparks wrote:
On Wed, 2 Apr 2008 02:41:24 -0700 (PDT)
Keith Dysart wrote:
On Apr 1, 8:08 am, Cecil Moore wrote:
Keith Dysart wrote:
Cecil Moore wrote:
If the average interference is zero, the average
reflected power is dissipated in the source resistor.
All of your unethical lies, innuendo, and hand-waving
will not change that fact of physics.
More precisely, the average value of the imputed reflected
power is numerically equal to the increase in the average
power dissipated in the source resistor.
More precisely, at the *exact* time of arrival of the
reflected wave, the average power dissipated in the
source resistor increases by *exactly* the magnitude
of the average reflected power. Do you think that is
just a coincidence?
Not quite, but close. And averages can not change instantly.
But your phraseology suggests another way to approach
the problem. Let us consider what happens just after
the arrival of the reflected wave. For computational
convenience, we'll use a frequency of 1/360 hertz so
that 1 degree of the wave is one second long.
The reflected wave arrives at 90 seconds (or 90 degrees)
after the source is turned on. If the line is
terminated in 50 ohms, then no reflection arrives and
during the one second between degree 90 and degree 91,
the source resistor dissipates 0.01523 J of energy. The
line also receives 0.01523 J of energy and the source
provides 0.03046 J.
* Esource.50[90..91] = 0.03046 J
* Ers.50[90..91] * * = 0.01523 J
* Eline.50[90..91] * = 0.01523 J
* Esource = Ers + Eline
as expected.
* Efor = Eline
since Eref is 0.
OK, Power at the source is found from (V^2)/50 = 0.03046w. *
Not quite.
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 at http://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. The source is acting like a resistor with an impedance of 2.468143/1.389317 = 1.776 ohms. As a result, the returning reflection does not truely see 50 ohms but sees 50 + 1.776 = 51.776 ohms. 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.
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.
--
73, Roger, W7WKB