On Mar 6, 11:41*am, Roger Sparks wrote:
On Thu, 6 Mar 2008 06:10:26 -0800 (PST)
Keith Dysart wrote:
On Mar 6, 1:17*am, Roger Sparks wrote:
On Wed, 5 Mar 2008 21:37:06 -0800 (PST)
[snip]
Thanks Keith. *I see what you are doing now, although I still don't understand your logic in faulting Cecil on the instantaneous values. *I agree with you that the instantaneous values can be tracked, but don't see a fault in Cecil's presentation. *
For the special situation described in
http://www.w5dxp.com/nointfr.htm
Cecil is attempting to show that the reflected energy is dissipated in
the source resistor.
The logic he employs is:
- before the reflection arrives back at the generator, the source
resistor
* is dissipating X watts.
- the reflected wave has an energy flow of Y watts.
- after the reflection arrives back at the generator, the source
resistor
* is dissipating Z watts.
- since Z is equal to X + Y, the energy in the reflected wave is being
* dissipated in the source resistor.
In other words, since the dissipation in the source resistor increases
by the same amount as the power in the reflected wave, the energy in
the reflected wave must be being dissipated in the source resistor.
Your logic and conclusion seem correct to me.
Cecil analyzes the circuit for a number of load resistances and
suggests
that the equality holds for any load resistance.
For example, with a load resistance of 12.5 ohms, the original
dissipation
in the source resistor is 50 W which increases to 68 W when the 18 W
reflected wave arrives back at the generator. That is, X = 50, Y = 18
and Z = 68, so Z is equal to X + Y.
Cecil does all of this analysis using average powers.
But we know that the power dissipation varies as a function of time
and
that the power in the reflected wave is a function of time. It is my
contention that if it is the energy in the reflected wave that is
increasing the dissipation in the source resistor, the dissipation in
source resistor should occur at the same time that the reflected wave
delivers the energy.
In other words, not only should Z.average = X.average + Y.average,
but Z.instantaneous should equal X.instantaneous + Y.instantaneous
for if the dissipation in the source resistor is not tracking the
energy in the reflected wave, it can not be the energy in the
reflected wave that is heating the resistor.
Hard to follow this long sentence/paragraph. *The dissipation in the source resistor should be the sum of instantaneous energy flows from both source(forward) and reflected energy flows. *
Agreed. And that is what I was trying to say.
We would not expect that the instantaneous energy flow would be equal from both source and reflection because of the sine wave nature of the energy flow.
So using the same 12.5 ohm example,
X.inst = 50 + 50 cos(2wt)
Y.inst = 18 + 18 cos(2wt)
X.inst + Y.inst = 68 + 68 cos(2wt)
but
Z.inst = 68 + 68 cos(2wt - 61.9degrees)
So Z.inst is not equal to Y.inst + X.inst.
I think you are mixing average energy flow (the 50 and 18 watts) with instantaneous flow (50 cos(2wt) and 18 cos(2wt)).
I do not think so.
The instantaneous energy flow for the reflected wave is
Pr.g(t) = 18 - 18 cos(2wt) (see correction in previous post)
This means that at t=0, 0 energy is flowing.
When 2wt is 180 degrees, 36 joules per second are flowing.
The average over a full cycle is 18 W.
So I am summing the instantaneous flows for the two contributors of
energy.
The correct sum, however, is 68 + 32cos(2wt).
The actual dissipation in the source resistor is
Vrs(t) = 82.46 cos(wt -30.96 degrees)
Irs(t) = 1.649 cos(wt -30.96 degrees)
Prs(t) = Vrs(t) * Irs(t)
= 68 + 68 cos(2wt -61.92 degrees)
Whenever cos(2wt-61.92degrees) is equal to 1, 132 joules per second
are being dissipated and whenever it is equal to -1, 0 joules are
being dissipated. This follows from the periodic nature of the energy
flow.
I think Z.inst = x.inst + y.inst = 50 cos(2wt) + 18 cos(2wt). *The two energy components arrive 90 degrees out of phase (in this example) so we should write z.inst = 50 cos(2wt) + 18 cos(2wt + 90).
This means that the dissipation in the resistor is not happening
at the same time as the energy is being delivered by the reflected
wave, which must mean that it is not the energy from the reflected
wave that is heating the source resistor.
So while analyzing average power dissipations suggests that the
energy from the reflected wave is dissipated in the source resistor,
analysis of the instantaneous power shows that it is not.
...Keith
Either your math or mine is not correct. *Which is incorrect?
Both. But I think I have now corrected mine.
...Keith