On Mar 6, 12:04*am, Roger Sparks wrote:
On Wed, 5 Mar 2008 06:06:04 -0800 (PST)

Keith Dysart wrote:
On Mar 5, 8:12*am, Roger Sparks wrote:
On Tue, 4 Mar 2008 17:00:31 -0800 (PST)

Keith Dysart wrote:
On Mar 4, 3:36*pm, Cecil Moore wrote:
After discovering the error on Roy's web page at:

http://eznec.com/misc/Food_for_thought.pdf

I have begun a series of articles that convey "The Rest
of the Story" (Apologies to Paul Harvey). Part 1 of
these articles can be found at:

http://www.w5dxp.com/nointfr.htm

Looks good. And well presented. There is only one small problem
with the analysis.

When the instantaneous energy flows are examined it can be seen
that Prs is not equal to 50 W plus Pref.

Taking just the second example (12.5 ohm load) for illustrative
purposes...

The power dissipated in Rs before the reflection arrives is
Prs.before = 50 + 50cos(2wt) watts

How do you justify this conclusion? *It seems to me that until the reflection arrives, the source will have a load of Rs = 50 ohms plus line = 50 ohms for a total of 100 ohms. *Prs.before = 50 watts.

The voltage on Rs, before the reflection returns is
Vrs(t) = 70.7cos(wt)
Prs(t) = Vrs(t)**2/50
* * * *= 50 + 50cos(2wt)

Prs.before.average = average(Prs(t))
* * * * * * * * * *= 50
since the average of cos is 0.

For the equation Vrs(t) = 70.7cos(wt), are you finding the 70.7 from sqrt(50^2 + 50^2)?

No. Cecil's circuit has a 100 V RMS source. For this source

Vs(t) = 141.4 cos(wt)

Before the reflection returns, the line acts as 50 ohms. This, in
series with
the 50 source resistor means that Vs(t)/2 is across Rs and Vs(t)/2
appears
across the line.

Vrs(t) = Vf.g(t) = Vs(t)/2 = 70.7 cos(wt)

The power dissipated in the resistor is

Prs(t) = Vrs(t)**2 / 50
= ((70.7 * 70.7) / 50) * cos(wt) * cos(wt)
= 100 * 0.5 * (cos(wt+wt) + cos(wt-wt))
= 50 (cos(2wt) + 1)
= 50 cos(2wt) + 50

If so, how do you justify that proceedure before the reflection returns? *I think the voltage across Rs is 50v until the reflection returns. *I also think the current would be 1 amp, for power of 50w, until the reflection returns.

That is the average power. But since the voltage is a sinusoid, the
instantaneous power as a function of time is
Prs(t) = 50 + 50 cos(2wt)
which does average to 50 W, thus agreeing with your calculations.


My apologies for leaving out the "(t)" everywhere which would have
made it clearer.

There would be no reflected power at the source until the reflection returns, making the following statements incorrect.

I should have been more clear. The below applies after the reflected
wave returns.
And I should have included the "(t)" for greater clarity.

The reflected power at the source is
Pref.s = 18 + 18cos(2wt) watts

But the power dissipated in Rs after the reflection arrives is
Prs.after = 68 + 68cos(2wt-61.9degrees) watts

Prs.after is not Prs.before + Pref, though the averages do sum.

And since the energy flows must be accounted for on a moment by
moment basis (or we violate conservation of energy), it is the
instantaneous energy flows that provide the most detail and
allow us to conclude with certainty that Prs.after is not
Prs.before + Pref.

The same inequality holds for all the examples except those
with Pref equal to 0.

Thus these examples do not demonstrate that the reflected power
is dissipated in the source resistor.

...Keith

Rs is shown as a resistance on only one side of the line. *It would simplify and focus the discussion if Rs were broken into two resistors, each placed on one side of the source to make the circuit balanced. *

I am not sure how this would help. It would make the arithmetic
somewhat more
complex.

...Keith

Well, as drawn, the circuit is unbalanced. *Reflections are about time and distance traveled during that time, so we really don't know where the far center point is located for finding instantaneous values. *We are making an assumption that the length of the source and resistor rs is zero which is OK, but we need to be aware that we are making that assumption. *

I find it valuable to understand how such an ideal circuit would
operate, then
extend the solution if necessary.

In practice, many circuits are small compared to the wavelength and
these assumptions
do not materially affect the answer.

Of course, it is important to know when they do, at which time more
complex analysis
becomes necessary.

By splitting the resistor into two parts, and then adding them together to make the calculations, we can see that a balanced circuit is intended, but the source resistors are combined for ease of calculation.

...Keith