On Sun, 23 Mar 2008 09:03:31 -0500
Cecil Moore wrote:

Roger Sparks wrote:
How can we justify calling the +/- 2*SQRT[Pf.g(t)*Pr.g(t) term the "interference term"?

Page 388, "Optics", by Eugene Hecht, 4th edition:
"The interference term becomes I12 = 2*SQRT(I1*I2)cos(A)"
where 'I' is the Irradiance (power density)[NOT Current]
Later Hecht says +2*SQRT(I1*I2) is the total constructive
interference term and -2*SQRT(I1*I2) is the total
destructive interference term. Chapter 9 is titled
"Interference" - recommended reading.

Am I correct in assuming that this equation describes the instantaneous power delivered to Rs?

Yes, if Tom, K7ITM, is correct about the equation working
for instantaneous power densities, not just for average
power densities as I had first assumed.

Let's say the instantaneous forward voltage dropped across
the source resistor is +50 volts and the instantaneous
reflected voltage across the source resistor is -30 volts.
The source resistor is 50 ohms.
Pf.rs(t) = (+50v)^2/50 = 50w
Pr.rs(t) = (-30v)^2/50 = 18w
Prs(t) = Pf.rs(t) + Pr.rs(t) - interference
Prs(t) = 50w + 18w - 2*SQRT(50*18) = 8 watts

If Tom is correct, that should be the actual dissipation
in the source resistor at that time which includes 60 watts
of destructive interference that will be dissipated 90 degrees
later when 2*SQRT(50*18) = +60 watts.
--
73, Cecil http://www.w5dxp.com

Thanks for your thoughtful reply.

TanH(30/50) = 30.96 degrees. This takes us back to the 12.5 ohm load example.

Is it possible that in your example here, the reflected voltage acts in series with Rs but arrives 90 degrees out of phase with the forward voltage? If so, then

Vrs = sqrt(50^2 + 30^2) (the reflected voltage should ADD to the source voltage)

= sqrt(3400) = 58.31v

The power to Rs would be

Prs = (V^2)/50 = 3400/50 = 68w

We previously found that 32w was used at the 12.5 ohm load, so 32 + 68 = 100w. The entire output from the source is accounted for.

If this is the case, we have here an example of constructive interference, and complete accounting for the power.

You might wonder why I would consider this alternative. If the destructive interference included a 90 degree delay, how would I know whether the 30v was the delayed voltage or exactly in phase with the source? It must be delayed by 90 degrees because the forward voltage is always 90 degrees ahead of the reflected wave (in 45 degree line length example).

Your example certainly works as written, but it also introduces a dilemma. Where is the power stored for 90 degrees?

To answer that question, I see two possiblities: The source voltage causes a reflection so the 60w is stored as an additional reflected wave on the transmission line.

Or second, the 60w is stored in the source.

--
73, Roger, W7WKB