Richard Harrison wrote:
Keith Dsart wrote:
"Therefore, the forward and reverse waves can not be transferring energy
across these points."

Waves in motion are transporting energy no matter how their constituents
seem to add at a particular point.

We can make Keith's assertion true by the addition of one
word.

"Therefore, the forward and reverse waves cannot be transferring
*net* energy across these points. As Ramo and Whinnery say about
the forward and reflected Poynting vectors:

If Pz+ = Pz- then Pz+ - Pz- = 0
--
73, Cecil http://www.w5dxp.com

On Dec 24, 12:01*am, Cecil Moore wrote:
Richard Harrison wrote:
Keith Dsart wrote:
"Therefore, the forward and reverse waves can not be transferring energy
across these points."

Waves in motion are transporting energy no matter how their constituents
seem to add at a particular point.

We can make Keith's assertion true by the addition of one
word.

"Therefore, the forward and reverse waves cannot be transferring
*net* energy across these points. As Ramo and Whinnery say about
the forward and reflected Poynting vectors:

With an open circuited line, I agree that there is no
net energy transfer at any point on the line. At most
points on the line, there is energy sloshing back and
forth, but netting to zero.

My statement about those 90 degree points where the
voltage or current is always 0 is much stronger: NO
energy transfer.

This follows inexorably from P(x,t) = V(x,t) * I(x,t).

If you disagree with the general applicability of this
equation, please indicate when it can and can not be
applied.

...Keith

Keith Dysart wrote:
With an open circuited line, I agree that there is no
net energy transfer at any point on the line. At most
points on the line, there is energy sloshing back and
forth, but netting to zero.

It is not "sloshing" back and forth unless you consider
the speed of light to be "sloshing". That's EM photonic
energy. If it's not traveling at the speed of light, it
doesn't exist. I would describe the movement of EM wave
energy as "racing" back and forth.

My statement about those 90 degree points where the
voltage or current is always 0 is much stronger: NO
energy transfer.

No *NET* energy transfer because the energy being transfered
in either direction is equal to the other direction. This
would be true of incoherent EM waves and even applies to
any and every type of wave energy.

This follows inexorably from P(x,t) = V(x,t) * I(x,t).

Yes, it certainly does for *NET* power.
Pnet = Pfor - Pref = 0 for ideal standing waves.

If you disagree with the general applicability of this
equation, please indicate when it can and can not be
applied.

It can be correctly applied when the person doing the
applying understands what it really means in reality.

I suggest you take a look at HP's s-parameter AP 95-1.
If you square the normalized s-parameter voltage equations,
you get power directly. The s-parameter equations are
designed that way.

b1 = s11*a1 + s12*a2 standard s-parameter voltage equation

a1^2 = Forward Power incident upon the impedance discontinuity

b1^2 = Reflected Power = (s11*a1 + s12*a2)^2
Power reflected from the impedance discontinuity

When a1^2 - b1^2 = 0

what do you think is the physical meaning of Reflected
Power being equal to:

Pref = (s11*a1)^2 + 2(s11*a1)(s12*a2) + (s12*a2)^2 ???

Do you recognize the irradiance equation from optical
engineering?

Ptot = P1 + P2 + 2*SQRT(P1*P2)cos(A)
--
73, Cecil http://www.w5dxp.com