"Cecil Moore" wrote in message
.. .
Roy Lewallen wrote:
Superposition means the following: If f(x) is the result of excitation x
and f(y) is the result of excitation y, then the result of excitation (x
+ y) is f(x) + f(y). . .

Now the big question is: Is superposition always reversible?
If not, it implies interaction between f(x) and f(y).
--
73, Cecil http://www.w5dxp.com

as long as everything is linear, yes.

Dave wrote:
Now the big question is: Is superposition always reversible?
If not, it implies interaction between f(x) and f(y).

as long as everything is linear, yes.

This is really interesting. Given the following:

b1 = s11(a1) + s12(a2) = 0

Let P1 = |s11(a1)|^2 = 1 joule/sec

Let P2 = |s12(a2)|^2 = 1 joule/sec

Therefore, Ptot = |b1|^2 = 0 joules/sec

Ptot = P1 + P2 + 2*SQRT(P1*P2)cos(180)

Ptot = 1 + 1 - 2 = 0 joules/sec = |b1|^2

Can one reverse the superposition whose result is
zero to recover the original two component waves?
If not, isn't that proof that the two original
component waves interacted?
--
73, Cecil http://www.w5dxp.com

"Cecil Moore" wrote in message
...
Dave wrote:
Now the big question is: Is superposition always reversible?
If not, it implies interaction between f(x) and f(y).

as long as everything is linear, yes.

This is really interesting. Given the following:

b1 = s11(a1) + s12(a2) = 0

Let P1 = |s11(a1)|^2 = 1 joule/sec

Let P2 = |s12(a2)|^2 = 1 joule/sec

Therefore, Ptot = |b1|^2 = 0 joules/sec

Ptot = P1 + P2 + 2*SQRT(P1*P2)cos(180)

Ptot = 1 + 1 - 2 = 0 joules/sec = |b1|^2

Can one reverse the superposition whose result is
zero to recover the original two component waves?
If not, isn't that proof that the two original
component waves interacted?
--
73, Cecil http://www.w5dxp.com

no, because you have done a non-linear operation on them by converting to
powers. obviously at the start 'a1' and 'a2' are separate.

"Dave" wrote in message news:7RoWh.109$Zm.79@trndny03...

"Cecil Moore" wrote in message
...
Dave wrote:
Now the big question is: Is superposition always reversible?
If not, it implies interaction between f(x) and f(y).

as long as everything is linear, yes.

This is really interesting. Given the following:

b1 = s11(a1) + s12(a2) = 0

Let P1 = |s11(a1)|^2 = 1 joule/sec

Let P2 = |s12(a2)|^2 = 1 joule/sec

Therefore, Ptot = |b1|^2 = 0 joules/sec

Ptot = P1 + P2 + 2*SQRT(P1*P2)cos(180)

Ptot = 1 + 1 - 2 = 0 joules/sec = |b1|^2

Can one reverse the superposition whose result is
zero to recover the original two component waves?
If not, isn't that proof that the two original
component waves interacted?
--
73, Cecil http://www.w5dxp.com

no, because you have done a non-linear operation on them by converting to
powers. obviously at the start 'a1' and 'a2' are separate.


i should expand a bit more. all your equations above have done is shown
that at the point where you are doing your analysis s11(a1) and s12(a2),
which add up to 0... also produce a net 0 power at that point. this is as
expected for destructive interference AT THAT POINT. as such your s
parameter analysis is insufficient to separate the individual components
after you combine them into a power. however, at the begining they are
obviously separate waves since you have represented them with separate input
values, and given a linear transfer function for your point on the wire, or
in space, they can always be kept separate. it is only your act of
calculating the power at that point that combines them.

Dave wrote:
i should expand a bit more. all your equations above have done is shown
that at the point where you are doing your analysis s11(a1) and s12(a2),
which add up to 0... also produce a net 0 power at that point. this is as
expected for destructive interference AT THAT POINT. as such your s
parameter analysis is insufficient to separate the individual components
after you combine them into a power. however, at the begining they are
obviously separate waves since you have represented them with separate input
values, and given a linear transfer function for your point on the wire, or
in space, they can always be kept separate. it is only your act of
calculating the power at that point that combines them.

True, but we also know that the power is zero at every point
between the Z0-match and the source. That gives us an
infinite number of points with which to work and we still
cannot reverse superpose the two original superposed waves.
--
73, Cecil http://www.w5dxp.com

Dave wrote:
"Cecil Moore" wrote:
Can one reverse the superposition whose result is
zero to recover the original two component waves?
If not, isn't that proof that the two original
component waves interacted?

no, because you have done a non-linear operation on them by converting to
powers. obviously at the start 'a1' and 'a2' are separate.

If V^2/Z0 and I^2*R cannot be reverse superposed
in reality, doesn't that imply than neither can V
and I be reverse superposed in reality?

Doesn't a real world EM wave require ExB joules/sec
for it to exist and to have an associated voltage
and current?
--
73, Cecil http://www.w5dxp.com

Dave wrote:
"Cecil Moore" wrote in message
. . .

I'm very sorry to see that Cecil has arrived to divert what was an
interesting and informative discussion to yet another one of his endless
argumentative junk science threads. Oh, well, it was nice while it lasted.

Roy Lewallen, W7EL

Roy Lewallen wrote:
I'm very sorry to see that Cecil has arrived to divert what was an
interesting and informative discussion to yet another one of his endless
argumentative junk science threads. Oh, well, it was nice while it lasted.

This from the person who used standing-wave current to
measure the phase shift through a loading coil knowing
all the while that standing-wave current has an
unvarying phase. I cannot think of any worse junk
science than trying to hoodwink the uninitiated.
--
73, Cecil http://www.w5dxp.com