On Mar 18, 12:16*pm, Cecil Moore wrote:
Roger Sparks wrote:
Keith Dysart wrote:
Pr.g(t) = -50 + 50 *cos(2wt)
* * * * = -50 -50
* * * * = -100
which would appear to be the 100 watts needed to
heat the source resistor.

Thanks for a comprehensive analysis Keith. *It was helpful to me to see how you step by step analyzed the circuit.

Note that his analysis proves that destructive interference
energy is indeed stored in the reactance of the network and
dissipated later in the cycle as constructive interference
in the source resistor at a time when the instantaneous
source power equals zero.

I've made no claims about destructive or other interference
so I am intrigued by what you say my analysis proves.

Could you expand on how my analysis "proves" that
"destructive interference energy is indeed stored in the
reactance of the network and dissipated later in the cycle
as constructive interference in the source resistor at a
time when the instantaneous source power equals zero."

I am not sure why you focus on the infinitesimally small
times when "the instantaneous source power equals zero".
Are you saying that at other times, an explanation other
than interference is needed?
Or does interference explain the flows of the energy
in the reflected waves for all times?

It would be good if you could provide the equations that
describe these flows, especially the ones describing the
storage of some of the energy from the reflected wave in
the reactance of the network.

...Keith

Keith Dysart wrote:
I am not sure why you focus on the infinitesimally small
times when "the instantaneous source power equals zero".

:-) :-) That was my exact reaction when you wanted to
focus on the infinitesimally small times associated
with instantaneous powers. :-) :-) You are to blame
for that focus, not I.

It only takes one example to prove your claims wrong.
I didn't want to focus on instantaneous power at all
but you insisted. Now that your own techniques are
being used to prove you wrong, you are objecting.

At the time when the instantaneous source power is zero,
the only other sources of energy in the entire network
is reflected energy and interference energy. That's
a fact of physics.

You have gotten caught superposing powers, something
that every sophomore EE knows not to do. I repeat:
The only time that power can be directly added is
when there is *ZERO INTERFERENCE*. The test for
zero interference is: (V1^2 + V2^2) = (V1 + V2)^2
--
73, Cecil http://www.w5dxp.com

On Mar 19, 12:00*pm, Cecil Moore wrote:
Keith Dysart wrote:
I am not sure why you focus on the infinitesimally small
times when "the instantaneous source power equals zero".

:-) :-) That was my exact reaction when you wanted to
focus on the infinitesimally small times associated
with instantaneous powers. :-) :-) You are to blame
for that focus, not I.

It only takes one example to prove your claims wrong.

Could you state the claim that you think you are proving
wrong?

I didn't want to focus on instantaneous power at all
but you insisted. Now that your own techniques are
being used to prove you wrong, you are objecting.

If you wish to have your equalities apply at only selected
points within the cycle, that works for me.

It only fails when you make the assertion that
for the circuit under consideration, the energy in
the reflected wave is dissipated in the source
resistor.

At the time when the instantaneous source power is zero,
the only other sources of energy in the entire network
is reflected energy and interference energy. That's
a fact of physics.

You have gotten caught superposing powers, something
that every sophomore EE knows not to do.

Could you kindly show me where?

The two expressions in which I have summed power are
Ps(t) = Prs(t) + Pg(t)
and
Pg(t) = Pf.g(t) + Pr.g(t)

You have not previously indicated that either of these
are incorrect. Please take this opportunity.

I repeat:
The only time that power can be directly added is
when there is *ZERO INTERFERENCE*. The test for
zero interference is: (V1^2 + V2^2) = (V1 + V2)^2

Surely you overstate. Powers can be directly added
whenever you have a system with ports adding and
removing energy from the system. Conservation of
energy says that the sum of the energy being stored
in the system is equal to the sums of all the ins
and outs at the ports. There are no conditions on
the ports adding or removing energy.

...Keith

Keith Dysart wrote:
If you wish to have your equalities apply at only selected
points within the cycle, that works for me.

Not only at selected points within the cycle but
also for average values. If zero average interference
exists then 100% of the average reflected energy is
dissipated in the source resistor which is the subject
of my Part 1 article. If the instantaneous interference
is zero, 100% of the instantaneous reflected power is
dissipated in the source resistor.

I repeat: When zero interference exists, 100% of the
reflected energy is dissipated in the source resistor.
--
73, Cecil http://www.w5dxp.com