On Mar 13, 10:28*pm, Cecil Moore wrote:
Keith Dysart wrote:
Show me the reactance and its power function of time such that
it stores and releases the energy at the correct time.

You've got to be kidding - EE201.

I see no reactance that performs this function.

Have you no ideal what the +j44.1 ohms means?

This is the third opportunity you have had to clearly
identify the component and show that its energy flow
as a function of time is exactly that needed to store
and release the energy in the reflected wave.

Since you have not done so, I conclude that you can't
find it either.

...Keith

Keith Dysart wrote:
This is the third opportunity you have had to clearly
identify the component and show that its energy flow
as a function of time is exactly that needed to store
and release the energy in the reflected wave.

P(t).reactance = [V(t).reactance][I(t).reactance]

Where is that term in your equations?
--
73, Cecil http://www.w5dxp.com

On Mar 14, 7:59*am, Cecil Moore wrote:
Keith Dysart wrote:
This is the third opportunity you have had to clearly
identify the component and show that its energy flow
as a function of time is exactly that needed to store
and release the energy in the reflected wave.

P(t).reactance = [V(t).reactance][I(t).reactance]

Where is that term in your equations?

It is unnecessary. But if you believe me wrong, show me
where it goes, compute the values, and show how it
accounts for the energy that is not dissipated in the
source resistor.

...Keith

Keith Dysart wrote:
On Mar 14, 7:59 am, Cecil Moore wrote:
P(t).reactance = [V(t).reactance][I(t).reactance]
Where is that term in your equations?

It is unnecessary. But if you believe me wrong, show me
where it goes, compute the values, and show how it
accounts for the energy that is not dissipated in the
source resistor.

It is unnecessary to account for all of the instantaneous
power???? Your problem is greater than just a simple
misunderstanding of the laws of physics by which we must
all abide.

The DC energy is stored in your vehicle's battery until
it is needed to start your vehicle. That delay between
stored energy and needed energy is related to the
(undefined) wavelength. Think about it.

In an AC circuit, the reactance has no say as to when
to store the energy and when it is delivered back to the
system. It is also related to wavelength which is defined.

When the source voltage is zero at its zero crossing
point/time, the instantaneous power dissipation in
the source resistor is NOT zero! Doesn't that give
you pause to wonder where the instantaneous power
is coming from when the instantaneous power delivered
by the source is zero???? Why do you ignore that power
and try to sweep it under the rug?

You are making the mistakes that your EE 201 professor
warned you not to make. You are superposing powers,
something that all the gurus on this newsgroup will
condemn. Until you learn not to superpose powers, you
will remain forever ignorant. Richard C. made the same
mistake when he declared that the reflections from non-
reflective thin-film coatings on glass are "brighter
than the surface of the sun". If one ignores the laws
of physics, anything is possible.

There is a condition where it is valid to superpose
powers. That is when (V1^2 + V2^2) = (V1 + V2)^2
You have obviously not satisfied that condition and
are paying for your violations of those laws of physics.
Until you give up on your superposition of powers, I
cannot help you.
--
73, Cecil http://www.w5dxp.com

On Mar 14, 5:51*pm, Cecil Moore wrote:
Keith Dysart wrote:
On Mar 14, 7:59 am, Cecil Moore wrote:
P(t).reactance = [V(t).reactance][I(t).reactance]
Where is that term in your equations?

It is unnecessary. But if you believe me wrong, show me
where it goes, compute the values, and show how it
accounts for the energy that is not dissipated in the
source resistor.

It is unnecessary to account for all of the instantaneous
power???? Your problem is greater than just a simple
misunderstanding of the laws of physics by which we must
all abide.

The DC energy is stored in your vehicle's battery until
it is needed to start your vehicle. That delay between
stored energy and needed energy is related to the
(undefined) wavelength. Think about it.

In an AC circuit, the reactance has no say as to when
to store the energy and when it is delivered back to the
system. It is also related to wavelength which is defined.

When the source voltage is zero at its zero crossing
point/time, the instantaneous power dissipation in
the source resistor is NOT zero! Doesn't that give
you pause to wonder where the instantaneous power
is coming from when the instantaneous power delivered
by the source is zero????

Previously explained, as you may have forgotten, with
the energy cyclically returned from the line.

For your convenience, I copy from a previous post:
----
Recall that
Pg(t) = 32 + 68cos(2wt)

For some of the cycle, energy flow is from the line
towards the resistor and the voltage source.

But this is not the energy in the reflected wave
which has the function
Pr.g(t) = -18 + cos(2wt)
and only flows in one direction, towards the source.
And it is this supposed energy that can not be
accounted for in the dissipation of the source
resistor.
----

And, as expected
Ps(t) = Prs(t) + Pg(t)

Energy is conserved.

...Keith

"Keith Dysart" wrote in message
...
On Mar 14, 5:51 pm, Cecil Moore wrote:
Energy is conserved.

qed.

now, can we get back to antennas?? its much more fun mocking art than the
endless discussions of electrons and holes sloshing back and forth.

Keith Dysart wrote:
And, as expected
Ps(t) = Prs(t) + Pg(t)

Would you please explain how energy is conserved in the
following example at the zero-crossing point for Vs?

Rs Vg Vl
+----/\/\/-----+----------------------+
| 50 ohm |
| |
Vs 45 degrees | Shorted
100v RMS 50 ohm line | Stub
| |
| |
+--------------+----------------------+
gnd

At the zero-crossing of Vs, Ps(t) = 0, i.e. the source
is supplying zero watts at that time but Prs(t) = 100w.
Where is the 100 watts coming from?
--
73, Cecil http://www.w5dxp.com

On Mar 16, 10:21*am, Cecil Moore wrote:
Keith Dysart wrote:
And, as expected
Ps(t) = Prs(t) + Pg(t)

Would you please explain how energy is conserved in the
following example at the zero-crossing point for Vs?

* * * * * * * *Rs * * * Vg * * * * * * * * * * Vl
* * * * *+----/\/\/-----+----------------------+
* * * * *| * *50 ohm * * * * * * * * * * * * * |
* * * * *| * * * * * * * * * * * * * * * * * * |
* * * * Vs * * * * * * * * 45 degrees * * * * *| Shorted
* * *100v RMS * * * * * * *50 ohm line * * * * | Stub
* * * * *| * * * * * * * * * * * * * * * * * * |
* * * * *| * * * * * * * * * * * * * * * * * * |
* * * * *+--------------+----------------------+
* * * * gnd

At the zero-crossing of Vs, Ps(t) = 0, i.e. the source
is supplying zero watts at that time but Prs(t) = 100w.
Where is the 100 watts coming from?

Vs(t) = 1.414cos(wt)

After settling, from some circuit analysis...

Vg(t) = 100 cos(wt+45degrees)
Ig(t) = 2 cos(wt-45degrees)

Vrs(t) = Vs(t) - Vg(t)
= 100 cos(wt-45degrees)
Irs(t) = Ig(t)
= 2 cos(wt-45degrees)

Is(t) = Ig(t)
= 2 cos(wt-45degrees)

Ps(t) = Vs(t) * Is(t)
= 100 + 141.4213562 cos(2wt-45degrees)

Prs(t) = Vrs(t) * Irs(t)
= 100 + 100 cos(2wt-90degrees)

Pg(t) = Vg(t) * Ig(t)
= 0 + 100 cos(2wt)

For confirmation of conservation of energy, the above
is in agreement with
Ps(t) = Prs(t) + Pg(t)

Ps(t) = 100 + 141.4213562 cos(2wt-45degrees)
so, Ps(t) = 0, occurs whenever
141.4213562 cos(2wt-45degrees)
is equal to
-100
which, for example, would happen when 2wt = 180 degrees
or wt = 90 degrees.

At this time, again for example, the energy being
dissipated in the source resistor
Prs(t) = 100 + 100 cos(2wt-90degrees)
= 100 + 0
= 100

Since no energy is being delivered from the source,
(Ps(t) is 0), then given
Ps(t) = Prs(t) + Pg(t)
the energy must be coming from the line. Let us
check the energy flow at the point Vg at this
time
Pg(t) = 0 + 100 cos(2wt)
= -100
as required.

So the energy being dissipated in the source
resistor at this time is being returned from
the line.

Just for completeness we can compute the line
state at the point Vg in terms of forward and
reverse waves...

Vf.g(t) = 70.71067812 cos(wt)
Vr.g(t) = 70.71067812 cos(wt+90degrees)
Vg(t) = Vf.g(t) + Vr.g(t)
= 100 cos(wt+45degrees)

If.g(t) = 1.414213562 cos(wt)
Ir.g(t) = 1.414213562 cos(wt-90degrees)
Ig(t) = If.g(t) + Ir.g(t)
= 2 cos(wt-45degrees)

Pf.g(t) = Vf.g(t) * If.g(t)
= 50 + 50 cos(2wt)

Pr.g(t) = Vr.g(t) * Ir.g(t)
= -50 + 50 cos(2wt)

And since
Pg(t) = Pf.g(t) + Pr.g(t)
= 0 + 100 cos(2wt)
confirming the previously computed values.

It is valuable to examine Pr.g(t) at the time when
Ps(t) is zero. Substituting wt = 90degrees
into Pr.g(t)...

Pr.g(t) = -50 + 50 cos(2wt)
= -50 -50
= -100
which would appear to be the 100 watts needed to
heat the source resistor. This is misleading.
When all the values of t are examined it will be
seen that only the sum of Pf.g(t) and Pr.g(t),
that is, Pg(t), provides the energy not provided
from the source that heats the source resistor.
wt=90 is a special case in that Pf.g(t) is 0 at
this particular time.

The line input impedance does have a reactive
component and it is this reactive component that
can store and return energy. The energy flow
into and out of this impedance (pure reactance
for the example under consideration) is described
by Pg(t).

In summary,
Ps(t) = Prs(t) + Pg(t)

and by substitution, if the solution is preferred
in terms of Pf and Pr,
Ps(t) = Prs(t) + Pf.g(t) + Pr.g(t)

Ps and Pr alone are insufficient to explain the
heating of the source resistor.

...Keith