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