On Sun, 09 Mar 2008 14:35:25 GMT
Cecil Moore wrote:

Keith Dysart wrote:
My issue was that you seemed, in that sentence, to be saying
that the reflected energy was dissipated in the source
resistor. But earlier you had stated that was not your claim.

My earlier claim was that the average power in a reflected
wave is dissipated in the source resistor when the forward
wave is 90 degrees out of phase with the reflected wave at
the source resistor. In that earlier claim, I didn't care
to discuss instantaneous power and thus excluded instantaneous
power from that claim.

For instantaneous values, it will be helpful to change the
example while leaving the conditions at the source resistor
unchanged. Here's the earlier example:

Rs Vg Vl
+----/\/\/-----+----------------------+
| 50 ohm |
| 1/8 WL |
Vs 45 degrees 12.5 ohm
100v RMS 50 ohm line Load
| |
| |
+--------------+----------------------+
gnd

Here's the present example:

Rs Vg Vl
+----/\/\/-----+----------------------+
| 50 ohm |
| 1 WL |
Vs 360 degrees 23.5+j44.1
100v RMS 50 ohm line ohm Load
| |
| |
+--------------+----------------------+
gnd

If I haven't made some stupid mistake, the conditions at
the source resistor are identical in both examples. But
in the second example, it is obvious that energy can
be stored in the transmission line during part of a cycle
(thus avoiding dissipation at that instant in time) and be
delivered back to the source resistor during another part
of the cycle (to be dissipated at a later instant in time).
That is the nature of interference energy and is exactly
equal to the difference between the two powers that you
calculated. You neglected to take into account the ability
of the network reactance to temporarily store energy and
dissipate it later in time.

All of the reflected energy is dissipated in the source
resistor, just not at the time you thought it should be.

I think we can all agree to what would happen in the turn off situation following a long period of stable power flow. When the source voltage steps from 100v to 0v, the power on the transmission line reflects to the 50 ohm source resistor and is all absorbed.

On the other hand, in the case of steady power flow, the source is presented with a reactive load of 73.5 + J44.1 ohms. The reactive part of this load results from returning power from the transmission line.

For the resistor Rs, it will have power applied from two sources, the source and the reflected power from the transmission line, i.e., the source power and reflected power are in series when considered in relationship with the resistor Rs. The problem is that 'what looks like two sources to resistor Rs, is really only one source, Vs'.

I think you are looking for solutions that show how power to Rs peaks at a different time from when power into the transmission line peaks, which is yet a different time from when power from Vs peaks. You are looking for the instantaneous timing of 3 peaks.


While performing this analysis, you will discover the ability
of superposing waves to redistribute energy (even in the
complete absence of ordinary reflections). Not possible in
the above special case examples, but under other conditions,
the redistribution of energy can attain perpetual steady-state
status (even in the complete absence of ordinary reflections).

There are two mechanisms involved in the re-routing of
steady-state reflected energy back toward the load.

1. Ordinary reflection of single EM waves governed by the
rules of the wave reflection model.

2. A redistribution of reflected energy back toward the
load as a result of superposition of waves accompanied
by an average level of steady-state interference.
--
73, Cecil http://www.w5dxp.com


--
73, Roger, W7WKB