On Jun 9, 2:04*am, Keith Dysart wrote:
On Jun 8, 7:39*pm, K1TTT wrote:



On Jun 8, 10:57*am, Keith Dysart wrote:

On Jun 8, 3:47*am, Richard Clark wrote:

On Sun, 6 Jun 2010 14:21:16 -0700 (PDT), Keith Dysart

wrote:
The 100W forward and 50W reflected have no relation to actual powers

From MECA, makers of Isolators - their application:

The isolator is placed in the measurement path of a test bench between
a signal source and the device under test (DUT) so that any
reflections caused by any mismatches will end up at the termination of
the isolator and not back into the signal source. This example also
clearly illustrates the need to be certain that the termination at the
isolated port is sufficient to handle 100% of the reflected power
should the DUT be disconnected while the signal source is at full
power. If the termination is damaged due to excessive power levels,
the reflected signals will be directed back to the receiver because of
the circular signal flow.
...
MECA offers twenty-four models of isolators and circulators in both N
and SMA-female connectors with average power ratings from 2 - 250
watts.

Good day Richard,

You have located several examples from reputable vendors where the
behaviour of directional couplers is described in terms of power
in a forward and reflected wave. This model of behaviour works within
its limits and allows for convenient computation and prediction of
the behaviour. But all of these papers have the appropriate
discipline and do not ask the question "where does the reflected
energy go?" which is good, for this exceeds the limits of the
model.

As soon as one assigns tangible energy to the reflected wave, it
becomes reasonable to ask for an accounting of this energy and
the model is incapable of properly accounting for the energy.

Following this weakness back through the model, the root cause
is the attempt to assign tangible energy to the reflected wave.
Think of it as a reflected voltage or current wave and all will
be well, but assign power to it and eventually incorrect
conclusions will be drawn.

For those who understand this, and know that "where did the
reflected energy go?" is an invalid question, using the power
model within its limits will not cause difficulties. But for
those who are not careful, great difficulties arise and a lot
of fancy dancing is offered to work around the difficulties,
unsuccessfully.

Just for fun, here is a simple example.

100V DC source, connected to a 50 ohm source resistor,
connected to 50 ohm transmission line, connected to a 50 ohm
load resistor. Turn on the source. A voltage step propagates
down the line to the load. The impedances are matched, so
there are no reflections. The source provides 100W. 100W is
dissipated in the source resistor. 100W is dissipated in the
load resistor. Energy moves along the transmission line
from the source to the load at the rate of 50W.
All is well.

lets see, when steady state is reached you have 100v through 2 50ohm
resistors in series i assume since you say the source provides 100w...
that would mean the current is 1a and each resistor is dissipating
50w.

Ouch. I can't even get the arithmetic right. You are, of course,
correct. Fortunately, the rest of the examples do not attempt
to compute actual values.



Disconnect the load. A voltage step propagates back along
the line from the load to the source. In front of this
step current continues to flow. Behind the step, the
current is 0. When the step reaches the source there is
no longer any current flowing. The source is no longer
providing energy, the source resistor is dissipating
nothing, and neither is the load resistor.

Proponensts of the power model claim that energy is
still flowing down the line, being reflected from the
open end and flowing back to the source. Since the
source is clearly no longer providing energy, great
machinations are required to explain why the reflected
'energy' is re-reflected to provide the forward 'energy'.

why would there be power flowing down the line,

Exactly. And yet, were you to attach a DC-couple directional
wattmeter to the line, it would indicate 50W forward and 50W
reflected (using your corrected numbers).

only as you monitored the initial transient. once the transient has
passed there is no power flowing in the line. to drive power requires
a potential difference, without potential difference there is no
current and no power.


in this simple dc case
once the voltage in the line equals the source voltage, which in this
perfect case would be after one transient down the line, all is in
equilibrium... and art is happy.

HOWEVER!!! *replace that dc source with an AC one and everything
changes,

Not at all. The measurements and computations for forward and
reflected
power that work for AC, work perfectly well for DC, pulses, steps,
or pick your waveform. There is no magic related to AC.

ah, but there is... in a sine wave the voltage never stops changing so
there is always a potential difference driving the wave along the
line. with DC you can reach equilibrium in one transit of the cable,
assuming it is lossless and the source and/or load is matched to the
line. otherwise you have to do the infinite sum to approximate the
steady state as for any other transient. complex waveforms get ugly
since they are infinite sums of sine waves, though for lossless and
dispersionless cables they aren't too hard to handle. The magic of a
single frequency AC source is it lets us use simplified equations that
make use of the sinusoidal steady state approximations... this is what
lets us do things like s parameters, phasors, E=IZ, and cecil's simple
power addition equation, without the need for messy summations and
other more detailed calculations.


now there are forward and reflected waves continuously going
down the line and back. *your DC analysis is no longer valid as it is
a very special case. *

Not so. And if it were, at exactly what very low frequency does
the AC analysis begin to fail?

none, there is no magic cutoff, its just how close you want to look at
it.


now you have to go back to the general equations
and use the length of the line to transform the open circuit back to
the source based on the length of the line in wavelengths... for the
specific case of the line being 1/4 wave long the open

The transformation properties apply only to sinusoidal AC (note, for
example, that the line length is expressed in wavelengths),

true, another simplification made possible by the sinusoidal steady
state approximation.

but
reflection coefficients and the the computation of forward and
reflected voltages, currents and powers are waveshape independant.

only as long as the loads are perfectly resistive and linear. real
loads often change impedance with frequency and so distort the
reflection of complex waveforms. use a scope with a good risetime and
a decent fast rise time pulse and you'll see it.


When analyzing in the time domain, use the delay of the line to decide
the values of the signals to superposed. Doing this for sinusoids
will lead to exactly the same results that are expected from, for
example, 1/4 wave lines.

...Keith