On Jun 14, 12:09*pm, Keith Dysart wrote:
On Jun 10, 5:52 pm, K1TTT wrote:

On Jun 9, 2:04 am, Keith Dysart wrote:

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).

now remember, this is AFTER you disconnected the load resistor.


only as you monitored the initial transient. *once the transient has
passed there is no power flowing in the line.

I agree completely with the second sentence. But the first is in
error. A directional wattmeter will indeed show 50W forward and 50W
reflected on a line with constant DC voltage and 0 current. I’ll
start by providing support for your second sentence and then get
back to reflected power.


no, it won't... as i will explain below.

Choose an arbitrary point on the line and observe the voltage. The
observed voltage will be a function of time. Let us call it V(t).
At the same point on the line, observe the current and call it I(t).


if you can choose an arbitrary point, i can choose an arbitrary
time... call it a VERY long time after the load is disconnected.

The energy flowing (i.e. power) at this point on the line will
be P(t)=V(t)I(t). If the voltage is measured in volts and the current
in amps, then the power will be in joules/s or watts.
We can also compute the average energy flow (power) over some
interval by integrating P(t) over the interval and dividing by
the length of the interval. Call this Pavg. For repetitive
signals it is convenient to use one period as the interval.

Note that V(t) and I(t) are arbitrary functions. The analysis
applies regardless of whether V(t) is DC, a sinusoid, a step,
a pulse, etc.; any arbitrary function.

To provide some examples, consider the following simple setup:
1) 50 ohm, Thevenin equivalent circuit provides 100 V in to an
* *open circuit.

this is the case we were discussing... the others are irrelevant at
this point


2) The generator is connected to a length of 50 ohm ideal
* *transmission line. We will stick with ideal lines for the
* *moment, because, until ideal lines are understood, there
* *is no hope of understanding more complex models.
3) The ideal line is terminated in a 50 ohm resistor.

Set the generator to DC:
* V(t)=50V, wherever it is measured on the line
* I(t)=1A, wherever it is measured on the line
* P(t)=50W, wherever it is measured on the line
* Pavg=50W
No disagreement, I hope.

only if this is a case WITH a load... this is not what we were
discussing above.



Set the generator to generate a sinusoid:
* V(t)=50sin(wt)
* I(t)=1sin(wt)
* P(t)=25+25sin(2wt)
* Pavg=25W
describes the observations at any point on the line.
Measurements at different points will have a different
phase relationship to each due to the delay in the line.
Note carefully the power function. The power at any point
on the line varies from 0 to 50W with a sinusoidal
pattern at twice the frequency of the voltage sinusoid.
The power is always positive; that is, energy is always
flowing towards the load.

Now let us remove the terminating resistor. The line is
now open circuit.

OK, now we are back to where i had objections above.


Again, set the generator to DC. After one round trip
time, the observations at any point on the line will
be:
* V(t)=100V
* I(t)=0A
* P(t)=0W
* Pavg=0W


ok, p(t)=0w, i=0a, i connect my directional wattmeter at some very
long time after the load is disconnected so the transients can be
ignored and i measure 0w forever. there is no power flowing in the
line.


The story for a sinusoid is more interesting...
The observations now depend on the location on the
line where the measurements are performed.
At a current 0:
* V(t)=100sin(Wt)
* I(t)=0A
* P(t)=0W
* Pavg=0W
At a voltage 0:
* V(t)=0V
* I(t)=2sin(wt)
* P(t)=0W
* Pavg=0W
Halfway between the current 0 at the load and the
preceding voltage 0 (that is, 45 degrees back
from the load:
* V(t)=70.7sin(wt)
* I(t)=1.414cos(wt)
* P(t)=50sin(2wt)
* Pavg=0W
This tells us that for the first quarter cycle of the
voltage (V(t)), energy is flowing towards the load,
with a peak of 50W, for the next quarter cycle, energy
flows towards the source (peak –50W), and so on, for
an average of 0, as expected.

For a sinusoid, when the load is other than an open
or a short, energy flows forward for a greater
period than it flows backwards which results in a
net transfer of energy towards the load.

At the location of current or voltage minima on
the line, energy flow is either 0 or flows towards the
load.

All of the above was described without reference to forward
or reflected voltages, currents or power. It is the basic
circuit theory description of what happens between two
networks. It works for DC, 60 HZ, RF, sinusoids, steps,
pulses, you name it. If there is any disagreement with the
above, please do not read further until it is resolved.

so you have derived voltages and currents at a couple of special
locations where superposition causes maximum or minimum voltages or
currents... this of course ONLY works after sinusoidal steady state
has been achieved, so your cases of connecting and disconnecting the
load must be thrown out. and how do you find a maximum or minimum
when the load is matched to the line impedance? standing waves are
always a trap... step back from the light and come back to the truth
and learn how to properly account for forward and reflected waves with
voltages or currents and all will be revealed.



Now it is time to introduce forward and reflected waves to
the discussion. Forward an reflected waves are simply a
mathematical transformation of V(t) and I(t) to another
representation which can make solving problems simpler.

The transformation begins with assuming that there is
a forward wave defined by Vf(t)=If(t)*Z0, a reflected
wave defined by Vr(t)=Ir(t)*Z0 and that at any point
on the line V(t)=Vf(t)+Vr(t) and I(t)=If(t)-Ir(t).

Since we are considering an ideal line, Z0 simplifies
to R0.

Simple algebra yields:
* Vf(t)=(V(t)+R0*I(t))/2
* Vr(t)=(V(t)-R0*I(t))/2
* If(t)=Vf(t)/R0
* Ir(t)=Vr(t)/R0

Now these algebraically transformed expressions have all
the capabilities of the original so they work for DC,
60 HZ, RF, sinusoids, steps, pulses, you name it.

hmmm, they work for all waveforms you say... even dc?? ok, open
circuit load, z0=50, dc source of 100v with 50ohm thevenin
resistance...

If(t) = 100/50 = 2a
Ir(t) = 100/50 = 2a

so there is a reflected 2a current you say? where does that go when
it gets back to the source? we know that when the 100v source is open
circuited as it must be in this case that it develops 100v across its
terminals... sure doesn't seem like any way that 2a can go back into
the source. it can't be reflected since the source impedance matches
the line impedance. so where does it go? maybe those simple
algebraic expressions have to be reconsidered a bit.

well, maybe i missed something... lets look at the voltages:

Vf(t)=(100+50*2)/2 = 100
Vr(t)=(100-50*2)/2 = 0

hmm, you do seem to get the right constant voltage... but it kind of
messes up your power flow since Vr=0 there can be no power in that
reflected wave, so it really doesn't make it much of a wave.

but wait, the equations must be consistent so

Ir(t)=0/50

So there really is no reflected current... but there is forward
current? so where does that forward current go if there is no load?
it doesn't reflect back if Ir=0... does it just build up at the open
end? going to be lots of trons sitting there after a while.





Some people look at the expression If(t)=Vf(t)/R0 and
think it looks a lot like a travelling wave so they decide
to compute the energy being moved by this wave:
* Pf(t)=Vf(t)*Vf(t)/R0
* Pr(t)=Vr(t)*Vr(t)/R0

So let us explore the results when applied to the DC examples from
above.

Terminated with 50 ohms:
* Vf(t)=50V
* If(t)=1A
* Vr(t)=0V
* Ir(t)=0A

* Pf(t)=50W
* Pr(t)=0W

This all looks good, no reflected wave and the appropriate
amount of energy is being transported in the forward wave.

Let us try the open ended line:
* Vf(t)=50V
* If(t)=1A
* Vr(t)=50V
* Ir(t)=1A

which yield
* V(t)=Vf(t)+Vr(t)=100V
* I(t)=If(t)-Ir(t)=0A

which is in agreement with the observations. But there is
definitely a forward voltage wave and a reflected voltage
wave. Carrying on to compute powers:

sorry, you can't have a 'voltage wave' without a 'current wave'...
just doesn't work i'm afraid. to have one you must have the other.

* Pf(t)=50W
* Pr(t)=50W
and
* P(t)=Pf(t)-Pr(t)=0W

So there you have it. The interpretation that the forward
and reflected wave are real and transport energy leads to
the conclusion that on a line charged to a constant DC
voltage, energy is flowing in the forward direction and
energy is flowing in the reflected direction.

obviously incorrect, by reductio ad absurdum if i remember the latin
correctly.


Because it is obvious that there is no energy flowing on
the line, this interpretation makes me quite uncomfortable.

Now there are several ways out of this dilemma.

Some say they are only interested in RF (and optics), and
that nothing can be learned from DC, steps, pulses or other
functions. They simply ignore the data and continue with
their beliefs.

Some claim that DC is special, but looking at the equations,
there is no reason to expect the analysis to collapse with
DC.

The best choice is to give up on the interpretation that the
forward and reflected voltage waves necessarily transport
energy. Consider them to be a figment of the analysis technique;
very convenient, but not necessarily representing anything real.

As an added bonus, giving up on this interpretation will remove
the need to search for “where the reflected power goes” and
free the mind for investment in activities that might yield
meaningful results.

...Keith