On Jun 17, 5:53*pm, K1TTT wrote:
On Jun 17, 10:31*am, Keith Dysart wrote:





On Jun 14, 7:04 pm, K1TTT wrote:

On Jun 14, 12:09 pm, Keith Dysart wrote:
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.

That works for me, as long as the time is at least twice the
propagation
time down the line.

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.

snip the experiments you do not consider relevent

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.

Not correct. See the equations for a directional wattmeter below.

there is no power flowing in the line.

I agree with this, though a directional wattmeter indicates
otherwise. See below.

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

The above equations are used in a directional wattmeter.

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...

You have made some arithmetic errors below. I have re-arranged
your prose a bit to allow correction, then comment.

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

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

You made a substitution error here. On the transmission line
with a constant DC voltage:
* V(t)=100V
* I(t)=0A

substituting in to
* Vf(t)=(V(t)+R0*I(t))/2
* Vr(t)=(V(t)-R0*I(t))/2

yields
* Vf(t)=50V
* Vr(t)=50V

and
* If(t) = 50/50 = 1a
* Ir(t) = 50/50 = 1a

Recalling that the definition of forward and reflected is
* V(t)=Vf(t)+Vr(t)
* I(t)=If(t)-Ir(t)
we can check our arithmetic
* V(t)= 50+50 = 100V
* I(t)= 1-1 * = 0A
Exactly as measured.
If you are disagree with the results, please point out
my error in the definitions, derivations or arithmetic.

except i can measure If and Ir separately and can see they are both
zero. *

How would you measure them? The standard definition of forward
and reflected waves are as I provided above. I suppose you could
construct your own definitions, but such a new definition would
be confusing and unlikely to have the nice properties of the
standard definitions.

this is where your argument about putting a resistor between
two batteries falls apart... there is a length of cable with a finite
time delay in the picture here. *so it is possible to actually watch
the waves reflect back and forth and see what really happens. *in the
case where the line is open at the far end we can measure the voltage
at both ends of the line and see the constant 100v, so where is the
voltage difference needed to support the forward and reverse current
waves? *

Well, as I suggested, the forward and reflected wave are somewhat
fictitious. They are very useful as intermediate results in solving
problems but, as you note above, quite problematic if one starts
to assign too much reality to them.

That is why they are quite analogous to the two currents in the two
battery example.

you can't have current without a voltage difference. *

You can, though, have an arbitrary current as an intermediate
result in solving a problem. With the two battery example, reduce
the resistor to 0 and an infinitely large current is computed in
both directions, though we know in reality that no current is
actually flowing.

plus i
can substitute in a lossy line and measure the line heating and
measure no i^2r loss that would exist if there were currents flowing
in the line, so there are none.

I certainly agree that there is no real current flowing, but the
fictititious If and Ir are indeed present (or pick a better word
if you like).

try this case, which is something we do at work... take a 14ga piece
of copper wire maybe 50' above ground, its characteristic impedance is
probably a few hundred ohms, and put 100kv on it... how much power is
it dissipating from your If and Ir heating losses?? *lets say 200 ohms
and 100kv gives 100,000/200a=500a, shouldn't take long to burn up that
wire should it? *and yet it never does... so why doesn't it have your
circulating currents

I really like this example. I thought of using something similar once
to convince Cecil, but never did. He is unreedemable. But your
example aptly demonstrates why If and Ir are not real, though still
useful because If-Ir is the actual current flowing.

Just as an aside, your comment above about requiring a voltage to
cause a current to flow is not quite correct. An ideal conductor has
zero resistance, so current can flow without voltage in an ideal
conductor. Strangely, this can even be achieved in the real world
with superconductors.

So I offer another example for your consideration. The experiment
is the same as above (100V, 50 ohm, ideal source and conductors)
but this time short the end of the transmission line. After waiting
for the line to settle, we find 0 volts everywhere and 2A flowing.
Recalling the coversion equations:
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

we obtain:
Vf(t)=(0+100)/2 = 50V
Vr(t)=(0-100)/2 = -50V
If(t)=1A
Ir(t)=-1A

Substituting to verify
V(t)=50+(-50)=0V
I(t)=1-(-1)=2A

Exactly as measured.

I will caution again, do not assign too much reality to these
forward and reflected waves; very useful for solving problems,
but trouble if carried too far.

....Keith