On Jun 18, 12:00*am, Keith Dysart wrote:
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?

you said you had a directional wattmeter, so do i. mine reads zero,
what does yours read?

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.

fictitious?? i can measure them, watch them flow from here to there,
and see physical effects of them, doesn't sound very fictitious to
me. weren't you the one who teaches tdr use? what are you seeing in
the steps of the tdr if not for reflected waves coming back to your
scope?


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

no, that is why they aren't analogous. lumped components can't
represent the traveling waves.


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.

no it isn't. in the limit at R-0 0V/R still looks like 0 to me.



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).

now you say they are fictitious, yet you have formulas for them and
can measure them... until you come to your senses and admit that
reflections are real there isn't much we can do for you so the rest of
your post is based on an incorrect premise.

snip

On Jun 18, 7:04*am, K1TTT wrote:
On Jun 18, 12:00*am, Keith Dysart wrote:
except i can measure If and Ir separately and can see they are both
zero. *

How would you measure them?

you said you had a directional wattmeter, so do i. *mine reads zero,
what does yours read?

That would most likely be because your wattmeter is AC coupled.
Working
with DC, you need a DC-coupled meter, in which case it would read as
I predicted.

A highly instructive exercise is to examine the schematic for your
directional wattmeter and work out how it implements the expressions
Vf(t)=(V(t)+R0*I(t))/2
Vr(t)=(V(t)-R0*I(t))/2

Pf=avg(Vf(t)*Vf(t)/R0)
Pr=avg(Vr(t)*Vr(t)/R0)
or some equivalent variant using If and Ir.

You will find a voltage sampler, a current sampler, some scaling,
and an addition (or subtraction). If your meter is analog, the
squaring function is usually implemented in the non-linear meter
scale. And the averaging function might be peak-reading or rely
on meter inertia.

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.

no it isn't. *in the limit at R-0 *0V/R still looks like 0 to me.

When computing the intermediate results using superposition, you
have the full battery voltage in to 0 ohms. First the battery on
the left, then the one on the right, and then you add the two
currents to get the total current. 0 is always the total, but
the intermediate results can go to infinity if you connect
the two batteries directly in parallel.

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).

now you say they are fictitious, yet you have formulas for them and
can measure them... until you come to your senses and admit that
reflections are real there isn't much we can do for you so the rest of
your post is based on an incorrect premise.

Then what are the equations for computing forward and reflected waves
and 'power'? That is where I started. Is that the premise you consider
to be incorrect?

Do follow through the arithmetic in the previously provided examples
and see if you can find any faults. Point me to any errors and I will
gladly reconsider my position.

If not, perhaps it is time for you to start questioning your
assumptions.

....Keith

On Jun 18, 11:26 pm, Keith Dysart wrote:
On Jun 18, 7:04 am, K1TTT wrote:

On Jun 18, 12:00 am, Keith Dysart wrote:
except i can measure If and Ir separately and can see they are both
zero.

How would you measure them?

you said you had a directional wattmeter, so do i. mine reads zero,
what does yours read?

That would most likely be because your wattmeter is AC coupled.
Working
with DC, you need a DC-coupled meter, in which case it would read as
I predicted.

mine is dc coupled... still nothing. please provide the schematic or
model number for the one you use to get such a reading.


A highly instructive exercise is to examine the schematic for your
directional wattmeter and work out how it implements the expressions
Vf(t)=(V(t)+R0*I(t))/2
Vr(t)=(V(t)-R0*I(t))/2

Pf=avg(Vf(t)*Vf(t)/R0)
Pr=avg(Vr(t)*Vr(t)/R0)
or some equivalent variant using If and Ir.

You will find a voltage sampler, a current sampler, some scaling,
and an addition (or subtraction). If your meter is analog, the
squaring function is usually implemented in the non-linear meter
scale. And the averaging function might be peak-reading or rely
on meter inertia.

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.

no it isn't. in the limit at R-0 0V/R still looks like 0 to me.

When computing the intermediate results using superposition, you
have the full battery voltage in to 0 ohms. First the battery on
the left, then the one on the right, and then you add the two
currents to get the total current. 0 is always the total, but
the intermediate results can go to infinity if you connect
the two batteries directly in parallel.

there is no battery on the right, unless you have changed the case
again. i thought we were doing the open circuit coax with the dc
source on one end.

but even if you have the case above and connect another 100v battery
on the open end of the line there will be no current flowing back as
there is no voltage difference to drive it.



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).

now you say they are fictitious, yet you have formulas for them and
can measure them... until you come to your senses and admit that
reflections are real there isn't much we can do for you so the rest of
your post is based on an incorrect premise.

Then what are the equations for computing forward and reflected waves
and 'power'? That is where I started. Is that the premise you consider
to be incorrect?

yep, your 'waves' don't include the proper term to make them
propagating waves that satisfy maxwell's equations... so they can't be
real... so your initial premise is incorrect.


Do follow through the arithmetic in the previously provided examples
and see if you can find any faults. Point me to any errors and I will
gladly reconsider my position.

If not, perhaps it is time for you to start questioning your
assumptions.

...Keith

my assumptions are perfectly valid... it is yours that are flawed from
the very beginning.

On Jun 18, 7:55*pm, K1TTT wrote:
On Jun 18, 11:26 pm, Keith Dysart wrote:

On Jun 18, 7:04 am, K1TTT wrote:

On Jun 18, 12:00 am, Keith Dysart wrote:
except i can measure If and Ir separately and can see they are both
zero.

How would you measure them?

you said you had a directional wattmeter, so do i. *mine reads zero,
what does yours read?

That would most likely be because your wattmeter is AC coupled.
Working
with DC, you need a DC-coupled meter, in which case it would read as
I predicted.

mine is dc coupled... still nothing. *please provide the schematic or
model number for the one you use to get such a reading.

Oh my. Could you kindly provide the schematic or a reference? Before
posting I spent an hour searching my library and the web for the
circuit
for a dc-coupled directional wattmeter and was unsuccessful.

Nothing prevents one from being constructed. That one is not readily
available should not prevent you from using the equations that they
implement and then predicting the DC behaviour of the line.

A highly instructive exercise is to examine the schematic for your
directional wattmeter and work out how it implements the expressions
* Vf(t)=(V(t)+R0*I(t))/2
* Vr(t)=(V(t)-R0*I(t))/2

* Pf=avg(Vf(t)*Vf(t)/R0)
* Pr=avg(Vr(t)*Vr(t)/R0)
or some equivalent variant using If and Ir.

You will find a voltage sampler, a current sampler, some scaling,
and an addition (or subtraction). If your meter is analog, the
squaring function is usually implemented in the non-linear meter
scale. And the averaging function might be peak-reading or rely
on meter inertia.

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.

no it isn't. *in the limit at R-0 *0V/R still looks like 0 to me..

When computing the intermediate results using superposition, you
have the full battery voltage in to 0 ohms. First the battery on
the left, then the one on the right, and then you add the two
currents to get the total current. 0 is always the total, but
the intermediate results can go to infinity if you connect
the two batteries directly in parallel.

there is no battery on the right, unless you have changed the case
again. *i thought we were doing the open circuit coax with the dc
source on one end.

No. I was referring to the simple DC circuit with two batteries and
a resistor. I was pretty sure that was your referent when you used
"R-0".

I understand that you have not found any flaws in my exposition,
except that you are still uncomfortable with the results. It is
possible to overcome this.

And I re-iterate the value of studying your directional wattmeter's
schematic, especially if it is DC-coupled.

....Keith