what happens to reflected energy ?
 

On Jun 14, 11:04*pm, K1TTT wrote:
read more »

no, i've read enough thank you.

I think there are 2 basic things you are missing... V and I are
functions not only of time, but of distance along the line, both for
the forward and reflected waves. looking at them at specific points
can lead to incorrect assumptions and over simplified conclusions.
Also, there are very good reasons why the sinusoidal steady state
condition is used to simplify the equations they way they are normally
shown, and why your over simplifications won't hack it in the long
run. to do the full analysis of a distributed system (except in very
special cases) for an arbitrary waveform requires summations of
reflections both ways on the line over many reflection periods.. the
results of this for step functions can be seen with simple tdr's on
lines with multiple discontinuities where you can see the reflections
ringing down.

what happens to reflected energy ?
 

Keith Dysart wrote:

. . .

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.

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

Well done, Keith. Thanks.

Roy Lewallen, W7EL

what happens to reflected energy ?
 

On Jun 16, 4:11*pm, Roy Lewallen wrote:
Well done, Keith. Thanks.

Two ignorant people patting each other on the back. If DC methods
worked on distributed networks, there would never have been any need
to invent distributed network analysis - yet it was invented because
DC methods failed miserably for distributed network analysis. How does
a 1/4WL impedance transformer perform at DC? How does a DC analysis
work for light waves? Good grief!
--
73, Cecil, w5dxp.com

what happens to reflected energy ?
 

On Jun 16, 5:26*pm, Cecil Moore wrote:
On Jun 16, 4:11*pm, Roy Lewallen wrote:

Well done, Keith. Thanks.

Two ignorant people patting each other on the back. If DC methods
worked on distributed networks, there would never have been any need
to invent distributed network analysis - yet it was invented because
DC methods failed miserably for distributed network analysis. How does
a 1/4WL impedance transformer perform at DC? How does a DC analysis
work for light waves? Good grief!

Ahhh. Your standard answer. You are an RF dude and there is nothing
that can possibly be learned from simple DC circuits.
Especially when the results challenge your beliefs.

Still, I observe that you have discovered no errors in the exposition.
Which is good.

....Keith

what happens to reflected energy ?
 

On Jun 17, 5:38*am, Keith Dysart wrote:
You are an RF dude and there is nothing
that can possibly be learned from simple DC circuits.

We studied DC circuits in EE101. Later we were forced to expand our
knowledge into distributed networks because the DC circuit model
failed at RF frequencies.

Still, I observe that you have discovered no errors in the exposition.

I have not wasted my time time trying to discover any errors. In a
nutshell, what new knowledge have you presented?
--
73, Cecil, w5dxp.com

what happens to reflected energy ?
 

Keith Dysart wrote:
. . .
By substitution, it is easy to show that the rules used to
derive Vf and Vr results in
P(t)=Pf(t)-Pr(t)

So, while superposing powers, does not work in general, it does
for this case. Directional wattmeters take advantage of this
ideosyncracy to let the user compute the power being delivered
to the load.
. . .

And it works only if Z0 is purely real. When it's not, P(t) also
includes Vf*Ir and Vr*If terms which are neither "forward" nor "reverse"
power. It's unfortunate in a way that Z0 is close enough to being real
for decent lines at Amateur frequencies that these terms are small.
Otherwise they'd have to be confronted and some mistaken assumptions
abandoned.

Roy Lewallen, W7EL

what happens to reflected energy ?
 

Roy Lewallen wrote in
:

Keith Dysart wrote:
. . .
By substitution, it is easy to show that the rules used to
derive Vf and Vr results in
P(t)=Pf(t)-Pr(t)

So, while superposing powers, does not work in general, it does
for this case. Directional wattmeters take advantage of this
ideosyncracy to let the user compute the power being delivered
to the load.
. . .

And it works only if Z0 is purely real. When it's not, P(t) also
includes Vf*Ir and Vr*If terms which are neither "forward" nor
"reverse" power. It's unfortunate in a way that Z0 is close enough to
being real for decent lines at Amateur frequencies that these terms
are small. Otherwise they'd have to be confronted and some mistaken
assumptions abandoned.

While Zo of transmission lines might not be purely real, if the sampler
element is calibrated for V/I being real, then the power is given by
'forward power' - 'reflected power'. This is true even if the
calibration impedance is different to the transmission line in which the
measurements are made, though significant departure will impact
measurement uncertainty.

For example, if I insert a 50 ohm Bird 43 in a 75 ohm line at 1.8MHz,
and measure Pf=150W and Pr=50W, then the power is 150-50=100W. The
insertion VSWR due to the Bird 43 is trivial in this case, so it hardly
disturbs the thing being measured. Each of the power measurements is of
little value alone, no inference can be made (in this case) of the
actual line VSWR, but the difference of the Pf and Pr readings does give
the power at that point.

I discuss this in my article entitled "http://vk1od.net/blog/?p=1004" at
http://vk1od.net/blog/?p=1004 .

Owen

what happens to reflected energy ?
 

Correction:

I discuss this in my article entitled "Power in a mismatched transmission
line" at http://vk1od.net/blog/?p=1004 .

Owen

what happens to reflected energy ?
 

Owen Duffy wrote:

While Zo of transmission lines might not be purely real, if the sampler
element is calibrated for V/I being real, then the power is given by
'forward power' - 'reflected power'. This is true even if the
calibration impedance is different to the transmission line in which the
measurements are made, though significant departure will impact
measurement uncertainty.

For example, if I insert a 50 ohm Bird 43 in a 75 ohm line at 1.8MHz,
and measure Pf=150W and Pr=50W, then the power is 150-50=100W. The
insertion VSWR due to the Bird 43 is trivial in this case, so it hardly
disturbs the thing being measured. Each of the power measurements is of
little value alone, no inference can be made (in this case) of the
actual line VSWR, but the difference of the Pf and Pr readings does give
the power at that point.

I discuss this in my article entitled "http://vk1od.net/blog/?p=1004" at
http://vk1od.net/blog/?p=1004 .

Owen

Adding a directional wattmeter to the mix raises an interesting issue.

Recalling my earlier example of a 100 watt transmitter connected to a
half wavelength 200 ohm transmission line and then to a 50 ohm resistive
load, the "forward power" on the line was 156.25 watts and the "reverse
power" or "reflected power" 56.25 watts. The subject of this topic asks
where the "reflected energy" goes. Does it go back into the transmitter?
Well, let's put a 50 ohm directional wattmeter between the transmitter
and the line. It measures 100 watts forward and 0 watts reverse. So what
happened to the "reverse power" on the transmission line? If it was
going back into the transmitter, we've now fixed the problem just by
adding the wattmeter. In fact, we could replace the wattmeter with a
piece of 50 ohm transmission line of any length, as short or long as we
want, and the "forward power" on that line will be 100 watts and
"reverse power" zero. So we've protected the transmitter from this
horrible, damaging "reverse power" just by adding a couple of inches of
50 ohm line -- or a directional wattmeter. Is this cool or what?

Or we can put the wattmeter at the far end of the line and read 100
watts forward and zero watts reverse, eliminating "reverse power" at the
load -- although we've lost 56.25 watts of "forward power". Or put it at
the center of the line, where we'd read a whopping 451.6 watts forward
and 351.6 reverse(*).

These are the "forward power" and "reverse power" on the short 50 ohm
transmission line (or lumped equivalent) inside the wattmeter. So we can
create "forward power" and "reverse power" just by moving the wattmeter
around. And most importantly, we can use it to isolate the transmitter
from that bad "reverse power". What a powerful tool!

I should point out that the example doesn't even mention, or need to
mention, the transmitter output impedance. The entire analysis holds for
any transmitter impedance, whether "dissipative", "non-dissipative",
linear, or nonlinear. Wherever the "forward power" and "reverse power"
come from and do to, it doesn't depend on any particular value or kind
of transmitter impedance.

(*) This brings up a great idea. Drop down to Radio Shack
or HRO and pick up one of those circulator things that separates forward
and reverse power. Put it into the middle of the line where you have
451.6 watts of "forward power" and 351.6 watts of "reverse power". Take
the 351.6 watts of "reverse power", rectify it, send it to an inverter,
and use it to run the transmitter -- you'll have plenty, even with poor
efficiency, and you can unplug the transmitter from the mains. There
should even be enough power left over to run your cooler and keep a
six-pack cold. Then work DX with your 451.6 watts of forward power while
you enjoy a cool one. Goodbye to electric bills! Hello to DX, free
power, and a tall cool one!

Roy Lewallen, W7EL

what happens to reflected energy ?
 

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

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?