wrote:
W5DXP wrote:

wrote:
Do not be afraid to admit that you have changed the definition of P = V
x I and therefore do not accept the standard definition.

Well, I was taking 'x' as a multiplication sign. Did you mean it as
a cross product sign?

No, simple multiply.

Well, then your equation is wrong. It should contain phase. The
power is V*I*cos(theta) which is known to be *AVERAGE* power since
V and I are RMS values. When the total voltage is zero and the total
current is 2*I, the phase angle is still 90 degrees for a lossless
stub.

In general in circuit analysis, one can
use superposition to evaluate voltages and currents, but it is quite
unusual for powers to sum properly at the same time.

They sum perfectly if interference term is taken into account. The
equation becomes something like, e.g.

P1 + P2 + constructive interference = total forward power

P3 + P4 - destructive interference = total reflected power.

This is all explained in _Optics_, by Hecht.

Oh, the waves are there all right, if you choose to model it so. But,
they are waves of voltage and current, not power.

How do those waves ever come into existence without a power
source? During the first two seconds when a 100w transmitter is
driving a one second long lossless line, it shoves 200 joules
into the line. What happens to that 200 joules?

A transmission line with distributed capacitance and inductance is
somewhat more complex than a voltage source and a capacitor. The
distributed capacitance and inductance form resonant circuits which
transfer energy between capacitance and inductance. Just like a
circuit, if you short the line at any point where the voltage is
always zero, or open it at any point where the current is always
zero, nothing is changed.

Of course, something is changed dramatically. You have introduced an
extreme impedance discontinuity where there was none before. How you
can assert that nothing is changed when something has changed as much
as it possibly can change is pretty strange. You have gone from 100%
transmission of waves to 100% reflection of waves.

Here is an example.

source-------------50 ohm coax--------------+----1/4WL stub-----open

What mechanism of physics keeps energy from crossing the '+' point? Note
that there is no physical impedance discontinuity at point '+'.

The voltage is always 0 so from the most basic of circuit theory,
the power is zero, so the energy flow is 0. I have some difficulty
articulating what happens at a current maximum. But if you will permit
me to move the question back another quarter wavelength to a voltage
maximum and current minimum, the answer is easy. The voltage on both
sides of the current minimum are always the same so that no current
flows, hence no energy flows.

Exactly what mechanism of physics allows a non-changing characteristic
impedance to cause reflections? There is no impedance discontinuity.
Therefore, there are no reflections. What you need to realize is that
no *NET* energy flows but the forward wave has absolutely nothing to
interfere with its flow. The reflected wave has absolutely nothing to
interfere with its flow. The total voltage and total current is *caused*
by the waves. Therefore, the ratio of the total voltage to total current
cannot affect their own cause. One more example of confusing cause and
effect. All this is obvious when observing ghosting during a TV experiment.

Rather I disagree with the view that in an open
or shorted ideal line, energy flows to the end, is reflected back,
flows back to the start and is reflected again. Repeat. To be even a
bit stronger, ENERGY does not cross any point where the voltage or
current is always 0. This is so that the expression P = V x I can
remain true.

You are again confusing the energy carriers with the wave energy. They
are not the same thing. The energy carriers don't cross the boundary
(except for random occurrences). But there is no physics mechanism to stop
the wave energy from crossing the boundary in both directions. There is
nothing there that can reverse the momentum of the waves.

The easiest way to view your mistake is by putting a directional wattmeter
half way down a stub. Guess what you will read? We had this discussion on
this newsgroup a few months ago. No one could prove that the mouth of the
stub has any effect at all on the forward wave and the reflected wave.

As to your question on standing waves, when a resonant circuit is exited
with an impulse function, a sinusoidal voltage and current occur in the
circuit. It does not seem much of a stretch to find a sinusoidal spatial
distribution of voltage and current in circuit constructed of
distributed capacitance and inductance (i.e. a transmission line).

The stretch is finding a physics mechanism that will create and sustain
standing waves without a forward wave and a reflected wave. Please describe
in detail how it is done. You need a physics mechanism capable of reversing
the momentum of waves in a constant Z0 environment. Good luck on that one.
--
73, Cecil http://www.qsl.net/w5dxp
"One thing I have learned in a long life: that all our science, measured
against reality, is primitive and childlike ..." Albert Einstein



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Dr. Slick wrote:
You're right, but we are talking about a one-port network, the
antenna and transmission line.

Hmmmmmm, I could have sworn that you were talking about the
source impedance. :-)
--
73, Cecil http://www.qsl.net/w5dxp



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wrote:
Let's apply these expressions to some simple examples.

Connect a length of 50 Ohm transmission line to a 9 Volt battery and
wait for the transient to die:

Why do you need to obfuscate the discussion by resorting to DC? Why
can't you explain things using RF? One previous poster refused to
discuss a single source system for some reason. Since reactance
disappears during DC steady-state, the relevance of a DC example is
certainly questionable. Hint: The Z0 of a piece of coax at DC is
NOT 50 ohms.
--
73, Cecil http://www.qsl.net/w5dxp



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W5DXP wrote:

wrote:
Let's apply these expressions to some simple examples.

Connect a length of 50 Ohm transmission line to a 9 Volt battery and
wait for the transient to die:

Why do you need to obfuscate the discussion by resorting to DC?

Mostly because a sensible answer should be able to explain from
DC to RF. If it only works at RF but collapses with the much
simply DC, then the answer seems suspect.

Why is there so much protestation about having the answer explain
DC behaviour as well?

Why can't you explain things using RF?

I like my answers to explain the more complex (RF) as well as the
simpler (DC). As well, applying the answer to the simpler system
sometimes points out absurdities in the answer.

One previous poster refused to discuss a single source system
for some reason.

I fully agree. If the explanation works for the more complex system,
it should also work for the simpler.

Since reactance disappears during DC steady-state, the relevance
of a DC example is certainly questionable.

Perhaps, but we are dealing with simple circuit theory here using
ideal components. All seems to work.

Hint: The Z0 of a piece of coax at DC is NOT 50 ohms.

I've seen nothing in the derivations for ideal transmission lines that
indicate frequency dependence. Care to elaborate?

In any case, if you wish to stay away from DC, we could just use an
arbitrarily low AC signal. How about 10**-100 Hertz? Indistinguishable
from DC for any engineering purpose? Or should we go lower? Just pick
the number.

....Keith

W5DXP wrote:

wrote:

W5DXP wrote:
Unfortunately, the analogy is not a good one. In a transmission line,
there must exist a discontinuity to cause a reversal of momentum of
the waves. No such discontinuity exists so there is nothing to reverse
the momentum of the forward and reflected waves.

But then I was not talking waves, but charge. The waves are a
manifestation of the changes in the charge distribution.

Yes, I know that. When are you going to stop confusing the charge
carriers with the waves? :-) The charges have exactly as much to do
with the waves as water molecules have to do with water waves. Water
molecules don't cross certain boundaries but that doesn't keep tsunamis
from wiping out an island.

Well, charge does seem to be key. Current is charge per time. Power
is voltage times current. I fail to understand the opposition to
using charge in the discussion. It is a basic element of circuit
theory.

Correct. I hope I have not given any other impression. For the most
part in my discussions I mean instantaneous power, since conversion to
average loses too much information to enable understanding.

Instantaneous power is essentially useless to this discussion
according to Hecht in _Optics_. Here's a quote: "Since the power
arriving cannot be measured instantaneously, the detector must
integrate the energy flux over some finite time." Irradiance is
*average* power.

This may be a challenge in optics but it is not at the lower AC
frequencies especially in a circuit. My power supplier does it all
the time. And sends me a bill to prove it.

But for the most part, greater understanding will arise from sticking
with instantaneous energy flow.

I hope, for your mental health, that you don't really believe that. :-)
If it doesn't work in optics, there is no reason to believe that it
will work for other photonic waves, including RF.

Actually, I suspect that optics has more to learn from RF than vice
versa
since, with RF, one has the luxury of measuring both voltage and current
(and simultaneously at lower frequencies). Optics seems constrained to
pure average power measurements.

But at certain points (1/2 wavelength apart), the voltage IS always 0
so the energy flow IS always 0. Unless you are prepared to discard
Pinst = Vinst * Iinst.

The *NET* energy flow is zero. The component energy flow encounters absolutely
nothing that can change its momentum. There are NO impedance discontinuities
to cause reflections. It would take a Transmission Line God to do that.

It appears that you are prepared to ignore Pinst = Vinst * Iinst. Could
you expand on when Pinst IS equal to Vinst * Iinst and when it isn't?

Simply convince me that Pinst is not always equal to Vinst * Iinst and
my difficultes will disappear.

....Keith

p(t) = v(t) * i(t). Period. No phase, no vectors, no cross products.

You can't calculate instantaneous power from phasor quantities, because
phasors assume a frequency that's the same for all quantities, and the
frequency of the power wave is twice that of the voltage and current waves.

Of course, a lot of people mean "average power" when they say "power",
so you have to be very careful when reading anything written by those folks.

Roy Lewallen, W7EL

W5DXP wrote:
wrote:

W5DXP wrote:

wrote:

Do not be afraid to admit that you have changed the definition of P = V
x I and therefore do not accept the standard definition.


Well, I was taking 'x' as a multiplication sign. Did you mean it as
a cross product sign?


No, simple multiply.


Well, then your equation is wrong. It should contain phase. The
power is V*I*cos(theta) which is known to be *AVERAGE* power since
V and I are RMS values. When the total voltage is zero and the total
current is 2*I, the phase angle is still 90 degrees for a lossless
stub.
. . .

W5DXP wrote:
wrote:

Let's apply these expressions to some simple examples.

Connect a length of 50 Ohm transmission line to a 9 Volt battery and
wait for the transient to die:


Why do you need to obfuscate the discussion by resorting to DC? Why
can't you explain things using RF? One previous poster refused to
discuss a single source system for some reason. Since reactance
disappears during DC steady-state, the relevance of a DC example is
certainly questionable. Hint: The Z0 of a piece of coax at DC is
NOT 50 ohms.
. . .

It is for the time it takes for the transient to die. Y'see, when you
connect a battery to a transmission line, you don't have DC. You only
have DC after you wait around forever or, for practical purposes, until
everything settles to its final value within some small error bound.

It's also wise to remember that, likewise, when you connect a source to
a line, you're not dealing with a single frequency. The system settles
down to a single frequency only after all transients have died out.
That's why a time domain analysis is necessary, or at least highly
preferable, to analyze transient conditions.

But I do agree that, although it's instructive and can help
understanding of transmission line phenomena, we have to be careful when
extrapolating pulse or step results to sine wave situations --
especially steady state.

Roy Lewallen, W7EL

wrote:
I see your confusion and apologize for not being completely clear.
When I say P = V * I, P, V and I are instantaneous values, the only
ones worth exploring if an understanding is desired.

Instantaneous power is useless if an understanding is desired.
I agree with Hecht and all my other references. Kraus, Jasik,
Balanis, Hecht, Ramo & Whinnery all agree that dealing with
instantaneous power is a waste of time. Instantaneous power
is essentially meaningless since the definition of power
requires a length of time. There is simply no such thing as
instantaneous power. Power is always the energy passing a
point during a slot of time. Zero energy passes a point in
dt of time as delta-t approaches zero, by definition.

Well, if you do a bit of fudging you can always make it work. But
I do not observe these fudge terms in the expressions related to
forward and reverse power in transmission lines.

Then you haven't read Dr. Best's QEX article. The interference is
there and takes the form of 2*Sqrt(P1)*Sqrt(P2) for 100% constructive
interference.

Of circuit theory I have a reasonable grasp, optics I leave to others.

Too bad. The field of optics has already solved the problems with which
you are wrestling. Take a look at this web page to figure out how Z0-
matching works.

http://www.mellesgriot.com/products/optics/oc_2_1.htm

As mentioned previously: stored in the line and moving back and
forth between quarter wave boundaries.

Mentioned, but not explained. How does the energy's momentum change
direction in a constant Z0 environment? What causes the back and
forth movement?

Back in basic circuit theory, some years ago, it was permitted to
short points of equal voltage and open conductors with no current.
Has this changed in the intervening years?

Ahhhh, I see your confusion. Circuit theory and transmission line
theory are not the same thing. Many have tried to mix the two
models and fallen on their faces. Transmission line theory is
simply more complicated than basic circuit theory. If you don't
believe it, replace a transmission line with ghosting with an
equivalent circuit - the ghosting disappears. Are they really
equivalent?

Because the observed voltages and currents are the same. The circuit
has been replaced by one which is indistinguishable from the first.

BS! I can certainly distinguish between a short and a non-shorted
transmission line. Please try again.

You have gone from 100%
transmission of waves to 100% reflection of waves.

An intriguing assertion, but one which can not be demonstrated through
any observations made on the circuit.

Of course it can by observing ghosting in a TV signal. With the line
shorted, there is no ghosting. With the line not shorted, there is
ghosting. Why is that so difficult to understand?

I am not sure why you are looking so hard for reflections. I am not
attempting to claim that there are any. I simply claim that no energy
can cross the boundaries because the current or voltage is always 0.

In the real world, if no RF energy crosses a boundary, then that energy
is reflected since it cannot stand still. Of course, the supernatural
world is an entirely different matter.

Basic electriciy. This says nothing about the presence or absence of
reflections or whether there is a mechanism to prevent the energy
crossing the boundary or it just happens through the dynamics of the
distributed capacitance and inductance.

Unfortunately, this is beyond the ability of basic electricity to
handle. It takes the wave reflection model (or quantum physics) to
handle it. Einstein said a model should be as simple as possible,
but not too simple. Once again, you seem to be trying to force
reality to obey your model instead of vice versa.

Oh, I fully understand what a 'directional watt' meter will indicate.

Then why are you so confused? :-)

May I suggest that you perform the same experiment with a real
instantaneous watt meter:

This is obviously a diversionary tactic. Instantaneous power is
essentially meaningless according to all my references and I
have a bunch of them. Would you care to provide a reference that
seriously deals with instantaneous power in transmission lines?

I'm sorry, I do not grasp what you are attempting to say here.

That's more than obvious.
--
73, Cecil http://www.qsl.net/w5dxp
"One thing I have learned in a long life: that all our science, measured
against reality, is primitive and childlike ..." Albert Einstein



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wrote:
Well, charge does seem to be key.

Sorry, charge is not the key. The charges are simply the medium that
makes the waves possible.

Actually, I suspect that optics has more to learn from RF than vice
versa ...

You have got to be kidding. Optics is at the leading edge. RF is at
the trailing edge. Things that are common knowledge in optics are
dismissed by this newsgroup (including you?).

It appears that you are prepared to ignore Pinst = Vinst * Iinst.

I will go with all my references - instantaneous power is essentially
useless for any kind of serious analysis. Average power is the only thing
that matters at RF and optical frequencies. Please provide a transmission
line reference that extols the virtues of instantaneous RF power.
--
73, Cecil http://www.qsl.net/w5dxp



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Roy Lewallen wrote:
p(t) = v(t) * i(t). Period. No phase, no vectors, no cross products.

Millions of power engineers have been taught that V*I*cos(theta) is power.
Are you disagreeing with that teaching? p(t) = v(t) * i(t) is virtually
worthless to a power engineer since a generator is a giant heat sink.

Most of my references say that instantaneous power is virtually meaningless
for RF and optics work. Can you provide a reference that extols the virtues
of instantaneous power applied to RF? Can you list any benefit for considering
instantaneous RF power?
--
73, Cecil http://www.qsl.net/w5dxp



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