Keith Dysart wrote:
Computing Pf and Pr will yield 50 W forward and 50 W reflected.
And yet no current is flowing anywhere. The voltage on the line
is completely static.

Why would you compute Pf and Pr when no DC current is
flowing? It is an invalid thing to do and unrelated to
reality.

And yet some will claim that 50 W is flowing forward and 50 W
is flowing backwards.

I know of no one who will claim that for static DC.
There are obviously no photons being emitted and
therefore, no waves.

Your example is unrelated to standing waves on an
RF transmission line where energy is in motion,
photons are continuously being emitted and absorbed,
and current and voltage loops are active.

One must realize the limitations of one's model.
The wave model obviously fails where there are no
waves.
--
73, Cecil http://www.w5dxp.com

On Jan 26, 9:07*am, Cecil Moore wrote:
Keith Dysart wrote:
Computing Pf and Pr will yield 50 W forward and 50 W reflected.
And yet no current is flowing anywhere. The voltage on the line
is completely static.

Why would you compute Pf and Pr when no DC current is
flowing?

To facilitate learning about how the equations work and
what they may mean.

It is an invalid thing to do

Not at all. The equations don't just stop working at 0 frequency.

In their general form F(t), there is no hint at all that F can
not be a constant. Or, if you prefer, a square wave with a width
several times longer than the length of the line.

and unrelated to reality.

Anything unreal will also be unreal for the specific case of
sinusoids.

And yet some will claim that 50 W is flowing forward and 50 W
is flowing backwards.

I know of no one who will claim that for static DC.
There are obviously no photons being emitted and
therefore, no waves.

You really should try to stop thinking about photons
for just a short while. All the behaviours of a transmission
line can be understood and characterized without reference
to photons.

Analysis using classic circuit principles works quite fine
and has no difficulty at low frequencies.

Your example is unrelated to standing waves on an
RF transmission line where energy is in motion,
photons are continuously being emitted and absorbed,
and current and voltage loops are active.

There is no standing wave, but the example is quite
valid none-the-less. If you like, consider it as a
long pulse. And if you only want sinusoids, Fourier
will convert the pulse to sinusoids which you can,
using superposition, use to solve the problem.

The simplicity of the constant voltage makes it
easy to check your results.

One must realize the limitations of one's model.
The wave model obviously fails where there are no
waves.

Think of it as a long pulse. That should satisfy your
need to have 'waves'.

...Keith

Keith Dysart wrote:
Not at all. The equations don't just stop working at 0 frequency.

As a matter of fact, EM waves cannot exist without
photons. There is zero wave activity at DC. Therefore,
the forward power and reflected power is zero at DC
steady-state.

You really should try to stop thinking about photons
for just a short while.

Yep, you guys would like to sweep the technical knowledge
from the field of optical physics and quantum electro-
dynamics under the rug. One wonders why.

Think of it as a long pulse. That should satisfy your
need to have 'waves'.

No, only photons will satisfy the definition of EM
waves. There are no photons. There are no waves. There
are no forward and reflected powers. There are no
changing E-fields or H-fields. There is not even any
movement of electrons associated with the source
voltage.
--
73, Cecil http://www.w5dxp.com

Cecil Moore wrote in news:mO0nj.4140$J41.257
@newssvr14.news.prodigy.net:

Keith Dysart wrote:
Not at all. The equations don't just stop working at 0 frequency.

As a matter of fact, EM waves cannot exist without
photons. There is zero wave activity at DC. Therefore,
the forward power and reflected power is zero at DC
steady-state.

You really should try to stop thinking about photons
for just a short while.

Yep, you guys would like to sweep the technical knowledge
from the field of optical physics and quantum electro-
dynamics under the rug. One wonders why.

Think of it as a long pulse. That should satisfy your
need to have 'waves'.

No, only photons will satisfy the definition of EM
waves. There are no photons. There are no waves. There
are no forward and reflected powers. There are no
changing E-fields or H-fields. There is not even any
movement of electrons associated with the source
voltage.

Food for thought!

There is no such thing as DC! How’s that you say? You have to turn it on
sometime and someday you may turn it off. There for, DC is just very low
frequency AC.

As another example of physics over thinking, I once had a physics
professor describe in detail how an electric motor works using Quark
physics. Very interesting but it has no practical value in using electric
motors. Much like this long thread has nothing much to do with antennas.


John Passaneau W3JXP

"JOHN PASSANEAU" wrote in message
...

There is no such thing as DC! How's that you say? You have to turn it on
sometime and someday you may turn it off. There for, DC is just very low
frequency AC.


John Passaneau W3JXP

That's false reasoning OM! Alternating current is not the same as
discontinuous current. In the example you provide, a DC supply is either off
or on; it does not reverse polarity! You do not make AC by switching on and
off DC, even at 50 Hz (or your 60 Hz)

Suzy wrote:
"You do not make AC by switching on and off DC, even at 50 Hz (or your
60 Hz)"

Not without inductance to provide the missing half cycle. Remember
vibrator supplies and their solid-state equivalents?

Best regards, Richard Harrison, KB5WZI

Suzy wrote:
"JOHN PASSANEAU" wrote in message
...
There is no such thing as DC! How's that you say? You have to turn it on
sometime and someday you may turn it off. There for, DC is just very low
frequency AC.


John Passaneau W3JXP

That's false reasoning OM! Alternating current is not the same as
discontinuous current. In the example you provide, a DC supply is either off
or on; it does not reverse polarity! You do not make AC by switching on and
off DC, even at 50 Hz (or your 60 Hz)

But through the techniques of linear circuit analysis, we can split the
pulsed DC into two components, a pure steady DC component and a
symmetrical AC component. We can do two separate analyses with the two
(AC and DC) excitations, and sum the results. The answer will be exactly
the same as if we had done the calculations directly. One can then
reasonably claim that the switched DC is the sum of an AC waveform and a
pure DC component.

Roy Lewallen, W7EL

On Jan 27, 9:45*am, Cecil Moore wrote:
Keith Dysart wrote:
Not at all. The equations don't just stop working at 0 frequency.

As a matter of fact, EM waves cannot exist without
photons. There is zero wave activity at DC. Therefore,
the forward power and reflected power is zero at DC
steady-state.

And yet all the circuit theory derivations of reflection
coefficient, power, voltage and current distributions
work just fine for DC (or, if you prefer, low rate pulses).

Digital designers use exactly that for solving real world
problems. They do not refuse to solve problems when the
conditions approach DC. The equations all work.

You really should try to stop thinking about photons
for just a short while.

Yep, you guys would like to sweep the technical knowledge
from the field of optical physics and quantum electro-
dynamics under the rug. One wonders why.

An intriguing accusation. Transmission lines can be
understood well, and real world problems solved without
reference to photons and EM waves. Just use the well
known circuit theory based equations. And they extend
all the way to DC.

For those who like EM waves and photons... Why do you
want to limit yourselves? Why won't you use the circuit
theory bases equations to solve problems for which they
work? Just because EM waves and photons do not?

Think of it as a long pulse. That should satisfy your
need to have 'waves'.

No, only photons will satisfy the definition of EM
waves. There are no photons. There are no waves. There
are no forward and reflected powers. There are no
changing E-fields or H-fields. There is not even any
movement of electrons associated with the source
voltage.

May be true. But why do you want to use that as an excuse
not to solve solveable problems?

...Keith

Keith Dysart wrote:
Transmission lines can be
understood well, and real world problems solved without
reference to photons and EM waves.

If that were true, we would not be having this
argument. The real world problem is - where does
the reflected EM energy go? Since you have not "solved
that real world problem", your methods are suspect.

OTOH, optical physicists solved that same problem
long before any of us were born.

Why won't you use the circuit
theory bases equations to solve problems for which they
work?

The main reason not to use your methods is that you use
them to arrive at wrong concepts. EM waves cannot exist
without energy. If there exists no EM wave energy, there
are no EM waves. If EM waves exist, they are necessarily
associated with energy and momentum, both of which must
be conserved. The amount of that energy flowing past a
measurement point/plane in a unit-time is the power
(density) associated with the reflected wave. Even the
energy and momentum of a single photon can be calculated.

Any length of transmission line with reflections contains
exactly the amount of energy necessary to support the
forward and reflected waves. That amount of energy exists
in the transmission line and is not delivered to the load
during steady-state.

Because your model doesn't tell you where the reflected
energy goes, you assume there is zero energy in reflected
waves. Again, I challenge you to use your fingers to
replace the reflected power circulator load in a system
with a KW source driving an open-circuit. That shock therapy
will, no doubt, change your mind about the non-existence of
reflected power.
--
73, Cecil http://www.w5dxp.com