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

On Jan 26, 12:15*pm, Roger Sparks wrote:
On Fri, 25 Jan 2008 19:13:31 -0800 (PST)





Keith Dysart wrote:
On Jan 24, 10:33*pm, Roger Sparks wrote:
[snip]
By examining this derivation, the reader can see that power and energy
is reflected when a wave encounters a discontinuity. *The reader can
also see that more power is present on the transmission line than is
delivered to the load.

This is the conventional phraseology for describing the behaviour at
the impedance discontinuity.

Allow me to offer a specific example for which this phraseology is
inappropriate.

Consider a 50 V step function generator with an output impedance of
50 ohms driving a 50 ohm line that is 1 second long terminated in an
open circuit.

Turn on the generator. A 50 V step propagates down the line. The
generator is putting 50 J/s into the line. One second later it
reaches the open end and begins propagating backwards.
After two seconds it reaches the generator. The voltage at the
generator is now 100 V and no current is flowing from the
generator into the line. In the 2 seconds, the generator put
100 joules into the line which is now stored in the line.
The line is at a constant 100 V and the current is zero everywhere.

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.

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

Does this seem like a reasonable claim for an open circuited
transmission line with constant voltage along its length and
no current anywhere?

I do not find it so.

...Keith

This is a reasonable observation for a static situation where energy is stored on a transmission line. *

If the example contained an ongoing consideration, like "Where does the power move to?", then it would be reasonable to consider that the wave continued to move, simply to avoid the complication of what EXACTLY happens when a wave starts and stops.

So have you thought about "where does the power go?"

When the generator is matched to the line so that
the reflected wave does not encounter an impedance
discontinuity when it arrives back at the generator
(and therefore is not reflected), where does the
reflected power go?
Does it enter the generator?
Is it dissipated somewhere?

Answers to these questions will quickly lead to
doubts about the *reality* of "reflected power".

...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

Keith Dysart wrote:
When the generator is matched to the line so that
the reflected wave does not encounter an impedance
discontinuity when it arrives back at the generator
(and therefore is not reflected), ...

On the contrary, it is redistributed back toward
the load in the process of destructive interference
and becomes constructive interference associated
with the forward wave. Whether you call that a
reflection or not, the fact that the forward
power equals the source power plus the reflected
power tells us that reflected power being dissipated
in the source would violate the conservation of
energy principle.

where does the reflected power go?

Power doesn't flow so
reflected power doesn't "go" anywhere. It is the
reflected energy that is flowing, i.e. going
somewhere. If the power were flowing, its
dimensions would be joules/sec/sec. Maybe
you would like to try to explain the physical
meaning of those dimensions?

Does it enter the generator?

forward power = source power + reflected power

A component of the forward energy is equal in
magnitude to the reflected energy so no, it
doesn't enter the source.

Is it dissipated somewhere?

Not during steady-state. During steady-state it
is being used for impedance transformation. After
steady-state, it is dissipated either in the
source or in the load as a traveling wave.

Answers to these questions will quickly lead to
doubts about the *reality* of "reflected power".

Reflected power is the Poynting vector associated
with the reflected wave. It exists at a point. It's
average magnitude (indirectly measured by a Bird) is
Re(ExH*)/2. It's direction is the direction of the
flow of reflected energy.

If you don't believe in reflected energy, let your
fingers become the circulator load resistor for
an open-circuit stub being driven by a KW source.
--
73, Cecil http://www.w5dxp.com

On Sat, 26 Jan 2008 19:24:22 -0800 (PST)
Keith Dysart wrote:

On Jan 26, 12:15*pm, Roger Sparks wrote:
On Fri, 25 Jan 2008 19:13:31 -0800 (PST)





Keith Dysart wrote:
On Jan 24, 10:33*pm, Roger Sparks wrote:
[snip]
By examining this derivation, the reader can see that power and energy
is reflected when a wave encounters a discontinuity. *The reader can
also see that more power is present on the transmission line than is
delivered to the load.

This is the conventional phraseology for describing the behaviour at
the impedance discontinuity.

Allow me to offer a specific example for which this phraseology is
inappropriate.

Consider a 50 V step function generator with an output impedance of
50 ohms driving a 50 ohm line that is 1 second long terminated in an
open circuit.

Turn on the generator. A 50 V step propagates down the line. The
generator is putting 50 J/s into the line. One second later it
reaches the open end and begins propagating backwards.
After two seconds it reaches the generator. The voltage at the
generator is now 100 V and no current is flowing from the
generator into the line. In the 2 seconds, the generator put
100 joules into the line which is now stored in the line.
The line is at a constant 100 V and the current is zero everywhere.

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.

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

Does this seem like a reasonable claim for an open circuited
transmission line with constant voltage along its length and
no current anywhere?

I do not find it so.

...Keith

This is a reasonable observation for a static situation where energy is stored on a transmission line. *

If the example contained an ongoing consideration, like "Where does the power move to?", then it would be reasonable to consider that the wave continued to move, simply to avoid the complication of what EXACTLY happens when a wave starts and stops.

So have you thought about "where does the power go?"

Yes, only the model I use substitutes a battery for the signal generator that you are using. The returning wave can recharge the battery, but how does the current stop? Or does it ever stop? From a practical aspect, the current must stop, but I can not explain how except by resistance that absorbs the circulating power.

We must keep the limitations of our models in mind.


When the generator is matched to the line so that
the reflected wave does not encounter an impedance
discontinuity when it arrives back at the generator
(and therefore is not reflected), where does the
reflected power go?
Does it enter the generator?
Is it dissipated somewhere?

Answers to these questions will quickly lead to
doubts about the *reality* of "reflected power".

...Keith

The reflected wave that does not encounter an impedance at the generator must be on an infinitely long line, and therefore the conditions must have changed between the time of launch and return.

It makes more sense to think that the reflected voltage of (using your example) 50v meets 50v of forward voltage, and therefore finds an infinite resistance, coming to a stop.

If that happens, doesn't the same condition occur as soon as the reflected wave is first generated at the line end? The current really stops as soon as the end is reached, with the energy contained in the magnetic field converted to electric field energy visible as voltage. If this happened, the reflected wave could be better described as an electric "jump", similar to a hydraulic jump found in open channel flow of liquids, where kinetic energy is converted to potential energy.

This idea of an electric "jump" requires not a reflection occuring without a discontinuity, but a moving wave front that absorbs the traveling wave, bringing it to a stop (in this case).

--
73, Roger, W7WKB

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

Roy Lewallen wrote:
"A better reason to avoid "stored" is that power isn`t stored at all,
anywhere. Anyone who believes so should be able to tell us how many
watts are stored in a 50 Ah, 12 volt battery."

The 50 Ah of energy stored in a battery may be withdrawn at a rate
determined by the load, so it is a variable.

Velocity in a transmission line is a constant determined by
constructtion of the line. Thus, energy stored in the line is determined
by its length, voltage, current, and phase angle. These predict the rate
of energy transfer (power).

Best regards, Richard Harrison, KB5WZI

"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

Richard Harrison wrote:
Roy Lewallen wrote:
"A better reason to avoid "stored" is that power isn`t stored at all,
anywhere. Anyone who believes so should be able to tell us how many
watts are stored in a 50 Ah, 12 volt battery."

The 50 Ah of energy stored in a battery may be withdrawn at a rate
determined by the load, so it is a variable.

Ah is a unit of charge, not energy. The battery, making the simplifying
(and invalid) assumption of constant voltage during discharge, contains
2.16 Mj of energy.

Velocity in a transmission line is a constant determined by
constructtion of the line. Thus, energy stored in the line is determined
by its length, voltage, current, and phase angle. These predict the rate
of energy transfer (power).

So the rate at which a transmission line transfers energy depends on its
length? If I put 100 watts into a one wavelength cable and get 100 watts
out, will I get 200 watts out if I extend the cable to two wavelengths?

Roy Lewallen, W7EL