RadioBanter - Re-Normalizing the Smith Chart (Changing the SWR into the

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Richard Harrison wrote:

I wrote:
"As energy can`t be destroyed it had to be reflected by a hard short or
open." Keith replied:
"Or just stopped and stored."

Wave energy is energy in motion. No motion, no waves.

There is no doubt that energy moves. The point of disagreement is on
how far it moves.

Keith wrote:
"I would strongly suggest that no energy crosses these points in the
line where the voltage and current are always zero since p(t) is always
zero."

See my comment above on power as a function of time. Keith erred in
saying "points in the line where the voltage and current are always
zero", as where SWR volts are zero, amps are max, and vice versa.

This last is true, but p(t) = v(t) * i(t); volts and amps must be
present simultaneously for there to be power.

I wrote:
"If energy were turned around before it reached the end of the line,
nulls more distant from the source than the turnaround point would not
exist."

Keith wrote:
"Not so,---."

There is no argument that can make wave interference where there are no
waves. In a lossless line, pre-existing waves could circulate forever.
But, our discussion relates to effects on actual lines.

There are many assumptions in this discussion which means it only
applies
to ideal lines. The extension to real lines, retains the fundamentals
but the details need tuning.

As a simple example, on a real line, the nulls are never 0.
But including this in the discussion would just make it more difficult
to locate the points of disagreement.

Keith wrote:
"Try visualizing how a step function charges the line."

Totally irrelevant.

Understanding a step will help with understanding line
behaviour. This knowledge can then assist in understanding
sinusoidal steady state.

....Keith

Keith wrote:
"The last is true, but p(t) = v(t)*i(t); volts and amps must be present
simultaneously for there to be power."

By the same token, a-c flow is discontinuous at all zero crossings! I
don`t think so.

Best regards, Richard Harrison, KB5WZI

Richard Harrison wrote:

Proof is that a directional sensor finds the same power flow, forward
or reflected, at a standing wave zero point as it does at a standing
wave maximum, or at any point in between.

Richard's statement about what the directional sensor "finds" is
perfectly correct - but unfortunately it cannot be used as proof.

The reason is that so-called "directional wattmeters" don't physically
sense directional power flow. All they sense from the transmission line
are the current and the voltage, as two separate samples. Then they add
or subtract these samples to give the sensor its directional properties.
All the meter reads is a detected RF *voltage*, which changes to a
different value when the sensor is reversed.

If you want to know what those meter readings mean, you need
transmission-line theory in order to understand them. You can then
calibrate the meter to read forward and reverse power - but you cannot
do that without using transmission-line theory to do it, and that theory
is the subject of this entire discussion.

Therefore the readings of a "directional wattmeter" cannot be used as
evidence for either side, because that argument would be circular - you
cannot use any theory to prove itself!

However, this attempt to use inadmissible evidence doesn't necessarily
affect Richard's wider argument about power flow. If that argument is
correct, there definitely *will* be other physical evidence to prove it.

--
73 from Ian G3SEK 'In Practice' columnist for RadCom (RSGB)
Editor, 'The VHF/UHF DX Book'
http://www.ifwtech.co.uk/g3sek

Ian White, G3SEK wrote:
However, this attempt to use inadmissible evidence doesn't necessarily
affect Richard's wider argument about power flow. If that argument is
correct, there definitely *will* be other physical evidence to prove it.

The biggest clue that I have noticed is that nobody has been able to
generate standing waves in a single source, single feedline, single
load system without the existence of a reflected wave.
--
73, Cecil, W5DXP

Richard Harrison wrote:

Keith wrote:
"The last is true, but p(t) = v(t)*i(t); volts and amps must be present
simultaneously for there to be power."

By the same token, a-c flow is discontinuous at all zero crossings! I
don`t think so.

There is certainly no power at the zero crossings. This variation in
the rate of energy flow is why the power dudes really prefer 3 phase;
energy flow is constant.

....Keith