On Jun 25, 1:18*pm, Cecil Moore wrote:
On Jun 25, 5:53*am, dave wrote:

try it! *you will read the exact same voltage on either side of that
connection point!

I already told you that only applies to the total voltage and total
current. You will NOT read the same forward voltage on either side,
you will NOT read the same forward current on either side, you will
NOT read the same reflected voltage on either side, and you will NOT
read the same reflected current on either side.

when someone gives me a voltmeter i can touch to that connection point
and measure the 4 components then we can talk. as far as designing
anything i need i can do it without giving those s parameters or your
4 components a second thought. it is very easy to transform and
combine the impedances to tell me what the load seen by the
transmitter is, or to figure out the needed stub for providing a
proper match without all that stuff.

On Jun 25, 5:25*pm, John S wrote:
But, it does. First, it causes the 50 ohms line (looking into the 291.4
ohms line to see a match due to the reflection. Second, the
re-reflection from that discontinuity is half of what maintains the
circulating energy on the line. The other half is the discontinuity of
the non-virtual load.

You are confusing reflection with wave cancellation (destructive
interference). I suggest that you study the separate sections on
reflections vs interference in "Optics", by Hecht. Nowhere does any
optical textbook indicate that superposition and reflection are the
same thing (and they are indeed NOT the same thing).

Superposition/interference applies to two or more waves. Reflection
applies to a single wave. When a reflected wave is re-reflected, it is
always a single wave event. Take a look at one of the s-parameter
equations:

b1 = s11(a1) + s12(a2) = 0

b1 is the reflected voltage toward the source which is, of course,
zero when looking into a Z0-match.

s11(a1) is the forward source wave reflected from the physical
impedance discontinuity at the Z0-match point. Note: This is a single
reflected component of a single forward wave.

s12(a2) is the part of the reflected wave from the load that is
transmitted back through the impedance discontinuity at the Z0-match
point. Note: This is a single transmitted component of a single
reflected wave.

Reflections are now over and done with. What happens next is
superposition/interference, i.e. the phasor addition of the two
component reflections. Note: The results of the phasor addition of two
component waves is NOT a reflection!!!

The zero result of the addition of those two phasors is associated
with destructive interference toward the load. That's what causes the
50 ohm Z0-match, not the component reflections. The redistribution of
the destructive interference energy back toward the load is NOT a re-
reflection and here's why:

A re-reflection preserves the modulation content of the re-reflected
wave which can be proved by TV ghosting experiments. Wave cancellation
due to destructive interference does NOT preserve the modulation. In
fact, any differences in modulation between the two superposed wave
components would wind up incident upon the source.
--
73, Cecil, w5dxp.com

On Jun 26, 6:43*am, dave wrote:
when someone gives me a voltmeter i can touch to that connection point
and measure the 4 components then we can talk.

I could design an expensive device that will do exactly that but it is
a lot easier to just calculate the values using Mathcad.
--
73, Cecil, w5dxp.com

On 6/29/2011 11:04 AM, Cecil Moore wrote:
On Jun 25, 5:25 pm, John wrote:
But, it does. First, it causes the 50 ohms line (looking into the 291.4
ohms line to see a match due to the reflection. Second, the
re-reflection from that discontinuity is half of what maintains the
circulating energy on the line. The other half is the discontinuity of
the non-virtual load.

You are confusing reflection with wave cancellation (destructive
interference). I suggest that you study the separate sections on
reflections vs interference in "Optics", by Hecht. Nowhere does any
optical textbook indicate that superposition and reflection are the
same thing (and they are indeed NOT the same thing).

You always fall back on the optics thing, don't you?

On Jun 24, 1:52*pm, Cecil Moore wrote:
If the reflected wave is re-reflected, it must be by an impedance
other than the virtual impedance generated by the reflected wave
itself. If the reflected wave is being used to generate a virtual
impedance, it cannot at the same time be being re-reflected.


I think i have finally seen the confusion point. you can not at the
same time talk about reflected waves and re-reflected waves and also
use s parameters or other steady state reflection parameters or
impedances... this means you can not talk about individual reflections
from the stub connection point without also analyzing the traveling
wave inside the stub, nor can you track and sum up all the reflections
using the steady state impedance of the stub as if it were a lumped
value. to do so violates the conditions where the s parameters and
reflection coefficients are valid, the sinusoidal steady state realm.

On Jun 25, 5:25*pm, John S wrote:
But, it does. First, it causes the 50 ohms line (looking into the 291.4
ohms line to see a match due to the reflection. Second, the
re-reflection from that discontinuity is half of what maintains the
circulating energy on the line. The other half is the discontinuity of
the non-virtual load.

John, EM wave reflection is what happens to a single wave. EM wave
superposition/interference is what happens between two (or more)
waves. They are not the same phenomenon and do NOT obey the same rules
of physics. Wave cancellation between two waves is *NOT a reflection*
but it does *redistribute* reflected energy back toward the load at
the Z0-match point. The destructive interference energy toward the
source is redistributed as constructive interference energy toward the
load but one can tell from the resulting phase and magnitude that it
was NOT a reflection.

A good reference on the differences between wave reflection and wave
superposition/interference is "Optics", by Eugene Hecht. The
international 4th edition is available in paperback for around $20
from Abebooks.com.
--
73, Cecil, w5dxp.com

On Jun 29, 2:13*pm, John S wrote:
You always fall back on the optics thing, don't you?

That's because optical physicists seem to be the only technical
people who understand interference effects. However, most of what one
needs to understand about the subject is contained in my Worldradio
energy article:

http://www.w5dxp.com/energy.htm
--
73, Cecil, w5dxp.com

On Jun 29, 5:41*pm, K1TTT wrote:
I think i have finally seen the confusion point. you can not at the
same time talk about reflected waves and re-reflected waves and also
use s parameters or other steady state reflection parameters or
impedances...

Someone said it a long time ago: You can't have your cake and eat it
too.

One cannot be using the reflected wave to establish a dynamic virtual
short while, at the same time, the virtual short is causing a dynamic
re-reflection of the reflected wave.

The virtual impedance is (Vfor+Vref)/(Ifor+Iref) but if that is the
impedance doing the re-reflection of that same Vref and Iref, then it
cannot also be being used to establish the dynamic virtual impedance,
i.e. if a virtual impedance is re-reflecting all of the reflected
energy, it will necessarily disappear from existence because there is
no longer a Vref and Iref to cause it to exist.
--
73, Cecil, w5dxp.com

On Jul 1, 7:56*am, W5XP wrote:

My apologies to JAMES W GRIFFITH, W5XP. I apparently made a typo and
left out a 'D' from my call.
--
73, Cecil, w5dxp.com

"W5DXP" wrote in message
...

On Jul 1, 7:56 am, W5XP wrote:

My apologies to JAMES W GRIFFITH, W5XP. I apparently made a typo and
left out a 'D' from my call.
--
73, Cecil, w5dxp.com
****************
-
LOL....I saw the callsign, knew something was wrong, but didn't know what.
......that happens at my age
--Wayne
W5GIE (exiled to W6)