Analyzing Stub Matching with Reflection Coefficients
 

Cecil Moore wrote:
Richard Harrison wrote:
Obviously, where vectors are in-phase they add and where they are
out-of-phase they subtract.

In fact, Jim Kelley's assertion that there is no interaction
between waves would result in isotropic radiation in the
far field of every antenna if one went out far enough to
measure the waves after they are propagating free of each
other. I wonder if NASA knows that?

Cecil,

You don't believe in superposition, do you? It is discussed in lots of
books if you want to understand.

8-)

73,
Gene
W4SZ

Analyzing Stub Matching with Reflection Coefficients
 

Jim Kelley wrote:
Cecil Moore wrote:
But they can act upon one another, Jim. The Florida State web
page says so. The Melles-Groit web page says so.

No they don't. If the waves themselves changed, then their resultant
superposition would also change. It's a completely unfounded notion,

If what you say is true, then if we measure field strengths
far enough away from an antenna to get outside the range
of interference, then all antennas are isotropic. Why don't
you call up NASA and tell them that permanent constructive
interference doesn't exist and they might as well be
using isotropic antennas?

It says their
energy components are redistributed.

Which is not the same as saying waves have an effect on other waves. I
said I didn't expect you to understand, and clearly you don't.

Well then, please explain it to me.
Here's the s-parameter equation for wave cancellation
in the b1 direction.

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

s11, a1, s12, and a2 are all real measured values. b1 is a
real measured value. All of the measured values are perfectly
consistent. Exactly how did b1 get to be 0 without s11(a1) and
s12(a2) canceling each other out?

What do *you* get when you add one volt at 0 deg to one volt
at 180 deg when they are coherent and traveling in a collinear
path in a transmission line? Assuming EM waves, a value of zero
tells us that wave cancellation has occurred. So what value do
you get?

As I said, I don't expect you to understand, and clearly here you don't.

You are a broken record, Jim, mindlessly uttering mantras to
cover up your inability to comprehend reality. I'm beginning
to understand what Roy meant by his "academic" statement.

Just as illustrated on the Florida State web page, when coherent
waves are also collinear, as they are in a transmission line, they
merge into the total wave and cease to exist as separate wave
components.

Yes, it very effectively shows how 1 + -1 = 0. Very profound, Cecil.

That, my friend, is permanent destructive interference, in
the flesh, as it were. One joule/second at 0 degrees plus
one joule/second at 180 degrees is indeed 0 joules/sec in
the direction of original travel of those two waves in a
transmission line. What happens after that is a two
joule/second reflection in the opposite direction away from
the impedance discontinuity that is causing the reflections
and permanent interference.
--
73, Cecil http://www.w5dxp.com

Analyzing Stub Matching with Reflection Coefficients
 

Jim, AC6XC wrote:
"Virtual impedance discontinuities do not cause reflections."

Reflection is a change in direction. On a transmission line a complete
reflection is caused by a physical open or short on the line.

Current is interrupted at an open circuit. Energy in the magnetic field
at the interruption collapses generating a voltage which doubles the
line voltage at the open. This causes the current direction to reverse
(a reverse in its phase) while not changing the phase of the voltage.

Conditions necessary for the reversal in travel direction of an EM wave
are to reverse either the magnetic or electric wave`s polarization, but
not to change both.

At an actual short on a line, volts are forced to zero at the short.
Current doubles at the short, and the voltage wave reverses its
polarization. The EM wave reverses its travel direction at the short as
it did in the case of an open circuit.

When volts or amps compete against an opponent of half their magnitude,
the stronger opponent wins. So, it`s volts or amps which determine which
direction a wave travels on a transmission line at a discontinuity.

At a short or an open on a line , it is the current or voltage the
discontinuity generates which turns the wave around. The line doesn`t
care how the amps or volts came to suddenly appear at the turnaround
point. If a virtual condition can generate the energy surge or
escalation needed for a reversal in direction, it is as acceptable as a
real discontinuity, in my opinion.

Best regards, Richard Harrison, KB5WZI

Analyzing Stub Matching with Reflection Coefficients
 

Roy Lewallen wrote:
Any of the proponents of "virtual discontinuity" reflections up to it?

Would you please explain if there is any energy associated
with your "inverted reflected pulse" and your "non-inverted
non-reflected pulse" or do they exist devoid of energy?
--
73, Cecil http://www.w5dxp.com

Analyzing Stub Matching with Reflection Coefficients
 

Alan Peake wrote:
"I was trying to underline that a stub puts a physical discontinuity on
the TL which will give you a real reflection."

Short-circuited 1/4-wave stubs have long been used as "metal insulators"
to support transmission lines. This would likely be impractical if the
stubs produced a discontinuity on the line at the operatibg frequency.
The number of stubs seems without limit also. Put enough together and
you`ve constructed rectangular waveguide.

Best regards, Richard Harrison, KB5WZI

Analyzing Stub Matching with Reflection Coefficients
 

Gene Fuller wrote:
You don't believe in superposition, do you? It is discussed in lots of
books if you want to understand.

Do you believe Jim's argument that two coherent EM
waves of equal magnitudes and opposite phases traveling
collinearly in the same direction in a transmission line
can never be canceled? If Jim is right, we can toss the
s-parameter analysis in the garbage can and join Roy in
calling it gobbledigook (sic).
--
73, Cecil http://www.w5dxp.com

Analyzing Stub Matching with Reflection Coefficients
 

Hi Alan -

The point of using pulses is that their width is short, ....


Hi Jim,
While the individual pulses are short, for them to simulate CW, they
must be right next to each other. Obviously, the pulses that cancel or
reinforce aren't the same ones, as Roy points out, and the CW which
appears to reflect from the virtual discontinuity is the sum of all
pulses, whether they are from the stub point, the other end of the stub
or the load end of the TL.

Alan

Analyzing Stub Matching with Reflection Coefficients
 

If you think of CW as a series of pulses, a "virtual short" occurs only
when an inverted reflected pulse arrives at the same point at the same
time as a non-inverted non-reflected pulse, causing the two to add to
zero. (Or, of course, more complex combinations of multiple pulses
arriving at the same point.) It isn't the same pulse which appears twice
to interfere with itself; it's different pulses of the pulse string,
sent at different times but arriving at the same point simultaneously
due to one being delayed by reflection and the other not. So you see,
the interval between those pulses is critical; if it changes, then the
location of the "virtual short" changes. This is analogous to the steady
state CW situation where the location of the "virtual short" changes
with frequency.

Hi Roy,
No disagreement with that. A single pulse will have reflections from the
open end of the stub, the load end of the TL (if it's not Zo) and the
stub attachment point. In a TDR, the first return will be from the
latter point but it won't be of the same magnitude and sign as it would
from a short. That requires all the pulses which are simulating (if you
like), the CW. It will be whatever you get when the pulse, which has
been happily travelling along a TL of Zo, meets a point where there are
now two TLs of Zo in parallel. I consider this to be a discontinuity and
will produce a real reflection albeit not, as I said, the same as that
from a virtual short etc. So, in the sense that the quarter wave stub
appears as a virtual short, I agree that the actual reflection to cause
this comes from the open end of the stub.

In theory, you could prove that reflection isn't occurring from a
"virtual discontinuity" by making an abrupt change in the excitation,
for example abruptly changing its level, then noting that the effect of
the change isn't seen back at the input until it propagates through the
"virtual discontinuity", on to physical discontinuities where reflection
actually takes place, and back. This might be difficult to do in
practice, though, except with some fairly sophisticated equipment or
very long lines because of the time intervals involved.

Yes, you'd have to have a stub of 1/4 wave plus N 1/2 waves to see this
effect. If N was large enough to see the excitation change (you could
probably use a "CW pulse" a la radar to do this) I would expect to see
some return from the stub attachment point, then the reflection from the
open end of the stub.

But let's suppose that you did somehow prove that a "virtual
discontinuity" reflects waves. Then you have to explain the mechanism by
which waves alter each other in a linear medium.

Well, I'm a believer in superposition so I'll leave that one to others :)

Analyzing Stub Matching with Reflection Coefficients
 

Alan Peake wrote:
[I wrote]
But let's suppose that you did somehow prove that a "virtual
discontinuity" reflects waves. Then you have to explain the mechanism
by which waves alter each other in a linear medium.

Well, I'm a believer in superposition so I'll leave that one to others :)

Those others have made, I'm sure, over a thousand postings so far, and
yet in them there's a complete lack of any explanation of the mechanism.

Roy Lewallen, W7EL

Analyzing Stub Matching with Reflection Coefficients
 

Jim Kelley wrote:
Cecil Moore wrote:

Jim Kelley wrote:
I said it because waves do not, according to the definition of the
word, 'act upon one another'.

But they can act upon one another, Jim. The Florida State web
page says so. The Melles-Groit web page says so.

No they don't. If the waves themselves changed, then their resultant
superposition would also change. It's a completely unfounded notion,
Cecil.

Here's an example of that "unfounded notion". Please
point out my error.

In the following example, the 100W signal generator
is equipped with a circulator load. The system is
Z0-matched during steady-state so b1 = 0 during
steady-state.

100W
SGCL--50 ohm line--x--1/2WL 291.4 ohm line--50 ohm load
a1-- b2--
--b1 --a2

b1 = s11(a1) + s12(a2) b2 = s21(a1) + s22(a2)

Let t0 be the time at which the 100W forward wave
reaches point 'x' for the first time.

Just after after t0, the source signal has split
into two parts. There are as yet, no reflections,
so a2=0. Every one of these three voltages can be
measured as real. These values remain constant
throughout steady-state.

x
a1=10----|
|----s21(a1)=5 toward the load
s11(a1)=5----|

Just after t0, b1=5. During steady-state, b1=0.

Please explain how b1 goes from 5 to 0 during the
transient build-up state without having s11(a1)
interact with anything. There, in a nutshell,
is your technical and logical contradiction.
--
73, Cecil http://www.w5dxp.com