Superposition
 

K7ITM wrote:
On Nov 17, 4:03 pm, Cecil Moore wrote:
The waves are launched by the external reflection from a
Z0-match and the internal reflection from the load.

So the waves are going opposite directions along the line??

No, all reflections travel toward the source and
therefore, are traveling in the same direction.
Their Poynting vectors are all toward the source.
Given the following Z0-match impedance discontinuity
in a transmission line with the source to the left
and the load to the right:

Z0-match
------Z01---+---Z02------
Pfor1-- Pfor2--
--Pref1=0 --Pref2

The power reflection coefficient is
rho^2 = [(Z02-Z01)/(Z02+Z01)]^2

Pref1 is a combination of two reflected waves

1. P1 = Pfor1(rho^2) "the external reflection from the
Z0-match"

2. P2 = Pref2(1-rho^2) "the internal reflection from
the load"

Pref1 = P1 + P2 + 2*SQRT(P1*P2)cos(A)

Pref1 equals zero at a Z0-match so P1+P2 and A=180 deg.
--
73, Cecil http://www.w5dxp.com

Superposition
 

Antonio Vernucci wrote:
The extra power measured at the receiver is obviously "created" at the
expense of power taken away from other regions of the space (according
to the transmit antenna pattern).

Too fundamental to deserve further discussions!

Almost everyone knows what occurs in free space -
constructive interference in the direction of
greater gain and destructive interference in the
direction of lesser gain. But my posting was not
about free space. I thank you for your input so
far but please now extend those EM wave concepts
to transmission lines.

Everyone doesn't agree that constructive and
destructive interference also happens at a Z0-
match point in a transmission line with reflections.
That is the topic that needs "further discussions".

Just as constructive interference functions to
increase antenna gain in one direction while
destructive interference functions to decrease
antenna gain in another direction, in a transmission
line at a Z0-match point, constructive interference
functions to increase the energy flow toward the
load while destructive interference functions to
decrease the energy flow toward the source.

Antonio, please don't bow out now. You are apparently
one of the few posters who fully understands interference.
--
73, Cecil http://www.w5dxp.com

Superposition
 

On Sun, 18 Nov 2007 12:05:05 -0500, "Stefan Wolfe"
wrote:


"Richard Clark" wrote in message
.. .
On Sun, 18 Nov 2007 00:16:47 -0500, "Stefan Wolfe"
wrote:

Let's rewind up that list of charades to revisit:
Well, power is a vector quanity subject to the rules of vector
math.
and ponder the implication of a negative power, for simplicity:
P1 = 50W @ 90deg
P2 = 50W @ 270deg
what does the math reveal?

I am not trying to define power as negative with respect to zero
Negative with respect to zero? What about with respect to positive?
Or even a smaller negative! How about half negative (only 90 degrees
shift instead of 180)?
What happened to:
Well, power is a vector quanity subject to the rules of vector
math.

I was so wanting to see your solution, much less how:
Ptotal = P1 + P2 + 2*SQRT(P1*P2)cos(A)
migrated into the negated sine in:
= (50 - 0) + (50 - 0) - 2SQRT P1*P2 sine(A)


Oh God this is getting frustrating!

You are lagging the community by at least 8 to 10 postings.

http://math2.org/math/integrals/more/restrig.htm

1. See top equation. THE INTEGRAL OF A COSINE IS A SINE.

What you published above is not the integral of the system and it has
problems even in isolation.

I added a minus
sign, I shouldn't have BUT IT DOESN'T MATTER.

As so many previous mistakes don't matter either. I also note (and I
am full aware of the math) that YOUR reference (see top equation)
solves with a constant added - as it should if this were a legitimate
exercise.

My students didn't get full credit for discarding constants,
especially when they didn't have the vaguest notion of how these
constants contribute to the outcome.

Let me guess, you will say "it doesn't matter" with emphasis. :-)

I think you are marginalizing your credibility with each succeeding post in
which you dispute freshman caluclus fundamentals. Depending upon your
education base, this could be quite a blow to you. I thought you knew
math...maybe it was just me getting the wrong impression. Are you pulling
our legs?

Well, I did think I had mined this troll to completion, but another
round may yet bring more fascinating entries.

73's
Richard Clark, KB7QHC

Superposition
 

Just as constructive interference functions to
increase antenna gain in one direction while
destructive interference functions to decrease
antenna gain in another direction, in a transmission
line at a Z0-match point, constructive interference
functions to increase the energy flow toward the
load while destructive interference functions to
decrease the energy flow toward the source.

I cannot follow your reasoning as I cannot understand what is a "Z0-match
point".

In my understanding:
- if the transmission line end is mismatched, in no point impedance can be equal
to Z0.
- conversely if the transmission line end is matched, impedance is equal to Z0
in all points. But there is no reflected wave and hence no energy flow toward
the source

Then, your "Z0-match point" must be something else which I cannot figure out.

73
Tony I0JX

Superposition
 

Tony, I0JX wrote:
"Then, your "Zo-matchpoint" must be comething else which I cannot figure
out."

I`ll guess with Tony that in a made-up case of a 50-ohm antenna and a
50-ohm transmitter connected by a 1/2-wavelength of 300-ohm twinlead, we
have a Zo-match to 50-ohms because the twin-lead, mismatched at both
ends, still looks to source and load like 50-ohms.

Best regards, Richard Harrison, KB5WZI

Superposition
 

Antonio Vernucci wrote:
I cannot follow your reasoning as I cannot understand what is a
"Z0-match point".

A Z0-match is defined in my ARRL Antenna Book.

Then, your "Z0-match point" must be something else which I cannot figure
out.

Here is an example of a Z0-match to 50 ohms at point '+':

XMTR--50 ohm coax---+---1/2 WL 300 ohm feedline---50 ohm load

The SWR on the 300 ohm feedline is 6:1. The SWR on the 50
ohm coax is 1:1. What happens to the energy and momentum
of the reflected waves on the 300 ohm feedline? Seems
obvious that there is destructive interference toward
the XMTR and constructive interference toward the load.
--
73, Cecil http://www.w5dxp.com

Superposition
 

On Sun, 18 Nov 2007 14:28:09 -0500, "Stefan Wolfe"
wrote:

BTW the constant you saw has
nothing to do with the equation at hand since we assume zerop power for that
component at t = 0.

An unemphatic "it doesn't matter!" :-)

I could have made C =100,

Hmmm, 0 is as good as 100?

what must the value of C
be in this equation? Ha!

You are right, you have been away from math class too long. The value
of the added C as the result of integration is unknown by definition.
This constant is, after all, from your own reference supplied by you.
Plugging in any value (including your fudge factor of 0) plucked from
the air is invalid. Hardly the best way to demonstrate the
conservation of energy (power, karma, the trade balance, or
what-have-you).

73's
Richard Clark, KB7QHC

Superposition
 

A Z0-match is defined in my ARRL Antenna Book.

Sorry, I have that book, but I do not remember where it is.

Here is an example of a Z0-match to 50 ohms at point '+':

XMTR--50 ohm coax---+---1/2 WL 300 ohm feedline---50 ohm load


OK, understood.

The SWR on the 300 ohm feedline is 6:1. The SWR on the 50
ohm coax is 1:1. What happens to the energy and momentum
of the reflected waves on the 300 ohm feedline? Seems
obvious that there is destructive interference toward
the XMTR and constructive interference toward the load.

I am not sure on whether I am able to correctly interpret your statement. My
understanding is:

- reflected power does not reach the transmitter, as it is fully reflected back
toward the load
- such re-reflected power reaches the load, where it is partially absorbed (thus
contributing to the total power delivered to the load) and partially reflected
back once more

Probably this is what you call destructive interference at the trasmitter and
constructive interference at the load.

The fact that reflected power is fully re-reflected to the load does not seem to
be appreciated by everyone. Many people still attribute their transmitter power
transistors failure to reflected power burning them. The failure is instead
clearly due to malfunctioning or poor design of the SWR protection circuit that:
- does not keep the collector voltage within its maximum allowable value when
load impedance is too high
- does not keep the collector current and the junction temperature within their
maximum allowable values when load impedance is too low

73

Tony I0JX

Superposition
 

Cecil, W5DXP wrote:
"Antonio, please don`t bow out now."

Why is a principle so trivial as superposition worth a thread in this
newsgroup?

Best regards, Richard Harrison, KB5WZI

Superposition
 

Richard Harrison wrote:
Why is a principle so trivial as superposition worth a thread in this
newsgroup?

Because most of the posters to this newsgroup do not
know what happens to the energy in the waves during
superposition inside a transmission line. They seem
to understand superposition in free space but not
inside a transmission line. Maxwell's laws are the
same for EM waves in free space and inside a
transmission line.
--
73, Cecil http://www.w5dxp.com