Constructive interference in radiowave propagation
 

On Apr 9, 11:49 pm, Cecil Moore wrote:
Jim Kelley wrote:
You can't get 400 watts to a load using two 100 watt PA's no matter how
much interfering and averaging you do, Cecil. How can you even make
that claim with a straight face?

I didn't make that claim. I said that Keith would have to prove
it was possible before his assertions made any sense. I challenged
Keith to prove that was possible. I certainly have never made that
claim. You usual lack of ethics is showing. If you are forced to
lie to make your point, your point is not worth making.

It would be highly valuable if you would point to any step in
my solutions that require that 100 + 100 equal 400. I certainly
have never stated this.

I know I have previously offered solutions that you have declared
incorrect but you have been unwilling to point to the error.

I have also offered to check your solution but you have never
offered one, apparently because your analysis techniques are
inadequate to solve the problem.

....Keith

PS Should you wish to try, recall that the problem was a signal
generator with a 450 Ohm output impedance driving a 450 Ohm line
connected to a 75 Ohm load. What is the magnitude of the re-reflected
signal at the generator? That is, what is the magnitude of the ghost
created by the reflected signal? (And you can't make changes to the
problem to solve it.)

Constructive interference in radiowave propagation
 

On Apr 10, 12:26 am, Cecil Moore wrote:
Jim Kelley wrote:
You can't get 400 watts to a load using two 100 watt PA's no matter how
much interfering and averaging you do, Cecil. How can you even make
that claim with a straight face?

I didn't make that claim. I challenged Keith to prove it
was possible. Only then could he claim that sources
obey the rules of superposition.

But let's talk about superposing two 100 watt waves
in a 50 ohm transmission line. Never mind exactly
how it is accomplished for now.

Wave1 = 70.7V at 0 deg and is 70.7^2/50 = 100 joules/sec
Wave2 = 70.7V at 0 deg and is 70.7^2/50 = 100 joules/sec

Please superpose those two coherent waves within a 50
ohm transmission line and see what answer is obtained
for total voltage and total power.

Hint: Power = V^2/Z0

Oh yes, let us do this.

70.7 + 70.7 - 141.4
141.4**2/50 - 400

100 + 100 - 200

But 200 is not equal to 400.

So which was the invalid step?

Adding the voltages together? (Might mean superposition is invalid?)
Or adding the powers together? (Might mean no conservation of
energy?)
Or perhaps V**2/Z0 is invalid? (Maybe this is the best choice?)
Or 200 is equal to 400? (This would be messy!)

For full marks, explain your answer.

....Keith

Constructive interference in radiowave propagation
 

On Apr 9, 9:50 pm, Cecil Moore wrote:
K7ITM wrote:
So, for example, if I send 50 watts of a sinusoid down a 50 ohm line,
and there's a transition to a 291.4 ohm line that's half a wave long
at the sinusoid's frequency, terminated in 50 ohms, there's no
reflected power on the 50 ohm line. Cool. I knew that.

Make that no *NET* reflected power. Two reflected waves had
to engage in wave cancellation for there to be no net
reflected power. RF waves respond to real-world physical
impedance discontinuities.

In re-
reading what Roy wrote, I see NO disagreement with that. But in the
291.4 ohm line, there's 100 watts forward and 50 watts reverse. At
the interface between the two lines, there's a total of 100 watts
coming in: 50 from the 50 ohm line and 50 from the 291 ohm line. And
wonder of wonder, there's 100 watts going out; it happens to all be in
the 291 ohm line.

Let's analyze that example:

50W--50 ohm line--+--1/2WL 291.4 ohm line---50 ohm load
Pfor1=50w-- Pfor2=100w--
--Pref1=0 --Pref2=50w

You are very comfortable with this style of example, you analyze it
well,
and the numbers confirm your expectation.

But a good theory works for more than just one example.

Why don't you analyze my example:

Generator with
291.4 Ohm ------------ 291.4 Ohm line ---- 50 Ohm load
Source Impedance Pfor2=100w--
--Pref2=50w

What is the power emitted by the generator into the line?
Where does Pref2=50W go?
Are there ghosts?
What is the magnitude?
What power would the generator emit if the line was terminated
with 291.4 Ohms?

Please do not modify the example for analysis since this may change
the results.

There is much to be learned by trying examples that may challenge
your expectations.

....Keith

Constructive interference in radiowave propagation
 

This thread has been a hoot - I just about snorted my coffee out my
nose a couple of times, reading this in the early AM...

Just to muddy the waters nicely, here is an excellent demonstration of
something for nothing, i.e. faster than light propagation by
constructive wave interference called "supraluminal propagation"..

http://gregegan.customer.netspace.ne...ETS/20/20.html

Lets see, for starters assume that the constructively interfered wave
is propagationg at 1.2C it means that E = ( MC^2 )^1.2 right????

denny
gawd, I can't wait to see how you guys handle this - worm holes,
anyone?

Constructive interference in radiowave propagation
 

Keith Dysart wrote:
It would be highly valuable if you would point to any step in
my solutions that require that 100 + 100 equal 400. I certainly
have never stated this.

I went over this before. In order for a source to
obey the rules for superposition, it must be linear
enough to engage in constructive interference. That
means it must be able to double its voltage and double
its current at the same time in order to supply the
total constructive interference energy requirements
required by superposition of coherent waves. No
commercially available amateur radio transmitter that
I know of will do that. Therefore, your simple proposed
method will not work on the average amateur radio
transmitter because it is NOT LINEAR. All you can do
is invent sources in your head. I'll bet you can even
leap tall buildings in a single bound, in your head.

I have also offered to check your solution but you have never
offered one, apparently because your analysis techniques are
inadequate to solve the problem.

Yes, they are as are everyone else's, including yours.
That's why there has *never been a solution* and the
argument continues to rage after decades. If anyone
had ever figured it out, we wouldn't be having this
discussion.

You and I are looking at the moon. You assert that it
is 1000 miles away. I say that's wrong. You ask me how
far away is it? I answer that I don't know but I know
it is not 1000 miles away. You say that since I don't
know how far away the moon is then it has to be 1000
miles away. That's where our present argument stands.
--
73, Cecil http://www.w5dxp.com

Constructive interference in radiowave propagation
 

Keith Dysart wrote:
On Apr 10, 12:26 am, Cecil Moore wrote:
Jim Kelley wrote:
You can't get 400 watts to a load using two 100 watt PA's no matter how
much interfering and averaging you do, Cecil. How can you even make
that claim with a straight face?
I didn't make that claim. I challenged Keith to prove it
was possible. Only then could he claim that sources
obey the rules of superposition.

But let's talk about superposing two 100 watt waves
in a 50 ohm transmission line. Never mind exactly
how it is accomplished for now.

Wave1 = 70.7V at 0 deg and is 70.7^2/50 = 100 joules/sec
Wave2 = 70.7V at 0 deg and is 70.7^2/50 = 100 joules/sec

Please superpose those two coherent waves within a 50
ohm transmission line and see what answer is obtained
for total voltage and total power.

Hint: Power = V^2/Z0

Oh yes, let us do this.

70.7 + 70.7 - 141.4
141.4**2/50 - 400

100 + 100 - 200

But 200 is not equal to 400.

So which was the invalid step?

There was no invalid step. The two coherent waves engaged in
total constructive interference just as they do in a Z0-matched
transmission line in the direction of the source. The extra
200 watts is supplied by total destructive interference at
the Z0-match in the direction of the source. 200 watts of
reflections have been eliminated toward the source and join
the forward power wave toward the load. In ham terms, all
the reflected power has been re-reflected.

The S-Parameter equation yields the same results where
b2 is the normalized voltage flowing toward the load.

s21(a1) = 70.7V/SQRT(50) = 10 [s21(a1)]^2 = 100W

s22(a2) = 70.7V/SQRT(50) = 10 [s22(a2)]^2 = 100W

b2 = s21(a1) + s22(a2) = 141.4V/SQRT(50) = 20

|b2|^2 = 20^2 = 400W = the forward power

constructive & destructive interference is built right
into the S-Parameter analysis. Forward and reflected
waves are built right into the S-Parameter analysis.

There was no invalid step.
--
73, Cecil http://www.w5dxp.com

Constructive interference in radiowave propagation
 

Keith Dysart wrote:
Why don't you analyze my example:

Generator with
291.4 Ohm ------------ 291.4 Ohm line ---- 50 Ohm load
Source Impedance Pfor2=100w--
--Pref2=50w

Any example where reflected energy is allowed to
reach the source cannot be analyzed in any valid
real-world way. That is what the decades-old
argument between experts is all about. There has
never been a valid way to analyze such a real-world
system. Of course, you can analyze this system in
your mind but real-world results with real-world
ham transmitters will prove your analysis invalid.

If there existed a valid way to analyze real-world
ham transmitters with incident reflections, everyone
would be using that method but, so far, bench
measurements prove that no valid analysis exists.
And so the argument continues to wage on. You
are not going to resolve it with a simple-minded
model that exists only in your mind and not in
reality.

Replace the above purely imaginary generator with an
IC-706 and get back to us.
--
73, Cecil http://www.w5dxp.com

Constructive interference in radiowave propagation
 

Cecil Moore wrote:
There was no invalid step. The two coherent waves engaged in
total constructive interference just as they do in a Z0-matched
transmission line in the direction of the source.
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Darn, that should have been "in the direction
of the load".
--
73, Cecil http://www.w5dxp.com

Constructive interference in radiowave propagation
 

On Apr 10, 10:00 am, Cecil Moore wrote:
Keith Dysart wrote:
On Apr 10, 12:26 am, Cecil Moore wrote:
Jim Kelley wrote:
You can't get 400 watts to a load using two 100 watt PA's no matter how
much interfering and averaging you do, Cecil. How can you even make
that claim with a straight face?
I didn't make that claim. I challenged Keith to prove it
was possible. Only then could he claim that sources
obey the rules of superposition.

But let's talk about superposing two 100 watt waves
in a 50 ohm transmission line. Never mind exactly
how it is accomplished for now.

Wave1 = 70.7V at 0 deg and is 70.7^2/50 = 100 joules/sec
Wave2 = 70.7V at 0 deg and is 70.7^2/50 = 100 joules/sec

Please superpose those two coherent waves within a 50
ohm transmission line and see what answer is obtained
for total voltage and total power.

Hint: Power = V^2/Z0

Oh yes, let us do this.

70.7 + 70.7 - 141.4
141.4**2/50 - 400

100 + 100 - 200

But 200 is not equal to 400.

So which was the invalid step?

There was no invalid step. The two coherent waves engaged in
total constructive interference just as they do in a Z0-matched
transmission line in the direction of the source. The extra
200 watts is supplied by total destructive interference at
the Z0-match in the direction of the source. 200 watts of
reflections have been eliminated toward the source and join
the forward power wave toward the load. In ham terms, all
the reflected power has been re-reflected.

The S-Parameter equation yields the same results where
b2 is the normalized voltage flowing toward the load.

s21(a1) = 70.7V/SQRT(50) = 10 [s21(a1)]^2 = 100W

s22(a2) = 70.7V/SQRT(50) = 10 [s22(a2)]^2 = 100W

b2 = s21(a1) + s22(a2) = 141.4V/SQRT(50) = 20

|b2|^2 = 20^2 = 400W = the forward power

constructive & destructive interference is built right
into the S-Parameter analysis. Forward and reflected
waves are built right into the S-Parameter analysis.

There was no invalid step.

Now just above, you started with 100W + 100W and ended with
400W. And you wonder why readers think you advocate this
position.

Does this not cause you some discomfort? It clearly
violates conservation of energy.

For me, it is pretty clear. If you start with 200W, you can
not end up with 400W.

....Keith

Constructive interference in radiowave propagation
 

On Apr 10, 10:16 am, Cecil Moore wrote:
Keith Dysart wrote:
Why don't you analyze my example:

Generator with
291.4 Ohm ------------ 291.4 Ohm line ---- 50 Ohm load
Source Impedance Pfor2=100w--
--Pref2=50w

Any example where reflected energy is allowed to
reach the source cannot be analyzed in any valid
real-world way.

A strange assertion. Consider two wire phone lines; transmitter
and receiver at each end. Consider cable modems; ditto. Consider
computer busses; ditto.

I am sure you will find a "good" reason why either these are
special cases or amateur RF is a special case and therefore
something does not apply.

Oh well.

....Keith