Cecil, W5DXP wrote:
"Antenna gain over isotropic is an application of constructive
interference."

Yes. An often offered annalog is an inflated spherical balloon. It
contains the same amount of air no matter how it is squeezed. Sqeeze it
one place and it bulges elsewhere.

An isotropic antenna, could one be constructed, would radiate equally
well in all directions. As a radiation pattern becomes lopsided, the
bulge is filled with the energy squeezed from elsewhere.

Directive gain of an antenna is a power ratio. It`s the power that you
would have to put into an isotropic wersus the power you have to put
into the gain antenna to lay the same signal on a point in the preffered
direction.

Other things equal, if a gain antenna radiates twice the power in the
preferred direction as an isotropic, it has a gain of 3 dBi.

Best regards, Richard Harrison, KB5WZI

Cecil Moore wrote:
Jim Kelley wrote:
Then your observation was unrelated to the topic of discussion, which
was your claim that 2 Joules per second could be obtained from a 1 watt
laser.

You know that I never claimed that, Jim. As bright a guy
as you are, why do you have to stoop to falsifying what
I have said?

So now you start with the accusations, and here's why (from
yesterday):

"If it were total constructive interference, two 1/2W
beams would yield an intensity of 2 watts."

It was pointed out to you that that notion violates conservation of
energy, but rather than admit that you were incorrect, you attempt to
revise history. Then there was this:

Hint: (2 watts/in^2 + 0 watts/in^2)/2 in^2 = 1 watt

:-) Actually, it's a beam of light that subtends a solid angle of
less than one degree. Let's call it a degree just to be
conservative. That would leave 719 other solid degrees where there is
0 watts, so according to Cecil's theory of spacial power averaging,
the answer could also be 2 watts+0+0+0... / 719 = 2.78 milliwatts.
But apparently those values don't give the answer you were going for.

Others have made points along these lines quite eloquently. Excellent
comments in this thread from Roy, Keith, Owen, Gene, Tom, et al.

73, Jim AC6XG

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

100W + 100W is not all we started with. Somewhere
else is 100W + 100W of destructive interference
that adds up to zero.

This is very much like the equation in Born and Wolf.

Itotal = 4I1 where I1 = I2 = 100 watts/unit-area

Such is the nature of constructive interference. That
you are ignorant of such is noted.

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

Absolutely NOT! There is 200 watts of destructive
interference somewhere else. Here is a real-world
example:

---291.4 ohm line---+---1/2WL 50 ohm line---291.4 ohm load
Pfor1=200W-- Pfor2=400W--
--Pref1=0W --Pref2=200W

On the load side of point '+':
P1 = Pfor1(1-rho^2) = 100W
P2 = Pref2(rho^2) = 100W

P1 + P2 + 2*SQRT(P1*P2) = Pfor2 = 400W
(200W of constructive interference)

On the source side of point '+':
P3 = Pfor1(rho^2) = 100W
P4 = Pref2(1-rho^2) = 100W

P3 + P4 - 2*SQRT(P3*P4) = Pref1 = 0W
(200W of destructive interference)
--
73, Cecil http://www.w5dxp.com

Keith Dysart wrote:
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.

We are discussing ham radio sources, Keith, which are
none of the above. How many hams use cable modems for
their RF transmissions? Please get real.
--
73, Cecil http://www.w5dxp.com

Jim Kelley wrote:

w5dxp wrote:
If it were total constructive interference, two 1/2W
beams would yield an intensity of 2 watts.

That should have been 2 watts per unit-area. Note that
I did NOT say two 1/2W beams would yield a power of 2
watts. The correct dimensions of intensity would be
2 watts/unit-area, a power density, not a power. I have
already apologized for that goof in dimensions.

It was pointed out to you that that notion violates conservation of
energy, but rather than admit that you were incorrect, you attempt to
revise history. Then there was this:

No, that notion *DOES NOT* violate the conservation of energy
principle which cannot be violated. That notion relies upon
one watt per unit area of total destructive interference
occurring somewhere else. The one watt of destructive
interference causes a flat black area. The energy from that
flat black area is redistributed to the area of constructive
interference and causes that area to be twice as bright as
the average area. Born and Wolf's equation is valid.

Itotal = 4I1 where I1 = I2 = 0.5W

2 watts/unit-area = 4(0.5) watts/unit-area

Again, I ask you to please cease from trying to twist my
words into a violation of the conservation of energy
principle. If you think I have uttered such words, you
are mistaken. Next time you are confused, instead of your
arrogant ASSumptions, please just ask me what I meant.
--
73, Cecil http://www.w5dxp.com

"K7ITM" wrote in
ps.com:

Hi Owen,

I had a quick look at your article. Though I didn't try to proof-read
it for accuracy, I was reminded that the equations I posted those long
years ago said that if you know the _instantaneous_ voltage and
current at a point on a line, and know its impedance (as a frequency-
independent quantity), the equations apply, and you can resolve that
instantaneous pair of values into forward and reverse. That's
something that's not immediately obvious when people think only about
sine waves.

Tom,

Something else that follows from the derivation is that whilst the
indicated Pf and Pr do not have stand along meaning, Pf-Pr does have
meaning irrespective of the nominal R for which the instrument is
calculated.

For example, if we cascade a 100W source, 50 ohm directional wattmeter, a
75 ohm directional wattmeter and a 100+j0 load, the instrument readings
should be:
- 75 ohm: Pf=112.4, Pr=12.4, P=100
- 50 ohm: Pf=104.1, Pr=4.1, P=100

This of course assumes that the instruments do not significantly disturb
the thing they are measuring, in this case the V/I conditions at the 100
ohm load.

So, while you can nominate any reference Zo for a Pf or Pr value (and so
vary those values), the power passing the instrument (Pf-Pr) is indicated
correctly irrespective of the calibration R.

(The article explains that the result of Pf-Pr is only meaningful if the
calibration impedance is purely real.)

Owen

On Apr 10, 4:13 am, "Keith Dysart" wrote:



Why don't you analyze my example:

(Since the Grand PoohBah Power Master seems unwilling, here, for your
consideration, I give you...)


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?
-- Assuming lossless line, 50W of course; 50V@1A at the load.

Where does Pref2=50W go?
-- Into satisfying the boundary conditions at the generator
to line interface. We have way to little info to
determine if the generator dissipates more, or less, or
the same, as if it were terminated in 291.4 ohms.

Are there ghosts?
-- Some people believe in them. I've never seen on--Oh not
THAT kind of ghost. Well, not at the load, but certainly
at the generator, if the generator is putting out a signal
that's an interesting enough function of time. TDRs really
do work.

What is the magnitude?
-- Of what? We have enough info to resolve Vf, If, Vr and Ir
on the line, and if we knew the line length and
the excitation, we could be a lot more definitive about
things, but since steady-state excitation doesn't
produce ghosts... but the reflection coefficient at the
line-load interface is -1/sqrt(2), so that's the voltage
ratio we'll see between the forward and reverse waves.

What power would the generator emit if the line was terminated
with 291.4 Ohms?
-- "100 watts, of course." For those who don't immediately see
that, it's not difficult to go through some math to show that
the power delivered to a load is independent of the load
impedance so long as magnitude((Zload-Rgen)/(Zload+Rgen)) is
constant--that is, so long as the magnitude of the reflection
coefficient is constant--as it is along a lossless line...
and from that, find the Thevenin or Norton equivalent of the
source in this example ... and from that, figure the power
that source will deliver to a matched load.

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.

:-) This reminds me of some thoughts I posted a long time ago about
lines whose Zo is somewhat reactive. For example, if a linear
sinusoidal source of impedance Zo is connected to a line also of
impedance Zo, what load maximizes the power in the load? If you keep
magnitude((Zload-Zo)/(Zload+Zo)) constant, is the power dissipated in
the load independent of the phase angle of (Zload-Zo)/(Zload+Zo)?

...Keith

Cheers,
Tom

On Apr 10, 4:46 pm, Cecil Moore wrote:
Keith Dysart wrote:
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.

We are discussing ham radio sources, Keith, which are
none of the above. How many hams use cable modems for
their RF transmissions? Please get real.

Are you sure there are different rules for ham radio sources
than for all the other ones? Something different about them?
Something that makes them not amenable to the techniques used
for others?

If that is so, I am surprised at your claim that ham radio
sources have something in common with light bulbs and lasers
such that knowledge of those subjects can help analysis, while
any knowledge of the behaviour of other electrical circuits is
for naught.

Seriously though, it does all work. The problems are solvable.
You don't need to throw up your hands and say: "too tough."
And then rationalize this response by calling them "too
different" or saying "insufficient information".
Much can be learned by solving the problems set using more
controlled examples. Try it. Using the regular techniques will
produce the same answer for all the problems you can currently
solve, as well as allowing you to solve ones you currently
declare as unsolvable.

....Keith

On Apr 10, 5:30 pm, "K7ITM" wrote:
On Apr 10, 4:13 am, "Keith Dysart" wrote:

Why don't you analyze my example:

(Since the Grand PoohBah Power Master seems unwilling, here, for your
consideration, I give you...)



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?

-- Assuming lossless line, 50W of course; 50V@1A at the load.

Where does Pref2=50W go?

-- Into satisfying the boundary conditions at the generator
to line interface.

This is a different way of expressing it. But I like it.

We have way to little info to
determine if the generator dissipates more, or less, or
the same, as if it were terminated in 291.4 ohms.

Are there ghosts?

-- Some people believe in them. I've never seen on--Oh not
THAT kind of ghost. Well, not at the load, but certainly
at the generator, if the generator is putting out a signal
that's an interesting enough function of time. TDRs really
do work.

What is the magnitude?

-- Of what?

I was think of the ghost, for those who believe in them, mostly
to force some computation. If there is one, it must be quantifiable.

We have enough info to resolve Vf, If, Vr and Ir
on the line, and if we knew the line length and
the excitation, we could be a lot more definitive about
things, but since steady-state excitation doesn't
produce ghosts... but the reflection coefficient at the
line-load interface is -1/sqrt(2), so that's the voltage
ratio we'll see between the forward and reverse waves.

What power would the generator emit if the line was terminated
with 291.4 Ohms?

-- "100 watts, of course." For those who don't immediately see
that, it's not difficult to go through some math to show that
the power delivered to a load is independent of the load
impedance so long as magnitude((Zload-Rgen)/(Zload+Rgen)) is
constant--that is, so long as the magnitude of the reflection
coefficient is constant--as it is along a lossless line...
and from that, find the Thevenin or Norton equivalent of the
source in this example ... and from that, figure the power
that source will deliver to a matched load.

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

:-) This reminds me of some thoughts I posted a long time ago about
lines whose Zo is somewhat reactive. For example, if a linear
sinusoidal source of impedance Zo is connected to a line also of
impedance Zo, what load maximizes the power in the load? If you keep
magnitude((Zload-Zo)/(Zload+Zo)) constant, is the power dissipated in
the load independent of the phase angle of (Zload-Zo)/(Zload+Zo)?

Hmmmmmm. I'll have to think on this. Perhaps after I work out whether
the source impedance of a properly tuned amateur transmitter is the
complex conjugate of the load impedance.

....Keith

Keith Dysart wrote:
Are you sure there are different rules for ham radio sources
than for all the other ones? Something different about them?
Something that makes them not amenable to the techniques used
for others?

You are using the rules of superposition in your examples.
I don't know if the rules of superposition apply to those
other sources but I do know that a ham transmitter, like
my IC-706, is not linear enough to abide by the rules of
superposition. How is superposition supposed to handle
foldback?

Ham transmitters cannot willy-nilly be shorted and opened
in order to ascertain their linear model characteristics.

A signal generator equipped with a circulator load solves
all the experimental problems but doesn't act like a ham
transmitter.

Seriously though, it does all work. The problems are solvable.

If that is true, why hasn't anyone ever solved them,
published the results, and ended the arguments? The
"solutions" produce different results depending upon
whose brain is being used. Nobody has ever *solved*
the problem and therefore the argument still continues
to rage.
--
73, Cecil http://www.w5dxp.com