On Feb 20, 12:22*am, Cecil Moore wrote:
Keith Dysart wrote:
Steven Best's article does have an equation of similar form:

PFTOTAL = P1 + P2 + 2 * sqrt(P1) * sqrt(P2) * cos(theta)

Starting with superposition of two voltages, he derived an
equation based on powers for computing the total power.

The derivation proves the equation to be valid.
The redundancy inside a transmission line of Z0 characteristic
impedance means we don't have to deal with voltages at all

Not quite, since the angle 'theta' is the angle between the voltages,
is it not?

if
we know the length of the transmission line and the reflection
coefficient of the load. Pfor = Vfor^2/Z0 *Pref = Vref^2/Z0

Perhaps you could carry out the calculations for a slightly
different experiment. Use a 25 ohm source impedance and a
voltage oF 70.7 RMS.

Note you are cutting the source voltage and source resistance
in half while keeping the rest of the network the same. The
forward power cannot remain at 100w.

I am pretty sure that it is 100 W. See more below.

Forward and reverse power for the short circuited load will
still be 100 W, but only 200 W will be dissipated in the
source resistor. It is not obvious to me how to compute the
interference term for this case.

Nope, forward and reverse power will not still be 100w.
Forward and reverse power will be 50w. It will be
44.44w + 4.94w + 0.549w + 0.06w + ... = 50 watts
The source is now exhibiting a power reflection coefficient
of 0.3333^2 = 0.1111 so there is a multiple reflection
transient state before reaching steady-state.

I suggest that you forgot to use Steven Best's equation when
you added the powers. Consider just the first re-reflection
where 4.94 is added to 44.4.

Pftotal = 44.444444 + 4.938272 + 2 * sqrt(44.444444) * sqrt(4.938272)
* cos(0)
= 49.382716 + 9.599615
= 79.01 W

To verify, let us consider the voltages:

Original Vf is 66.666666 V (peak)
First re-reflection is -66.666666 V * -0.3333333 - 22.222222 V (peak)
giving a new Vf of 88.888888 V (peak)
which is 79.01 W into 50 ohms.

It does, indeed, converge to 100 W forward but you have to use the
proper equations when summing the powers of superposed voltages.
You can not just add them.

So forward and reverse powers each converge to 100 W and 200 W is
dissipated in the source resistor, which suggests that your
equation for computing dissipation in the source resistor is
incorrect.

To take the tedium out of these calculation you might like the
Excel spreadsheet at
http://keith.dysart.googlepages.com/...oad,reflection

...Keith

Keith Dysart wrote:
On Feb 20, 12:22 am, Cecil Moore wrote:
The derivation proves the equation to be valid.
The redundancy inside a transmission line of Z0 characteristic
impedance means we don't have to deal with voltages at all

Not quite, since the angle 'theta' is the angle between the voltages,
is it not?

That's part of the redundancy I was talking about. If one
calculates the voltage reflection coefficient at the load
and knows the length of the transmission line, one knows
the angle between those voltages without actually calculating
any voltages. For a Z0-matched system, one only needs to
know the forward and reflected powers in order to do
a complete analysis. No other information is required.

I am pretty sure that it is 100 W. See more below.

Good grief, you are right. I wrote that posting and made
a sophomoric mistake after a night out on the town. I
should have waited until after my first cup of coffee
this morning. Mea culpa.

It does, indeed, converge to 100 W forward but you have to use the
proper equations when summing the powers of superposed voltages.
You can not just add them.

You're right - 20 lashes for me. Some day I will learn not
to post anything while my left brain is in a pickled state.

So forward and reverse powers each converge to 100 W and 200 W is
dissipated in the source resistor, which suggests that your
equation for computing dissipation in the source resistor is
incorrect.

It happened to be correct in the earlier example because
the ratio Rs/Z0 = 1.0, so the equation was correct
*for those conditions*. Rs/Z0 needs to be included when
the source resistor and the Z0 of the feedline are different.
Now that I've had my first cup of coffee this morning, let's
develop the general case equation. Note that we will converge
on the equation that I posted earlier.

50w + 50w + 2*SQRT(50w*50w) = 200 watts

The interference is not between the forward wave and reflected
wave on the transmission line. The interference is actually
between the forward wave and the reflected wave inside the
source resistor where the magnitudes can be different from
the magnitudes on the transmission line.

Note that the forward RMS current is the same magnitude at
every point in the network and the reflected RMS current
is the same magnitude at every point in the network.

Considering the forward wave and reflected wave separately,
where Rs is the source resistance:

If only the forward wave existed, the dissipation in the
source resistor would be Rs/Z0 = 1/2 of the forward power,
i.e. (Ifor^2*Z0)(Rs/Z0) = Ifor^2*Rs = 50 watts.

If only the reflected wave existed, the dissipation in
the source resistor would be 1/2 of the reflected power,
i.e. (Iref^2*Z0)(Rs/Z0) = Iref^2*Rs = 50 watts.

We can ascertain from the length of the feedline and from
the Gamma angle at the load that the cos(A) is 1.0
This is rather obvious since we know the source "sees"
a load of zero ohms.

Now simply superpose the forward wave and reflected wave
and we arrive at the equation I posted last night which I
knew had to be correct (even in my pickled state).

P(Rs) = 50w + 50w + 2*SQRT(50w*50w) = 200 watts

For the general equation: P1=Pfor(Rs/Z0), P2=Pref(Rs/Z0)

P(Rs) = P1 + P2 + 2[SQRT(P1*P2)]cos(A)

And of course, the same thing can be done using voltages
which will yield identical results. What some posters
here don't seem to realize are the following concepts
from my energy analysis article at:

http://www.w5dxp.com/energy.htm

Given any two superposed coherent voltages, V1 and
V2, with a phase angle, A, between them:

If ( 0 = A 90 ) then there exists constructive
interference between V1 and V2, i.e. cos(A) is a
positive value.

If A = 90 deg, then cos(A) = 0, and there is no
destructive/constructive interference between
V1 and V2, i.e. cos(A) = 0

If (90 A = 180) then there exists destructive
interference between V1 and V2, i.e. cos(A) is a
negative value.

Dr. Best's article didn't even mention interference and
indeed, on this newsgroup, he denied interference even
exists as pertained to his QEX article.

The failure to recognize interference between two coherent
voltages is the crux of the problem.
--
73, Cecil http://www.w5dxp.com

Cecil Moore wrote:
Given any two superposed coherent voltages, V1 and
V2, with a phase angle, A, between them:

If ( 0 = A 90 ) then there exists constructive
interference between V1 and V2, i.e. cos(A) is a
positive value.

i.e. (V1^2 + V2^2) (V1 + V2)^2

If A = 90 deg, then cos(A) = 0, and there is no
destructive/constructive interference between
V1 and V2.

i.e. (V1^2 + V2^2) = (V1 + V2)^2

If (90 A = 180) then there exists destructive
interference between V1 and V2, i.e. cos(A) is a
negative value.

i.e. (V1^2 + V2^2) (V1 + V2)^2
--
73, Cecil http://www.w5dxp.com

Keith Dysart wrote:
So forward and reverse powers each converge to 100 W and 200 W is
dissipated in the source resistor, which suggests that your
equation for computing dissipation in the source resistor is
incorrect.

I remembered what I did when I used that equation. It is
the same technique that Dr. Best used in his article. If
we add 1WL of 25 ohm line to the example, we haven't
changed any steady-state conditions but we have made the
example a lot easier to understand.

Rs 50w-- 100w--
+--/\/\/--+------------------------+------------+
| 25 ohm --50w --100w |
| |
Vs 1WL 1/2WL |Short
70.7v 25 ohm 50 ohm |
| |
+---------+------------------------+------------+

Now the forward power at the source terminals is 50w and
the reflected power at the source terminals is 50w. So the
ratio of Rs/Z0 at the source terminals is 1.0. Therefo

P(Rs) = 50w + 50w + 2*SQRT(50w*50w) = 200w

To take the tedium out of these calculation you might like the
Excel spreadsheet at
http://keith.dysart.googlepages.com/...oad,reflection

My firewall/virus protection will not allow me to download
EXCEL files with macros. Apparently, it is super easy to
embed a virus or worm in EXCEL macros.
--
73, Cecil http://www.w5dxp.com

Keith Dysart wrote:

Keith, I am preparing a web page on this subject.
Here are a couple of the associated graphics for
the earlier simple example.

http://www.w5dxp.com/easis1.GIF
http://www.w5dxp.com/easis2.GIF
--
73, Cecil http://www.w5dxp.com

Keith Dysart wrote:

Hey Keith, how about this one?

Rs Pfor=50w--
+----/\/\/-----+----------------------+
| 50 ohm --Pref |
| |
Vs 45 degrees RLoad+j0
100v RMS 50 ohm line |
| |
| |
+--------------+----------------------+

The dissipation in the source resistor is:

P(Rs) = 50w + Pref

How can anyone possibly argue that reflected power
is *never* dissipated in the source resistor? :-)
--
73, Cecil http://www.w5dxp.com

On Feb 21, 9:53*pm, Cecil Moore wrote:
Keith Dysart wrote:

Hey Keith, how about this one?

* * * * * * * *Rs * * * * * * Pfor=50w--
* * * * *+----/\/\/-----+----------------------+
* * * * *| * *50 ohm * * * * * *--Pref * * * *|
* * * * *| * * * * * * * * * * * * * * * * * * |
* * * * Vs * * * * * * * * * 45 degrees * * RLoad+j0
* * * 100v RMS * * * * * * * 50 ohm line * * * |
* * * * *| * * * * * * * * * * * * * * * * * * |
* * * * *| * * * * * * * * * * * * * * * * * * |
* * * * *+--------------+----------------------+

The dissipation in the source resistor is:

P(Rs) = 50w + Pref

How can anyone possibly argue that reflected power
is *never* dissipated in the source resistor? :-)
--
73, Cecil *http://www.w5dxp.com

Two Saturdays ago I was on a road trip and used
9 litres/100km. How can anyone argue that the fuel
consumption is never equal to the day of the month?
Numerical coincidences can be much fun.

But you will like the generator below even better.
The power dissipated in the generator resistors is
always equal to 50 + Pref, regardless of the load
and line length, thereby always accounting for Pref.

Pfor=50w--
+--/\/\/---+----------------+-------------------+
| 100 ohm | --Pref |
| | 100 ohm |
| +--/\/\/--+ any length |
| ^ | 50 ohm line any load
Vs Is | |
100v RMS 1A RMS | |
| | | |
+----------+---------+------+-------------------+
The generator output impedance is 50 ohms.
Dissipation in the generator resistors is always
50 Watts plus Pref.
With a shorted or open load, the power dissipation
in the generator is 100 W.

Numerical coincidence as proof that Pref is always
dissipated in the generator. :-)

On a more serious note, how would you analyze this
generator using reflected power and constructive
and destructive interference?

...Keith

Keith Dysart wrote:
Numerical coincidence as proof that Pref is always
dissipated in the generator. :-)

It's no coincidence. The special case example was
carefully selected to make the angle 'A' between
the forward and reflected waves equal to 90 degrees.
Since cos(A) exists in the interference term and
cos(90) = 0, it makes the interference term equal
to zero.

When there is no interference at the source resistor,
100% of the reflected power is *always* dissipated in
the source resistor. This is a chosen special case
condition, NOT a coincidence. In the absence of
interference, there is simply no other place for the
reflected energy to go, i.e. it cannot be redistributed
back toward the load.

There are any number of special cases that will cause
the forward wave and reflected wave to be 90 degrees
out of phase at the source resistor. One of Roy's cases
was just such a case - 1/2WL of 50 ohm feedline with a
0 +/- j50 ohm load.

For any length feedline, there exists a load that
will cause the reflected wave to be 90 degrees out of
phase with the forward wave at the source resistor.
For any of those infinite number of cases, the
reflected energy will be dissipated in the source
resistor.

On a more serious note, how would you analyze this
generator using reflected power and constructive
and destructive interference?

Without analyzing it, based on our previous discussion
of voltage sources and current sources, I would say
that any constructive interference in the voltage
source is offset by an equal magnitude of destructive
interference in the current source and vice versa, i.e.
the net interference inside the source is always zero.
--
73, Cecil http://www.w5dxp.com

On Feb 22, 5:37 am, Keith Dysart wrote:
On Feb 21, 9:53 pm, Cecil Moore wrote:



Keith Dysart wrote:

Hey Keith, how about this one?

Rs Pfor=50w--
+----/\/\/-----+----------------------+
| 50 ohm --Pref |
| |
Vs 45 degrees RLoad+j0
100v RMS 50 ohm line |
| |
| |
+--------------+----------------------+

The dissipation in the source resistor is:

P(Rs) = 50w + Pref

How can anyone possibly argue that reflected power
is *never* dissipated in the source resistor? :-)
--
73, Cecil http://www.w5dxp.com

Two Saturdays ago I was on a road trip and used
9 litres/100km. How can anyone argue that the fuel
consumption is never equal to the day of the month?
Numerical coincidences can be much fun.

But you will like the generator below even better.
The power dissipated in the generator resistors is
always equal to 50 + Pref, regardless of the load
and line length, thereby always accounting for Pref.

Pfor=50w--
+--/\/\/---+----------------+-------------------+
| 100 ohm | --Pref |
| | 100 ohm |
| +--/\/\/--+ any length |
| ^ | 50 ohm line any load
Vs Is | |
100v RMS 1A RMS | |
| | | |
+----------+---------+------+-------------------+
The generator output impedance is 50 ohms.
Dissipation in the generator resistors is always
50 Watts plus Pref.
With a shorted or open load, the power dissipation
in the generator is 100 W.

Numerical coincidence as proof that Pref is always
dissipated in the generator. :-)

On a more serious note, how would you analyze this
generator using reflected power and constructive
and destructive interference?

...Keith

It's easy to lose sight of what's important when you get bogged down
in numerical coincidences and the like. To me, some things are
clearly important with respect to analyzing such systems:

1. If a generator is linear and matched to a line (Zgen = Zline, not
Zgen* = Zline), then no matter where a "reverse" signal comes from,
that signal does not reflect at the source:line junction. The
"reverse" signal can come from a reflection at a load, from another
generator at the other end of the line, from something feed in through
a coupler, from an electric eel biting the line--it doesn't matter.
There is no need for an analysis involving "constructive" or
"destructive" interference.

1a. Just because a "reverse" signal on the line does not reflect at
the generator:line junction, that does NOT mean that additional power
is dissipated inside the source.

2. You MUST have an accurate model of the inside of the source to
know how it will respond to some particular load and to signals that
impinge on its output port. With respect to figuring out what goes on
inside the source and what power may or may not be dissipate there,
there is NO advantage to knowing how the load or signals got there.

3. To correctly analyze conditions on a line that's fed only from one
end, with a load on the other end, there is NO NEED OR ADVANTAGE to
know what goes on inside the generator (beyond knowing the power it
delivers to that effective load, perhaps).

3a. There may be some advantage in knowing the source impedance of a
generator (or transmitter) in calculating the power delivered to a
load at the source's output port, but there is no advantage to knowing
it if you want to determine the standing wave ratio or reflection
coefficient on the line, or what net impedance that line+load presents
to the source; that is all determined solely by the line and the load.

The stuff about constructive/destructive interference with respect to
figuring out what happens inside a source is, to me, just so much
dancing on the head of pins. Welcome to dance if you so wish, but I'd
just as soon sit that one out.

Cheers,
Tom

K7ITM wrote:
It's easy to lose sight of what's important when you get bogged down
in numerical coincidences and the like. To me, some things are
clearly important with respect to analyzing such systems:

Is it a coincidence when one amp flows through a one ohm
resistor with one volt across it and dissipates one watt?
No, it is the laws of physics in action. The fact that
everything is a unity magnitude is because of the particular
values chosen for the example.

My example was NOT coincidence. I deliberately chose values
that would cause the forward wave and reflected wave to be
90 degrees out of phase at the source resistor. Under those
conditions, there is no interference present and all of the
reflected energy is dissipated in the source resistor. There
are an infinity of such examples and it is true for both
voltage sources and current sources.

The fact that there is even one example discredits the assertion
that reflected energy is *never* dissipated in the source.

1. If a generator is linear and matched to a line (Zgen = Zline, not
Zgen* = Zline), then no matter where a "reverse" signal comes from,
that signal does not reflect at the source:line junction. The
"reverse" signal can come from a reflection at a load, from another
generator at the other end of the line, from something feed in through
a coupler, from an electric eel biting the line--it doesn't matter.
There is no need for an analysis involving "constructive" or
"destructive" interference.

An interference analysis reveals exactly where all the energy is
going and that's what this discussion is all about. It may not
matter to you but it obviously matters to Keith and me.

1a. Just because a "reverse" signal on the line does not reflect at
the generator:line junction, that does NOT mean that additional power
is dissipated inside the source.

That's true. If total destructive interference exists at the
source resistor, then all of the reflected energy is redistributed
back toward the load. For the simple sources we have been using,
predicting how much reflected energy is dissipated in the source
resistor is a piece of cake.

I took Roy's chart and without calculating a single voltage or
current, not only matched Roy's correct results but I uncovered
an error he had made. That's a pretty good track record considering
that Roy's data went unchallenged for many years.

2. You MUST have an accurate model of the inside of the source to
know how it will respond to some particular load and to signals that
impinge on its output port. With respect to figuring out what goes on
inside the source and what power may or may not be dissipate there,
there is NO advantage to knowing how the load or signals got there.

We are discussing single-source, single transmission line, single
mismatched load systems. Where the energy components come from is
obvious.

3. To correctly analyze conditions on a line that's fed only from one
end, with a load on the other end, there is NO NEED OR ADVANTAGE to
know what goes on inside the generator (beyond knowing the power it
delivers to that effective load, perhaps).

This discussion is all about what is going on inside the source.
If you don't care to engage in that discussion, please feel free
not to.

3a. There may be some advantage in knowing the source impedance of a
generator (or transmitter) in calculating the power delivered to a
load at the source's output port, but there is no advantage to knowing
it if you want to determine the standing wave ratio or reflection
coefficient on the line, or what net impedance that line+load presents
to the source; that is all determined solely by the line and the load.

The argument about what happens inside a source is about 20 years
old now and is still raging. I'm simply trying to contribute
something to that argument.

The stuff about constructive/destructive interference with respect to
figuring out what happens inside a source is, to me, just so much
dancing on the head of pins. Welcome to dance if you so wish, but I'd
just as soon sit that one out.

That's your opinion and that's OK. If you choose not to attempt
to understand interference, you will forever remain ignorant of
its usefulness as an analysis tool. For the simple examples presented
so far, how much reflected power is dissipated in the source
resistor has been accurately predicted for all examples.

The interference phenomenon is well understood in the field of
optical physics and is a very useful tool in that field. The
principles are the same for RF waves. Why not use the tool?
Incidentally, optical physicists are NOT dancing on the head
of a pin when they calculate the irradiance of the bright rings
and dark rings.
--
73, Cecil http://www.w5dxp.com