On Feb 18, 7:58 pm, Cecil Moore wrote:
Roy Lewallen wrote:
For the last four entries, the SWR is infinite, and the reverse power is
a full 100 watts. The source is perfectly matched to the line for all
table entries. Yet the source resistor dissipation varies from 0 to 400
watts depending on the load impedance - despite no difference in source
match, or forward or reverse power for the four entries.

The last two entries are particularly interesting. When the line is open
circuited at the far end (last table entry), there is no power at all
dissipated in the source resistor. So none of the reverse power is
dissipated in the source resistor.

Such is the nature of *total destructive interference* as
described by Hecht in "Optics". All of the reflected energy
is redistributed back toward the load.

Yet when the line is short circuited
at the far end (next to last table entry), the source resistor
dissipates twice the sum of the forward and reverse powers.

Such is the nature of *total constructive interference* as
described by Hecht in "Optics". All of the reflected energy
plus some more supplied by the source is dissipated in the
source resistor.

From the last entry alone we can conclude that THE REVERSE POWER IS NOT
DISSIPATED IN OR ABSORBED BY THE SOURCE RESISTANCE.

Of course not from only the last entry when total destructive
interference is occurring. 100% of the reflected energy is
redistributed back toward the load.

OTOH, when total constructive interference is occurring, not
only is 100% of the reflected energy dissipated in the source
resistor but the source has to supply twice as much energy as
the forward power plus the reflected power combined.

Perhaps the following energy analysis will shed some light on
the misconceptions. "Shedding some light" seems appropriate
since these concepts are from the field of optical physics.

This posting will provide an energy analysis approach to the
same previous W7EL data specifically avoiding any reference
to voltage and current.

The example that Roy provided in "Food for Thought: Forward
and Reflected Power" is:

Rs
+----/\/\/-----+----------------------+
| 50 ohm |
| |
Vs 1/2 wavelength ZLoad
141.4v 50 ohm line |
| |
+--------------+----------------------+

http://eznec.com/misc/Food_for_thought.pdf

We will create a new chart, step by step, that doesn't use
voltages or currents. Note that the first two columns are
copied from W7EL's chart. The Gamma reflection coefficient
is calculated at the load and |Rho|^2 is the power reflection
coefficient. The reflected power is the forward power multiplied
by |Rho|^2. 'GA' is the reflection coefficient Gamma Angle.

Zl fPa Rho GA,deg |Rho|^2 rPa
1. 50 + j0 100 0.0 0 0.0 0
2. 100 + j0 100 0.3333 0 0.1111 11.1
3. 25 + j0 100 0.3333 180 0.1111 11.1
4. 37 +/-j28 100 0.3378 97.1 0.1141 11.4
5. 0 +/-j50 100 1.0 -90 1.0 100
6. 0 +/-j100 100 1.0 -53.2 1.0 100
7. 0 + j0 100 1.0 -180 1.0 100
8. infinite 100 1.0 0 1.0 100

So far, everything agrees with W7EL's chart. We will now use
the following power equation not only to predict the dissipation
in the source resistor but also to explain the redistribution of
energy associated with interference. The power equation is:

Pa(R0) = fPa + rPa + 2*(fPa*rPa)cos(180-GA)

Could you expand on why the expression on the right is equal to
the average power dissipated in R0(Rs)? How was the expression
derived?

As well, what would be the equivalent expression for the following
example?

+-------+-------------+----------------------+
| | |
^ | Rs |
Is +-/\/\/-+ 1/2 wavelength ZLoad
2.828A 50 ohm | 50 ohm line |
| | |
+---------------+-----+----------------------+

The forward power is the same, the source impedance is the same,
but the conditions which cause maximum dissipation in the source
resistor are completely different.

Why is it not the same expression as previous since the conditions
on the line are the same?

What is the expression that describes the power dissipated in the
source resistor?

How is the expression derived?

Where 'GA' is the reflection coefficient Gamma angle and the last
term, 2*SQRT(fPa*rPa)cos(180-GA), is known as the *INTERFERENCE TERM*.

fPa rPa (180-GA) Pa(R0) interference term
1. 100 0 180 100 0
2. 100 11.1 180 44.4 -66.7
3. 100 11.1 0 177.8 +66.7
4. 100 11.4 82.9 119.8 + 8.35
5. 100 100 270 200 0
6. 100 100 233.2 80.2 -119.8
7. 100 100 360 400 +200
8. 100 100 180 0 -200

Except for the error that W7EL made in the Pa(R0) for example
number 7, these values of Pa(R0) agree with W7EL's posted values.
Therefore, the power-interference equation works. Not only does
it work, but it tells us the magnitude of interference between
the forward wave and the reflected wave when they interact at
the source resistor. Line by line:

1. There is zero interference because there are no reflections.

2. There is 66.7 watts of destructive interference present.

3. There is 66.7 watts of constructive interference present.

4. There is 8.35 watts of constructive interference present.

5. There is zero interference because the forward wave and
reflected waves are 90 degrees apart.

6. There is 119.8 watts of destructive interference present.

7. There is 200 watts of constructive interference present.

8. There is 200 watts of destructive interference present.
All of the reflected energy is redistributed back toward the load.

Wonder no more where the power goes. Constructive interference
requires extra energy from the source. Destructive interference
redistributes some (or all) of the reflected energy back toward
the load. Under zero interference conditions, all of the reflected
power (if it is not zero) is dissipated in the source resistor.
--
73, Cecil http://www.w5dxp.com

Keith Dysart wrote:

w5dxp wrote:
Pa(R0) = fPa + rPa + 2*(fPa*rPa)cos(180-GA)

Opps, sorry - a typo. That equation should be:

Pa(R0) = fPa + rPa + 2*SQRT(fPa*rPa)cos(180-GA)

Could you expand on why the expression on the right is equal to
the average power dissipated in R0(Rs)? How was the expression
derived?

This is essentially the same as the irradiance-interference
equation from optical physics. It's derivation is covered
in detail in "Optics" by Hecht, 4th edition, pages 383-388.
It is also the same as the power equation explained in detail
by Dr. Steven Best in his QEX article, "Wave Mechanics of
Transmission Lines, Part 3", QEX, Nov/Dec 2001, Eq 12. It
can also be derived independently by squaring the s-parameter
equation: b1^2 = (s11*a1 + s12*a2)^2

As well, what would be the equivalent expression for the following
example?

Pa(R0) = fPa + rPa + 2*SQRT(fPa*rPa)cos(GA)

Note the 180 degree phase difference between the two
examples. Why that is so is explained below.

+-------+-------------+----------------------+
| | |
^ | Rs |
Is +-/\/\/-+ 1/2 wavelength ZLoad
2.828A 50 ohm | 50 ohm line |
| | |
+---------------+-----+----------------------+

The forward power is the same, the source impedance is the same,
but the conditions which cause maximum dissipation in the source
resistor are completely different.

If by "completely different", you mean 180 degrees different,
you are absolutely correct. The 1/2WL short-circuit and open-
circuit results are reversed when going from a voltage source
to a current source.

Why is it not the same expression as previous since the conditions
on the line are the same?

We are dealing with interference patterns between the forward
wave and the reflected wave. In the voltage source example,
the forward wave and reflected wave are flowing in opposite
directions through the resistor. In the current source example,
the forward wave and reflected wave are flowing in the same
direction through the resistor. That results in a 180 degree
difference in the cosine term above. I believe all your other
excellent questions are answered above.
--
73, Cecil http://www.w5dxp.com

On Feb 19, 3:30*pm, Cecil Moore wrote:
Keith Dysart wrote:
* w5dxp wrote:
Pa(R0) = fPa + rPa + 2*(fPa*rPa)cos(180-GA)

Opps, sorry - a typo. That equation should be:

Pa(R0) = fPa + rPa + 2*SQRT(fPa*rPa)cos(180-GA)

Could you expand on why the expression on the right is equal to
the average power dissipated in R0(Rs)? How was the expression
derived?

This is essentially the same as the irradiance-interference
equation from optical physics. It's derivation is covered
in detail in "Optics" by Hecht, 4th edition, pages 383-388.
It is also the same as the power equation explained in detail
by Dr. Steven Best in his QEX article, "Wave Mechanics of
Transmission Lines, Part 3", QEX, Nov/Dec 2001, Eq 12. It
can also be derived independently by squaring the s-parameter
equation: *b1^2 = (s11*a1 + s12*a2)^2

Steven Best's article does have an equation of similar form:

PFTOTAL = P1 + P2 + 2 * sqrt(P1) * sqrt(P2) * cosè

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

So that made sense. But why would adding in the forward
power in the line be useful for computing the power in the
source resistor? And the answer: Pure happenstance. The
value of the source resistor is the same as the value
of the line impedance, so identical currents produce
identical powers.

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.

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.

As well, what would be the equivalent expression for the following
example?

Pa(R0) = fPa + rPa + 2*SQRT(fPa*rPa)cos(GA)

Note the 180 degree phase difference between the two
examples. Why that is so is explained below.

* * * *+------+-------------+----------------------+
* * * *| * * * | * * * * * * * * * * * * * * * * * *|
* * * *^ * * * | * Rs * * * * * * * * * * * * * * * |
* * * Is * * * +-/\/\/-+ * * * *1/2 wavelength * *ZLoad
* * 2.828A * * *50 ohm | * * * * 50 ohm line * * * *|
* * * *| * * * * * * * | * * * * * * * * * * * * * *|
* * * *+---------------+-----+----------------------+

The forward power is the same, the source impedance is the same,
but the conditions which cause maximum dissipation in the source
resistor are completely different.

If by "completely different", you mean 180 degrees different,
you are absolutely correct. The 1/2WL short-circuit and open-
circuit results are reversed when going from a voltage source
to a current source.

Why is it not the same expression as previous since the conditions
on the line are the same?

We are dealing with interference patterns between the forward
wave and the reflected wave. In the voltage source example,
the forward wave and reflected wave are flowing in opposite
directions through the resistor. In the current source example,
the forward wave and reflected wave are flowing in the same
direction through the resistor.

I see that now. It works because you were effectively using
superposition to compute the voltage across the source resistor
and then converting this to power using the expression derived
by Steven Best.

But it was happenstance that the voltage across the resistor
was the same as the forward voltage on the line. A very misleading
coincidence.

That results in a 180 degree
difference in the cosine term above. I believe all your other
excellent questions are answered above.
--
73, Cecil *http://www.w5dxp.com

Keith Dysart wrote:
Steven Best's article does have an equation of similar form:

PFTOTAL = P1 + P2 + 2 * sqrt(P1) * sqrt(P2) * cosè

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 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.

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.

50w + 50w + 2*SQRT(50w*50w) = 200 watts
--
73, Cecil http://www.w5dxp.com

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