While working on an energy-based presentation of W7EL's
data from the following web page, I came across an
instance where my energy analysis differed from W7EL's
results under the "Food for Thought: Forward and Reverse
Power" section. Assuming Roy was correct, I attempted to
find my error and failed to do so.

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

**********begin quote**********

Zl fPa rPa Pa(tx) Pa(src) Pa(R0) Pa(Rl) frac R0 frac Rl
0 + j0 100 100 0 0 0 0 - -
infinite 100 100 0 0 0 0 - -

Not only that, but notice the last two cases. Here, the reverse power is
a full 100 watts. The source match is 1:1. Yet *none* of this reverse
power is dissipated in the source resistance. In fact, no power at all
is dissipated in the source resistance.

ANY MODEL PRESENTED TO ACCOUNT FOR WHAT HAPPENS TO "FORWARD" AND
"REVERSE" POWER AT TRANSMISSION LINE ENDS HAD BETTER GIVE RESULTS THAT
AGREE WITH THE ABOVE TABLE.

**********end quote**********

Unfortunately, my results do not agree.

In the line where ZL is zero, i.e. a short-circuit, the
dissipation in the source resistance is 400 watts, i.e.
all of the forward power and reflected power is dissipated
in the source resistor plus an additional 200 watts associated
with constructive interference. All 400 watts must be supplied
by the source so Pa(src) must also be 400 watts. It should read:

Zl fPa rPa Pa(tx) Pa(src) Pa(R0) Pa(Rl) frac R0 frac Rl
0 + j0 100 100 0 400 400 0 1.0 -
infinite 100 100 0 0 0 0 - -

For the ZL=0 case:
Pa(R0) = fPa + rPa + 2*SQRT(fPa*rPa) = 400 watts
This is *total constructive interference* as defined by
Hecht in "Optics".

For the ZL=infinite case:
Pa(R0) = fPa + rPa - 2*SQRT(fPa*rPa) = 0 watts
This is *total destructive interference* as defined by
Hecht in "Optics".

Since Roy doesn't read my postings or emails, could someone
please pass this information on to him.
--
73, Cecil http://www.w5dxp.com

On Feb 18, 7:13 am, Cecil Moore wrote:
While working on an energy-based presentation of W7EL's
data from the following web page, I came across an
instance where my energy analysis differed from W7EL's
results under the "Food for Thought: Forward and Reverse
Power" section. Assuming Roy was correct, I attempted to
find my error and failed to do so.

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

**********begin quote**********

Zl fPa rPa Pa(tx) Pa(src) Pa(R0) Pa(Rl) frac R0 frac Rl
0 + j0 100 100 0 0 0 0 - -
infinite 100 100 0 0 0 0 - -

Not only that, but notice the last two cases. Here, the reverse power is
a full 100 watts. The source match is 1:1. Yet *none* of this reverse
power is dissipated in the source resistance. In fact, no power at all
is dissipated in the source resistance.

ANY MODEL PRESENTED TO ACCOUNT FOR WHAT HAPPENS TO "FORWARD" AND
"REVERSE" POWER AT TRANSMISSION LINE ENDS HAD BETTER GIVE RESULTS THAT
AGREE WITH THE ABOVE TABLE.

**********end quote**********

Unfortunately, my results do not agree.

In the line where ZL is zero, i.e. a short-circuit, the
dissipation in the source resistance is 400 watts, i.e.
all of the forward power and reflected power is dissipated
in the source resistor plus an additional 200 watts associated
with constructive interference. All 400 watts must be supplied
by the source so Pa(src) must also be 400 watts. It should read:

Zl fPa rPa Pa(tx) Pa(src) Pa(R0) Pa(Rl) frac R0 frac Rl
0 + j0 100 100 0 400 400 0 1.0 -
infinite 100 100 0 0 0 0 - -

For the ZL=0 case:
Pa(R0) = fPa + rPa + 2*SQRT(fPa*rPa) = 400 watts
This is *total constructive interference* as defined by
Hecht in "Optics".

For the ZL=infinite case:
Pa(R0) = fPa + rPa - 2*SQRT(fPa*rPa) = 0 watts
This is *total destructive interference* as defined by
Hecht in "Optics".

Since Roy doesn't read my postings or emails, could someone
please pass this information on to him.
--
73, Cecil http://www.w5dxp.com

Why did you bother with Hecht? It's simple enough to go back to the
second equation above the line you disagree with, the one at the
bottom of page 7 in that pdf, and see it does not agree there either.
It's obviously a typo and should be corrected. I'll drop Roy a line
about it, in case he doesn't see this.

Cheers,
Tom

K7ITM wrote:
It's obviously a typo and should be corrected. I'll drop Roy a line
about it, in case he doesn't see this.

So much for this statement screamed at us by Roy. :-)

"ANY MODEL PRESENTED TO ACCOUNT FOR WHAT HAPPENS TO 'FORWARD' AND
'REVERSE' POWER AT TRANSMISSION LINE ENDS HAD BETTER GIVE RESULTS THAT
AGREE WITH THE ABOVE TABLE. Otherwise, it's wrong. The values in the
above table can be measured and confirmed."
--
73, Cecil http://www.w5dxp.com

K7ITM wrote:
It's obviously a typo and should be corrected.

Tom, do you think this is also a typo? :-)

"Yet *none* of this reverse power is dissipated in the
source resistance. In fact, no power at all is dissipated
in the source resistance. THE REVERSE POWER IS NOT DISSIPATED
IN OR ABSORBED BY THE SOURCE RESISTANCE."
--
73, Cecil http://www.w5dxp.com

K7ITM wrote:
On Feb 18, 7:13 am, Cecil Moore wrote:
While working on an energy-based presentation of W7EL's
data from the following web page, I came across an
instance where my energy analysis differed from W7EL's
results under the "Food for Thought: Forward and Reverse
Power" section. Assuming Roy was correct, I attempted to
find my error and failed to do so. . .

Why did you bother with Hecht? It's simple enough to go back to the
second equation above the line you disagree with, the one at the
bottom of page 7 in that pdf, and see it does not agree there either.
It's obviously a typo and should be corrected. I'll drop Roy a line
about it, in case he doesn't see this.

Thanks very much to both Cecil, for finding the error, and Tom, for
passing it along.

Tom is correct, that the information in the table should follow directly
from the equations at the bottom of the preceding page. The table entry
was in error, but not the equations or underlying principles. For Rl = 0
+ j0 the equation at the bottom of page 7

Pa(R0) = |Ilrms|^2 * R0 = (Vrms^2 * R0) / [(R0 + Rl)^2 + Xl^2]

gives the correct result of 400 watts, not 0 as shown in the table. The
table has been corrected, and the comments following it have been
modified to reflect the corrected value. Here's the corrected table and
text:

****************************

Zl fPa rPa Pa(tx) Pa(src) Pa(R0) Pa(Rl) frac R0 frac Rl
50 + j0 100 0 100 200 100 100 0.50 0.50
100 + j0 100 11.1 88.9 133 44.4 88.9 0.33 0.67
25 + j0 100 11.1 88.9 267 178 88.9 0.67 0.33
37 +/-j28(*)100 11.4 88.6 209 120 88.6 0.58 0.42
0 +/-j50 100 100 0 200 200 0 1.00 0
0 +/-j100 100 100 0 80.0 80.0 0 1.00 0
0 + j0 100 100 0 400 400 0 1.00 0
infinite 100 100 0 0 0 0 - -

(*) For any Zl that causes exactly a 2:1 SWR, rPa will equal 11.1 and
Pa(Rl) = 88.9. The values shown for 37 +/-j28 are slightly different
because this impedance doesn’t result in quite exactly a 2:1 SWR.

For the second, third, and fourth entries, the SWR is 2:1. The forward
and reverse powers are the same for all three, and the source impedance
(50 ohms) is the same for all the above cases. So here we have three
cases where the reverse powers are the same, and the impedance match
looking back toward the source is the same (1:1), yet the dissipation in
the source resistor Pa(R0) is very different. The obvious conclusion is
that THE POWER DISSIPATED IN THE SOURCE RESISTANCE ISN’T DETERMINED
DIRECTLY BY THE SOURCE MATCH, THE SWR, OR THE REVERSE POWER. Otherwise
it would be the same in all three cases, since all these quantities are
the same for all three.

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. 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.

From the last entry alone we can conclude that THE REVERSE POWER IS NOT
DISSIPATED IN OR ABSORBED BY THE SOURCE RESISTANCE. And the table
clearly shows that the source resistor dissipation bears no relationship
to the amount of reverse power.

**************************

The corrected essay has been uploaded to replace the previous one at
http://eznec.com/misc/Food_for_thought.pdf. Please note the uppercase
"F" -- it has to be entered exactly as shown.

Again, thanks very much for the corrections. It's my sincere intention
to present material that's accurate, and I appreciate the help in
finding and correcting errors I've made.

Roy Lewallen, W7EL

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)

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

Cecil Moore wrote:

Sorry about the misalignment of the numbers. Here they are
properly aligned in fixed font.

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

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
--
73, Cecil http://www.w5dxp.com

Cecil Moore wrote:

I apparently used some tabs in the previous posting that
caused the columns not to be aligned. Hopefully, this
posting remedies the problem.

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

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
--
73, Cecil http://www.w5dxp.com

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