W7EL's Food for Thought: Forward and Reverse Power
 

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

W7EL's Food for Thought: Forward and Reverse Power
 

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

W7EL's Food for Thought: Forward and Reverse Power
 

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

W7EL's Food for Thought: Forward and Reverse Power
 

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

W7EL's Food for Thought: Forward and Reverse Power
 

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

W7EL's Food for Thought: Forward and Reverse Power
 

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

W7EL's Food for Thought: Forward and Reverse Power
 

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

W7EL's Food for Thought: Forward and Reverse Power
 

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

W7EL's Food for Thought: Forward and Reverse Power
 

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

W7EL's Food for Thought: Forward and Reverse Power
 

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

W7EL's Food for Thought: Forward and Reverse Power
 

On Feb 19, 11:07 am, Keith Dysart wrote:
....
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.

Which, of course, illustrates the point that I believe Roy made in the
article, that the load conditions (whether they are the result of
something going on involving a transmission line, or just a lumped
load) tell us nothing about what's going on inside a source, ideal or
otherwise. For that, you MUST know the characteristics of the source
itself, and for that you do NOT need to know anything about the load
beyond the impedance it presents to the generator output terminals
(possibly as a function of time, frequency, amplitude and other
factors). For example, in the case of the signal generator on my
bench when 100dB of attenuation is cranked in, the change in
dissipation inside the generator versus load impedance is
inconsequential: at least 99.999 percent of the generator's available
RF output is dissipated in the attenuator when the load is matched (50
ohms). The additional maximum possible 0.001 percent increase
depending on load would be difficult to detect: about 0.00004dB
change. On the other hand, the change in dissipation inside my 450MHz
transmitter versus load impedance is substantial, BUT bears no
resemblance to either the current-source-with-shunt-resistor or the
voltage-source-with-series-resistor model, for multiple reasons.

Cheers,
Tom

W7EL's Food for Thought: Forward and Reverse Power
 

K7ITM wrote:
Keith Dysart wrote:
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.

Which, of course, illustrates the point that I believe Roy made in the
article, that the load conditions (whether they are the result of
something going on involving a transmission line, or just a lumped
load) tell us nothing about what's going on inside a source, ideal or
otherwise.

That's true, but Roy specified exactly what was inside his source
so we are free to analyze it based on his specifications.
--
73, Cecil http://www.w5dxp.com

W7EL's Food for Thought: Forward and Reverse Power
 

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

W7EL's Food for Thought: Forward and Reverse Power
 

K7ITM wrote:

Which, of course, illustrates the point that I believe Roy made in the
article, that the load conditions (whether they are the result of
something going on involving a transmission line, or just a lumped
load) tell us nothing about what's going on inside a source, ideal or
otherwise. For that, you MUST know the characteristics of the source
itself, and for that you do NOT need to know anything about the load
beyond the impedance it presents to the generator output terminals
(possibly as a function of time, frequency, amplitude and other
factors). For example, in the case of the signal generator on my
bench when 100dB of attenuation is cranked in, the change in
dissipation inside the generator versus load impedance is
inconsequential: at least 99.999 percent of the generator's available
RF output is dissipated in the attenuator when the load is matched (50
ohms). The additional maximum possible 0.001 percent increase
depending on load would be difficult to detect: about 0.00004dB
change. On the other hand, the change in dissipation inside my 450MHz
transmitter versus load impedance is substantial, BUT bears no
resemblance to either the current-source-with-shunt-resistor or the
voltage-source-with-series-resistor model, for multiple reasons.

Yes. In all the cases, you can replace the transmission line and load
with any other combination of transmission line and load which present
the same impedance to the source, and the source resistor dissipation
will be exactly the same.

For example, look at the first entry, 50 + j0:

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

If we used a 100 ohm transmission line instead of a 50 ohm transmission
line, we'd have

Zl fPa rPa Pa(tx) Pa(src) Pa(R0) Pa(Rl) frac R0 frac Rl
50 + j0 112.5 12.5 100 200 100 100 0.50 0.50

There's no change in source or load dissipation even though the forward
and reverse powers in the transmission line are different. If we do away
with the transmission line altogether and connect the load directly to
the source resistor, completely eliminating traveling waves, the result
is exactly the same.

A literally infinite number of other examples, using various line
impedances, load resistances, and line lengths, can be created. You'll
find that the *only* factor (other than source voltage and source
resistance value) which determines source resistor dissipation is the
impedance which it sees - regardless of how that impedance is created.
It has nothing to do with constructive or destructive interference,
traveling waves of voltage, current, or power, or anything else
happening on the line, or even if there is a line at all. And the exact
amount of source resistor dissipation can be immediately calculated by
analyzing a simple circuit of three components: the voltage source, the
source resistance, and the impedance seen by the source resistance. No
other information is required.

Roy Lewallen, W7EL

W7EL's Food for Thought: Forward and Reverse Power
 

Roy Lewallen wrote:
No other information is required.

No other information is required if, as you assert, you
don't care where the power goes. You said: "I personally
don't have a compulsion to understand where this power
'goes'." That's perfectly acceptable, but don't turn
around and present yourself as an expert on "where the
power goes". Please choose either not to care or to engage
in a technical discussion of where the power goes. It
is not fair play for you to try to have it both ways.

If it is applied properly, the power-interference equation
tells us exactly where the power goes so your following
statement is false: "While the nature of the voltage and
current waves when encountering an impedance discontinuity
is well understood, we're lacking a model of what happens
to this 'reverse power' we've calculated."

We are NOT lacking a model of what happens to the "reverse
power". You have just chosen not to understand the model
and are trying to use your guru status to belittle and
discredit anyone attempting to use that model. Are you
afraid it might turn out to be valid?
--
73, Cecil http://www.w5dxp.com

W7EL's Food for Thought: Forward and Reverse Power
 

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

W7EL's Food for Thought: Forward and Reverse Power
 

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

W7EL's Food for Thought: Forward and Reverse Power
 

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

W7EL's Food for Thought: Forward and Reverse Power
 

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

W7EL's Food for Thought: Forward and Reverse Power
 

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

W7EL's Food for Thought: Forward and Reverse Power
 

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

Forward and Reverse Power
 

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

Forward and Reverse Power
 

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

Forward and Reverse Power
 

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

Forward and Reverse Power
 

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

Forward and Reverse Power
 

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

Forward and Reverse Power
 

Cecil Moore wrote:


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.

Cecil,

I don't recall just how you became such an expert on optics, but your
proposed use of constructive and destructive interference is not the way
to calculate bright and dark rings. Great for handwaving explanations,
or textbook explanations, but close to useless for detailed calculations.

In the real world most most problems of interest are not simple one
dimensional set-ups with ideal lossless components. All of your nice
power equations with cosine cross-terms get completely unwieldy in the
real world.

Do you even wonder why you seem to be the pioneer in trying to apply
constructive and destructive interference to HF problems? Do you suppose
that no other smart folks ever thought along the same path? Do you
suppose there is a reason why essentially all of the textbooks and
scholarly writings on transmission lines virtually ignore constructive
and destructive interference for detailed calculations?

You have recently demonstrated that you can get exactly the same answers
as Keith, Roy, and others. However, beyond satisfying your own needs,
you have demonstrated exactly nothing in addition to the results
available from conventional analysis. That is dancing on the head of a pin.

There is really no particular need to discover "where the power goes".
The equations for ordinary classical physics are self-consistent. If one
gets the fields analyzed correctly, or equivalently the voltages and
currents, then energy and power will take care of themselves. You simply
will not find a case where all of the forces or fields are worked out
correctly but the energy is not conserved.

73,
Gene
W4SZ

Forward and Reverse Power
 

Gene Fuller wrote:
You have recently demonstrated that you can get exactly the same answers
as Keith, Roy, and others.

Which means, that contrary to all the earlier assertions,
an energy+interference analysis works. The analysis that
Roy called "gobbledegook" caught an error in his own data
that he had overlooked for many years and about which he
had made some false technical assumptions which he promoted
on his web page.

If an energy analysis provided nothing more than uncovering
the errors in Roy's math and concepts, it was more than worth
the effort.

There is really no particular need to discover "where the power goes".

Maxwell and Bruene have been arguing about such for 20
years now. The need to discover "where the power goes"
apparently exists for them and others.

If you, like Roy and Tom, don't care where the power
goes, that OK, but please don't try to present yourselves
as experts on a subject you don't care enough about
enough to have ever studied it in detail.
--
73, Cecil http://www.w5dxp.com

Forward and Reverse Power
 

On Feb 23, 2:32 pm, Cecil Moore wrote:
....
If you, like Roy and Tom, don't care where the power
goes,


I really do wish you'd learn to read. I did NOT say I don't care
where the power goes.

Forward and Reverse Power
 

K7ITM wrote:
Cecil Moore wrote:
If you, like Roy and Tom, don't care where the power
goes,

I really do wish you'd learn to read. I did NOT say I don't care
where the power goes.

I apologize, Tom. To me, the following quote sounded like
you don't care.

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.

If you had to chose the correct implication which would it be?

A. I care about what happens inside a source.

B. I don't care what happens inside a source.
--
73, Cecil http://www.w5dxp.com

Forward and Reverse Power
 

K7ITM wrote:
On Feb 23, 2:32 pm, Cecil Moore wrote:
...
If you, like Roy and Tom, don't care where the power
goes,


I really do wish you'd learn to read. I did NOT say I don't care
where the power goes.

I care where the power goes, too. And I know where it goes: from my
transmitter through the transmission line to the antenna.

Roy Lewallen, W7EL

Forward and Reverse Power
 

Roy Lewallen wrote:
I care where the power goes, too. And I know where it goes: from my
transmitter through the transmission line to the antenna.

Roy, here's a quote from your web page:
"I personally don't have a compulsion to understand
where this power 'goes'."

How can you care where the power goes when you don't care
to understand where the power goes?

You also scream out:
"THE REVERSE POWER IS NOT DISSIPATED IN OR ABSORBED BY THE
SOURCE RESISTANCE."

I just presented a special case example where all of the
reverse energy is dissipated (converted to heat) in the
source resistance. All it takes is one example to prove
your above statement to be false.

Your web page even contains one of those special case
examples. When your load is 0 +/- j50 at the end of
a 1/2WL piece of 50 ohm line, the forward wave and
reflected wave are 90 degrees out of phase at the
50 ohm source resistor. Under that special condition,
all of the reflected energy is dissipated in the
source resistor. Your own example proves your statement
to be false.
--
73, Cecil http://www.w5dxp.com

Forward and Reverse Power
 

Cecil Moore wrote:


Maxwell and Bruene have been arguing about such for 20
years now. The need to discover "where the power goes"
apparently exists for them and others.

If you, like Roy and Tom, don't care where the power
goes, that OK, but please don't try to present yourselves
as experts on a subject you don't care enough about
enough to have ever studied it in detail.


Cecil,

I must have missed the messages in which you resolved this 20 year
battle through your pioneering analysis technique.

Who won?

8-)

73,
Gene
W4SZ

Forward and Reverse Power
 

On Sun, 24 Feb 2008 16:23:49 GMT, Gene Fuller
wrote:

Who won?

Gad Gene,

Now we will get postings signed Cecil Balboa! (Older, pudgier, but
still swinging after all these years.)

73's
Richard Clark, KB7QHC

Forward and Reverse Power
 

Gene Fuller wrote:
I must have missed the messages in which you resolved this 20 year
battle through your pioneering analysis technique.

I didn't say it had been resolved. Those guys argue
about real-world class-C amplifiers. I didn't want
to discuss sources at all but Keith led me down the
primrose path. :-)

The small part of the argument that I can resolve is
the ideal class-A voltage and current sources. But
perhaps that resolution will spill over into the
more complicated configurations. I don't recall
anyone else considering what effect interference
has on energy redistribution inside an RF source.

But I indeed do have it figured out for ideal linear
voltage and current sources. Not only does it agree
with the results of other valid analysis methods,
but it answers the question: What happens to the
reflected energy when it is incident upon an ideal
linear source?
--
73, Cecil http://www.w5dxp.com

Forward and Reverse Power
 

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? :-)

It is time for a more complete analysis. For the sake of an
example let us set Rload to 150 ohms. That makes a nice
voltage RC at the load of 0.5.

Vs = 100 V rms

At the generator end
Vf.g = 50 V rms
Pf.g = 50 W avg

Vr.g = 25 V rms
Pr.g = 12.5 W avg

At the source resistor, before the reflection returns, the
dissipation in the source resistor is
Prs.br = 50 W avg
and after the reflection returns it is
Prs.ar = 62.5 W avg

Yup, it sure looks like that reflected wave is dissipated in the
source resistor since 62.5 = 50 + 12.5, does it not?

But let us examine in more detail...

Vs = 100 V rms
Vs(t) = 141 cos(wt)

Vf.g = 50 V rms
Vf.g(t) = 70.7 cos(wt)

Pf.g = 50 W avg
Pf.g(t) = 50 cos(2wt) + 50

Vr.g = 25 V rms
Vr.t(t) = 35.35 cos(wt-90degrees)

Pr.g = 12.5 W avg
Pr.g(t) = 12.5 cos(2wt) + 12.5

The power in the source resistor, before the reflection returns is
Prs.br = 50 W avg
Prs.br(t) = 50 cos(2wt) + 50

and after the reflection returns
Prs.ar = 62.5 W avg
Prs.ar(t) = 62.5 cos(2wt-53.13degrees) + 62.5

So, while the average powers seem to sum nicely to support the
claim, when the actual power as a function of time is examined
it can be seen that the power in the source resistor is NOT the
sum of its dissipation pre-reflection plus the power from the
reflection.

This solidly disproves the claim that the reflected power is
dissipated in the source resistor.

You may find the spreadsheet at
http://keith.dysart.googlepages.com/...oad,reflection
useful in taking some of the tedium out of the calculations.
And yes, if your anti-virus tools do not like Excel macros, you may
have to invoke your authority over your tools. (They do know who is
boss, do they not?)

...Keith

Forward and Reverse Power
 

Keith Dysart wrote:
This solidly disproves the claim that the reflected power is
dissipated in the source resistor.

I will respond to your posted data after awhile. In the
meantime, let's correct the misconception contained in
your above statement.

No one is asserting that reflected power is *always*
dissipated in the source resistor so that discussion
is a dead horse. The assertion that I am making is:

Reflected RF energy is *sometimes* dissipated (converted
to heat) in the source resistor and sometimes it is not.
With an ideal voltage or current source, I can tell you
when it happens and when it doesn't happen along with
the quantitative values involved.

What is being questioned is Roy's assertion in his
"Food for Thought" paper. Quoting:

"THE REVERSE POWER IS NOT DISSIPATED IN OR ABSORBED BY
THE SOURCE RESISTANCE."

It seems clear that he is saying that reverse power is
*never* dissipated in the source resistance. That is
simply not true. Sometimes, there is nowhere else for
the reflected energy to go.

And yes, if your anti-virus tools do not like Excel macros, you may
have to invoke your authority over your tools. (They do know who is
boss, do they not?)

I am reluctant to turn off my firewall and virus protection
while I download the EXCEL file. I once had a virus/worm
that even survived a reformatting of my hard drive. Apparently,
it was stored somewhere besides the hard drive.
--
73, Cecil http://www.w5dxp.com

Forward and Reverse Power
 

Keith Dysart wrote:
So, while the average powers seem to sum nicely to support the
claim, when the actual power as a function of time is examined
it can be seen that the power in the source resistor is NOT the
sum of its dissipation pre-reflection plus the power from the
reflection.

Whoa Keith, Dr. Best's power equation that is being used
does not work for instantaneous powers. It is adapted from
the irradiance equation from optical physics which is a
*time averaged power density*. The irradiance equation
does NOT work for instantaneous powers even in the
field of optical physics and it is not supposed to.

Trying to use Dr. Best's power equation on instantaneous
powers is akin to trying to measure the feedpoint impedance
of an antenna with a DC ohm-meter. It is the misuse of a
tool. All powers or power densities appearing in the
power equation *must* be integrated over at least one
complete cycle. Instantaneous powers are simply excluded
from this energy model that we have been discussing.

Note that every voltage, current, and power in Dr. Best's
QEX article is an average value.

I have made absolutely no assertions about instantaneous
powers so your instantaneous power data is irrelevant to
my assertions. When I say "power", I am always talking about
time averaged power - never about instantaneous power.

You are free to make assertions about instantaneous power
but those assertions do not apply to my statements
about average power. They may be interesting to you but
have nothing to do with anything that I have been saying.
--
73, Cecil http://www.w5dxp.com

Forward and Reverse Power
 

Keith Dysart wrote:
So, while the average powers seem to sum nicely to support the
claim, when the actual power as a function of time is examined
it can be seen that the power in the source resistor is NOT the
sum of its dissipation pre-reflection plus the power from the
reflection.

Here's a quote from "Optics", by Hecht, concerning power
density (irradiance). "Furthermore, since the power
arriving cannot be measured instantaneously, the detector
must integrate the energy flux over some finite time, 'T'.
If the quantity to be measured is the net energy per unit
area received, it depends on 'T' and is therefore of limited
utility. If however, the 'T' is now divided out, a highly
practical quantity results, one that corresponds to the
average energy per unit area per unit time, namely 'I'."
'I' is the irradiance (*AVERAGE* power density).

i.e. The irradiance/interference equation does not work
for instantaneous powers which are "of limited utility".
--
73, Cecil http://www.w5dxp.com