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