The Rest of the Story
 

Keith Dysart wrote:
Ps(t) = 100 + 116.6190379cos(2wt-30.96375653)
Prs(t) = 68 + 68cos(2wt-61.92751306)
Pg(t) = 32 + 68cos(2wt)
Pf.g(t) = 50 + 50cos(2wt)
Pr.g(t) = -18 + 18cos(2wt)

Now it is time for you to explain exactly why you
believe in a conservation of power principle. Do
you demand that the instantaneous power delivered
by a battery charger be instantaneously dissipated
in the battery being charged? If not, why do you
require such for the example under discussion?

The correct equation for adding the powers above is?

Prs(t) = Pf.g(t) + Pr.g(t) +/- 2*SQRT[Pf.g(t)*Pr.g(t)]

The last term is the interference term. The sign of
the interference term is negative if Vf.g(t) and
Vr.g(t) are out of phase. The sign of the interference
term is positive if Vf.g(t) and Vr.g(t) are in phase.
Vf.g(t) and Vr.g(t) are in phase for half of the cycle.
Vf.g(t) and Vr.g(t) are out of phase for the other half
of the cycle. The "excess" energy from the destructive
interference is dissipated in the source resistor as
constructive interference after being delayed by 90
degrees.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Keith Dysart wrote:
To me it was a complete
surprise that summing the voltages produces the correct
total voltages and, at the same time, summing the powers
(which are a squared function of the voltage) also
produce the correct result.

Why is that a surprise to you? I have been telling you
for many days about that special case whe

(V1^2 + V2^2) = (V1 + V2)^2 for zero interference

This applies equally well to phasors or instantaneous
values of voltage. The above special case is what my
Part 1 article is all about.

Since a 45 degree long transmission line forces the
above RMS voltage equation to be true for all values of
resistive loads, the *average* reflected power based on
RMS voltage values is *always* dissipated in the source
resistor. Since a conservation of power principle does
not exist, it is perfectly acceptable for destructive
interference energy to be stored for part of the cycle
and be dissipated later in the cycle.

So
Pg(t) = Pf.g(t) + Pr.g(t)
is always true.

Of course it is true but it says absolutely nothing about
the dissipation of the reflected power in the source resistor
which is the subject of the discussion. The above equation
remains always true while the dissipation of the reflected
power in the source resistor varies from 0% to 100%.
--
73, Cecil http://www.w5dxp.com

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On Fri, 21 Mar 2008 17:08:39 -0700 (PDT)
Keith Dysart wrote:

On Mar 21, 9:12*am, Roger Sparks wrote:
On Fri, 21 Mar 2008 05:10:01 -0700 (PDT)

Keith Dysart wrote:
On Mar 20, 1:07*pm, Roger Sparks wrote:
Keith Dysart wrote:
On Mar 16, 10:21 am, Cecil Moore wrote:
snip
Would you please explain how energy is conserved in the
following example at the zero-crossing point for Vs?

* * * * * * * *Rs * * * Vg * * * * * * * * * * Vl
* * * * *+----/\/\/-----+----------------------+
* * * * *| * *50 ohm * * * * * * * * * * * * * |
* * * * *| * * * * * * * * * * * * * * * * * * |
* * * * Vs * * * * * * * * 45 degrees * * * * *| Shorted
* * *100v RMS * * * * * * *50 ohm line * * * * | Stub
* * * * *| * * * * * * * * * * * * * * * * * * |
* * * * *| * * * * * * * * * * * * * * * * * * |
* * * * *+--------------+----------------------+
* * * * gnd

At the zero-crossing of Vs, Ps(t) = 0, i.e. the source
is supplying zero watts at that time but Prs(t) = 100w.
Where is the 100 watts coming from?

For the first 90 degrees of time, the circuit can be represented as

* * * * * * * * *Rs * * * Vg * * * * * * * * * * Vl
* * * * * *+----/\/\/-----+----------------------+
* * * * * *| * *50 ohm * * * * * * * * * * * * * |
* * * * * *| * * * * * * * * * * * * * * * * * * /
* * * * * Vs * * * * * * * * * * * * * * * * * * \ 50 ohm resistor
* * * *100v RMS * * * * * * * * * * * * * * * * */
* * * * * *| * * * * * * * * * * * * * * * * * * \
* * * * * *| * * * * * * * * * * * * * * * * * * |
* * * * * *+--------------+----------------------+
* * * * * gnd

After 90 degrees of time has passed, the circuit can be represented as

* * * * * * * * *Rs * * * Vg * * * * * * * * * * Vl
* * * * * *+----/\/\/-----+----------------------+
* * * * * *| * *50 ohm * * * * * * * * * * * * * |
* * * * * *| * * * * * * * * * * * * * * * * * * |
* * * * * Vs * * * * * * * * * * * * * * * * * *---
* * * *100v RMS * * * * * * * * * * * * * * * * --- 50 ohm inductive
* * * * * *| * * * * * * * * * * * * * * * * * * |
* * * * * *| * * * * * * * * * * * * * * * * * * |
* * * * * *+--------------+----------------------+
* * * * * gnd
HEAVILY CLIPPED
We differ on how to utilize the term Pg(t). *I think it describes the standing wave, not the voltage across Rs. *

I describe Vg(t) as the actual voltage that would be measured by an
oscilloscope at the point g; i.e. the beginning of the transmission
line. It is certainly not the voltage across Rs. Perhaps it IS what
you mean by standing wave. In any case the voltage across the resistor
is
Vs(t) = Vrs(t) + Vg(t)
Vrs(t) = Vs(t) - Vg(t)

Pg(t) is the power being delivered to the transmission line at any
instance t.
Pg(t) = Vg(t) * Ig(t)

...Keith

You gave a very thoughtful reply Keith. It looks to me like two perspectives, both right so far as the analysis has proceeded.
--
73, Roger, W7WKB

The Rest of the Story
 

Keith Dysart wrote:
Cecil Moore wrote:
The net total voltage at point g is the standing wave
voltage that is present at that point.

I suppose that is a name you could use, but do you still
call it "standing wave voltage" when there is no
reflected wave, i.e. Vr is 0 for all time. There is
no standing wave when Vr is always 0.

We are not discussing matched systems. If the total
RMS voltage differs from the RMS forward voltage,
then Vr is NOT zero.
--
73, Cecil http://www.w5dxp.com

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On Sat, 22 Mar 2008 03:48:51 -0700 (PDT)
Keith Dysart wrote:

On Mar 21, 9:32*pm, Roy Lewallen wrote:
Keith Dysart wrote:
. . .
The equation
* Pg(t) = Pf.g(t) + Pr.g(t)
is more interesting. The basis for this is superposition.
The forward and reverse voltage and current are superposed
to derive the actual voltage and current. It would seem
invalid to also sum the powers. To me it was a complete
surprise that summing the voltages produces the correct
total voltages and, at the same time, summing the powers
(which are a squared function of the voltage) also
produce the correct result.

But by starting with the equations used to derive forward
and reverse voltage and current, it can be easily shown
with appropriate substitution that Ptot is always equal
to Pforward + Preverse (Or Pf - Pr if you use the other
convention for the direction of the energy flow)s. It
simply falls out from the way that Vf and Vr are derived
from Vactual and Iactual.

It only holds true when Z0 is purely real. Of course, when it isn't,
time domain analysis becomes very much more cumbersome. But it's not
hard to show the problem using steady state sinusoidal analysis, and
that's where the cos term appears and is appropriate.

*
* So
* * Pg(t) = Pf.g(t) + Pr.g(t)
* is always true. For any arbitrary waveforms. Inclusion
* of cos(theta) terms would be incorrect.

Thanks for providing the limitation.

But I am having difficulty articulating where the math in the
following
derivation fails.

Starting by measuring the actual voltage and current at a single
point on the line, and wishing to derive Vf and Vr we have the
following four equations:

V = Vf + Vr
I = If - Ir
Zo = Vf / If
Zo = Vr / Ir

rearranging and substituting
Vf = V - Vr
= V - Zo * Ir
= V - Zo * (If - I)
= V - Zo * (Vf/Zo - I)
= V - Vf + Zo * I
= (V + Zo * I)/2
similarly
Vr = (V - Zo * I)/2

Pf = Vf * If
= Vf**2 / Zo
= ((V + Zo * I)(V + Zo * I)/4)/Zo
= (V**2 + 2(V * Zo * I) + Zo**2 * I**2) / (4 * Zo)

Pr = Vr * Ir
= (V**2 - 2(V * Zo * I) + Zo**2 * I**2) / (4 * Zo)

So, comtemplating that
P = Pf - Pr
and substituting
P = (V**2 + 2(V * Zo * I) + Zo**2 * I**2) / (4 * Zo)
- (V**2 - 2(V * Zo * I) + Zo**2 * I**2) / (4 * Zo)
= 4(V * Zo * I) / (4 * Zo)
= V * I
as required.

So when Zo is real, i.e. can be represented by R, it is
clear that P always equals Pf - Pr. And it does not even
matter which value of R is used for R. It does not have
to be the characteristic impedance of the transmission
line, the subtraction of powers still produces the correct
answer.

But when Zo has a reactive component, it still cancels
out of the equations. So why is this not a proof that
also holds for complex Zo.

I suspect it has to do with complex Zo being a concept
that only works for single frequency sinusoids, but am
having difficulty discovering exactly where it fails.

And if it is related to Zo and single frequency sinusoids,
does that mean that P = Pf - Pr also always works for
single frequency sinusoids?

...Keith

I am very impressed with this series of equations/relationships. These equations clarify your previous postings and provide a basis for future enrichment.

I think that a complex Zo would not be a transmission line, but would be an end point. Any complex end point could be represented by a length of transmission line with a resistive termination. Once that substitution was made, the problem should come back to the basic equations you presented here.
--
73, Roger, W7WKB

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On Sat, 22 Mar 2008 14:14:35 GMT
Cecil Moore wrote:

Keith Dysart wrote:
Ps(t) = 100 + 116.6190379cos(2wt-30.96375653)
Prs(t) = 68 + 68cos(2wt-61.92751306)
Pg(t) = 32 + 68cos(2wt)
Pf.g(t) = 50 + 50cos(2wt)
Pr.g(t) = -18 + 18cos(2wt)

Now it is time for you to explain exactly why you
believe in a conservation of power principle. Do
you demand that the instantaneous power delivered
by a battery charger be instantaneously dissipated
in the battery being charged? If not, why do you
require such for the example under discussion?

The correct equation for adding the powers above is?

Prs(t) = Pf.g(t) + Pr.g(t) +/- 2*SQRT[Pf.g(t)*Pr.g(t)]

Prs(t) = Pf.g(t) + Pr.g(t) +/- 2*SQRT[Pf.g(t)*Pr.g(t)]

= [SQRT(pf.g(t) + SQRT(pr.g(t)]^2

and/or= [SQRT(pf.g(t) - SQRT(pr.g(t)]^2

How can we justify calling the +/- 2*SQRT[Pf.g(t)*Pr.g(t) term the "interference term"?

Am I correct in assuming that this equation describes the instantaneous power delivered to Rs?

The last term is the interference term. The sign of
the interference term is negative if Vf.g(t) and
Vr.g(t) are out of phase. The sign of the interference
term is positive if Vf.g(t) and Vr.g(t) are in phase.
Vf.g(t) and Vr.g(t) are in phase for half of the cycle.
Vf.g(t) and Vr.g(t) are out of phase for the other half
of the cycle. The "excess" energy from the destructive
interference is dissipated in the source resistor as
constructive interference after being delayed by 90
degrees.
--
73, Cecil http://www.w5dxp.com


--
73, Roger, W7WKB

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On Mar 22, 10:14*am, Cecil Moore wrote:
Keith Dysart wrote:
Ps(t) = 100 + 116.6190379cos(2wt-30.96375653)
Prs(t) = 68 + 68cos(2wt-61.92751306)
Pg(t) = 32 + 68cos(2wt)
Pf.g(t) = 50 + 50cos(2wt)
Pr.g(t) = -18 + 18cos(2wt)

Now it is time for you to explain exactly why you
believe in a conservation of power principle. Do
you demand that the instantaneous power delivered
by a battery charger be instantaneously dissipated
in the battery being charged?

Are you making this a trick question by using the
word "dissipated"?

If not, then yes. Consider my laptop which has an
external power supply connected by a cord to the
laptop.

Conservation of energy means that the instantaneous
energy flow (i.e. power) along this cord into the
laptop is always exactly equal to the sum of
- the sum of energy being dissipated as heat
in each individual component
- the increase in energy being stored in
components such as capacitors, inductors
and batteries, minus any stored energy
being returned from components such as
capacitors, inductors and batteries
- energy being emitted such as
- light (e.g. display)
- sound (e.g. fans, speakers)
- RF (e.g. Wifi antennas, RFI)
- etc.

So yes, the phrase "conservation of power" is
appropriately descriptive and follows from
conservation of energy.

If not, why do you
require such for the example under discussion?

I require it for both.

The correct equation for adding the powers above is?

Prs(t) = Pf.g(t) + Pr.g(t) +/- 2*SQRT[Pf.g(t)*Pr.g(t)]

The last term is the interference term. The sign of
the interference term is negative if Vf.g(t) and
Vr.g(t) are out of phase. The sign of the interference
term is positive if Vf.g(t) and Vr.g(t) are in phase.
Vf.g(t) and Vr.g(t) are in phase for half of the cycle.
Vf.g(t) and Vr.g(t) are out of phase for the other half
of the cycle. The "excess" energy from the destructive
interference is dissipated in the source resistor as
constructive interference after being delayed by 90
degrees.

Where is "cos(theta)" in this?
And what "theta" is to be used?
Is it the same theta that was used to conclude that
this was a "no interference" example?
See "A Simple Voltage Source - No Interference" at
http://www.w5dxp.com/intfr.htm

So the plus/minus in the equation aboves means that
for part of the cycle we should add and part of the
cycle we should subtract.
This equation is ambiguous and is therefore incomplete
since it does not tell us when we should add and when
we should subtract.
Could you provide a mathematically precise description
of when we should do each?

Where is this destructive interference energy stored
while waiting to be dissipated constructively later?
If it is in a capacitance, which one? Does the
voltage on the capacitance increase appropriately
to account for the energy being stored?
Similarly if it is stored in an inductance.

...Keith

The Rest of the Story
 

On Mar 22, 10:55*am, Cecil Moore wrote:
Keith Dysart wrote:
Cecil Moore wrote:
The net total voltage at point g is the standing wave
voltage that is present at that point.

I suppose that is a name you could use, but do you still
call it "standing wave voltage" when there is no
reflected wave, i.e. Vr is 0 for all time. There is
no standing wave when Vr is always 0.

We are not discussing matched systems. If the total
RMS voltage differs from the RMS forward voltage,
then Vr is NOT zero.

True for this example. But it is generally unwise to pick
names or descriptions that will have to change when
the component values are changed.

...Keith

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On Mar 22, 11:17*am, Roger Sparks wrote:
I think that a complex Zo would not be a transmission line, but would be an end point. *Any complex end point could be represented by a length of transmission line with a resistive termination. *Once that substitution was made, the problem should come back to the basic equations you presented here..

The characteristic impedance for a transmission line is
Zo = sqrt( (R + jwL) / (G + jwC) )

For a lossline (no resistance in the conductors, and no
conductance between the conductors), this simplifies to
Zo = sqrt( L / C )

So real lines actually have complex impedances. But the
math is simpler for ideal (lossless) lines and there is
much to be learned from studying the simplified examples.

But caution is needed when taking these results to
the real world of lines with loss.

...Keith

The Rest of the Story
 

On Sun, 23 Mar 2008 03:42:36 -0700 (PDT)
Keith Dysart wrote:

On Mar 22, 11:17*am, Roger Sparks wrote:
I think that a complex Zo would not be a transmission line, but would be an end point. *Any complex end point could be represented by a length of transmission line with a resistive termination. *Once that substitution was made, the problem should come back to the basic equations you presented here.

The characteristic impedance for a transmission line is
Zo = sqrt( (R + jwL) / (G + jwC) )

For a lossline (no resistance in the conductors, and no
conductance between the conductors), this simplifies to
Zo = sqrt( L / C )

So real lines actually have complex impedances. But the
math is simpler for ideal (lossless) lines and there is
much to be learned from studying the simplified examples.

But caution is needed when taking these results to
the real world of lines with loss.

...Keith

Yes, I concur with these comments.

The characteristic impedance can also be found from

Zo = 1/(C*Vel)

where C is the capacitance of the line per unit distance and Vel is the velocity of the wave.

This second solution for Zo demonstrates the power storage capabilities of the transmission line over time.

But as you say, real lines also have resistance losses and other losses so use great care when taking these results into the real world

--
73, Roger, W7WKB