On Mar 21, 5:03*pm, Roger Sparks wrote:
On Fri, 21 Mar 2008 19:43:12 GMT

Cecil Moore wrote:
Roger Sparks wrote:

Cecil Moore wrote:
Roger, do you understand why EE201 professors admonish
their students not to try to superpose powers?

No, I really don't. *

It is because (V1 + V2)^2 usually doesn't equal V1^2+V2^2
because of interference. Keith's addition of powers
without taking interference into account is exactly
the mistake that the EE201 professors were talking about.
One cannot validly just willy-nilly add powers. It is
an ignorant/sophomoric thing to do.

If we add two one watt coherent waves, do we get a two
watt wave? Only in a very special case. For the great
majority of cases, we do *NOT* get a two watt wave. In
fact, the resultant wave can be anywhere between zero
watts and four watts. The concepts behind Keith's
calculations are invalid. If you are also trying to
willy-nilly add powers associated with coherent waves,
your calculations are also invalid.
--
73, Cecil *http://www.w5dxp.com

OK, yes, I agree. *It is OK to add powers when you are adding the power used by light bulbs. *It is not OK to willy nilly multiply the voltage or current by the number of bulbs to learn the power used. *You must carefully consider the circuit that connects the bulbs before selecting the proper method of calculating power, especially the possibility that the bulbs may be connected to phased power as in 3 phase or in traveling waves.

My analysis used voltages, currents and impedances to compute
all the voltages and currents within the circuit. Some were
derived using superposition of voltages and currents but most
were derived using basic circuit theory (E=IR, Ztot=Z1+Z2, etc.)

Having done that, the powers for the three components (the
voltage source, resistor, and entrance to the transmission line)
in the circuit were computed. These powers were not derived using
superposition but by multiplying the current through the component
by the voltage across it. This is universally accepted as a valid
operation.

Having the power functions for each of the component, we can
then turn to the conservation of energy principle: The energy
in a closed system is conserved.

This is the basis for the equation
Ps(t) = Prs(t) + Pg(t)
This equation says that for the system under consideration
(Fig 1-1), the energy delivered by the source is equal to
the energy dissipated in the resistor plus the energy
delivered to the line. This is extremely basic and satisfies
the conservation of energy principle. This is not
superposition and any inclusion of cos(theta) terms would
be incorrect.

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.

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

...Keith

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.

...Keith

Roy Lewallen, W7EL

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

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

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

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

On Sun, 23 Mar 2008 05:10:26 -0700, Roger Sparks
wrote:

Zo =3D 1/(C*Vel)

where C is the capacitance of the line per unit distance and Vel is the vel=
ocity of the wave.

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

What is old is new again.

Hasn't this "formula" been put in the ground once before? For one,
what is velocity but something that has to be computed first only to
find us refilling all the terms back into this shortcut? For two,
"power storage capabilities... over time?" Apparently the stake
missed the heart of this one.

= A0 =A 0 = A0 = A 0 Vs = A 0 =A 0 = A0 = A0 =A 0 = A0 = A 0

73's
Richard Clark, KB7QHC

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