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