Roger Sparks wrote:
Vrs(t) = Vs(t) - Vg(t) + Vref(t)

Good job, Roger. Let's simplify the equation through
substitution.

Let V1(t) = [Vs(t) - Vg(t)]

Let V2(t) = Vref(t)

Vrs(t) = V1(t) + V2(t)

The power to Rs would be (Vrs^2)/50 = (70.7^2)/50 = 100w, all coming
from the reflection.

Yes, because when Vs(t)=0, there is zero instantaneous
interference.

In summary, power to the resistor Rs comes via two paths, one longer
than the other by 90 degrees (in this example). The short path is the
series path of two resistors composed of the source resistor Rs and the
input to the 50 ohm transmission line measured by Vg. The long path is
the series path of one resistor Rs and one capacitor composed of the
shorted transmission line. Both paths are available at all times. Power
flows through both paths to Rs at all times, but because of the time
differential in arrival timing, at some point Rs will receive power only
from Vs, and at another point, receive power only from Vref.

What Keith is missing is that:

Vrs(t)^2 = [V1(t) + V2(t)]^2 NOT= V1(t)^2 + V2(t)^2

In his math, Keith is asserting that since

Prs(t) NOT= P1(t) + P2(t), then the reflected power is not
being dissipated in the source resistor. But every sophomore
EE knows NOT to try to superpose powers like that. Roger,
I'll bet you know not to try to superpose powers?

Since Keith doesn't listen to me, would you pass that
technical fact on to him?

When Keith uses the correct equation:

Prs(t) = P1(t) + P2(t) +/- 2*SQRT[P1(t)*P2(t)]

he will see that the equation balances and therefore
100% of the reflected energy is dissipated in the
source resistor since the interference term averages
out to zero over each cycle.
--
73, Cecil http://www.w5dxp.com

Cecil Moore wrote:
Roger Sparks wrote:
Vrs(t) = Vs(t) - Vg(t) + Vref(t)

Good job, Roger. Let's simplify the equation through
substitution.

Let V1(t) = [Vs(t) - Vg(t)]

Let V2(t) = Vref(t)

Vrs(t) = V1(t) + V2(t)

This works. It certainly helps to link my work with yours.

The power to Rs would be (Vrs^2)/50 = (70.7^2)/50 = 100w, all coming
from the reflection.

Yes, because when Vs(t)=0, there is zero instantaneous
interference.

In summary, power to the resistor Rs comes via two paths, one longer
than the other by 90 degrees (in this example). The short path is the
series path of two resistors composed of the source resistor Rs and
the input to the 50 ohm transmission line measured by Vg. The long
path is the series path of one resistor Rs and one capacitor composed
of the shorted transmission line. Both paths are available at all
times. Power flows through both paths to Rs at all times, but because
of the time differential in arrival timing, at some point Rs will
receive power only from Vs, and at another point, receive power only
from Vref.

What Keith is missing is that:

Vrs(t)^2 = [V1(t) + V2(t)]^2 NOT= V1(t)^2 + V2(t)^2

In his math, Keith is asserting that since

Prs(t) NOT= P1(t) + P2(t), then the reflected power is not
being dissipated in the source resistor. But every sophomore
EE knows NOT to try to superpose powers like that. Roger,
I'll bet you know not to try to superpose powers?

Using circuit theory, at the peak under steady conditions, we have
141.4v applied to 70.7 ohms which gives a current of 2a. The 70.7 ohms
is sqrt(50^2 + 50^2). Two amps flowing through Rs = 50 ohms, the power
to Rs is (2^2) * 50 = 4 * 50 = 200w.

Using your simplified equation for the voltage across Rsj

Vrs(t) = V1(t) + V2(t)

Vrs(t) = V1*sin(wt + 90) + V2*sin(wt)

In our case, V1 = V2 because both voltages are developed over a 50 ohm
load. As a result, in our case, 100w will be delivered to Rs both at wt
= 0 and wt = -90, two points 90 degrees apart. If we are looking for
the total power delivered to Rs, then it seems to me like we SHOULD add
the two powers in this case. This recognizes that power is delivered to
Rs via two paths, each carrying 100w. Alternatively, we could add the
two voltages together to find the peak voltage, and then square that
number and divide by the resistance of Rs. Both methods should give the
same result.

So it looks to me like Keith is right in his method, at least in this
case.


Since Keith doesn't listen to me, would you pass that
technical fact on to him?

When Keith uses the correct equation:

Prs(t) = P1(t) + P2(t) +/- 2*SQRT[P1(t)*P2(t)]

he will see that the equation balances and therefore
100% of the reflected energy is dissipated in the
source resistor since the interference term averages
out to zero over each cycle.

I am still uncomfortable with my summary statement that both paths, the
long and short, are available and active at all times. On the other
hand, it is consistent with your advice that the waves never reflect
except at a discontinuity. The conclusion was that 100w is delivered to
Rs via each path, with the path peaks occurring 90 degrees apart in
time. Surprisingly, 100 percent of the reflected energy is ALWAYS
absorbed by Rs. There is no further reflection from Vs or Rs.

This is counter intuitive to me. I like to resolve the circuit into one
path, one wave form. We do that with circuit analysis. With traveling
waves, we frequently have two or more paths the exist independently, so
we must add the powers carried on each path, just like we add the
voltages or currents. (But watch out and avoid using standing wave
voltages or currents to calculate power!)

Most surprising to me is the observation that my beginning statement
(from my previous post) about the circuit evolving from a circuit with
two resistive loads, into a circuit with a resistive load and
capacitive load, is really incorrect. Once steady state is reached,
BOTH circuits are active at the same time, forming the two paths
bringing power to RS. Using that assumption, we can use the traveling
waves to analyze the circuit.

73, Roger, W7WKB

Roger Sparks wrote:
So it looks to me like Keith is right in his method, at least in this case.

Roger, do you understand why EE201 professors admonish
their students not to try to superpose powers?
--
73, Cecil http://www.w5dxp.com

On Fri, 21 Mar 2008 11:22:22 -0500
Cecil Moore wrote:

Roger Sparks wrote:
So it looks to me like Keith is right in his method, at least in this case.

Roger, do you understand why EE201 professors admonish
their students not to try to superpose powers?
--
73, Cecil http://www.w5dxp.com

No, I really don't.

I do understand that you had better not try to find the power flowing from superposed voltages because they may be the sum of apples and oranges. That is, the superposed voltages may be flowing in opposite directions so they will each be carrying power but the power will be available/reachable only for the direction toward which the wave is traveling. An SWR meter comes to mind here.

--
73, Roger, W7WKB

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

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.

--
73, Roger, W7WKB

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

On Mar 20, 2:07*pm, Cecil Moore wrote:
Roger Sparks wrote:
Vrs(t) = Vs(t) - Vg(t) + Vref(t)

Good job, Roger. Let's simplify the equation through
substitution.

Let V1(t) = [Vs(t) - Vg(t)]

Let V2(t) = Vref(t)

Vrs(t) = V1(t) + V2(t)

The power to Rs would be (Vrs^2)/50 = (70.7^2)/50 = 100w, all coming
from the reflection.

Yes, because when Vs(t)=0, there is zero instantaneous
interference.

In summary, power to the resistor Rs comes via two paths, one longer
than the other by 90 degrees (in this example). *The short path is the
series path of two resistors composed of the source resistor Rs and the
input to the 50 ohm transmission line measured by Vg. *The long path is
the series path of one resistor Rs and one capacitor composed of the
shorted transmission line. *Both paths are available at all times. Power
flows through both paths to Rs at all times, but because of the time
differential in arrival timing, at some point Rs will receive power only
from Vs, and at another point, receive power only from Vref.

What Keith is missing is that:

Vrs(t)^2 = [V1(t) + V2(t)]^2 NOT= V1(t)^2 + V2(t)^2

In his math, Keith is asserting that since

Prs(t) NOT= P1(t) + P2(t), then the reflected power is not
being dissipated in the source resistor. But every sophomore
EE knows NOT to try to superpose powers like that. Roger,
I'll bet you know not to try to superpose powers?

Since Keith doesn't listen to me, would you pass that
technical fact on to him?

When Keith uses the correct equation:

Prs(t) = P1(t) + P2(t) +/- 2*SQRT[P1(t)*P2(t)]

he will see that the equation balances and therefore
100% of the reflected energy is dissipated in the
source resistor since the interference term averages
out to zero over each cycle.

It would be excellent if you would correct my exposition
using the correct formulae. We could then see if your
proposal actually provides accurate results.

...Keith

PS. It won't.

Keith Dysart wrote:
It would be excellent if you would correct my exposition
using the correct formulae. We could then see if your
proposal actually provides accurate results.

Roger didn't understand your terms and subscripts and
neither do I. I doubt that anyone understands your
formulas well enough to discuss them.

What is the equation for the forward voltage component
dropped across Rs?

What is the equation for the reflected voltage
component dropped across Rs?

Given valid equations for those two voltages, I can
prove everything I have been saying.
--
73, Cecil http://www.w5dxp.com

On Mar 21, 1:00*pm, Cecil Moore wrote:
Keith Dysart wrote:
It would be excellent if you would correct my exposition
using the correct formulae. We could then see if your
proposal actually provides accurate results.

Roger didn't understand your terms and subscripts and
neither do I. I doubt that anyone understands your
formulas well enough to discuss them.

What is the equation for the forward voltage component
dropped across Rs?

What is the equation for the reflected voltage
component dropped across Rs?

Given valid equations for those two voltages, I can
prove everything I have been saying.

Since I am not sure exactly what your are looking for,
and you could not be sure that I did them right,
it would probably be best if you were to compute these
equations which you need.

...Keith