On Dec 27, 10:53*am, Cecil Moore wrote:
Keith Dysart wrote:
It is worth doing to convince yourself. Then examine
P(t) to understand how the instaneous energy transfer
varies with time.

Oh, I know how to integrate P(t).

But did you know that your favourite V * I * cos(theta)
was derived from the function describing instantaneous
power?

But I don't
comprehend the utility of the following:

The instantaneous value of voltage is 10 volts.
The instantaneous value of current is 1 amp.
The voltage and current are in phase.

The instantaneous power is 10 joules per 0 sec?

There is definitely a problem with that.

But an instantaneous value of 10 joules/sec; that
is useful. With the function describing the
intantaneous values with respect to time, you
can integrate. You can find the total energy
transfered. You can find when it is transferred.
Is it steady? Or does it vary? There is lots
to learn.

You can even learn that when the instantaneous
power at some point is 0 for all instances,
then no energy is transferred. That would be
a useful learning.

...Keith

Roy Lewallen wrote:
Roger wrote:

I think we are in sync here, but something is missing. When I think
of a traveling sine wave, it must have a beginning as a point of
beginning discussion. I pick a point which is the zero voltage point
between wave halves. It follows that the maximum voltage point will
be 90 degrees later. I think you are doing the same thing, but maybe
not.

Next I imagine the whole wave moving down the transmission line as an
intact physical object, with the peak always 90 degrees behind the
leading edge. In our example 1/2 wave line, the leading edge would
reach the open end 180 degrees in time after entering the example. We
can see then, that the current peak will be at the center of the
transmission line when the leading edge reaches the end.

Yes, you're describing some of the properties of a sinusoidal traveling
wave. I generally describe them mathematically.

Again, we seem to be in complete agreement except for the statement
"In those situations, infinite currents or voltages occur during
runup". For many years I thought that "initial current into a
transmission line at startup" would be very high, limited only by the
inductive characteristics of the line. With this understanding, I
thought that voltage would lead current at runup. It was not until I
saw the formula Zo = 1/cC that I realized that a transmission line
presents a true resistive load at startup. Current and voltage are
always in phase at startup.

They are provided that Z0 is purely resistive. That follows from the
simplifying assumption that loss is zero or in the special case of a
distortionless line, and it's often a reasonable approximation. But it's
generally not strictly true.

But that doesn't have anything to do with my statement, which deals with
theoretical cases where neither end of the line has loss. For example,
look at a half wavelength short circuited line driven by a voltage
source. Everything is fine until the initial traveling wave reaches the
end and returns to the source end.

If we agree that voltage and current are always in phase in the
traveling wave, then we should find that in our example, the system
comes to complete stability after one whole wave (two half cycles) is
applied to the system.

Assuming you're talking about the half wavelength open circuited line
driven by a voltage source -- please do the math and show the magnitude
and phase of the initial forward wave, the reflected wave, the wave
re-reflected from the source, and so forth for a few cycles, to show
that what you say is true. My calculations show it is not. I'd do it,
but I find that the effort of showing anything mathematically is pretty
much a waste of effort here, since it's generally ignored. It appears
that the general reader isn't comfortable with high school level
trigonometry and basic complex arithmetic, which is a good explanation
of why this is such fertile ground for pseudo-science. But I promise
I'll read your mathematical analysis of the transmission line run-up.

Roy Lewallen, W7EL

OK. I think I should tweak the example just a little to clarify that
our source voltage will change when the reflected wave arrives back at
the source end. To do this, I suggest that we increase our transmission
line to one wavelength long. This so we can see what happens to the
source if we pretended that we had not moved it all.

We pick our lead edge at wt-0 and define it to be positive voltage. The
next positive leading edge will occur at wt-360. Of course, a half
cycle of positive voltage will follow for 180 degrees following points
wt-0 and wt-360.

Initiate the wave and let it travel 540 degrees down the transmission
line. At this point, the leading edge wt-0 has reflected and has
reached a point 180 degrees from the full wave source. This is the point
that was originally our source point on the 1/2 wave line.
Mathematically, wt-0 is parallel/matched with wt-360, but because the
wt-0 has been reflected, the current has been reversed but the voltage
has not been changed.

Lets move to wave point wt-1 and wt-361 so that we will have non-zero
voltage and current.

vt = vr(wt-1) + vf(wt-361) = 2*.5*sin(1)

where 1v p-p has been originally applied and vf(t) = vr(t) and vt(t) is
total voltage at any time point. Notice that the total voltage is now
2vr(t) = 2vf(t). This doubling continues as the wave moves forward, with

vt(t) = vr(wt-2) + vf(wt-362) ......... = 2vf(t) = 2vr(t)

The current is similar except, very important, the current was reversed
when reflected from the open end.

it(t) = vf(wt-362) - vr(wt-2) ...... = vf(t) - vr(t) = 0


The effect on the old source point is to make the impedance infinitely
high for all ongoing wave forward motion, which is not stopping the
wave, only indicating that power no longer moves past this point.

At time 720 degrees, the reflected wave (wt-0) reaches our revised
source point where it matches with wt-720 and begins raising the source
impedance, stopping power movement into the system. From this time on,
no further power enters the revised system because of the high impedance.

We should notice here that no power leaves the system after this time as
well. The high impedance works both ways, for forward and reflected
wave. There is no need for additional reflection analysis because both
source and full wave systems are stable and self contained after this
720 degree point. The source is effectively "turned off", and the full
wave system isolated.

If the source was parallel with a 50 ohm resistor (assuming a 50 ohm
transmission line), then the reflected wave would be matched with the
resistor and absorbed. Power would be continually moved through the
transmission line, giving hot spots on the line at 90 and 270 degree
points. We would still see the doubled voltage points at the end (360
degrees) and 180 degree points.

If we move to the shorted transmission line case, the math is identical
except that the voltages are reversed but currents add. The result at
the source is a low impedance where power can no longer be applied
because the voltage is always zero. Could we make current flow through
a resistor in parallel to the transmission line at the source in this
case? I think not.

So I think.

Thanks for taking time to consider these words. It takes real time to
carefully consider the arguments (and to present them).

73, Roger, W7WKB

Roger wrote:
There is a third possibility. The interaction of the two waves can
establish a very high resistance, so high that no current flows-zero.

This is a common confusion of cause and effect. Take a
short-circuit 1/4WL stub for instance. A very high
resistance is established at the mouth of the stub but
that high resistance has zero effect on the forward current
which keeps on flowing into the stub. Without the forward
current flowing uninhibited into the stub, the very high
resistance could not be maintained. In fact, the current
is a maximum at the shorted end of a 1/4WL stub. If you
don't believe it, measure it. Where did that current come
from if current cannot flow into the stub?

The stub impedance is the result of the ratio of voltage to
current. It is a virtual impedance and since it is an effect,
it cannot be the cause of anything. The people who say that
a virtual impedance is the same thing as an impedor have not
read the definitions of those things in the IEEE Dictionary.
--
73, Cecil http://www.w5dxp.com

Dave Heil wrote:
Please note that I now say and have
previously written that I do not believe that the Sun rises, ...

Good, that is all I was trying to get you to admit -
that the "rising of the sun" is an illusion caused by
the rotation of the earth.
--
73, Cecil http://www.w5dxp.com

Keith Dysart wrote:
Cecil Moore wrote:
The instantaneous power is 10 joules per 0 sec?

There is definitely a problem with that.

But an instantaneous value of 10 joules/sec; that
is useful.

But that instantaneous instant is NOT one second
long. Exactly how long is that instantaneous
instant? 1 ms? 1 us? 1 ns? more? less?
--
73, Cecil http://www.w5dxp.com

Dave Heil wrote:
Please note that I now say and have
previously written that I do not believe that the Sun rises, sets or
travels across the sky.

Very good, that is all I was trying to get you to admit -
that the "rising of the sun" is an illusion which was
my original point.

Incidentally, Webster's says the sun does actually rise. :-)
--
73, Cecil http://www.w5dxp.com

Cecil Moore wrote:

[...]

OK Cecil, good job well done ... again, you logic is flawless--I BELIEVE!

Time to come home now. (to reality)

Heils' got a "hardon", let him be--time'll fix it ... maybe ... ;-)

He still ain't done nothin' I ain't ever done.

Regards,
JS

Cecil Moore wrote:

...
But that instantaneous instant is NOT one second
long. Exactly how long is that instantaneous
instant? 1 ms? 1 us? 1 ns? more? less?

TIME?

I thought we already dealt with that; ain't no such thing--there IS
movement ... you will argue these points forever--until you STOP
believing in time (like Santa.)

Regards,
JS

Cecil Moore wrote:
Roger wrote:
There is a third possibility. The interaction of the two waves can
establish a very high resistance, so high that no current flows-zero.

This is a common confusion of cause and effect. Take a
short-circuit 1/4WL stub for instance. A very high
resistance is established at the mouth of the stub but
that high resistance has zero effect on the forward current
which keeps on flowing into the stub. Without the forward
current flowing uninhibited into the stub, the very high
resistance could not be maintained. In fact, the current
is a maximum at the shorted end of a 1/4WL stub. If you
don't believe it, measure it. Where did that current come
from if current cannot flow into the stub?

Stored in the 1/4 WL between the short and mouth. No more current
needed once stability is reached.


The stub impedance is the result of the ratio of voltage to
current. It is a virtual impedance and since it is an effect,
it cannot be the cause of anything. The people who say that
a virtual impedance is the same thing as an impedor have not
read the definitions of those things in the IEEE Dictionary.

Measured in ohms, virtual, and impedance is not the same as impedor. OK
An effect caused by earlier events. Agreed.

73, Roger, W7WKB

On Dec 27, 6:28*pm, Cecil Moore wrote:
Keith Dysart wrote:
Cecil Moore wrote:
The instantaneous power is 10 joules per 0 sec?

There is definitely a problem with that.

But an instantaneous value of 10 joules/sec; that
is useful.

But that instantaneous instant is NOT one second
long. Exactly how long is that instantaneous
instant? 1 ms? 1 us? 1 ns? more? less?

Time for some calculus review. Look up differentiation.

Perhaps try googling "in the limit as t approaches 0..."

But at least you now see the utility.

...Keith