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

The sudden switch in circuit design at time 90 degrees is not unique to
start up, but is true for any adjustment made to Vs and any returning
wave from the shorted stub,. *As a result, a true stable circuit will
never be found unless some voltage adjustment is allowed for the 90
timing shift caused by the shorted stub. *Keith (in his analysis of the
circuit) recognizes that Vs drives into a reactive circuit.

If we want to understand how constructive and destructive interference
act to cause a 50 ohm resistor to evolve into a 50 ohm capacitor, then
we need to examine how traveling waves might do this. *It would be nice
to have a formula or wave sequence that fully addressed this evolution..

*From circuit theory, we have the inductive reactance of a
short-circuited line less than 1/4 wavelength long is
* * *XL = Zo * tan (length degrees)
* * * * = 50 * tan(45)
* * * * = 50 ohms
*From traveling wave theory, we would have the applied wave from the
source arriving 90 degrees late to the stub side of resistor Rs. *This
ignores the fact that current must already be passing through resistor
Rs because voltage has been applied to Rs from the Vs side 90 degrees
earlier. *Whoa! *Things are not adding up correctly this way!

We need to treat the wave going down the 50 ohm line as a single wave
front. *The wave reverses at the short-circuit, reversing both direction
of travel and sign of voltage. *When the wave front reaches the input
end, 90 degrees after entering (for this case), the voltage/current
ratio is identical to the starting ratio (the line was 45 degrees long,
tan(45) = 1), and the returning voltage directly adds to the voltage
applied from Vs. *As a result, the current flowing through Rs will
increase, and Vs will see a changed (decreased) impedance.

After 90 degrees of signal application, we should be able to express the
voltage across Rs as

* * *Vrs(t) = Vs(t) - Vg(t) + Vref(t)

Just to ensure clarity, you are using a different convention for terms
and signs than I do.

For me, Vg(t) is the actual voltage at point Vg. In terms of forward and
reverse: Vg(t) = Vf.g(t) = Vr.g(t), meaning the actual voltage at point
g is sum of the forward and reverse voltage at point g.

In my terms, this leads to
* Vrs(t) = Vs(t) - Vg(t)
* * * * *= Vs(t) - Vf.g(t) - Vr.g(t)

Your Vg seems to be same as my Vf.g and you seem to be using a
different nsign for Vref than I do for Vr.

You must be carrying these differences through the equations correctly
because the final results seem to agree.

Thanks for looking at my posting very carefully, because I think you are drawing the same conclusions as I, at least so far. *

I think you are correct in saying that For me, Vg(t) is the actual voltage at point Vg. In terms of forward
and
reverse: Vg(t) = Vf.g(t) = Vr.g(t), meaning the actual voltage at
point g is sum of the forward and reverse voltage at point g. *

Ooooppps. That should have been Vg(t) = Vf.g(t) + Vr.g(t)

but doesn't this describe the standing wave? *

Not by itself. It is simply the function describing the voltage at
point g.

If you were to compute Vx(g) = Vf.x(t) + Vr.x(t) for many points
x, then you might end up with something akin to a standing wave.
Note that Vg(t) = Vf.g(t) + Vr.g(t) holds true for any function
while standing waves only really appear when the excitation
function is sinusoidal.

At point g, the forward wave has passed through resistor Rs,

This is not a way that I would recommend thinking about what is
happening. The voltage source and source resistor are lumped
circuit elements and thinking that wave is passing through
them is likely to lead to confusion.

It might be valuable to study the circuit completely with a
lumped voltage source, source resistor and reactance. Once
that circuit is understood, replace the reactance with the
transmission line. When studying the circuit with the
transmission line instead of the reactance, always check
to see whether you would use the same words if the TL were
replaced with the reactance. If not, question why.

loosing part of the original wave energy, and is inroute beginning the long path to the resistor Rs. *The reflected wave has arrived back to one side of Rs and the voltage is already in series with the source voltage acting on Rs. *I think we need to subtract the two terms so that we can separate the apples and oranges.

Vs(t) = Vs(wt + 90), Vg(t) = Vs(wt + 90)/2, Vref(t) = Vs(wt)/2

Substitute so

Vrs(t) = Vs(wt + 90) - Vs(wt + 90)/2 + Vs(wt)/2

* * *= Vs(wt + 90)/2 + Vs(wt)/2

Allow Vs to be represented by a sine wave, we have

A small aside. It is conventional to use cos for sinusoidal waves
because
it maps more readily into phasor notation.

This is an important observation. *cos(0) is 1 so to me it seems like we are saying that we begin at time zero with peak voltage or peak current. *I wanted to begin with zero current at time zero so I used the sine. *Am I missing something important here?

It does not alter the outcome, so is not that important.

* * 2Vrs(t) = Vs*sin(wt + 90) + Vs*sin(wt)
* * * * * * = 2Vs*(sin(wt + 45)(cos(45))
* * Vrs(t) = Vs(sin(wt + 45)(cos(45))

Vs is defined as 100v RMS, which equals 100 * 1.414 = 141.4v Peak. *The
maximum voltage would occur when the sin term was 90 degrees and equals
1, which would occur at wt = 45 degrees. * We would have

* * *Vrs(45) = 141.42 * sin(90)(cos(45)
* * * * * * *= 141.42 * 1 * 0.7071
* * * * * * *= 100v

Now consider the current. *After the same 90 degrees of signal
application, we should be able to express the current through Rs as

* * *Irs(t) = Is(t) + Iref(t)

Is(t) = Is(wt + 90), Iref(t) = Is(wt)

The reflected current has been shifted by 90 degrees due to the
reflection so we must rewrite Iref(t) to read

* * *Iref(t) = Is(wt + 90)

Substitute,

* * * Irs(t) = Is(wt + 90) + Is(wt)

Allow Is to be represented by a sine wave, we have

* * * Irs(t) = Is*sin(wt + 90) + Is*sin(wt)

* * * * * * *= 2*Is(sin(wt + 45)(cos(45))

How do we find Is? *Is is the initial current found by dividing the
applied voltage at peak (141.42v) by the initial resistance (100 ohms)..

* * *Is = 141.42/100 = 1.4142a

The maximum current would occur when the sin term was 90 degrees and
equals 1, which would occur at wt = 45 degrees. * We would have

* * * Irs(t) = 2*Is(sin(wt + 45)(cos(45))

* * * * * * *= 2 * 1.4142 * 1 * 0.7071

* * * * * * *= 2a

These results agree with the results from Keith and from circuit theory.

We have a theory and at least the peaks found from the theory agree with
the results from others. * How about Cecil's initial question which is
* 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?

We will use the equation

* *Vrs(t) = Vs(t) - Vg(t) + Vref(t)

Vs(t) = Vs(wt + 90), Vg(t) = Vs(wt + 90)/2, Vref(t) = Vs(wt)/2

The challenge looks at the time when Vs(t) = Vs*sin(wt + 90) = 0, which
occurs when wt = -90. *When wt = -90, all the power to the source
resistor Rs is from the reflected portion described as Vref(t) =
Vs(wt)/2. *The voltage across Rs would be

* *Vrs(-90) = Vref(-90)
* * * * * * = Vs*sin(-90)/2
* * * * * * = 141.4/2
* * * * * * = 70.7v

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

In summary, power to the resistor Rs comes via two paths, one longer
than the other by 90 degrees (in this example). *

This is not quite complete. The power to Rs does come from two places
but they are the voltage source and the line.

* Ps(t) = Prs(t) + Pg(t)

where Pg(t) is the power at point g flowing out of the generator into
the line and Ps(t) is the power flowing from the source to the source
resistor and the line.

The power into the line can be seaparated into a forward component and
a reflected component:

* Pg(t) = Pf(t) + Pr(t)

(Sometimes this is written as P = Pf - Pr, but that is just a question
of how the sign is used to represent the direction of energy flow.)

Substituting we have

* Ps(t) = Prs(t) + Pf.g(t) + Pr.g(t)
or
* Prs = Ps(t) - Pf.g(t) - Pr.g(t)

Whenever Ps(t) equal 0, then Pf.g(t) also equals 0 so at those times,
this simplifies to
* Prs = - Pr.g(t)
which is the same conclusion you came to.

No, I think my Pr.g(t) was positive, not negative. *

I think that is just because your reference direction was different.
I use the same reference direction for both Pf and Pr. It is often
done that the reference direction is different for Pf and Pr.

The difference is important because of the timing of the energy flows. *Here is where it is important to observe that the forward wave entering the transmission line will have no further effect on resistor Rs until it has traveled 90 degrees of time. *On the other hand, the reflected wave Pr.g(t) is only an instant of time away from entering the resistor Rs, but has not actually entered it because you are measuring it as being present at Vg, not as voltage Vrs.

But it is wise not to forget the Pf.g(t) term because at times when
Ps(t) is not equal to zero, the Pf.g(t) term will be needed to
balance the energy flows.

Pf.g(t) is the remaining power after the wave generated by Ps(t) has passed through Rs and left behind Prs(t). *Another Prs(t + x) will be generated by the returning reflected wave.

Again, I don't think it wise to think of the wave as passing through
Rs.

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.

And at still other times, the power in the source resistor will depend
on
all of Ps(t), Pf.g(t) and Pr.g(t), or, more simply, it will depend on
Ps(t) and Pg(t).

...Keith

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