On Wed, 2 Apr 2008 06:33:46 -0700 (PDT)
Keith Dysart wrote:

On Apr 2, 9:17*am, Roger Sparks wrote:
On Wed, 2 Apr 2008 02:41:24 -0700 (PDT)

Keith Dysart wrote:

clip



But your phraseology suggests another way to approach
the problem. Let us consider what happens just after
the arrival of the reflected wave. For computational
convenience, we'll use a frequency of 1/360 hertz so
that 1 degree of the wave is one second long.

The reflected wave arrives at 90 seconds (or 90 degrees)
after the source is turned on. If the line is
terminated in 50 ohms, then no reflection arrives and
during the one second between degree 90 and degree 91,
the source resistor dissipates 0.01523 J of energy. The
line also receives 0.01523 J of energy and the source
provides 0.03046 J.
* Esource.50[90..91] = 0.03046 J
* Ers.50[90..91] * * = 0.01523 J
* Eline.50[90..91] * = 0.01523 J
* Esource = Ers + Eline
as expected.
* Efor = Eline
since Eref is 0.

I come up with 141.4v across 50 plus 50 ohms. *The current should be 1.414a. *Power to each 50 ohm resistor would be *50*1.4142^2 = 200w. *For the 1 degree interval of 1 second, that would be 200 joules.

Right? *Peak current flows at 90 degrees?
--
73, Roger, W7WKB- Hide quoted text -

- Show quoted text -

We may be using different sources. My Vs is 141.4cos(wt)
so that between 90 degrees and 91 degrees, my source
voltage is going from 0 to -2.468 V.

And an ooopppps. I actually did the calculations for
a shorted load rather than 12.5 ohms as stated.
With the reflection coefficient of -1 and a 90 degree
delay, the reflected voltage between 90 and 91 degrees
changes from -70.711 V to -70.700 V.

Hoping these details resolve the disparity,

...Keith

I am sorry Keith, but I still can not duplicate your work. I wonder if the difference is in how we each apply the recharge voltage to the transmission line.

I redid my table showing the power in the source and reflected waves. I now have the power to resistor Rs coming from voltage agreeing with the expected power calculated from current.

The table can be found at http://www.fairpoint.net/~rsparks/Sm...Reflection.pdf.

-*/---------
--
73, Roger, W7WKB

On Apr 3, 9:47*am, Roger Sparks wrote:
On Wed, 2 Apr 2008 06:33:46 -0700 (PDT)
I am sorry Keith, but I still can not duplicate your work. *I wonder if the difference is in how we each apply the recharge voltage to the transmission line.

I do not know. The spreadsheet that did the work is now available from
this page:
http://keith.dysart.googlepages.com/radio6

Perhaps with the extra detail, you can find the discrepancy.

I redid my table showing the *power in the source and reflected waves. *I now have the power to resistor Rs coming from voltage agreeing with the expected power calculated from current. *

My sheet agrees with yours for the samples I checked.

...Keith

Keith Dysart wrote:
My sheet agrees with yours for the samples I checked.

Here's an example that changes nothing but allows
isolation of the component energy flows.

50 ohms
+---1WL-50-ohm---\/\/\/\---1WL-50-ohm---+
| |
Source 141.4*cos(wt) volts Load +j50
| |
+---------------------------------------+
GND GND

It appears to me that the confusion arises from
the *series* resistor.
--
73, Cecil http://www.w5dxp.com

On Thu, 03 Apr 2008 15:49:15 -0500
Cecil Moore wrote:

Keith Dysart wrote:
My sheet agrees with yours for the samples I checked.

Here's an example that changes nothing but allows
isolation of the component energy flows.

50 ohms
+---1WL-50-ohm---\/\/\/\---1WL-50-ohm---+
| |
Source 141.4*cos(wt) volts Load +j50
| |
+---------------------------------------+
GND GND

It appears to me that the confusion arises from
the *series* resistor.
--
73, Cecil http://www.w5dxp.com

This circuit has a reflection at the 50 ohm resistor because the impedance transitions from 50 ohms to 100 ohms, then back to 50 ohms.

--
73, Roger, W7WKB

Roger Sparks wrote:
This circuit has a reflection at the 50 ohm resistor because the impedance transitions from 50 ohms to 100 ohms, then back to 50 ohms.

Yet conditions are identical to the earlier example.
That should tell us something.
--
73, Cecil http://www.w5dxp.com

On Apr 3, 4:49*pm, Cecil Moore wrote:
Keith Dysart wrote:
My sheet agrees with yours for the samples I checked.

Here's an example that changes nothing but allows
isolation of the component energy flows.

* * * * * * * * * * *50 ohms
* * +---1WL-50-ohm---\/\/\/\---1WL-50-ohm---+
* * | * * * * * * * * * * * * * * * * * * * |
Source 141.4*cos(wt) volts * * * * * * * Load +j50
* * | * * * * * * * * * * * * * * * * * * * |
* * +---------------------------------------+
* *GND * * * * * * * * * * * * * * * * * * *GND

It appears to me that the confusion arises from
the *series* resistor.

It seems to me that this circuit is quite different.
There is a reflection where the left TL connects
to the 50 ohm resistor. There is a reflection when
the reflected wave on the left TL arrives back at the
source and there is a reflection when the reflected
wave on the right TL hits the 50 ohm resistor.

This circuit takes an infinitely long time to settle.
The previous circuit only had a reflection at the load
and settled after one round trip.

Quite different.

...Keith

Keith Dysart wrote:
It seems to me that this circuit is quite different.

Yet, steady-state conditions are identical. That
should tell us something about what is wrong
with the analysis so far. Wave reflection theory
works if the transmission line is a multiple of
one wavelength even if the multiple is zero.
--
73, Cecil http://www.w5dxp.com

On Apr 4, 12:01*am, Cecil Moore wrote:
Keith Dysart wrote:
It seems to me that this circuit is quite different.

Yet, steady-state conditions are identical. That
should tell us something about what is wrong
with the analysis so far.

Please expand on what it tells us "about what is wrong
with the analysis so far".

...Keith

Keith Dysart wrote:
Please expand on what it tells us "about what is wrong
with the analysis so far".

You have not been able to tell where the instantaneous
reflected energy goes. This new example should help
you solve your problem. If your steady-state equations
for the new example are not identical to the steady-
state equations for the previous example, then something
is wrong with your previous analysis. IMO, something
is obviously wrong with your previous analysis since
it requires reflected waves to contain something other
than ExH joules/sec, a violation of the laws of EM
wave physics.

feedline1 feedline2
source---1WL 50 ohm---Rs---1WL 50 ohm---+j50

The beauty of this example is that conditions at
the source resistor, Rs, are isolated from any
source of energy other than the ExH forward wave
energy in feedline1 and the ExH reflected wave
energy in feedline2. That's all the energy there
is available at Rs and that energy cannot be denied.
--
73, Cecil http://www.w5dxp.com

On Thu, 3 Apr 2008 10:35:17 -0700 (PDT)
Keith Dysart wrote:

On Apr 3, 9:47*am, Roger Sparks wrote:
On Wed, 2 Apr 2008 06:33:46 -0700 (PDT)
I am sorry Keith, but I still can not duplicate your work. *I wonder if the difference is in how we each apply the recharge voltage to the transmission line.

I do not know. The spreadsheet that did the work is now available from
this page:
http://keith.dysart.googlepages.com/radio6

Perhaps with the extra detail, you can find the discrepancy.

I redid my table showing the *power in the source and reflected waves.. *I now have the power to resistor Rs coming from voltage agreeing with the expected power calculated from current. *

My sheet agrees with yours for the samples I checked.

...Keith

I think that at 91 degrees, the sign of the returning wave should be positive, not negative. Thus, it should be +1.234v. My reasoning is that the voltage from the line side of the resistor adds to the voltage from the source side of the resistor, but in this case it takes a little longer to get around the loop. The 90 degrees is timing, the lead edge of the wave is not reversed by turning the corner at what we call the short.

Another way of thinking about it is that if the transmission line was shorter, the added voltage to the resistor from the line side would arrive even earlier, and add. If the line were zero length, there would be no question but that the voltage would add.

Lastly, your current peaks at 135 degrees, not 45 degrees, which should be the delay.

It is interesting to notice that the current is delayed 45 degrees, but the voltage is delayed 90 degrees. It is as if the current does not take the entire trip around the loop, but the voltage does. The current apparently travels an average of half the distance around the loop.

I know that the convention for calculating the reflection coefficient is to make it negative for a shorted transmission line. The negative sign makes sense because the voltages at the short cancel so we have a zero voltage across the short. That convention sure messes up our spread sheet calculations however.

--
73, Roger, W7WKB