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

On Apr 3, 6:47*pm, Roger Sparks wrote:
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. *

I do believe that -1.234 is correct. The reflection coefficient is -1,
so at the
short, Vr(t) is always equal to -Vf(t), which, when summed, give a
Vtot(t) of 0
which is what is expected at a short. Vr.g(t) is delayed by 90 degrees
from
Vf.g(t) so at 91 degrees, we should expect Vr.g(91) to be (-1 *
Vf.g(1)) or
-1.234. This is also how the spreadsheet computes it.

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.

Vg(t) is the voltage across the line, so Vg(t) is the sum of Vf.g(t)
and Vr.g(t),
recalling that Vtot = Vf + Vr.

And
Vrs(t) = Vs(t) - Vg(t)
Since Vf.g(t) = Vs(t)/2, and Vg(t) = Vf.g(t) + Vr.g(t) we have, by
substitution,
Vrs(t) = 2 * Vf.g(t) - Vf - Vr.g(t)
= Vf.g(t) - Vr.g(t)

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. *

Be careful if you are comparing things before and after the 90 degree
point. At 90
degrees, the circuit conditions change dramatically.

At the entrance to the line, Ig is delayed 90 degrees from Vg which is
to be expected
since the line is behaving as a pure inductor from any time after 90
degrees. Before
that the line is behaving as a resistance.

At the resistor, the offset is only half this because the resistor and
the line have
the same magnitude of impedance.

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.

One does have to be careful to treat the signs consistently. In this
sheet, the convention
is that flows (current and energy) to the right (assuming the load is
to the right) are
positive. Some conventions use positive to mean flows to the right for
the forward wave
and flows to the left for the reflected wave.

On this sheet, Vtot = Vf+Vr, Itot=If+Ir and Ptot=Pf+Pr.
I hope no gremlins snuck in.

...Keith

On Thu, 3 Apr 2008 17:50:05 -0700 (PDT)
Keith Dysart wrote:

On Apr 3, 6:47*pm, Roger Sparks wrote:
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. *

I do believe that -1.234 is correct. The reflection coefficient is -1,
so at the
short, Vr(t) is always equal to -Vf(t), which, when summed, give a
Vtot(t) of 0
which is what is expected at a short. Vr.g(t) is delayed by 90 degrees
from
Vf.g(t) so at 91 degrees, we should expect Vr.g(91) to be (-1 *
Vf.g(1)) or
-1.234. This is also how the spreadsheet computes it.

OK, I see now. You reversed the sign to calculate Vrs so once I get to the Vrs column, we are the same. You used the -1 reversal to calculate Vg, but plus one to calculate Vrs.

clip

On this sheet, Vtot = Vf+Vr, Itot=If+Ir and Ptot=Pf+Pr.
I hope no gremlins snuck in.

...Keith

I will go back and look some more at the power relationships from your earlier posting.


--
73, Roger, W7WKB