Walter Maxwell wrote:
Well, Cecil, so far I haven't been able to follow the logic in the lines above.
Perhaps they are correct for that particular example, but I'm not so sure.
Please pick any example of your choice of matched systems. Dr. Best's equation
13 will be valid for any matched system. VFtotal = V1 + V2 will be valid for
any system, matched or not.
For
example, if 100 w is available only 75 w will enter the 150-ohm line initially,
so initially only 56.25 w is absorbed in the 50-ohm load and 18.75 w is
reflected, which then sees a 3:1 mismatch on return to the 50-ohm line, making
the re-reflected power 4 .69 w . I haven't had time to work through the
remaining steps to the steady state.
When you do, you will get the same values as I.
So before I do I notice that you state that
Eqs 10 thru 15 are valid for any matched system. So let's see if this is so by
using the example Steve used earlier in his Part 3, the one he took from
Reflections and then called it a 'fallacy'.
As I have said before, Steve's equations are correct but he simply drew the
wrong conclusions from them. His "fallacy" is a fallacy but has nothing to
do with his equations. He simply drew the wrong conclusions from valid
equations.
This is that example: 50-ohm lossles line terminated with a 150 + j0 load. 100 w
available from the source at 70.71 v. Matched at the line input, in the steady
state the total forward power Pfwd = 133.33 w and power reflected is Pref =
33.333 w. With these steady state powers the forward voltage is 81.65 v and
reflected voltage is 40.82 v. Due to the integration of the reflected waves the
source voltage 70.71 increased 10.94 v to 81.65 v. Therefore, in the steady
state V1 = 81.65 v and V2 = 40.82 v.
Nope, you simply misunderstood what Steve said. V1 and V2 are *NOT* on the 50 ohm
line. In fact, the 1 WL 50 ohm line is irrelevant and just serves to confuse. V1
and V2 are voltages existing *at the match point* at the INPUT of the tuner. Steve
chose an example that is virtually impossible to explain or understand.
Let's take it step by step starting with Steve's example:
100W XMTR---50 ohm line---tuner---1WL 50 ohm line---150 ohm load
The 1WL 50 ohm lossless line is irrelevant except for power measurements so
eliminate it.
100W XMTR---50 ohm line---tuner---150 ohm load
The tuner can now be replaced by 1/4WL of 86.6 ohm feedline.
100W XMTR---50 ohm line---x---1/4WL 86.6 ohm line---150 ohm load
100W-- 107.76W--
--0W --7.76W
V1--
V2--
Now we have an example that is understandable and we haven't changed
any of the conditions.
The magnitude of the reflection coefficient is 0.268. That makes the magnitude
of the transmission coefficient equal to 1.268 (Rule of thumb for matched
systems with single step-function impedance discontinuities)
So V1 = 70.7 * 1.268 = 89.6V V2 = VF2 * 0.268 = 6.95V
VFtotal = V1 + V2 = 89.6V + 6.95V = 96.6V
PFtotal = 96.6V^2/86.6 = 107.76W
So, Cecil, do you still believe these equations are valid for every matched
situation?
Yes, they are, Walt, once one understands them. I don't blame you for being
confused about Steve's example. He chose the worst example possible and
didn't explain it very well at all.
I still maintain that you two are two inches apart and neither one of you
will budge an inch.
--
73, Cecil
http://www.qsl.net/w5dxp
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