On Sat, 05 Jun 2004 23:36:07 -0500, Cecil Moore wrote:
Walter Maxwell wrote:
I'm trying to locate my Hecht paper, but can't at the moment. Let's make sure
we're talking about the same set of of equations. The ones I'm saying are
invalid as stated in his article appear on Page 46, Col 2.
Walt, I'm sorry, all I have is the QEX CD and the pages are not the same as
they were in QEX magazine. Dr. Best's equations 13 and 15 are the classical
physics interference equations virtually identical to the irradiance equations
in _Optics_, by Hecht.
But Cecil, you CAN'T add V1 and V2 in any manner to obtain forward power,
because adding V1 and V2 does not yield forward voltage.
Yes, it does, Walt. Consider the following matched system similar to the
example in Dr. Best's article.
100w XMTR-----50 ohm line---x---1/2WL 150 ohm line---50 ohm load
P1 = 100W(1-rho^2) = 100(1-.25) = 75W
P2 = 33.33W(rho^2) = 33.33W*.25 = 8.33W
P1 = 75W, P2 = 8.33W, PFtotal = 133.33W
V1 = 106.07V, V2 = 35.35V, VFtotal = 141.4V
V1 + V2 = VFtotal, 106.07V + 35.35V = 141.4V
How many times do I have to explain that V1 + V2 does NOT equal VF total?
But it does, Walt. See above.
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. 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. 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'.
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.
Now using V1 and V2 in Eqs 7 and 8 we get P1 = 133.33 w and P2 = 33.333 w. So
far so good, these Eqs are correct and valid.
But now let's plug V1 and V2 into Eq 9:
(V1 + V2) = 122.47 v. Then 122.47^2/50 = 300 w. Something's amiss here. So let's
go a step further and plug P1 and P2 into Eq 13:
P1 = 133.33 w and P2 = 33.333 w.
Then using Eq 13, PF total = (sqrt 133.33 + sqrt 33.333)^2 = 300 w.
We get the same incorrect value as with Eq 9. Now why can this be?
For starters, P1 derived from Eq 7 is already the total forward power. So why is
the forward power value P1 plugged into Eq 13 do derive the total foward power
PFtotal?
Do you see where this is going, Cecil?
So now let's go one step backward to use Eq 11:
Plugging V1 and V2 into Eq 11 we get VFtotal^2 = 14 ,999.6983 (call it 15,000).
Therefore, VFtotal = sqrt VFtotal^2 = 122.4733 v.
This is the same incorrect value obtained using Eq 9.
So, Cecil, do you still believe these equations are valid for every matched
situation?
Walt
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