On Tue, 08 Jun 2004 12:58:17 -0500, Cecil Moore wrote:
Walter Maxwell wrote:
On Mon, 07 Jun 2004 22:22:49 -0500, Cecil Moore wrote:
snip.
Please consider the following matched system with a 1/2 second long lossless
transmission line. The physical rho is 0.5. Assume VF1 is a constant 70.7V.
100W XMTR---50 ohm line---x---1/2 second long lossless 150 ohm line---50 ohm load
VF1=70.7V-- VF2--
--VR1 --VR2
V1 is equal to 70.7(1.5) = 106.06V and is a constant value
V2 starts out at zero and builds up to 35.35V
VF2=V1+V2 assuming they are in phase
Here is (conceptually simplified) how VF2 builds up to its steady-state value.
VR2(t+1) is 1/2 of VF2(t) and V2(t+1) is 1/4 of VF2(t)
Time in
Seconds V1 VR2 V2 VF2
0 106.06 0 0 106.06
1 106.06 53.03 26.5 132.56
2 106.06 66.26 33.13 139.19
3 106.06 69.58 34.79 140.85
4 106.06 70.4 35.2 141.26
5 106.06 70.63 35.32 141.38
6 106.06 70.69 35.34 141.39
7 106.06 70.7 35.35 141.4 very close to steady-state
My rounding is a little off but above is a close approximation to what happens.
Note that VF2 is equal to V1 + V2 and converges on the proper value of 141.4V.
Sorry to spoil your fun, Cecil, but 141.4 v is not the forward voltage, it's the
max voltage of the standing wave. This is the same mistake that Steve made. I
asked you to rethink what the forward voltage and you have come up wrong.
OK Cecil, I now understand why 141.4 v is the forward voltage. But there is more
below.
Nope, I haven't, Walt. The maximum voltage of the standing wave equals the
maximum forward voltage plus the maximum reflected voltage. The forward
voltage is 141.4V. The reflected voltage is 70.7V. The maximum standing-wave
voltage is 141.4V + 70.7V = 212.1V. The minimum standing-wave voltage is
141.4V - 70.7V = 70.7V. The VSWR = 212.1V/70.7V = 3:1. Rho = 0.5
SWR = (1+Rho)/(1-Rho) = 1.5/0.5 = 3:1 Everything is perfectly consistent.
So the maximum standing-wave voltage is 212.1V, not 141.4V.
And yes, Cecil I now concur that the standing wave voltage is 212.1 v. There is
still more.
Please reconsider - here's the steady-state powers for the above
matched example:
100W XMTR-----50 ohm line---x---150 ohm line---50 ohm load
PF1=100W-- PF2=133.33W--
--PR1=0 --PR2=33.33W
The forward power is 133.33W and Z0=150 ohms. Last time I looked,
the square root of [133.33W(150 ohms)] was 141.4V.
141.4V^2/150 ohms equals 133.33W. The forward voltage is indeed 141.4V.
Yep, I agree, but as I said above, there is still more.
Until you discover your mental block, whatever it is, this discussion is
not going to progress. The above example is exceptionally simple so your
mistake must also be simple. You, yourself, have used the above example
and always agreed that 100W + 33.33W = 133.33W, i.e. forward power equals
generated power plus reflected power.
I assure you, Cecil, there is no mental block.
Returning to Steve's paper, and as I said in the preceding post, Steve precedes
his Eq 13 stating this equation works under the conditions where V1 and V2 are
in phase. This means that the system is matched and thus the re-reflected
voltage equals the reflected voltage. With forward voltage V1 = 141.4 and
reflected voltage V2 = 70.71, and using these values in his Eqs 7 and 8 and then
plugging the resulting values of P1 and P2 in Eq 13, we get PFtotal = 300 w.
This again proves that Eq 13 is invalid, because P1 from Eq 7 already derived
the total forward power, yet he uses P2 in Eq 13 to obtain the total forward
power.
Do you not see what's wrong here?
Walt
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