Walter Maxwell wrote:
On Mon, 07 Jun 2004 22:22:49 -0500, Cecil Moore wrote:
Walter Maxwell wrote:
The equation he misunderstands is Eq 4.23 on Page 100 in Johnson, which is
explained and derived on Pages 98 and 99, to derive the voltage E of the
standing wave for any position along a mismatched transmission line. Steve's
misunderstanding is that he believes the equation expresses the value of the
forward voltage on the line. Consequently, his Eq 6 in Part 1 says that 'Vfwd =
the terms on the right-hand side of the equation copied from Johnson', while the
correct version is 'E = the terms on the righ-hand side'.
Johnson's equation contains all possible components - both forward and
reflected, so Johnson's equation predicts the total net standing-wave
voltage. (Note that Johnson's equation contains both positive and negative
exponents.)
Dr. Best's equation contains only the forward components of Johnson's
equation. Therefore, Dr. Best's equation predicts *only* the forward-
traveling voltage. (Note that Dr. Best's equation contains only negative
exponents and represents the E+ half of Johnson's equation while omitting
the E- half of Johnson's equation.)
In other words, the two equations do *not* predict the same quantity
and they are *not* supposed to be the same equations. As I said before,
Dr. Best made a lot of conceptual errors but his equations seem to be
valid. If you break Johnson's equation into two parts representing E+
and E-, the E+ part will be Dr. Best's Eq 6.
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.
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.
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.
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.
--
73, Cecil
http://www.qsl.net/w5dxp
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