Richard Clark wrote:
On Sun, 06 Jun 2004 23:18:37 -0500, Cecil Moore
wrote:
I'm sorry, Walt,
Your belief that V2 is a reflected wave is the root of the misunderstanding.
V2 is a re-reflected wave
Another way of saying "You are right, Walt, you are wrong."
It's a very minor mistake, Richard, and one easily made. If you have
a copy of Dr. Best's QEX article, please feel free to express your
take on this discussion.
Let's see if I can present superposition in ASCII graphics. rho
is the voltage reflection coefficient and tau is the voltage
transmission coefficient. Assume VF1 has a phase angle of zero
degrees. Phase angles are important in the following but since
the system is matched, all phase angles are either at zero degrees
or at 180 degrees at the match point. So a sign change is equivalent
to a 180 degree phase shift.
100W XMTR---50 ohm line---x---1/2WL 150 ohm line---50 ohm load
VF1=70.7V-- VF2=141.4V--
--VR1=0V --VR2=70.7V
According to the rules of superposition, the two voltages incident
upon 'x', VF1 and VR2, can be considered separately and then added.
----------------------------------------------------------------
Breaking VF1 down into its two superposition components yields:
x
|
VF1=70.7V (100W)--|
|-- V1=106.06V (75W)
V3=35.35V (25W)--|
|
VF1 = 70.7V at zero degrees (100W)
V1 = VF1(forward-tau) = 70.7(1.5) = 106.06V at zero degrees (75W)
V3 = VF1(forward-rho) = 70.7(0.5) = 35.35V at zero degrees (25W)
Note that PF1 = 100W = P1 + P3 = 75W + 25W
-----------------------------------------------------------------
Breaking VR2 down into its two superposition components yields.
x
|
|-- VR2=70.7V (33.33W)
V4=35.35V (25W)--|
|-- V2=35.35V (8.33W)
|
VR2 = 70.7V at 180 degrees (33.33W)
V2 = VR2(reverse-rho) = 70.7(-0.5) = 35.35V at zero degrees (8.33W)
V4 = VR2(reverse-tau) = 70.7(0.5) = 35.35V at 180 degrees (25W)
Note that PR2 = 33.33W = P2 + P4 = 25W + 8.33W
-----------------------------------------------------------------
Now, following the rules of superposition:
To get the total forward voltage, add V1 + V2
VF2 = V1 + V2 = 106.06V + 35.35V = 141.4V (133.33W)
To get the total reflected voltage, add V3 and V4
VR1 = V3 + V4 = 35.35V - 35.35V = zero volts (0W)
Note: Dr. Best neglected to mention P3 and P4 in his QEX article.
P3+P4 is the interference joules/sec. They are scalar values.
All voltages are consistent and all powers are consistent.
So, in this matched system, all reflected power is re-reflected:
PF2 = 133.33W = P1 + P2 + P3 + P4 = 75W + 8.33W + 25W + 25W
PF2 = P1 + P2 + (complete constructive interference) = 133.33W
PR1 = P3 + P4 - (complete destructive interference) = zero watts
Note that the voltage forward-rho = (150-50)/(150+50) = +0.5
reverse-rho = (50-150)/(50+150) = -0.5 (180 deg phase shift)
--
73, Cecil
http://www.qsl.net/w5dxp
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