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From: Bitrex on 6 Aug 2010 02:34
In Wes Hayward's book "Introduction to Radio Frequency Design" on page
116, the voltage reflection coefficient is given as gamma = V-/V+ =
(z-1)/(z+1) where z is the normalized impedance, Zin/Z_o. As an
example, the author then takes a circuit comprised of a 2 volt source
into a voltage divider to calculate the output voltage based on
reflected voltage waves. For example, with a source resistance of 1 ohm
and a load resistance of 0.5 ohm, gamma = (0.5 -1)/(0.5 +1) = -0.333.
The forward wave is said to be 1 and the reverse wave is -0.333 leaving
a sum of 0.667, which is the same thing you would get using the voltage
divider equation: (2*0.5)/(0.5+1) = 0.667.

My point of confusion is that this example only seems to work with a
source voltage of 2 volts. Why does a 2 volt supply cause the forward
wave to be 1? I'm probably missing something obvious, but could use some
insight.
From: Andrew Holme on 7 Aug 2010 17:19

"Bitrex" <bitrex(a)de.lete.earthlink.net> wrote in message
news:gPudnRJ1dJ5-M8bRnZ2dnUVZ_rydnZ2d(a)earthlink.com...
> In Wes Hayward's book "Introduction to Radio Frequency Design" on page
> 116, the voltage reflection coefficient is given as gamma = V-/V+ =
> (z-1)/(z+1) where z is the normalized impedance, Zin/Z_o. As an example,
> the author then takes a circuit comprised of a 2 volt source into a
> voltage divider to calculate the output voltage based on reflected voltage
> waves. For example, with a source resistance of 1 ohm and a load
> resistance of 0.5 ohm, gamma = (0.5 -1)/(0.5 +1) = -0.333. The forward
> wave is said to be 1 and the reverse wave is -0.333 leaving a sum of
> 0.667, which is the same thing you would get using the voltage divider
> equation: (2*0.5)/(0.5+1) = 0.667.
>
> My point of confusion is that this example only seems to work with a
> source voltage of 2 volts. Why does a 2 volt supply cause the forward
> wave to be 1? I'm probably missing something obvious, but could use some
> insight.
>

If there's a voltage V across a matched resistive load R connected via a
transmission line of characteristic impedance R to a generator with a source
impedance R, the Thevenin equivalent of the generator is a voltage soure of
e.m.f. 2*V in series with a source resistance R. The open-circuit voltage
across the generator output would be 2*V if you disconnected the load.



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