| Ken Kundert 
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			Jonathan,Your question seems to be "how do I convert from Sphi to L?". Let me show you how its done assuming a sinusoidal signal. Let,
 (1)    v(t) = A sin(2 pi f0 t + phi(t))
 Assume phi is small, then
 (2)    v(t) = A sin(2 pi f0 t) + A cos(2 pi f0 t) phi(t)
 (3)    Sv(f) = (A2/4)(Sphi(f - f0) + Sphi(f + f0))
 Let f = f0 + df and assume that the phase noise is bandlimited so that Sphi(f + f0) is negligible. Then
 (4)    Sv(f0 + df) = (A2/4)Sphi(df)
 Finally,
 (5)    L(f0 + df) = (A2/4)Sphi(df)/(A2/2)
 (6)    L(f0 + df) = Sphi(df)/2
 
 Clearly from (1), Sphi has units of rads2/Hz (as it does everywhere in my paper, including eqn 19 on page 12). But as you said, L has units of 1/Hz.  So, from all appearances, the units on (6) do not balance. That is because factor of 1/2 appears to be a unitless number. In fact, it has units of 1/rads2. It got those units when the Taylor series was applied to (1).  To see how, let
 x(phi) = 2 pi f0 t + phi
 y(x) = A sin(x)
 Then
 dy/d phi = dy/dx dx/d phi
 Notice that dy/dx has units of V/rad and dx/d phi has units of rads/rad. So dy/d phi has units of V/rad. And so
 y(x) = A sin(2 pi f0 t + phi)    [V]
 dy/d phi = A cos(2 pi f0 t)      [V/rad]
 Thus, the last term in (2) appears to have units of [V*rads], but it really has units of [V/rad][rads] = [V].
 
 I believe it was these subtleties in the units that Jitter Man was trying to warn you about.
 
 -Ken
 
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