Ken Kundert

Jonathan, Your question seems to be "how do I convert from S_{phi} to L?". Let me show you how its done assuming a sinusoidal signal. Let, (1) v(t) = A sin(2 pi f_{0} t + phi(t)) Assume phi is small, then (2) v(t) = A sin(2 pi f_{0} t) + A cos(2 pi f_{0} t) phi(t) (3) S_{v}(f) = (A^{2}/4)(S_{phi}(f  f_{0}) + S_{phi}(f + f_{0})) Let f = f_{0} + df and assume that the phase noise is bandlimited so that S_{phi}(f + f_{0}) is negligible. Then (4) S_{v}(f_{0} + df) = (A^{2}/4)S_{phi}(df) Finally, (5) L(f_{0} + df) = (A^{2}/4)S_{phi}(df)/(A^{2}/2) (6) L(f_{0} + df) = S_{phi}(df)/2
Clearly from (1), S_{phi} has units of rads^{2}/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/rads^{2}. It got those units when the Taylor series was applied to (1). To see how, let x(phi) = 2 pi f_{0} 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 f_{0} t + phi) [V] dy/d phi = A cos(2 pi f_{0} 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
