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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