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I'm trying to get the jitter for an oscillator. The nominal frequency \$f_0\$ is 10 MHz, I'm integrating from 1 Hz to 100 kHz. The specifications are given as "RF Output Phase Noise (SSB)", in dBc/Hz.

The formula I'm using is:

\$x_{RMS}=\frac{1}{2 \pi f_0} \sqrt{\int_{1 Hz}^{10^5 Hz} 2 \mathscr{L} (f) df}\$

The corresponding python code is:

import numpy as np

f  = np.array([1e0,1e1, 1e2, 1e3, 1e4, 1e5])
Lf = np.array([-50,-70,-113,-128,-135,-140])
f0 = 10e6

Sphi = 2*10.0**(Lf/10.0)

jitter = 1.0/2/np.pi/f0 * np.sqrt(np.trapz(Sphi,f))

print jitter * 1e12,
print 'ps'

raw_input('done')

The result is:

159.083754236 ps
done

Two different online tools gave me an identical results of 68.66 ps, and this is also closer to what I'm expecting from lab data.

Online jitter tool

Am I using the formula incorrectly? Or maybe I made a mistake in my code?

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It turns out that the trapezoidal method is inaccurate when used on the linear data. The piece-wise linear approximation is only accurate on the log-log graph. Since I havn't found a nice detailed source on the subject, I did the calulations.

Function \$\mathscr{L}(f)\$ passing through \$(f_1,\mathscr{L}_1)\$ and \$(f_2,\mathscr{L}_2)\$, as a line on a log-log figure: \begin{equation} log_{10}(\mathscr{L}(f))=\frac{1}{10}\frac{\mathscr{L}_{2|dBc}-\mathscr{L}_{1|dBc}}{log_{10}(f_2)-log_{10}(f_1)}log_{10}(f)+\frac{1}{10}\frac{log_{10}(f_2)\mathscr{L}_{1|dBc}-log_{10}(f_1)\mathscr{L}_{2|dBc}}{log_{10}(f_2)-log_{10}(f_1)} \end{equation} This can be rewritten as: \begin{equation} \begin{aligned} \mathscr{L}(f) &= 10^{a \cdot log_{10}(f)+b}=10^{log_{10}(f^a)}10^b=f^a10^b \\ a &= \frac{\mathscr{L}_{2|dBc}-\mathscr{L}_{1|dBc}}{10(log_{10}(f_2)-log_{10}(f_1))} \qquad b = \frac{log_{10}(f_2)\mathscr{L}_{1|dBc}-log_{10}(f_1)\mathscr{L}_{2|dBc}}{10(log_{10}(f_2)-log_{10}(f_1))} \end{aligned} \end{equation} Integration of \$\mathscr{L}(f)\$ from \$f_1\$ to \$f_2\$: \begin{equation} \int_{f_1}^{f_2} 2\mathscr{L}(f)df=2\cdot10^b\int_{f_1}^{f_2} f^a df=2\frac{f_2^{a+1}-f_1^{a+1}}{a+1}10^b \end{equation} With multiple segments: \begin{equation} \begin{aligned} &\int 2\mathscr{L}(f)df = 2 \sum \frac{f_{i+1}^{a_i+1}-f_i^{a_i+1}}{a_i+1}10^{b_i} \\ a_i &= \frac{\mathscr{L}_{i+1|dBc}-\mathscr{L}_{i|dBc}}{10(log_{10}(f_{i+1})-log_{10}(f_i))} \qquad b_i = \frac{log_{10}(f_{i+1})\mathscr{L}_{i|dBc}-log_{10}(f_i)\mathscr{L}_{i+1|dBc}}{10(log_{10}(f_{i+1})-log_{10}(f_i))} \end{aligned} \end{equation}

The python code is changed to:

import numpy as np

f  = np.array([1e0,1e1, 1e2, 1e3, 1e4, 1e5])
Lf = np.array([-50,-70,-113,-128,-135,-140])
f0 = 10e6

# f_{i}, f_{i+1}, L_{i}, L_{i+1}
fi = f[:-1]
fip1 = f[1:]
Li = Lf[:-1]
Lip1 = Lf[1:]

ai = (Lip1-Li) / (np.log10(fip1) - np.log10(fi)) /10.0
bi = (Li*np.log10(fip1)-Lip1*np.log10(fi)) / (np.log10(fip1) - np.log10(fi)) /10.0
Sphi = 2*np.sum(  10.0**(bi) * (fip1**(ai+1)-fi**(ai+1))/(ai+1) )

jitter = 1.0/2/np.pi/f0 * np.sqrt(Sphi)

print jitter * 1e12,
print 'ps'

raw_input('done')

The output is 68.66 ps, it matches the online tools.

I'm still interested if somebody have a more complete source on this subject or a simpler method.

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