imtired
Junior Member
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Posts: 23
Santa Rosa
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The paper doesn't give the function for the phase noise curve, but by eyeing the graph, this is what I calculated:
Assuming L(f) = k + a/f in the region of integration, (where a,k are constants, and f is frequency)
integral( k+a/f, from f1 to f2 ) = k*(f2 - f1) + a* ln( f2/f1 ), where k=10^-150/10, and a = 10^-140/10 * 12e3, f1=12e3, f2=20e6.
plugging in and convert to dBc: 10 * log( 1e-15*(20e6-12e3) + 1e-14*12e3*ln(20e6/12e3) ) ~ -76.8 dBc.
My answer is not the same as in the paper, but I think it's pretty close, given that I had to eyeball the function. But this should demonstrate how to go about integrating the area under a phase noise curve, as a step to calculating the jitter.
Notice that L(f) has units dBc/Hz. Once you integrate over frequency, you get [dBc/Hz] * [Hz] = dBc.
I hope this helps you.
Regards, Robert
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