= Two-line frequency separation
For equal line intensities, the correlation is proportional to $e^{-i\omega_1\tau}+e^{-i\omega_2\tau}=2e^{-i\bar\omega\tau}\cos(\Delta\omega\tau/2)$. The <two-line coherence beating> envelope is therefore $|\cos(\Delta\omega\tau/2)|$.
In the original first plot, the first visibility zeros are at approximately $\tau=\pm100\,\mathrm{ps}$; the envelope revives at $0$ and about $\pm200\,\mathrm{ps}$. Hence
$$
\frac{|\Delta\omega|}{2}(100\times10^{-12}\,\mathrm s)=\frac\pi2,
\qquad \boxed{|\Delta\omega|\simeq3.14\times10^{10}\,\mathrm{rad\,s^{-1}},\quad |\Delta\nu|\simeq5.0\,\mathrm{GHz}.}
$$
Use the envelope recurrence or its nulls, not the much faster carrier fringes. The visibility zeros indicate equal component intensities in the ideal model; unequal intensities would leave a nonzero minimum. The arbitrary vertical normalization does not affect the inferred separation.
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