= Atomic clock frequency lock
{title2=$E(\nu_c)=S(\nu_c+d)-S(\nu_c-d)$}
A peak has zero slope at its center, so a single peak-height reading does not tell <feedback control> which way to tune. Instead sample the two sides at ordinary <frequencies> $\nu_c\pm d$, where $d>0$ is a chosen modulation offset, and subtract:
$$
E(\nu_c)=S(\nu_c+d)-S(\nu_c-d).
$$
For an even signal around the true center $\nu_0$, write $\epsilon=\nu_c-\nu_0$. A <Taylor expansion> gives $E(\nu_0+\epsilon)=2S'(\nu_0+d)\epsilon+O(\epsilon^3)$. On the descending side of a positive central peak $S'(\nu_0+d)<0$, so the update $\nu_c\mapsto\nu_c+gE$ has restoring sign for $g>0$ and is locally stable in this ideal discrete update when $0<-2gS'(\nu_0+d)<2$. A detector dip requires the opposite sign. These conditions follow from linearizing the update, not from assuming any particular electronics.
For the ideal signal $S=A\cos^2[\pi(\nu-\nu_0)T]+S_{\mathrm{bg}}$, choose $d=1/(4T)$. Then $E\simeq-2\pi AT\epsilon$. With $T=5\,\mathrm{ms}$ and $A=1$, a center error of $+1\,\mathrm{Hz}$ gives exact $E=-\sin(0.01\pi)\simeq-0.03141$. The sign directs the update toward lower <frequency>. Constant background cancels, but background drift between samples or asymmetric neighboring <spectral lines> can bias the zero. An initial scan matters because other fringes can also supply zeros.
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