Atomic clock frequency lock ($E({\nu }_c)=S({\nu }_c+d)-S({\nu }_c-d)$)

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:For an even signal around the true center ${\nu }_0$, write $ϵ={\nu }_c-{\nu }_0$. A Taylor expansion gives $E({\nu }_0+ϵ)=2S^{\prime }({\nu }_0+d)ϵ+O({ϵ}^3)$. On the descending side of a positive central peak $S^{\prime }({\nu }_0+d)<0$, 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 $0<-2g{S}^{\prime }({\nu }_0+d)<2$. 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 $E\simeq -2\pi ATϵ$. 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.