Lock-in harmonic rejection

Consider this original constructed input electric potential difference, with $t$ in seconds and $f_0=1\mathrm{Hz}$:The unwanted component has three times the reference frequency and more than twice the wanted peak amplitude. Nevertheless, the lock-in quadratures over $T=4/f_0=4\mathrm{s}$ are exactlyso $R=0.4\mathrm{V}$ and $atan2(Y,X)=\pi /3=60^{\circ }$. This is a peak amplitude calculation, not an SR830 display prediction.

To see why the larger component vanishes, multiply $B\cos (3\omega t+\psi )$ by each reference:Each sinusoid has zero average value of a function over a whole reference cycle. More generally, a component $B\cos (n\omega t+\psi )$ with integer $n\ge 2$ gives products at $(n-1)\omega$ and $(n+1)\omega$, whose integrals vanish over any integer number of reference cycles. This is the orthogonality behind selecting a Fourier coefficient with sinusoidal references. A nonsinusoidal reference can contain additional frequencies and need not have the same rejection property.

An unwanted component at the reference frequency survives. If $u(t)=B\cos (\omega t+\psi )$ is added to the input, its coordinate vector adds to the wanted one:For example, adding $B=0.1\mathrm{V}$ with $\psi =0$ to the wanted component gives $X=0.3\mathrm{V}$, $Y=0.2\sqrt{3}\mathrm{V}$, $R=\sqrt{0.21}\mathrm{V}\approx 0.458258\mathrm{V}$, and phase difference $atan2(0.2\sqrt{3},0.3)\approx 49.1066^{\circ }$. Longer averaging cannot separate these two synchronized contributions.