Past exam of the physics A course of the University of Cambridge / 2014 / Paper 1 / Section D / D13 / b / Solution

A lock-in amplifier multiplies the measured signal by a synchronized reference and low-pass filters the product. For $s(t)=A\cos (\omega t+\varphi )+n(t)$ and reference $2\cos \omega t$, the signal product is $A\cos \varphi +A\cos (2\omega t+\varphi )$. Averaging away the second harmonic gives $X=A\cos \varphi$ in this peak-amplitude convention. A second reference $-2\sin \omega t$ gives $Y=A\sin \varphi$, soA single quadrature could read zero for a substantial signal $90^{\circ }$ from its reference. Two quadratures avoid that blind spot; the phase is undefined at zero amplitude. Instrument RMS calibrations can include a different constant normalization.

The reference shifts the wanted frequency to zero, where a narrow low-pass bandwidth can select it. Uncorrelated broadband noise averages down, whereas a simple amplifier would enlarge signal and noise together. Longer averaging improves narrow-band sensitivity but slows the response to changing amplitude. The signal must remain phase coherent with the reference; frequency mismatch creates a beat instead of a constant output. interference at the same reference frequency survives, so shielding and independent background measurements remain necessary.^[ref]