 Lock-in quadratures

Let the input electric potential difference be $s(t)=A\cos (\omega t+\varphi )$, with peak amplitude $A\ge 0$, angular frequency $\omega >0$, and phase difference $\varphi$ relative to $\cos \omega t$. For an observation interval $[0,T]$ containing an integer number $N\ge 1$ of cycles, each lasting one period of a function, $T=2\pi N/\omega$, define the lock-in quadratures in a peak convention:These integrals average the products with dimensionless references $2\cos \omega t$ and $-2\sin \omega t$; $X$ and $Y$ have the input's units. The minus sign makes positive $\varphi$ give positive $Y$. The trigonometric functions giveThe oscillating terms have zero integral over the stated interval, proving $X=A\cos \varphi$ and $Y=A\sin \varphi$. Their orthogonality turns an oscillating signal into two constant coordinates.

Represent those coordinates as a vector, or as the complex number $Z=X+\mathrm{i}Y$. Its modulus and complex argument recover the peak amplitude and phase difference:Here $atan2(Y,X)$ selects the angle with the correct signs of both coordinates, rather than applying the arctangent to $Y/X$ and losing the quadrant. When $X=Y=0$, the amplitude is zero and the phase difference is undefined. When $\varphi =\pi /2$, $X=0$ despite a nonzero signal; measuring $Y$ prevents this blind spot. Rotating both references by a fixed angle rotates the coordinate vector but preserves $R$.

The factors of two above deliberately recover peak amplitude. The SR830 instead reports the selected sinusoidal component in root mean square units: in a calibrated, zero-offset display, $R_{\mathrm{rms}}=R/\sqrt{2}$. For $R=0.4\mathrm{V}$ this is $0.282843\mathrm{V}$, while the peak-to-peak span of that component is $0.8\mathrm{V}$. Display normalization, reference sign, gain and angle units must be checked for the particular instrument.^[ref]