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An observation interval of finite duration is not an infinitely sharp frequency selector. Combine the lock-in quadratures intoFor unwanted electric potential difference $s(t)=B\cos ((\omega +\delta )t+\psi )$, where $\delta$ is an angular frequency offset and $\omega +\delta >0$, expansion into complex exponentials gives the exact responseThe sinc function formula follows by integrating the exponential function and factoring $e^{\mathrm{i}\nu T}-1=2\mathrm{i}{e}^{\mathrm{i}\nu T/2}\sin (\nu T/2)$; its value at zero follows from a limit. The first term changes slowly after multiplication; the second oscillates at the sum frequency. It vanishes at exact matching when $T$ is an integer number of reference cycles, each lasting one period of a function, but cannot generally be discarded in a short-window calculation.

For $|\delta |\ll \omega$ and many reference cycles, the first term usually dominates: its modulus is $B|\mathrm{sinc}(\delta T/2)|$ and its complex argument includes $\delta T/2$. Thus $|\delta |T\ll 1$ gives almost no rejection. A longer observation separates closer frequencies, with the first zeros of this term at $|\delta |=2\pi /T$. Those zeros are not a brick-wall cutoff: the sinc function has nonzero lobes beyond them. At $\delta =0$ a synchronized contribution remains however large $T$ becomes.

For a numerical check, use $f_0=1\mathrm{Hz}$, $T=4\mathrm{s}$, $\psi =0$ and an unwanted frequency $f=1.125\mathrm{Hz}$. Then $(f-f_0)T=1/2$ and the exact expression giveswhereas the difference-frequency term alone gives $2/\pi \approx 0.636620$. An interference amplitude $B=0.9\mathrm{V}$ therefore contributes $0.539254\mathrm{V}$ to $|Z_T|$ in this constructed example. A nearby frequency can survive strongly even though the three-times-frequency component in lock-in harmonic rejection cancels exactly.

A practical lock-in amplifier's low-pass filter weights past products differently from this rectangular observation interval. Its time constant and roll-off determine near-reference rejection and settling time; narrowing detection slows the response.^[ref]

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.

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]