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core24 min studyLesson 956 in Physics

Lock-in detection and phase-sensitive measurement

Derive two-quadrature demodulation with explicit amplitude and phase conventions, then connect output bandwidth to settling time.

Before you begin
On this page

The question for today

What you’ll understand

  • Calculate in-phase and quadrature outputs in a declared peak-amplitude convention.
  • Explain how mixing and low-pass filtering select a narrow frequency region.
  • Distinguish noise rejection from overload protection and settling.

A lock-in measurement uses a reference to ask how much of a signal oscillates at a selected frequency and phase. Multiplication moves that component to a slowly varying or constant output, and a low-pass filter rejects much of the remaining spectrum. This can reveal a small coherent response buried under broadband noise. The improvement comes from frequency-selective measurement with a declared averaging time, rather than from creating information with unlimited gain. Reference coherence, amplitude normalization, front-end range, and output settling remain essential parts of the result.

Fix the reference normalization and quadrature sign

Use a sinusoidal input v(t)=Apk cos(ωrt+φ), where Apk is a peak voltage, φ is the signal phase relative to the reference, and ωr is angular frequency. In this lesson, multiply by two cos(ωrt) for X and by minus two sin(ωrt) for Y. The factor two removes the half-amplitude produced by multiplying equal-frequency sinusoids. The negative sine choice makes a positive input phase produce a positive Y.

Low-pass filtering removes the sum-frequency terms when its bandwidth is sufficiently small compared with the reference frequency. The surviving components are the cosine and sine projections of the input phasor. Their magnitude recovers the peak amplitude. Real instruments often scale their displays to RMS sinusoidal amplitude instead; the corresponding factor of square root two must be established through the manual or a calibration signal. A phase convention cannot be inferred from the labels X and Y alone.

X=LPF⁡[2v(t)cos⁡ωrt]=Apkcos⁡ϕ,Y=LPF⁡[−2v(t)sin⁡ωrt]=Apksin⁡ϕ,R=X2+Y2=ApkX=\operatorname{LPF}[2v(t)\cos\omega_rt]=A_{\mathrm{pk}}\cos\phi,\quad Y=\operatorname{LPF}[-2v(t)\sin\omega_rt]=A_{\mathrm{pk}}\sin\phi,\quad R=\sqrt{X^2+Y^2}=A_{\mathrm{pk}}
The low-pass filter has unit DC gain and strongly rejects 2ωr. These are this lesson’s peak-voltage conventions, not a specification of every instrument. Recover phase with atan2(Y,X), using both signs to select the quadrant.↔ Scroll wide equations horizontally.

Mixing selects a frequency neighborhood, not one instantaneous value

A component offset from the reference by Δf becomes a beat at that difference frequency. Its survival depends on the output filter rather than on exact frequency equality in a finite observation. A reference phase that drifts changes X and Y even if the input amplitude is constant. Measuring both quadratures avoids losing a signal merely because it is ninety degrees from the selected X reference, although it does not remove all phase-related systematic effects.

For a one-pole output filter, reducing bandwidth also slows its response to changes. Specify whether a quoted noise density is before or after the mixer: the mixer normalization and its two sidebands affect the baseband density. If a white one-sided density eBB is already defined at the low-pass input, its output RMS is eBB times the square root of the filter's equivalent bandwidth. This avoids silently applying the same density to differently normalized signal points.

HLP(s)=11+sτ,BEN=14τ,X(t)=Xf+(Xi−Xf)e−t/τ,σX=eBBBENH_{\mathrm{LP}}(s)=\frac1{1+s\tau},\qquad B_{\mathrm{EN}}=\frac1{4\tau},\qquad X(t)=X_f+(X_i-X_f)e^{-t/\tau},\qquad \sigma_X=e_{\mathrm{BB}}\sqrt{B_{\mathrm{EN}}}
τ is in seconds, bandwidth is in hertz, and the step equation assumes one first-order filter initially at Xi. eBB is the white one-sided baseband voltage density after the declared demodulation normalization. Multi-pole or other output filters have different settling and bandwidth factors.↔ Scroll wide equations horizontally.

A coherent disturbance survives the same selection

A disturbance exactly at the reference frequency can be indistinguishable from the desired response in a single measurement. Optical chopping, electrical modulation, cable pickup, and ground currents can all generate such coherent backgrounds. Varying the stimulus, blocking the signal path, or measuring a control configuration helps identify them. A narrow output filter rejects broadband noise but cannot tell which physical mechanism produced a surviving coherent component.

The input stage must also tolerate the entire incoming waveform before narrow output filtering occurs. A large out-of-band disturbance can overload the amplifier or converter and generate distortion near the reference. Waiting longer cannot reverse that clipping. Practical measurements therefore allow settling after changes, monitor overload indications, verify the reference, and state the filter order and normalization. The small recovered number becomes meaningful only when those measurement conditions accompany it.

Work it through

Recover a phasor with two quadratures

In the declared peak-voltage convention, a coherent input has peak amplitude 10 µV and phase +60°. Assume exact frequency lock, adequate input range, and a settled ideal output filter. Compute X, Y, R, and the phase recovered from the pair. Use sin60°=√3/2.

  1. The in-phase projection is X=10 cos60°=5 µV. A single X output would therefore be smaller than the full signal amplitude.

  2. The quadrature projection is Y=10 sin60°=5√3 µV≈8.66025 µV. It is positive because the reference for Y was defined as minus two sine.

  3. The magnitude is R=52+(53)2=10R=\sqrt{5^2+(5\sqrt3)^2}=10 µV and atan2(Y,X)=+60°. These are peak-scaled results; an RMS-scaled display would report the corresponding amplitude divided by √2.

The result

The settled outputs are X=5 µV and Y≈8.660 µV, recovering 10 µV peak amplitude at +60°.

The living notebook

Explore the model

Interactive lab

Interactive lab · vectors

Quadratures are signed projections

For each fresh comparison, Reset and set BOTH Angle=60° and Magnitude. Compare magnitudes 1 and 2. Predict projections (0.50,√3/2) and (1,√3), while the magnitude doubles. Interpret these only as normalized peak-amplitude quadratures in the declared sign convention. The vector instrument contains no multiplier, output filter, reference tracking, noise spectrum, or overload model.

Quadratures are signed projectionsPlayback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Current readings: x component: 1.393; y component: 0.975; Magnitude: 1.7. All parameters are adjustable with the labeled controls below.−7.46−3.7303.737.46−3.5−1.7501.753.5xyvvₓ = 1.39
Vectorx / y projectionsTrace marker
x component
1.393
y component
0.975
Magnitude
1.7
t = 0 s

Playback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forTwo signed projections recover a phasor without requiring its phase to align with one axis.

Pause & reason

Check your understanding

What does a narrower lock-in output bandwidth generally require after a signal change?

Your turn

Try it yourself

A one-pole demodulated X output initially equals 4 µV and steps toward a new steady value of 10 µV with τ=0.10 s. Find X after 0.30 s. How long is required for the residual error to fall to one percent of the six-microvolt step? Use exp(−3)≈0.0497871 and ln100≈4.60517.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

The residual relative to the step amplitude is exp(−t/τ), rather than the residual divided by the final ten-microvolt reading.

Compare with the solution

At 0.30 s, X=10−6e−3≃9.70128X=10-6e^{-3}\simeq9.70128 µV. One percent of the step is 0.060 µV, so the condition is exp(−t/τ)≤0.01. It requires t≥τln⁡100≃0.460517t\geq\tau\ln100\simeq0.460517 s. The specified tolerance is relative to the change of six microvolts. A multi-stage output filter would have a different transient, so its settling time should not be inferred solely from this single-pole example.

Keep the idea

Take it with you

  • Reference normalization and quadrature sign determine the reported amplitude and phase.
  • Output bandwidth trades noise rejection against settling and tracking speed.
  • A lock-in does not reject coherent backgrounds or repair an overloaded input.

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