Someone designs a lock-in measurement with time constant and a per octave (four-pole)…

PHYSICS · Multi-part · Fermi University Physics · Electronics, Instrumentation & Communication · Feedback, operational amplifiers and analogue signal processing · Feedback, operational amplifiers and analogue signal processing

Inside the Instrument, Ch Low-noise front ends and noise measurement

Problem

Someone designs a lock-in measurement with time constant τ=200 ms\tau = 200\,\mathrm{ms} and a 24 dB24\,\mathrm{dB} per octave (four-pole) output filter, and argues: "the equivalent noise bandwidth of any RCRC-type lock-in filter is 1/2πτ1/2\pi\tau, so here Bnoise=1/(2π×0.2 s)=0.796 HzB_{\rm noise} = 1/(2\pi\times0.2\,\mathrm{s}) = 0.796\,\mathrm{Hz}." They then report the signal-to-noise ratio for a 2 μV2\,\mu\mathrm{V} signal in an output noise density of 5 nV/Hz5\,\mathrm{nV/\sqrt{Hz}} as SNR=2 μV/(5 nV/Hz×0.796 Hz)\mathrm{SNR} = 2\,\mu\mathrm{V}/(5\,\mathrm{nV/\sqrt{Hz}}\times\sqrt{0.796\,\mathrm{Hz}}). ((a)) Find the error in this reasoning. ((b)) Compute the correct equivalent noise bandwidth and the correct signal-to-noise ratio.

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