A signal is recorded for at a sampling rate

PHYSICS · Multi-part · Fermi University Physics · Oscillations, Waves & Acoustics · Electronics, Instrumentation & Communication · Coupled oscillators, normal modes and Fourier analysis

Modes, Packets and Media, Ch The continuum limit, Fourier series and the spectrum of a measured signal

Problem

A signal is recorded for T=2.0 sT = 2.0\,\mathrm{s} at a sampling rate νs=44.1 kHz\nu_{\rm s} = 44.1\,\mathrm{kHz}, giving N=νsT≈88200N = \nu_{\rm s}T \approx 88200 samples in its discrete transform, with resolution Δν=νs/N=0.5 Hz\Delta\nu = \nu_{\rm s}/N = 0.5\,\mathrm{Hz}. Someone argues: "If instead we sample the same 2.0 s2.0\,\mathrm{s} record at νs=88.2 kHz\nu_{\rm s} = 88.2\,\mathrm{kHz}, we get N=176400N = 176400 samples, twice as many. Keeping the original sampling rate in the ratio, the resolution improves to Δν=44100/176400=0.25 Hz\Delta\nu = 44100/176400 = 0.25\,\mathrm{Hz}: doubling the sampling rate has halved the resolvable bandwidth between lines." ((a)) Find the error in this reasoning. ((b)) State the correct frequency resolution obtained by sampling the same 2.0 s2.0\,\mathrm{s} record at 88.2 kHz88.2\,\mathrm{kHz}.

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