A 20 mV DC signal is superimposed with 10 mV RMS Gaussian noise whose spectrum is flat up to 5 kHz. An integrating voltmeter measures the DC signal. What minimum averaging time in seconds yields a 99% accurate result with 95% certainty?
A one-percent error on twenty millivolts is 0.2 mV. For approximately ninety-five-percent certainty, take two standard deviations ≤0.2 mV, so averaged RMS noise must be ≤0.1 mV. White-noise averaging over T seconds gives variance σ²/(2BT)=100/(10000T)=0.01/T mV². Thus 0.1/√T≤0.1 requires T≥1 s, B. Using 1.96 rather than two gives about 0.96 s, closest to one second.
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