A lowfrequency signal is to be measured by using an accelero

A low-frequency signal is to be measured by using an accelerometer. The signal is physically a displacement of the form 5 sin(0.2t). The noise floor of the accelerometer (i.e., the smallest magnitude signal it can detect) is 0.4 volt/g. (g = 9.81 m/s^2; acceleration due to gravity) The accelerometer is calibrated at 1 volt/g. Can the accelerometer measure this signal?

Solution

To find the acceleration produced by the signal:

let y = 5sin(.2t)

y\" = .2 cos(.2t) max value is 0.2 units ( m^2/s)

Now the calibration is for 1 vol to 9.81 m^2/s

Threshold of 0.4 volts means a threshold of .4*9.81 = 3.924 m^2/s

The disturbance is too small to pass the threshold. Hence would not be sensed.