SolSec7.2 - Problems and Solutions Section 7.2 (7.1-7.5)...

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Problems and Solutions Section 7.2 (7.1-7.5) 7.1 A low-frequency signal is to be measured by using an accelerometer. The signal is physically a displacement of the form 5 sin (0.2 t ). The noise floor of the accelerometer (i.e. the smallest magnitude signal it can detect) is 0.4 volt/g. The accelerometer is calibrated at 1 volt/g. Can the accelerometer measure this signal? Solution: From the problem statement: x ( t ) = 0.5sin(0.2 t )m x & ( t ) = 0.1cos(0.2 t ) m s x & & ( t ) = -0.02sin(0.2 t ) m s 2 The peak acceleration is: ± 02 1 98 0 0204 . . . m s m s 2 2 g g Accelerometer calibration is g V 1 , therefore the peak output of the accelerometer is: ± 0 0204 1 0 0204 .. g g V V Since the noise floor on the accelerometer is 0.4 V, then this acceleration cannot be measured.
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7.2 Referring to Chapter 2, calculate the response of a single-degree-of-freedom system to a unit impulse and then to a unit triangle input lasting T second.
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SolSec7.2 - Problems and Solutions Section 7.2 (7.1-7.5)...

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