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Unformatted text preview: 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: ± 0.2 m s 2 1 g 9.8 m s 2 é ë ê ù û ú = ±0.0204 g Accelerometer calibration is g V 1 , therefore the peak output of the accelerometer is: ± 0.0204 g 1V g é ë ê ù û ú = ±0.0204V Since the noise floor on the accelerometer is 0.4 V, then this acceleration cannot be measured. 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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This note was uploaded on 11/28/2010 for the course ME 4440 taught by Professor Hill during the Winter '09 term at Detroit Mercy.
- Winter '09