During a step function calibration, a first-order pressure gage is exposed to a

Question

During a step function calibration, a first-order pressure gage is exposed to a step change from 10psi to 55psi. The sensor is calibrated to static sensitivity of 75mV/psi. If after 1.4sec the instrument reports 3.15V, estimate the time constant of the sensor. Determine the pressure reading after 2.8sec. The same sensor is then inserted into an environment that oscillates periodically at 0.3Hz. The digital data acquisition measures a steady-state signal with a mean value of 1.575V that oscillates at the exact same frequency but amplitude of + or - 1.35V. Determine the mean, maximum, and minimum pressures of the input environment that the sensor is measuring along with the phase lag in units of seconds.

Would changing the sensitivity or the time constant be more effective at reducing the lag time of the sensor? What benefit or penalty occurs for the steady-state output of the periodic signal?

Explanation / Answer

The static sensitivity of the sensor is 75 mV/psi. There is a step change from 10 psi to 55 psi and after 1.4 seconds the voltage report is 3.15 V. We assume that at t= 0 seconds, the voltage is 0. Hence, the relationship between voltage and time is V = mt. We can calculate m from the data. At t=1.4 second, voltage V is 3.15 V. Therefore, m = 3.15/1.4 = 2.25 V/s. Time constant is defined as the time taken by the sensor to reach 63.2% of the step change in input. The step change in input is 45 psi. So, time constant is the time taken to reach 63.2% of 45 psi i.e. 28.44 psi.The change here from 10 psi is 18.44 psi. The sensitivity is 75 mV/psi. So, for a change of 18.44 psi, the change in voltage is 1.383 Volts. Substituting this in the equation V = mt, we get time constant as 0.615 s.

To determine pressure reading after 2.8 seconds, we need to find the change in voltage. Substituting t as 2.8 seconds, V = 2.25 * 2.8 V = 6.3 V. So, for a change of 6.3 V the change in pressure is 6.3/0.075 psi or 84 psi.The pressure is thus 94 psi as we measure the change from 10 psi.

Now, the sensor is taken outside and it measures the mean, maximum and minimum voltages of a oscillating signal. The mean voltage is 1.575 V. There is a change of (6.3 - 1.575) V = 4.725 V. Correspondingly, there is a change in pressure of 4.725/0.075 psi or 63 psi. Thus , the mean pressure is (94-63) psi or 31 psi.The max. voltage is 1.35 V. There is a change of (6.3 - 1.35) V = 4.95 V. Correspondingly, there is a change in pressure of 4.95/0.075 psi or 66 psi. Thus , the max. pressure is (94-66) psi or 28 psi. The min. voltage is -1.35 V. There is a change of (6.3 -(- 1.35)) V = 7.65 V. Correspondingly, there is a change in pressure of 7.65/0.075 psi or 102 psi. Thus , the mean pressure is (94-102) psi or -8 psi.

To determine the phase lag, we need to see how long it takes for the sensor to report a change in pressure. For the actual signal that oscillates periodically, the change from max. pressure to min pressure is instantaneous like a square wave oscillating from -A to A. Correspondingly, when the signal changes from max to min. the voltage of the sensor changes from 1.35 V to -1.35 V. There is a change of 2.7 volts. The time for this change is calculated by t =V/m = 2.7/2.25 = 1.2 seconds. Thus , there is a delay or phase lag of 1.2 seconds.

Increasing the sensitivity will reduce the lag time of the sensor. For steady state output, when sensitivity is increased, the sensor will report changes in pressure even for small perturbations which are negligible which can be unproductive.

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