Question

QUESTION 4 (20 marks) For any given experiment data of a sensor, explain how to determine a. to ench 63.2 2 ofHe Panes the settling time of a second-order system Tire to at 22 Tire the time constant of a first-order system (i) () b. A thermocouple with a time constant of 0.05s is considered to behave as a first-order system. For the thermocouple to measure the dynamic temperature with an error less than 5%, what is (8) the range of frequency and time? c. A pressure transducer behaves as a second-order system. If the undamped natural frequency is 2000 Hz and the damping is 75% of its critical value, determine the frequency ranges over (8) which the measurement error is less than 5% . 21

Answer 1

a) (i) the time-constant of a first-order system The system's step response can be used to measure the time constant. The response should have first-order characteristics like no overshoot and non-zero initial slope from the response, we measure the time-constant that is, time required for the amplitude to reach 63% of its final value (11) settling time of a second-order system The time required for the transient's damped oscillations to reach and stay within 72% of the steady-state value Sometimes, an approximation for the settling time is used as To son Damped natural frequency wga und 1-5². Damping ratio S can be calculated from percent overshout s- tn (20s/100) +²+ In? (%08/100) Time constant, 7= 0.05 s. Dynamic error < 5% implies Synonse = (M/W) - 1) x 100% = (1 2 2 4 - 1) X 10%

So, 0.95 x < 1.05 - 7.6272+ But greater than 1 272 = cannot be ( w²₂²1 0.95 < t <1 I w2z2ti o< w z < 0·33. This gives. osws 0.33 = 0.33 rad/s. 0.05 . 6.6 radis Orad/ Kw 6.6 rad/s. = 2000 Hz = 12566.37 rad/s. S = 0.75 M (W) = 0.95 MW) , 1.05. ogs & Mw) < 1.

> 5982 rad/s. Los 6283-19 rad/s.