Question

A chemical process unit exhibits an underdamped second order behavior as given by the transfer function: Y(s) U(s) 18 s2 + 3s + 9 Calculate the process gain, natural period of oscillation and damping factor for the system. (3m) b. If a step change in the input with the magnitude of 3 is introduced, i. Determine the step response (5m) ii. Estimate the new steady-state value of y. (2m)

Answer 1

Given :- Trans for function:- Y(s) UCS) 18 8?+38 + 9 Process gain (Kp) know that transfer function of We a second order system is Y U15) Ке T232+ 2z28+1 Where! Kp is process gain [ is time constant 3 is damping factor form as convert the given transfer in the standard follows Y C5) 18 UCS) 52 +3579 Divide numerator and denominater by g Y CS) UCS) = 18/9 (5+ 38 +99/9

YES) 2 UCS) +18+ 3 9 with stand and form comparing it gain , Kp = 2 Process 72 - I g =) == C-ve time constant not possible) Natural period of oscillations:- Wn= 누 13 = = 3 (radians /time) 2378=1/38 23 T = 1 Wh 2 2 15 3 13 | 3 = 1 Damping factor 3=½

a step change in the input with magnitude of 3 is introduced Y (5) 18 2+38 +9 UCS) UCS) - 3 3 S i Y (5)= 18 3 (52 +3 5+9) 5) Using inverse laplace transform L-S 18 y (t) (52+38+9) s y (t) = 6 6-(233 sin (2 037) + 6cos (233) 1") New steady state value applying final value theorem lim y(t) tto lim sY(s) S=0 3 27m rims 4 18 (52+38 +9) 54 lim sto 32+36 +9 E6 54 9

The inverse Laplace in the second part is mathematical part, that's why I wrote the final answer. You can do it using calculators and other software tools. Otherwise the calculation may be hectic.

Thankyou.