Question

Problem 1) (25 points) Suppose that a second order system has the pole pair of –2 jn and no zeros. (a) Determine the kind of the unit step time response (overdamped, undamped, etc.) and write its general form. (05 points) (b) Find the settling and peak times, T, and Tp (if appropriate), respectively. (05 points) (c) Write the system's transfer function that produces a unit step time response having the steady- state value Css of 10. (05 points) (d) Find the peak value Cmax of the response. (05 points) (e) Propose a new transfer function that produces a unit step response two times (2x) faster in terms of settling time while maintaining the same peak and steady-state values. (05 points)

Answer 1

S= -d+jit Pole pair at Charalleristi's equation : *--*2)[sc..) -0 $+45 +(4+7°) - $$45 +13.87 - 0 comparing it with $498W, S4 o 26 Wn=4 wna 13.87 = 3.724 sed/see 2x Ex3-124 = 4 Ep= 0:537 (a) as Ep<1 : unit step response is underdam bed. Generel eq.foxm of 2nd ordes unit step Jesponse : -Ewnt е Cce)= ta Sin (wat ++) VT-E2 where o= tant NER ६ ) . wx () 5+4,02] C(t) = -gt 1.e [ : Sin (2.65++ 0.711 0= 52.93° (6) Settling time (Ts). 4. EW To: 2 sec -

Peak time (Tp)= I wa T 2.65 To: 1.185 sec (C) T-sansfer function Tess kwo $+ 25W,stwo Steady state state value Ke 10 Tess= 138.7 $+45 + 13.87 (d) for max: ETT/J1-62 Maxim Owezeshoot : *100 (05)= e OS: 20.537x1T%0.711 9.3.24 - Coman 10+ 10 voor 9.30 Comax 10.932 for Fes= Ts = futs Tp = Tp. kalook come me, Elwin €w=4 1:185 7.0985 - W1-16 Twin-white will Wn: 93.0985 ED Ws 4.798 E = 0.833

Tear ansfer function (Tess) kon so ta twist win Tics) 13 x 23.0285 $ + QxLx S +293.0985 Tés) 930.985 5+85+23.09985