Question

Question 1 (20 points) As a control system engineer you have been asked to design a controller that would improve the error and the transient response for the unity feedback system below. The proposed solution must be cost-effective, so consider a passive network-based compensator. The transient response of the closed-loop transfer function to a ramp input has a 30% overshoot (%OS = 30) and a settling time Ts= 2.73 seconds. You need to decrease the peak time by a factor of 2, decrease the percentage overshoot by a factor of 2 and improve the steady-state error by a factor of 30 a. What type of compensator should you implement in this case? b. Find the gain K for the uncompensated system. c. Find the peak time, Tp, the velocity constant, Kv, and the steady-state error, for the uncompensated system. d. Assume that the compensator zero is placed at -5 and design your compensator accordingly. RS) + E(s) (1) K (5) + 11)

Answer 1

Carmen = 307 mpy Ts = 2.73 Specifications, need to dochase the peak the by a factor of 2, de chase the percentage one shoot by a fare for of 2 and improne the factor of 30 Steady-state error by ou E(S) Gwer system system Rls) to $(5+5)(s+1) Based an on the specifications the lay-kood compensates is used 301. ::. Stip-o. Comen mpy h. 0.3 3 37/11 - 32 In(0.3) =-12 37 11-32 2 1-44 1-22 = 0-35 3 1.44 T2 1044 3 0035

tone TS =4 Selthing zwr 4 own - 4 ZNTS Ou 35 X 273 Wn= 40186 hadise alep Determine the domluant pole, sd Dentuant pole, sd = -3 won Jun 11-32 me have sd =-4 =-0-3544-186 wna 40186 ?=0.35 * j 4.1864,-0.352 Sod=-1.46 £ 3.92 Sty 2 +3.927 Drawn the pole- zero plot The pole-zero plot of opon loop trasfer function is shown in graph .poles are rephesented by the Symbol "s'. The pole at point of is the comment pole isde

Step -3 by To find the angle to be contributed lead Natural Angle to be contestated by lead Network ato to make point, pas a point on hoot loves het & (Sum of angles Now & Sum of angles Contributed by poles uncompensated of uncompensated System Contulbuted by poles of uncompensated I System Contulbuted by th18a zuos of System = 0,+B2+03 = 110+102 +48.36. Sum of ongles 2 bo. 36 Swee there is no fuite zano in uncompensated angle contributton by zero susten , there is no 260. 36t niso set nai, B = 260.36 -180- 30.36 free the angle 8360, the lead compensator us healiza cascade of two compensators, contesbulay half of the requered arogle p = 80'3612 40018.

Jo as shown Darow adopt To fud the poles and and gends of the compensator a lina Ap pornalled to x-ants in figure The bisector pc is drawn to bised the angle Apo. The angles epp and Bpe are constiveles aus schon. Here 12cpp = LBPC = 8/2 = 40.187 2 20.09 pole of -2.5 = 0.4 the compuusatos / Pe = -5-8. zeho of the compensates, zes he know that Zc =YT ET me know that PL = ago = ->p = Y5.8* E-YAT 5.880.4 aZ0-43 2 Determine the trasfes fuction of lead compustar Trasfer, fuetion of ? love (s = (S+YT) ² lead Compensator alla stessa } 2 (لادی 2 2 Cely = (s+205) (+508)

open loop trasfes fuction 04 lead compensated Systen the valus of geon is к Golsta (s+2.5), (575-8)’s (5+5)(3+1) Here k guen by at the dominant pole; sd,on the host tows. product of netos lengths from all pelesto sesd paoduct of weeter lengths from all genasto sash 1, x 12 x 1₂ x 14 454,24:5-546.4 15 4.2 z 158 f. Gols = 158 (+25) (5+5-8923(5+5)/s+1) 1. K = 2 velocity erros constant 3 Galb) = $158 (5+2.5) ? (5 + 5.8² $(3+5)/SD sto -1580+25)? -5.87 Suce the velocity (5-8) 25%) errer is groake teot Salest the requirement, the design is acceptach.

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