Question

Consider a system with the following transfer function,

G(s) = 10/[ s(s + 2)(s + 8)]

Design a compensator according to the following design objectives:

• Mp ≤ 16.3% and tr ≤ 0.45 sec;

• Kv ≥ 80 sec^−1 .

Submit your answer regarding the detailed compensator design procedures, and the corresponding MATLAB code and figures to verify your design.

Answer 1

GLS) = 10 s(372) (318) coording to the specificators Design compensator . ky 2 Box) mphibicy. 4th 20.45 k -all-32 = 16.3%. to = t-o - Wn1-32 »-75 = (0,163) ta = 20) Il - 32 O=o4137 = 60° wn/1-32 > GET =-1-814– win = 2.1 041-(0.57 ozbox 3.14 3-29 180. ?.</ 3-297 3-29+ + 2 c.04 rockers un = 5, 38 .:. Demamant pole sd = - 3Wntjund 1-32 -0.585.3887 5.388 din 0.52 sd 2-2694_44rb6j|

Sum of angles, Contalbulto niso | Slip-2 & Dream the pole-zero phot - The pole-quo plot of open loep trasfes fuelto is shown in graph. poles are represented by the Symbol 71 me pole at posut p is the obminant pole, sh - Step 3 To find the angle to be contributed by I leod Network let f = Angle to be contributed by lead Network to make pont P, as a point on root hans. I dum of angles I I entertbulit kay peles of uncompensates ) (by zares of uncompensated } = (017ezt ®3) -(0)) -150 Systen = (20+100 +40) +180 To - Bor Suce the angle contes button is more tian to due lead compensantas is realised bas Cascade of tuo compusoters udin each conysuscator, conditbuty half of the regulred angle 7 P = 80% = 400

Sep & Jo fund the pales and gass of the compusert. beran a lone Ap paralls to v-ands. The Blsector pe iw drown to beat the angle Apo. The angles app and Bpc hare constructed as Shan in Graphe Here LUPO = (Bpc = /2 = 10% = 20° from groph Pole of the zero of the compensate compensate pe=-73 ze=-3.4. S-PC trasfer fuction of lead controlle Cac (3) = size R klol open loop trasfes noted Inele of lead compensated Gols) = (3+3.498 i Jusle of System k=product of weeter lengths from pole to sasd paoduct of meetos lengtus from zero to Sad k = 1, x dax da x244 = 6x5x7.6 x 2(A) = 319.2 265) zds

2 - Gols) = 319.2 (+3.4) (577.332 s (st2/st2) . Conclushe check for error equerent. la velocity error Coefficket 1. kuat s60(3) sto Ht (319.2) ( 5+3+) wo, sto (6+73) (3+2)(st.8) z 319.28 3.4 Xlo. 243.27. 7032x2x8. Srce kv 22 43-27 are the velocity error constant of the Compensated syster is sighty apot year is acceptable. The design

bus leod Compensat penlyn By Rool 1cm=und Scale sd (-242.646 50120 4 -02=1001 13_05-40 - A Biserbos (2-3 SUCH