Question

2. (25 pts) For the block diagram shown below. a) Let d(t) = 0 and r(t) =2t. Find K so that the steady state error e=0.3 b) Find the steady state errores when K-5, R(S)=1/s, D(s)=0. c) Find the steady state value of c(t), when D(s)=3/s” and R(s)0 D(s) C(s) R(5) 5+ 2 (s + 1)(s + 3)

Answer 1

² if dlt)=0 then transfer function S+2 -C1) RIS) (5+1) (5+3) ess203 dit1-0, 8CH = 2t, A = 2 (ramp input) for samp :- ess = A - input Ko nebo 3 = -0 TH1s=1 L unity feedback S70 5+2 - lim S70 velocity but ko=lim sals) HIS) Elinos que 3 1873) ko = 2x but ko value in egun o K=10 Cs Scanned with K-10] Ang Scanned with CamScanner

© K-5 ; RIS) - 5 lunit step] 015)=0 for step input - Kp = lim G15) HIS) 520 AR! Sledy na pe state enor It ke basitional lim GIS) HIS) S6 error Constant Kp = S +2 HIS) = 1 unity Kapa be lim k S0S sar x yang [muz i gjithcka feedback 5+1) (5+3) 0 kp=0 put the value in eann ② ess - Htoo =0 © R(S)=0 S+2 S+1)(8+3) DLS) Dis ostents cs Scanned with tæk Scanned with CamScanner

CCS) DIS) DIS (5+2)/ 6+1) (5+3) I + 154215552 s(s+2) C(s). Dis) S1S+1)(5+3) +K (5+2) ses+2)_ C(S) = SIS+1) 65+3) tkisiz? DIS) D(S) DES) = A s(572) CCS)= s(s+1)(5+3) +K (S+2)] clt = lim sto sces) Steaady state value = lim tas S2 Css = lim sis(5+2) 370 L. scsti) (5+3)+k1S+2)] - 2 eu - 2x3 - 2 Scanned with Caim Scanner