Question

**please provide a full and clear handwriting answer

** Double check your answer to make sure it's %100 correct

R(S) C(s) F(S) G(S) P(S) HS)

Answer 1

(+Gp) IG FGP (G) (It Gip)2 Stipe p. IG + GPFP FGFP (1+Gp)² Substitute T = F Gip It Gp Stip FG FGØ (l+ GpJ² It Gp Ste + Gp above equation with compare given equation. - s²+25 $2+25+20 It Gp

It GP + 25 26 كه دي 26 + و - 2 HGP 8²+28 وه ده +ع GP مه - 25 کر 20+کھ 8 = مG و بهد (65 45 s - 25 را از put P = 5 G = S 20 5ts 28 4 G .ع 4 (st 2) S+2

so, G(s) S+2 Now compare calculated 'T' with given't" EGP 200k !+ Gip (S+1 (5+3) (5+23+20) T. (1+ Gp) 200k (S+ )(5+3) Cs? +2 5+20) substitute G ce e in above equations 4 5 200k FE (8+2) s (5+) (s+3) (52+29+20) s² + 25 + 20 2ook SCS+2) (SD) (5+3) 65+25+20)

FCS) = 2+25720) rook s(s+1 (5+3) (S+2) (83+25+20) 200k F(S) s(s+l) (5+2) (5+3) 80, Result 4 G(S) sta 200k FCS) ہے SCS+1) (5+2) (5+3)

steady state exc08 ess ith .tke for unit step sig nol. where - Lt Kp = CCs) so Lt GCS) So 200 k TCS) = (SI) (+3) (82 +25+20) T(S) 200k (5+1) (5+3) (5+25+20) error is calculater for open loop Transfer fun&tion OLT. F = numerator denominator - numerator

o. L-т. с- Ооо (5+1) (s+3) (32+23++20) — 2оор GCS) So1 роб k L+ ke | | S»o (s+) (s+3) (s24 25+ go) – 20ok 2ook ke А (O) (3) (20) — 2ook 2ook Кр Go - 2ov k kp = 20 k 20 k G

substitute kp in ess ess 1+ kp 1+ 2ok 6 - zok But Given condition of Steady state error for step input ess=0 . V 1 + 2ok 6-2ok 6-2ok U 6-20k zok = ) 6 - 20% =0 6 - 20k to t zok to K = 6 20 = 0.3

Result! So, required values "k" to get Steady state error of zero for the Step input is 0.3 003

(+Gp) IG FGP (G) (It Gip)2 Stipe p. IG + GPFP FGFP (1+Gp)² Substitute T = F Gip It Gp Stip FG FGØ (l+ GpJ² It Gp Ste + Gp above equation with compare given equation. - s²+25 $2+25+20 It Gp