Question

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Question 1 1 pts What is the closed-loop time constant for the system shown? R(s) C(s) Question 2 1 pts What is the closed-loop gain for the system shown? Ris) cls) 5+5 10

Answer 1

Solution: Question ti Transfer functions connected in parallel (feed loop). (6) Tranfer fanction G(s) = C(s) R(S) Н. It HiCS) H) H21 For the given problem: C(S) GLS) = Ris) eka on It exa G(6)= S 6 St12 12 {2+1 12 Standard form for pst order transfer function Isti where K gain constant T = closed loop time T= = 0.0833 12 Questions, L(S) G(S) = (5+5+50 R(S) st5 1+15 xto 3+5 G(S) - heating S+55

0.0909 G(S) = 0.01818 S + Standard form for 1st order transfer function GGG) = 7 st1 = grain compare above equation! whese k= K=0.ogog Quotion 3: 344514 Trans for function! G(s) = 1+ 1x 10 4514 - G(S) = 3²+48 +4 +90 G(s)= R(S) s+45+84 For the steady state! R(t = 1 R(S)= substute this in above equation c(s) = R(s). 8445+84 8 (S) s(5445+24)

Questions! s²44544 G(s) R(S) 1+ Asta) x 10 844814 8 G(s) - 8445+ 34 Standard form of the second-order septem- wn G(6) = 8²728 wstwa where! wna natural frequeency 8 = Damping ratio Comparing (542&constien) with 8445+84 11-23 84 - wn= 9.165 radised & numeroter! Questions we=8 Con 2.828 rad see for wn=91165 28 n = 4 > 28x9.165 = 4 $= 0.21822 & for con=2.828 2x 21828 4 &= 0.7072