Question

3. Consider an LTI system with transfer function H(s). Pole-zero plot of H(s) is shown below. Im O--- Re (a) How many ROCs can be considered for this system? (b) Assume system is causal. Find ROC of H(S) (c) Assume y(t) is system output with step unit as input. Given lim yết) = 5 , Find H(s). (d) (optional) Find y(2) (y(t) for t = 2).

Answer 1

6 given LTI System. XT s KO: = n+1 Roc @ No. of where n = no of - poles. nou of ROC = 3+1 = 4. & Roc g2-2 -254-1 Hence 6 For ROC causal system must be right 2 Rided ha)=0 & Lo causal )

O from figure MB) = YG) XC) HS). X(5) YES) = HG) o xG) as input xG) - SCA) And X) - | H6 = YG from figure H(s) = K (S-1 + j) (S-1-j) (5+1) (5+2) given condition a linya) = 5 from Final value theorem has lm sys) = 5 od sto K(S-1ti) (5-1-f) as lim sys) = 5x S S [S+!) (5+2) وع K(-1+4) (-1 mg) s KLED + j)?) s K = 5

Kes Aster Hence 5 (stri) (5-1-j) s(s+15+2) JHS) = YO) = 5(st+j) (st-j) S(SH) (5+2) Sportial A fraction S+1 on solving ing partial fraction we get part Az 56-173)(--))(=5 / 5:02 ) 5x10 25 5(1-3 +j) (-3 -2 (-1)

Hence - 5 25 Y) = 8 + 25 5 + 5+2 25 - - - S+2 sti taking Inverse Laplace on both side we get. gas = 5 ult) + 25 é et ult) - 25 et ults 9 (2) = 5+ 2504 - 2562 - 5 - 2-1 ge) = 219 give positive Hope you understand this please do feedback if this is helpful.