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Consider the following controller in a unity feedback configuration: (s + 10) C(s) = k· (s + 5) (a) (by hand) Using an approx

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L(s) Rili) face pcs) 115) = CCS) PCS) it cup(s) K (stlo) (sts) (5+2) (57 55+25) It K(sto) 11 (5+5) (5+2) (3 75+25) L(S) - 1ks* t; 4.33 리 NORL RL RL Po (-2) -10 PG -j4-33 Number of branches terminating at infinity P-2 = 4-1 =3 EGE 29+1) 180 (29+1) 150Angle of departure for open Loop poles Ød = 1800 0 = EØp - {02 lo 34.33 ØP3 &p, =907 tamil tamos -96-59 8P3 = tant 90 4-33 =-It To find the stability values of t use Routh- Hurwitz Criterion. I+Gls) k (5+10)11 It G (3)=0 (5+2) (s+) (5755+25) (5+75+

From 1,2 &3 k<8.1652.

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