Question

Consider the unity feedback system is given below Ris) Cs) G(8) with transfer function: -K(s +1) G(S) 52 + 25 + 2 Considering that can have only positive values, the system is unstable when the value of K is ........

Answer 1

RIS) E(S) Gls) C(s) H(s) from Black diagram C(s) = E(S) GIS) and Els) = R(S) - C(s) HIS) ((s) = (RIS) - C(s) H(s)]G(s) c(s) = R(S) G(s) – C(s) G(S) HIS) C(s) + ((s) G (3) H(S) 2 R(S) G(S) C(s) [ 1+ G(s)H(s)] R(S) 6(S) C(s) R(S) G(s) I+G(s)H(S) closed Loop transfer function C(s) R(S) Gls) 17.(S)H(s)

Here unity feedback, .. HIS) = 1 - Closed loop transfer function ((s) R(3) DOWS G(S) t615) -K(SH)² 32+25+2 - klst)2 32test2 -k(st)? s²tasta-k (3²+1+25) -K(SH)² s2 +25+2 - KS2-K-2KS a 11 -K/st) (l-k)s?+(2-2K)s +2-k (cs) R(S) be K(SH)2 (K-1)32+(2k-2)5+ K-2 Characteristic equation will → (K-1)32+ QK-2)5+(K-2) = 0 for testing stability we will use Stability criterion Routh Hurwitz The Routh Hurwitz Array will be

3 S? (K-1) (K-2) s' (QK-2) so (2K-2) (K-2)-(K-1)(0) (2K-2) =(K-2) . The table becomes (K-2) s² (K-1) s' (2K-2) so (K-2) for system to be stable all the elements of the first column of the table are positive Kalzo » K> ak- 270 K> 1 K-270 K>2 and kyo Combining these results we would have the values of K for System to be Stable / K>Q .. off the value of k would be less than 2 but greater eater than o the system would be unstable, . System is system is unstable OLK< 2u unstable if