Question

(d) What is the denominator polynomial that would yield poles at s 0,1j,-1-j? (2 pnts) (e) For a standard second order transfer function, what pole locations would yield a settling time of (2 pnts) 2.3 milliseconds and a peak time of 6.2 millinoconds!

Answer 1

d) S = 0 S = -1 + j S = -1 - j Denominator polynomial is = (S - 0)(S - (-1 + j) (S - (-1 - j)) = S(S + 1 - j) (S + 1 + j) = S(S^2 + S + Sj + S + 1 + j - js - j - j^2) = S(S^2 + 2S + 2) [j^2 = -1] = S[S^2 + 2s + 2] = S^3 + 2S^2e) given setting time = 2.3 m sec Peak time = 6.2 m sec peak time, t_p = pi/w_d [w_d is damped frequency] 6.2 m = pi/w_d w_d = pi/6.2 m = 506.70 setting time t_s = 3/delta w_n (5% tolerance) 2.3m = 3/delta w_m delta w_n = 3/2.3m = 1304.34 delta = 1304.34/w_n - (1) \r\n we know w_d = w_n squareroot 1 - delta^2 506.70 = w_n squareroot 1 - (1304.34/w_n)^2 squaring on both sides 25644.89 = w^2_n (1 - 1701302.83/w^2_n) 256744.89 = w^2_n (w^2_n - 1701302.83/w^2_n) 256744.89 + 1701302.83 = w^2_n (or) (506.70)^2 + (1304.34)^2 = w^2_n