自动控制原理（双语教学）：9.10.3.5

Solution 9.10.3.5 S G a = a G @G @a = a ab (s + a)(s+ b) @ @a ab s 2 +(a + b)s + ab = s 2 +(a+ b)s + ab b  ;ab(s + b) (s 2 +(a + b)s + ab) 2 + ab s 2 +(a+ b)s+ ab  = s 2 +(a + b)s + ab  ;a(s + b) (s 2 +(a + b)s+ ab) 2 + a s 2 +(a+ b)s+ ab  = s 2 +(a + b)s + ab " ;as;ab+ s 2 +(a + b)s + ab (s 2 +(a+ b)s + ab) 2 # = s(s + b) s 2 +(a+ b)s + ab : T c (s) = ab s 2 +(a + b)s + ab 1+ ab s 2 +(a + b)s + ab = ab s 2 +(a+ b)s +2ab : S T c a = a T c @T c @a = a ab s 2 +(a+ b)s +2ab @ @a ab s 2 +(a + b)s+2ab = s 2 +(a+ b)s +2ab b  ;ab(s +2b) (s 2 +(a + b)s +2ab) 2 + b s 2 +(a + b)s +2ab  = s 2 +(a + b)s +2ab  ;a(s +2b) (s 2 +(a + b)s +2ab) 2 + 1 s 2 +(a + b)s +2ab  = s 2 +(a + b)s +2ab " ;as;2ab + s 2 +(a+ b)s +2ab (s 2 +(a + b)s +2ab) 2 # = s(s + b) s 2 +(a+ b)s +2ab : S G K = S T c K =0;; Since there is no dependence on K.