自动控制原理（双语教学）：9.10.3.7

Solution 9.10.3.7 S G a = a G @G @a = a s + a s(s + b) @ @a s + a s(s + b) = as(s + b) (s + a)  1 s(s + b)  = a (s + a) ;; T c (s) = s + a s(s + b) 1+ s + a s(s + b) = s + a s 2 +(b+1)s + a : S T c a = a T c @T c @a = a s + a s 2 +(b+1)s + a @ @a s + a s 2 +(b +1)s + a = a(s 2 +(b+1)s+ a) s + a  ;(s + a) [s 2 +(b+1)s + a] 2 + 1 s 2 +(b+1)s + a  = a(s 2 +(b+1)s+ a) s + a " ;(s + a)+s 2 +(b+1)s + a [s 2 +(b+1)s + a] 2 # = as(s + b) (s + a)(s 2 +(b+1)s + a) ;; S G b = b G @G @b = b s + a s(s + b) @ @b s + a s(s + b) = bs(s+ b) (s + a)  ;s(s + a) (s 2 + bs) 2  = ;bs s(s + b) ;; S T c b = b T c @T c @b = b s + a s 2 +(b+1)s + a @ @b s + a s 2 +(b+1)s+ a = b(s 2 +(b +1)s + a) s + a  ;s(s + a) [s 2 +(b+1)s + a] 2  = ;bs s 2 +(b+1)s + a S G K = S T c K =0;; since there is no dependence on K.