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九、解答：非线性和线性如图示

非线性：\(N = \dfrac{4}{\pi A} \Rightarrow \dfrac{-1}{N} = \dfrac{-\pi A}{4}\)

\(A = \dfrac{1}{\pi}\) 时，\(\dfrac{-1}{N} = \dfrac{-1}{4} = G(j\omega)\)

闭环特征方程：\(D = 1+(1+N)G = 0 \Rightarrow 1+G+NG=0 \Rightarrow 1+\dfrac{NG}{1+G}=0\)

故线性部分：\(G' = \dfrac{G}{1+G} = \dfrac{\dfrac{K}{s(5s+1)(10s+1)}}{1+\dfrac{K}{s(5s+1)(10s+1)}} = \dfrac{K}{s(5s+1)(10s+1)+K}\)

相位变化：\(0°\sim -270°\)