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\(\because E(s) = X_i(s) - X_0(s)\)　　　　BUFF控制考研

\(\therefore \dfrac{E(s)}{X_i(s)} = 1 - \dfrac{X_0(s)}{X_i(s)} = 1 - \dfrac{k_2 G_0(s)\cdot(T_1 s+1) + k_1 k_2}{s(T_1 s+1)(T_2 s+1) + k_1 k_2}\)

\(= \dfrac{s(T_1 s+1)(T_2 s+1) - k_2 G_0(s)(T_1 s+1)}{s(T_1 s+1)(T_2 s+1) + k_1 k_2}\)

\(\therefore e_{ss}(\infty) = \lim_{s\to0} sE(s) = \lim_{s\to0} s\cdot\dfrac{E(s)}{X_i(s)}\cdot X_1(s) = \lim_{s\to0} s\dfrac{s(T_1 s+1)(T_2 s+1) - k_2 G_0(s)(T_1 s+1)}{s(T_1 s+1)(T_2 s+1) + k_1 k_2}\cdot\dfrac{V_0}{s^2}\)

\(\therefore \diamond e_{ss}(\infty) = 0\)

即 \(\dfrac{\left[s(T_1 s+1)(T_2 s+1) - k_2 G_0(s)(T_1 s+1)\right]V_0}{s\left(s(T_1 s+1)(T_2 s+1) + k_1 k_2\right)} = 0 \therefore s(T_1 s+1)(T_2 s+1) - k_2 G_0(s)(T_1 s+1) = 0\)

\(\therefore G_0(s) = \dfrac{s(T_1 s+1)(T_2 s+1)}{k_2(T_1 s+1)} = \dfrac{s(T_2 s+1)}{k_2}\)

五、（1）\(\because G(s) = \dfrac{k(1-s)}{s(s+2)} = \dfrac{-k(s-1)}{s(s+2)}\)，且为单位负反馈

\(\therefore D(s) = 1 + G(s)H(s) = 1 - \dfrac{k(s-1)}{s(s+2)} = 0 \Rightarrow \therefore \dfrac{k(s-1)}{s(s+2)} = 1\)，\(\therefore\) 为0度根轨迹。

1）\(\because n=2\)，\(m=1\)，\(n-m=1\) \(\therefore\) 有2条根轨迹，1条渐进线起点分别为 \(P_1=0\)，\(P_2=-2\)，终点为 \(Z_1=1\) 和一个无穷零点BUFF控制考研

2）实轴上的根轨迹位于：\([-2,0]\) 和 \([1,+\infty]\)

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