3) 渐近线
\[
\begin{cases}
\sigma_a = \dfrac{\sum p_i - \sum z_j}{n-m} = \dfrac{0-2-1}{1} = -3 \\[2mm]
\varphi_a = \dfrac{2k\pi}{n-m} = 2\pi
\end{cases}
\]
4）分离点：\(\dfrac{1}{d-\sum p_i} = \dfrac{1}{d-\sum z_j} \Rightarrow \dfrac{1}{d} + \dfrac{1}{d+2} = \dfrac{1}{d-1}\)，解得 \(d=-0.732\)；\(d_2=2.732\)
5）与虚轴的交点：\(D(s) = s(s+2) - k(s-1) = s^2 + (2-k)s + k\)
劳斯表：
| \(s^2\) | \(1\) | \(k\) |
|---|---|---|
| \(s^1\) | \(2-k\) | |
| \(s^0\) | \(k\) |
\(\therefore\) 当 \(k=2\) 时，出现全零行，即 \(D(s) = s^2 + 2 = 0 \Rightarrow \therefore s_1 = +\sqrt{2}j\)；\(s_2 = -\sqrt{2}j\)
\(\therefore\) 交点为 \((0,\sqrt{2}j)\) \((0,-\sqrt{2}j)\)
（2）响应振荡收敛即为欠阻尼状态。
\(\therefore\) 由模值条件可知分离点 \(d_1\) 对应的 \(k\) 值：\(\dfrac{k(1-(-0.732))}{|-0.732|\cdot|2-0.732|} = 1\)
\(\therefore\) 解得 \(k=0.536\)