分离点：\(\dfrac{dG(s)}{ds} = 0 \Rightarrow s_1 = -1.27, s_2 = -4.73\)

渐近线：\(\sigma_a = 1, \phi_a = \dfrac{\pi}{2-1}\)

如图所示：

（图中红笔手写补充推导，位于图右侧空白处，内容如下：）

\(d = \sqrt{3^2 - (\sqrt{3})^2} = \sqrt{6}\)

\(\cos\beta = \dfrac{\sqrt{6}}{3} = \xi\)

\(\cos\beta = \dfrac{d_2}{d_0} = \dfrac{d_2}{d_1} = \dfrac{\sqrt{3}}{3}\)

\(\Rightarrow d_2 = d_2 = \sqrt{2}\)

四边 \((\sqrt{2}, h = \sqrt{(\sqrt{6})^2 - (\sqrt{2})^2} = 2\)

~~即该点为\((\sqrt{2}, j2)\)~~

~~即\(s_A = \sqrt{2} \pm 2j\)~~

即\(s_A = -\sqrt{2} + 2j\)

(2)

令 \(s = j\omega\) 代入：\(D(s) = 0 \Rightarrow (\sigma + 3)^2 + \omega^2 = 3\)

根轨迹是一个圆。上图中 A 点是震荡最大值

A 点坐标为\(s_A = -3 \pm \sqrt{3}j\)

利用模值条件，求得 \(k = 4, \xi = \arccos\beta = \dfrac{\sqrt{6}}{3}\)

四、

相角范围：\(-90° \rightarrow -270°\)

经过第三象限和第二象限

\(\omega = \dfrac{\sqrt{15}}{15}\)时，与实轴交于\((-\dfrac{15k}{8}, j0)\)

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