（接上页）
\[0.78 = \omega_2 \sqrt{1-2\zeta^2}, \quad 42.7 - 40\lg\frac{9.49}{\omega^2} = 20\lg\frac{1}{2\zeta\sqrt{1-\zeta^2}}, \quad 40\lg 9.49 = 20\lg\frac{K}{\omega_1}\]
解得 \(K=13.65\)，\(\omega_1=0.14\)，\(\omega_2=0.95\)，\(\zeta=0.4\)，\(\omega_3=3\)，\(\omega_4=83.33\)
故开环系统的传递函数为
\[G(s) = \frac{13.65\left(\dfrac{1}{0.14}s+1\right)\left(\dfrac{1}{3}s+1\right)}{s\left[\left(\dfrac{1}{0.95}s\right)^2 + \dfrac{2\times 0.4}{0.95}s + 1\right]\left(\dfrac{1}{83.33}s+1\right)}\]
图5-114 系统开环对数幅频渐近特性曲线
5-61 已知系统开环传递函数为 \(G(s)H(s) = \dfrac{K(T_1 s+1)}{s^2(T_2 s+1)}\)，\(T_1 > T_2 > 0\)，试求 \(K\) 变化下系统相角裕度的最大值。
解 系统的开环频率特性为
\[G(\mathrm{j}\omega)H(\mathrm{j}\omega) = \frac{K(1+\mathrm{j}T_1\omega)}{-\omega^2(1+\mathrm{j}T_2\omega)} = \frac{K\sqrt{1+T_1^2\omega^2}}{\omega^2\sqrt{1+T_2^2\omega^2}}\angle\left(\arctan T_1\omega - 180^\circ - \arctan T_2\omega\right)\]
由系统相角裕度的定义可知
\[\gamma = 180^\circ + \angle G(\mathrm{j}\omega_c)H(\mathrm{j}\omega_c) = \arctan T_1\omega_c - \arctan T_2\omega_c\]
且有
\[|G(\mathrm{j}\omega_c)H(\mathrm{j}\omega_c)| = \left|\frac{K\sqrt{1+T_1^2\omega_c^2}}{\omega_c^2\sqrt{1+T_2^2\omega_c^2}}\right| = 1\]
令 \(\dfrac{\mathrm{d}\gamma}{\mathrm{d}\omega_c} = 0\)，得
\[\frac{T_1}{1+T_1^2\omega_c^2} - \frac{T_2}{1+T_2^2\omega_c^2} = 0\]
求出
\[\omega_c = \frac{1}{\sqrt{T_1 T_2}}\]
而
\[\frac{\mathrm{d}^2\gamma}{\mathrm{d}\omega_c^2} = -\frac{T_1^3\omega_c}{(1+T_1^2\omega_c^2)^2} + \frac{T_2^3\omega_c}{(1+T_2^2\omega_c^2)^2}\]
（本段推导在本页末尾被截断，续见下一页）
・316・