则
\[\varphi(\omega_x) = -90° - \arctan T_1\omega_x - \arctan T_2\omega_x = -180°\]
即
\[\arctan T_1\omega_x + \arctan T_2\omega_x = 90°\]
等式两端取正切,得
\[\tan(\arctan T_1\omega_x + \arctan T_2\omega_x) = \infty\]
根据两角和的三角函数关系,得
\[\frac{\tan(\arctan T_1\omega_x) + \tan(\arctan T_2\omega_x)}{1 - \tan(\arctan T_1\omega_x)\tan(\arctan T_2\omega_x)} = \infty\]
表明应有
\[1 - T_1T_2\omega_x^2 = 0, \quad \omega_x = \sqrt{1/T_1T_2}\]
(2) 系统(2)的相频特性为
\[\varphi(\omega) = -90° - \arctan T_1\omega - \arctan T_2\omega - \arctan T_3\omega\]
则 \(\varphi(\omega_x) = -90° - \arctan T_1\omega_x - \arctan T_2\omega_x - \arctan T_3\omega_x = -180°\)
即
\[\arctan T_1\omega_x + \arctan T_2\omega_x = 90° - \arctan T_3\omega_x\]
等式两端取正切,得
\[\tan(\arctan T_1\omega_x + \arctan T_2\omega_x) = \tan(90° - \arctan T_3\omega_x)\]
根据两角和的三角函数关系,得
\[\frac{\tan(\arctan T_1\omega_x) + \tan(\arctan T_2\omega_x)}{1 - \tan(\arctan T_1\omega_x)\tan(\arctan T_2\omega_x)} = \tan(90° - \arctan T_3\omega_x)\]
表明应有
\[\frac{T_1\omega_x + T_2\omega_x}{1 - T_1T_2\omega_x^2} = \frac{1}{T_3\omega_x}\]
即
\[1 - (T_1T_2 + T_1T_3 + T_2T_3)\omega_x^2 = 0\]
解得
\[\omega_x = \frac{1}{\sqrt{T_1T_2 + T_1T_3 + T_2T_3}}\]
(3) 系统(3)的相频特性为
\[\varphi(\omega) = -90° - \arctan T_1\omega - \arctan T_2\omega - \arctan T_3\omega - \arctan T_4\omega\]
则 \(\varphi(\omega_x) = -90° - \arctan T_1\omega_x - \arctan T_2\omega_x - \arctan T_3\omega_x - \arctan T_4\omega_x = -180°\)
即
\[\arctan T_1\omega_x + \arctan T_2\omega_x = 90° - \arctan T_3\omega_x - \arctan T_4\omega_x\]
等式两端取正切,得
\[\tan(\arctan T_1\omega_x + \arctan T_2\omega_x) = \tan(90° - \arctan T_3\omega_x - \arctan T_4\omega_x)\]
根据两角和的三角函数关系,得
\[\frac{\tan(\arctan T_1\omega_x) + \tan(\arctan T_2\omega_x)}{1 - \tan(\arctan T_1\omega_x)\tan(\arctan T_2\omega_x)} = \frac{1 - \tan(\arctan T_3\omega_x)\tan(\arctan T_4\omega_x)}{\tan(\arctan T_3\omega_x) + \tan(\arctan T_4\omega_x)}\]
即
\[\frac{T_1\omega_x + T_2\omega_x}{1 - T_1T_2\omega_x^2} = \frac{1 - T_3T_4\omega_x^2}{T_3\omega_x + T_4\omega_x}\]
有
\[T_1T_2T_3T_4\omega_x^4 - (T_1T_2 + T_1T_3 + T_1T_4 + T_2T_3 + T_2T_4 + T_3T_4)\omega_x^2 + 1 = 0\]
解得
\[\omega_x = \left(\frac{B \pm \sqrt{B^2 - 4A}}{2A}\right)^{1/2}; \quad \forall \omega_x > 0, \ B^2 - 4A > 0\]
其中
\[A = T_1T_2T_3T_4, \quad B = T_1T_2 + T_1T_3 + T_1T_4 + T_2T_3 + T_2T_4 + T_3T_4\]
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