考研851 自动控制原理
真题 · 真题答案
\[20\lg\frac{10}{0.02} - 40\lg\frac{0.2}{0.02} - 20\lg\frac{0.8}{0.2} - 20\lg\frac{\omega_c}{1} = 0 \Rightarrow \omega_c = 1.25\]

图:客观索引

\[\gamma = 180 + \varphi(\omega_c) = 47°\]

四、

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由已知得

非线性:\(-\dfrac{1}{N(X)} = -\dfrac{\pi X}{4}\sqrt{1-\left(\dfrac{h}{X}\right)^2} - j\dfrac{\pi h}{4}\)

线性:\(G(j\omega) = \dfrac{20}{j\omega(0.1j\omega+1)} = \dfrac{-2}{1+0.01\omega^2} - j\dfrac{20}{\omega(1+0.01\omega^2)}\)

如图所示:

图:客观索引

由图可知,负倒特性曲线与G(jω)曲线有交点。所以存在自持振荡,并且是稳定的自持振荡。(由不稳定区→稳定区)

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