考研851 自动控制原理
题海 · pdf-page · p.292
\[ G_3(\mathrm{j}\omega) = \frac{1+\mathrm{j}K_d\omega}{\mathrm{j}\omega(\mathrm{j}\omega+0.4)} = \frac{\sqrt{1+(K_d\omega)^2}}{\omega\sqrt{0.4^2+\omega^2}} \Big/ \arctan K_d\omega - 90° - \arctan\frac{\omega}{0.4} \]

\(|G(\mathrm{j}\omega_c)| = \left|\dfrac{\sqrt{1+(K_d\omega_c)^2}}{\omega_c\sqrt{0.4^2+\omega_c^2}}\right| = 1\),解得

\[ \omega_c = \sqrt{\frac{K_d^2-0.16+\sqrt{(K_d^2-0.16)^2+4}}{2}} \]

由于 \(0.1 \leqslant K_d \leqslant 1.5\),则

\[ 0.96 \leqslant \omega_c \leqslant 1.58 \]

再由

\[ \gamma = 180° + \varphi(\omega_c) = 90° + \arctan K_d\omega_c - \arctan\frac{\omega_c}{0.4} \]

考虑到 \(0.1 \leqslant K_d \leqslant 1.5\)\(0.96 \leqslant \omega_c \leqslant 1.58\),则 \(28.10° \leqslant \gamma \leqslant 81.33°\)

MATLAB 验证:令 \(\omega_n=1\),\(\zeta=0.2\),\(K_d=0.1\)\(K_d=1.5\),作系统开环对数频率特性及单位阶跃响应,分别如图 5-68 和图 5-69 所示。测得 \(\omega_c\),\(\gamma\),\(\sigma\%\),\(t_p\),\(t_s\)(\(\Delta=2\%\))范围如下:

图:自控原理题海_p292_fig1

图:自控原理题海_p292_fig2

(a) \(G(s)=\dfrac{\omega_n^2(1+K_ds)}{s(s+2\zeta\omega_n)}\)(\(\omega_n=1\),\(\zeta=0.2\),\(K_d=0.1\))  (b) \(G(s)=\dfrac{\omega_n^2(1+K_ds)}{s(s+2\zeta\omega_n)}\)(\(\omega_n=1\),\(\zeta=0.2\),\(K_d=1.5\))

图 5-68 系统开环对数频率特性(MATLAB)

图:自控原理题海_p292_fig3

图:自控原理题海_p292_fig4

(a) \(G(s)=\dfrac{\omega_n^2(1+K_ds)}{s(s+2\zeta\omega_n)}\)(\(\omega_n=1\),\(\zeta=0.2\),\(K_d=0.1\))  (b) \(G(s)=\dfrac{\omega_n^2(1+K_ds)}{s(s+2\zeta\omega_n)}\)(\(\omega_n=1\),\(\zeta=0.2\),\(K_d=1.5\))

图 5-69 系统单位阶跃响应(MATLAB)

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