考研851 自动控制原理
真题 · image

(2) 校正装置:\(G_c = \dfrac{(10s+1)(2s+1)}{(100s+1)(0.2s+1)}\)

图:客观索引

转折频率为:0.01 0.1 0.5 5

计算 w=5 时的 L:

\[20*\lg\dfrac{1}{0.01^0} - 20*\lg\dfrac{0.1}{0.01} + 0*\lg\dfrac{0.5}{0.1} + 20*\lg\dfrac{5}{0.5} = 0\]

图:客观索引

(3) 校正后:\(G_cGH = \dfrac{10}{s(2s+1)} \cdot \dfrac{(10s+1)(2s+1)}{(100s+1)(0.2s+1)} = \dfrac{10(10s+1)}{s(100s+1)(0.2s+1)}\)

定义法计算截止频率:0.986

相位裕度:\(r = 180° - 90° - \operatorname{arctg}(100\omega_c) - \operatorname{arctg}(0.2\omega_c) + \operatorname{arctg}(10\omega_c) = 73.6°\)

转折频率:0.01 0.1 5