(1) 令 \(M(s) = 0\)
_fig1.png)
\[\dfrac{C(1)}{R(0)} = \dfrac{(G_d + G_1)\dfrac{G_3}{1+G_3H_2}}{1 + \dfrac{G_3G_1H_1}{1+G_3H_2}} = \dfrac{(G_c+G_1)G_2G_3}{1+G_3H_2+G_1G_2G_1H_1}\]
(2) 令 \(R(s) = 0\)
_fig2.png)
\[\dfrac{C(s)}{M(s)} = \dfrac{G_3}{1+G_3H_2+G_1G_2G_1H_1}\]
三、(12分) 已知单位反馈控制系统的开环传递函数为 \(\Phi(s) = \dfrac{3s+10}{5s^2-2s-10}\)。
(1) 试求系统的静态位置稳态系数、静态速度稳态系数。
(2) 若 \(r(t) = 1+t\)，求系统的稳态误差。
解：(1) \(\Phi(s) = \dfrac{C(s)}{R(s)}\)，\(G(s) = \dfrac{\Phi(s)}{1-\Phi(s)}\)
\[= \dfrac{\dfrac{3s+10}{5s^2-2s+10}}{1-\dfrac{3s+10}{5s^2-2s+10}}\]
\[= \dfrac{3s+10}{5s^2-s}\]
\[K_p = \lim_{s\to0}[G(s)+1] = \lim_{s\to0}\left[\dfrac{3s+10}{s(5s-1)}+1\right] = \infty\]
\[K_v = \lim_{s\to0} sG(s) = \lim_{s\to0} s \cdot \dfrac{3s+10}{s(5s-1)} = -10\]
(2) \(R(s) = \dfrac{1}{s}+\dfrac{1}{s^2}\)
\[e_{ss} = \dfrac{1}{K_p}+\dfrac{1}{K_v} = \dfrac{1}{\infty}-\dfrac{1}{10} = -0.1\]
\[\Phi(s) = 5s^2-2s+10 \quad \text{某知都为正,某成稳定}\]