\[
c_{ss}(\infty) = \lim_{s \to 0} sC(s) = \lim_{s \to 0} s\Phi(s)R(s)
\]
\[
= \lim_{s \to 0} s \cdot \frac{12}{2.9s^2 + (1+2.9a)s + (a+67.2)} \cdot \frac{5.8}{s}
\]
\[
= 1
\]
求得
\[
\frac{69.6}{a+67.2} = 1, \quad a = 2.4
\]
将 \(a=2.4\) 代入闭环传递函数，整理后得
\[
\Phi(s) = \frac{4.138}{s^2 + 2.745s + 24}
\]
故闭环特征方程为
\[
s^2 + 2.745s + 24 = s^2 + 2\zeta\omega_n s + \omega_n^2 = 0
\]
从而
\[
\omega_n = \sqrt{24} = 4.90, \quad \zeta = \frac{2.745}{2\omega_n} = 0.280
\]
根据求出的 \(\zeta\) 与 \(\omega_n\)，可以算得
\[
\sigma\% = 100e^{-\pi\zeta/\sqrt{1-\zeta^2}}\% = 40\%, \quad t_s = \frac{3.5}{\zeta\omega_n} = 2.55(\Delta=5\%)
\]
(2) 求 \(\omega_r\) 与 \(A_m\)。
将闭环传递函数改写为
\[
\Phi(s) = \frac{\Phi(0)}{\left(\dfrac{s}{\omega_n}\right)^2 + 2\zeta\left(\dfrac{s}{\omega_n}\right) + 1}
\]
其中 \(\Phi(0) = 4.138/\omega_n^2 = 0.172\)。
闭环频率特性
\[
\Phi(j\omega) = \Phi(\omega) \angle \Phi(j\omega)
\]
其中
\[
\Phi(\omega) = |\Phi(j\omega)| = \frac{\Phi(0)}{\sqrt{\left(1-\dfrac{\omega^2}{\omega_n^2}\right)^2 + 4\zeta^2\dfrac{\omega^2}{\omega_n^2}}}
\]
令
\[
\frac{d\Phi(\omega)}{d\omega} = \frac{-\Phi(0)\left[-\dfrac{2\omega}{\omega_n^2}\left(1-\dfrac{\omega^2}{\omega_n^2}\right) + 4\zeta^2\dfrac{\omega}{\omega_n^2}\right]}{\left[\left(1-\dfrac{\omega^2}{\omega_n^2}\right)^2 + 4\zeta^2\dfrac{\omega^2}{\omega_n^2}\right]^{3/2}} = 0
\]
求得
\[
\omega_r = \omega_n\sqrt{1-2\zeta^2} = 4.50
\]
将 \(\omega_r\) 代入 \(\Phi(\omega)\)，可得
\[
\Phi_m(\omega_r) = \frac{\Phi(0)}{2\zeta\sqrt{1-\zeta^2}} = 0.32
\]
因 \(r(t) = A_r\sin\omega t, A_r = 5\)，故稳态输出最大振幅
\[
A_m = A_r\Phi_m(\omega_r) = 1.60
\]
(3) 设计 \(G_n(s)\)。
将系统结构图6-70改画为图6-71(a)，并化简为图6-71(b)。令 \(G_n(s) = 2.9s+1\)，必
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