欠阻尼二阶系统动态分析

二阶系统的设计一般取 $ζ = 0.4 \sim 0.8$

\begin{aligned} c (t) & = 1 - \frac{1}{\sqrt{1 - ζ^{2}}} e^{- ζ ω_{n} t} (\sqrt{1 - ζ^{2}} \cos ω_{d} t + ζ \sin ω_{d} t) \\ = 1 - \frac{1}{\sqrt{1 - ζ^{2}}} e^{- ζ ω_{n} t} \sin (ω_{d} t + β) \\ β & = \arctan \frac{\sqrt{1 - ζ^{2}}}{ζ} \\ ζ & = \cos β \end{aligned}

$ω_{n} ↑ t_{d}, t_{r}, t_{p}, t_{s} ↓$
$ζ ↑ t_{d}, t_{r}, t_{p} ↑ σ, t_{s} ↓$

延迟时间：经验公式

\begin{array}{r} h (t_{d}) = 0.5 t_{d} = \frac{1 + 0.6 ζ + 0.2 ζ^{2}}{ω_{n}} \end{array}

上升时间：

\begin{array}{r} h (t_{r}) = 1 \sin (ω_{d} t_{r} + β) = 0 t_{r} = \frac{π - β}{ω_{d}} \end{array}

峰值时间：

\begin{array}{r} \frac{d h}{d t} ∣_{t = t_{p}} = 0 t_{p} = \frac{π}{ω_{d}} \end{array}

阻尼振荡周期的一半，与闭环极点虚部成反比；闭环极点距离负实轴距离越远，峰值时间越短

超调量：

\begin{array}{r} \sin (π + β) = - \sqrt{1 - ζ^{2}} \end{array}

\begin{aligned} h (t_{p}) & = 1 - \frac{1}{\sqrt{1 - ζ^{2}}} e^{- π ζ / \sqrt{1 - ζ^{2}}} \sin (π + β) \\ σ % & = \frac{h (t_{p}) - h (\infty)}{h (\infty)} = e^{- π ζ / \sqrt{1 - ζ^{2}}} \end{aligned}

调节时间：

\begin{array}{r} t_{s} \leq \frac{3.5}{ζ ω_{n}} Δ = 0.05 \\ t_{s} \leq \frac{4.4}{ζ ω_{n}} Δ = 0.02 \end{array}