欠阻尼二阶系统动态性能指标计算
二阶系统的经典结构
G ( s ) = w n 2 s ( s + 2 ξ w n ) Φ ( s ) = G ( s ) 1 + G ( s ) = w n 2 s 2 + 2 ξ w n s + w n 2 \begin{aligned} G(s)&=\frac{w_n^2}{s(s+2\xi w_n)}\\ \Phi(s)&=\frac{G(s)}{1+G(s)}=\frac{w_n^2}{s^2+2\xi w_ns+w_n^2}\\ \end{aligned} G(s)Φ(s)=s(s+2ξwn)wn2=1+G(s)G(s)=s2+2ξwns+wn2wn2
欠阻尼二阶系统极点的两种表示方法
直角坐标系表示
λ 1 , 2 = σ ± j w d = − ξ w n ± j 1 − ξ 2 w n \lambda_{1,2}=\sigma\pm jw_d=-\xi w_n\pm j\sqrt{1-\xi^2}w_n λ1,2=σ±jwd=−ξwn±j1−ξ2wn
- w d w_d wd: 阻尼振荡频率(也是特征根的虚部)
- w n w_n wn: 自然频率(无阻尼振荡频率)
- ξ \xi ξ: 阻尼比
- w d = 1 − ξ 2 w n w_d=\sqrt{1-\xi^2}w_n wd=1−ξ2wn
"极"坐标表示
{ ∣ λ ∣ = w n ∠ λ = β { cos β = ξ sin β = 1 − ξ 2 \begin{cases}|\lambda|=w_n\\\angle\lambda=\beta\end{cases}\qquad\qquad\begin{cases}\cos\beta=\xi\\\sin\beta=\sqrt{1-\xi^2}\end{cases} {∣λ∣=wn∠λ=β{cosβ=ξsinβ=1−ξ2
- β \beta β: 阻尼角
- 极点与原点的距离为 w n w_n wn
欠阻尼二阶系统的单位阶跃响应
C ( s ) = Φ ( s ) R ( s ) = w n 2 s 2 + 2 ξ w n s + w n 2 ⋅ 1 s h ( t ) = 1 − e − ξ w n t 1 − ξ 2 sin ( 1 − ξ 2 w n + β ) ( 0 ≤ ξ < 1 ) \begin{aligned} C(s)&=\Phi(s)R(s)=\frac{w_n^2}{s^2+2\xi w_ns+w_n^2}·\frac{1}{s}\\ h(t)&=1-\frac{e^{-\xi w_nt}}{\sqrt{1-\xi^2}}\sin{(\sqrt{1-\xi^2}w_n+\beta)}\qquad(0\le\xi<1)\\ \end{aligned} C(s)h(t)=Φ(s)R(s)=s2+2ξwns+wn2wn2⋅s1=1−1−ξ2e−ξwntsin(1−ξ2wn+β)(0≤ξ<1)
- h ( t ) = 1 − e − σ t sin β ⋅ sin ( w d + β ) h(t)=1-\frac{e^{-\sigma t}}{\sin\beta}·\sin{(w_d+\beta)} h(t)=1−sinβe−σt⋅sin(wd+β)
- σ \sigma σ为特征根的实部
- w d w_d wd为特征根的虚部
- β \beta β为阻尼角
ξ = 0 , β = 90 ° \xi=0,\beta=90° ξ=0,β=90°时
h ( t ) = 1 − cos w n t h(t)=1-\cos{w_nt} h(t)=1−coswnt
0 < ξ < 1 0<\xi<1 0<ξ<1时
- 系统响应在 ( 1 + e − σ t sin β ) (1+\frac{e^{-\sigma t}}{\sin\beta}) (1+sinβe−σt)和 ( 1 − e − σ t sin β ) (1-\frac{e^{-\sigma t}}{\sin\beta}) (1−sinβe−σt)之间(上下包络线之间)
欠阻尼二阶系统动态性能指标计算(重点)
峰值时间 t p t_p tp
t p = π 1 − ξ 2 w n t_p=\frac{\pi}{\sqrt{1-\xi^2}w_n} tp=1−ξ2wnπ
- 峰值时间的定义为 π \pi π比上极点的虚部
超调量 σ % \sigma\% σ%
σ % = e − ξ π 1 − ξ 2 × 100 % = e − cos β π sin β × 100 % = e − π cot β × 100 % \begin{aligned} \sigma\%&=e^{-\frac{\xi\pi}{\sqrt{1-\xi^2}}}\times100\%\\ &=e^{-\frac{\cos\beta\pi}{\sin\beta}}\times100\%\\ &=e^{-\pi\cot\beta}\times100\% \end{aligned} σ%=e−1−ξ2ξπ×100%=e−sinβcosβπ×100%=e−πcotβ×100%
- ξ ↑ cos β ↑ sin β ↓ cot β ↑ σ % ↓ \xi\uparrow\qquad\cos\beta\uparrow\qquad\sin\beta\downarrow\qquad \cot\beta\uparrow\qquad\sigma\%\downarrow ξ↑cosβ↑sinβ↓cotβ↑σ%↓
- 超调量只与阻尼比有关
- c o t β cot\beta cotβ也是实部比虚部
调节时间 t s t_s ts
t s = 3.5 ξ w n ( Δ % = 5 % ) t s = 4.5 ξ w n ( Δ % = 2 % ) \begin{aligned} t_s&=\frac{3.5}{\xi w_n}\quad(\Delta\%=5\%)\\ t_s&=\frac{4.5}{\xi w_n}\quad(\Delta\%=2\%)\\ \end{aligned} tsts=ξwn3.5(Δ%=5%)=ξwn4.5(Δ%=2%)
- 调节时间与极点的实部有关
- 实部绝对值越大，调节时间越短(负半轴)
- 调节时间根据包络线计算(不是严格意义上的调节时间)
“最佳阻尼比” ξ = 0.707 \xi=0.707 ξ=0.707时, w n w_n wn确定时， t s t_s ts实际最小
二阶系统动态性能随极点位置分布的变化规律
- 实部不变，虚部越大 t p ↓ β ↑ ξ ↓ σ % ↑ t s 不变 t_p\downarrow\quad\beta\uparrow\quad\xi\downarrow\quad\sigma\%\uparrow\quad t_s不变 tp↓β↑ξ↓σ%↑ts不变
- 虚部不变，实部变大(绝对值) t s ↓ β ↓ σ % ↓ t p 不变 t_s\downarrow\quad\beta\downarrow\quad\sigma\%\downarrow\quad t_p不变 ts↓β↓σ%↓tp不变
- β \beta β不变时, 实部虚部一起变大 ξ 不变 σ % 不变 t s ↓ t p ↓ \xi不变\quad\sigma\%不变\quad t_s\downarrow\quad t_p\downarrow ξ不变σ%不变ts↓tp↓
例题
例1 控制系统结构如图所示
(1) 开环增益K=10时，求系统的动态性能指标
G ( s ) = 10 s ( 0.1 s + 1 ) Φ ( s ) = G ( s ) 1 + G ( s ) = 10 0.1 s 2 + s + 10 = 100 10 s 2 + 10 s + 100 { w n 2 = 10 2 ξ w n = 100 { w n = 10 ξ = 0.5 t p = π 1 − ξ 2 w n = 0.363 σ % = e − π ξ 1 − ξ 2 = 16.3 % t s = 3.5 ξ w n = 0.7 ( Δ = 5 % ) \begin{aligned} G(s)&=\frac{10}{s(0.1s+1)}\\ \Phi(s)&=\frac{G(s)}{1+G(s)}=\frac{10}{0.1s^2+s+10}=\frac{100}{10s^2+10s+100}\\ &\begin{cases}w_n^2=10\\2\xi w_n=100\end{cases}\qquad\begin{cases}w_n=10\\\xi=0.5\end{cases}\\ t_p&=\frac{\pi}{\sqrt{1-\xi^2}w_n}=0.363\\ \sigma\%&=e^{-\frac{\pi\xi}{\sqrt{1-\xi^2}}}=16.3\%\\ t_s&=\frac{3.5}{\xi w_n}=0.7\quad(\Delta=5\%)\\ \end{aligned} G(s)Φ(s)tpσ%ts=s(0.1s+1)10=1+G(s)G(s)=0.1s2+s+1010=10s2+10s+100100{wn2=102ξwn=100{wn=10ξ=0.5=1−ξ2wnπ=0.363=e−1−ξ2πξ=16.3%=ξwn3.5=0.7(Δ=5%)
(2) 确定使系统阻尼比 ξ = 0.707 \xi=0.707 ξ=0.707的K值
G ( s ) = K s ( 0.1 s + 1 ) Φ ( s ) = G ( s ) 1 + G ( s ) = K 0.1 s 2 + s + K = 10 K s 2 + 10 s + 10 K { w n 2 = 10 K 2 ξ w n = 10 K = ( 5 ξ ) 2 ⋅ 1 10 = 25 10 ξ 2 = 5 \begin{aligned} G(s)&=\frac{K}{s(0.1s+1)}\\ \Phi(s)&=\frac{G(s)}{1+G(s)}=\frac{K}{0.1s^2+s+K}=\frac{10K}{s^2+10s+10K}\\ &\begin{cases}w_n^2=10K\\2\xi w_n=10\end{cases}\\ K&=(\frac{5}{\xi})^2·\frac{1}{10}=\frac{25}{10\xi^2}=5 \end{aligned} G(s)Φ(s)K=s(0.1s+1)K=1+G(s)G(s)=0.1s2+s+KK=s2+10s+10K10K{wn2=10K2ξwn=10=(ξ5)2⋅101=10ξ225=5
例2 二阶系统的结构图及单位阶跃响应分别如图(a),(b)所示。试确定系统参数 K 1 , K 2 , a K_1,K_2,a K1,K2,a的值。
Φ ( s ) = K 1 K 2 s ( s + a ) 1 + K 2 s ( s + a ) = K 1 K 2 s 2 + a s + K 2 { K 2 = w n 2 a = 2 ξ w n 由 单位阶跃响应曲线有 h ( ∞ ) = 2 = lim s → 0 s Φ ( s ) R ( s ) = lim s → 0 K 1 K 2 s 2 + a s + K 2 = K 1 { t p = π 1 − ξ 2 w n = 0.75 σ % = e − π ξ 1 − ξ 2 = 0.09 解得 { ξ = 0.608 w n = 5.278 { K 2 = 27.85 a = 6.42 因此有 K 1 = 2 K 2 = 27.85 a = 6.42 \begin{aligned} \Phi(s)&=\frac{K_1\frac{K_2}{s(s+a)}}{1+\frac{K_2}{s(s+a)}}=\frac{K_1K_2}{s^2+as+K_2}\\ &\begin{cases}K_2=w_n^2\\a=2\xi w_n\end{cases}\\ 由&单位阶跃响应曲线有\\ h(\infty)&=2=\lim_{s\rightarrow0}s\Phi(s)R(s)=\lim_{s\rightarrow0}\frac{K_1K_2}{s^2+as+K_2}=K_1\\ &\begin{cases}t_p=\frac{\pi}{\sqrt{1-\xi^2}w_n}=0.75\\\sigma\%=e^{-\frac{\pi\xi}{\sqrt{1-\xi^2}}}=0.09\end{cases}\\ 解得&\quad\begin{cases}\xi=0.608\\w_n=5.278\end{cases}\\ &\quad\begin{cases}K_2=27.85\\a=6.42\end{cases}\\ 因此有&\quad K_1=2\quad K_2=27.85\quad a=6.42 \end{aligned} Φ(s)由h(∞)解得因此有=1+s(s+a)K2K1s(s+a)K2=s2+as+K2K1K2{K2=wn2a=2ξwn单位阶跃响应曲线有=2=s→0limsΦ(s)R(s)=s→0lims2+as+K2K1K2=K1⎩ ⎨ ⎧tp=1−ξ2wnπ=0.75σ%=e−1−ξ2πξ=0.09{ξ=0.608wn=5.278{K2=27.85a=6.42K1=2K2=27.85a=6.42
