考研851 自动控制原理
题海 · 题海 · p.293
\[\omega_c = 0.96323 \sim 1.5784\text{rad/s}, \quad \gamma = 28.054° \sim 81.323°, \quad \sigma\% = 5\% \sim 44\%\]
\[t_p = 2.28 \sim 2.99\text{s}, \quad t_s = 4.32 \sim 14.0\text{s}(\Delta=2\%), \quad e_{ss}(\infty) = 0\]

MATLAB文本:exe542c.m

wn=1;keth=0.2;Kd1=0.1;Kd2=1.5;

G1=tf(wn^2[Kd1,1],conv([1,0],[1,2keth*wn]));

G2=tf(wn^2[Kd2,1],conv([1,0],[1,2keth*wn]));

cloop1=feedback(G1,1);

cloop2=feedback(G2,1);

figure(1);margin(G1);grid;

figure(2);margin(G2);grid;

figure(3);step(close1);grid;

figure(4);step(close2);grid;

综上可知,对于典型二阶系统,加入测速反馈校正装置,通过减小系统的截止频率来提高系统的相角裕度;而比例-微分校正装置,则是通过超前校正来提高系统的相角裕度,同时也提高了系统的截止频率。

5-43 已知单位反馈系统的开环传递函数如下。试用奈奎斯特判据或对数频率稳定判据,判断闭环系统的稳定性,并确定闭环稳定系统的相角裕度和幅值裕度。

(1) \(G(s)=\dfrac{100}{s(0.2s+1)}\);　(2) \(G(s)=\dfrac{100}{s(0.25s+1)(0.0625s+1)} \cdot \dfrac{0.2s^3}{0.8s+1}\);

(3) \(G(s)=\dfrac{50}{(0.2s+1)(s+2)(s+0.5)}\);　(4) \(G(s)=\dfrac{100}{s(0.8s+1)(0.25s+1)}\);

(5) \(G(s)=\dfrac{100(s+1)}{s(0.1s+1)(0.5s+1)(0.8s+1)}\);　(6) \(G(s)=\dfrac{-10}{2s(1-20s)}\);

(7) \(G(s)=\dfrac{10}{s(0.1s+1)(0.25s+1)}\);　(8) \(G(s)=\dfrac{10}{s(0.2s+1)(s-1)}\);

(9) \(G(s)=\dfrac{1000}{s(s^2+2s)(0.2s+1)}\);　(10) \(G(s)=\dfrac{5(1-0.5s)}{s(1+0.1s)(1-0.2s)}\)。

解 (1) 系统(1)的频率特性为

\[G(j\omega) = \frac{100}{j\omega(j0.2\omega+1)}\]

则系统(1)的开环对数幅频和相频特性为

\[L(\omega) = 20\lg100 - 20\lg\omega - 10\lg[1+(0.2\omega)^2]\]
\[\varphi(\omega) = -90° - \arctan0.2\omega\]

系统(1)的开环幅相特性为

\[G(j\omega) = -\frac{20}{1+0.04\omega^2} - j\frac{100}{\omega(1+0.04\omega^2)}\]

系统(1)的开环对数频率特性图如图5-70所示;开环幅相特性图如图5-71所示。

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