考研851 自动控制原理
题海 · 题解 · p.272
\[ G(s)=\dfrac{K\left[\left(\dfrac{s}{\omega_1}\right)^2+2\zeta_1\left(\dfrac{s}{\omega_1}\right)+1\right]}{\left[\left(\dfrac{s}{\omega_2}\right)^2+2\zeta_2\left(\dfrac{s}{\omega_2}\right)+1\right]\left(\dfrac{1}{400}s+1\right)} \]

③ 由给定条件确定开环传递函数参数。

由 \(20\lg K=-20\),解得 \(K=0.1\);

再由 \(40\lg\dfrac{10}{\omega_1}=20\),即 \(\omega_1=\sqrt{10}=3.162\);

再由在 \(\omega=\omega_1\) 处,谐振峰值为8dB,则

\[ 20\lg 2\zeta_1\sqrt{1-\zeta_1^2}=8,\quad 即\quad \zeta_1=0.203 \]

由 \(40\lg\dfrac{\omega_2}{10}=20\),即 \(\omega_2=\sqrt{1000}=31.62\);再由在 \(\omega=\omega_2\) 处,修正 \(20-14=6\text{dB}\),则

\[ 20\lg 2\zeta_2=6,\quad 即\quad \zeta_2=0.998 \]

于是,系统的开环传递函数为

\[ G(s)=\dfrac{0.1\left[\left(\dfrac{s}{\sqrt{10}}\right)^2+2\times0.203\left(\dfrac{s}{\sqrt{10}}\right)+1\right]}{\left[\left(\dfrac{s}{\sqrt{1000}}\right)^2+2\times0.998\left(\dfrac{s}{\sqrt{1000}}\right)+1\right]\left(\dfrac{1}{400}s+1\right)} \]

MATLAB 验证结果如图5-43所示。

图:自控原理题海_p272_fig1

图5-43 \(G(s)=\dfrac{0.1\left[\left(\dfrac{s}{\sqrt{10}}\right)^2+2\times0.203\left(\dfrac{s}{\sqrt{10}}\right)+1\right]}{\left[\left(\dfrac{s}{\sqrt{1000}}\right)^2+2\times0.998\left(\dfrac{s}{\sqrt{1000}}\right)+1\right]\left(\dfrac{1}{400}s+1\right)}\) 开环对数幅频特性曲线(MATLAB)

MATLAB 文本:exe523c.m

K=0.1;w1=3.162;w2=31.62;thema1=0.203;thema2=0.998;

G=tf(K[(1/w1)^2,2thema1/w1,1],conv([(1/w2)^2,2*thema2/w2,1],[1/400,1]));

bode(G);grid

5-24 已知单位反馈系统开环传递函数如下,其相应的概略幅相曲线如图5-44所示,试用奈奎斯特判据判别闭环系统的稳定性(\(\zeta,\omega_n,K,T_i(i=1,2,\cdots,6)\)皆大于零)。

(1) 图5-44(a) \(G(s)=\dfrac{K}{(T_1s-1)(T_2s+1)(T_3s+1)}\);

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