考研851 自动控制原理
题海 · 解答/仿真 · p.194

图:自控原理题海_p194_fig1

图:自控原理题海_p194_fig2

图 4-91 \(1+\dfrac{100bs}{(s+5-\mathrm{j}8.66)(s+5+\mathrm{j}8.66)}=0\)

概略参数根轨迹图

图 4-92 \(1+\dfrac{100bs}{(s+5-\mathrm{j}8.66)(s+5+\mathrm{j}8.66)}=0\)

参数根轨迹图(MATLAB)

\[ p_1=0,\quad p_2=-1.05+\mathrm{j}2.26 \]
\[ p_3=-1.05-\mathrm{j}2.26 \]

设闭环根为 \(s\),根据根轨迹的幅值条件

\[ K=4.4=|s-p_1|\cdot|s-p_2|\cdot|s-p_3| \]

应用MATLAB方法可解得

\[ s_1=-0.857,\quad s_2=-0.622+\mathrm{j}2.18, \]
\[ s_3=-0.622-\mathrm{j}2.18 \]

(2) 应用MATLAB方法可得多项式的等效开环传递函数为

\[ G(s)=\frac{K}{s^5+4s^4+4s^3+s^2+2s} \]
\[ =\frac{K}{s(s+2)(s+2.21)(s-0.103-\mathrm{j}0.665)(s-0.103+\mathrm{j}0.665)} \]

其中,\(K=1\)。系统的开环极点为

\[ p_1=0,\quad p_2=-2,\quad p_3=-2.21\quad p_4=0.103+\mathrm{j}0.665,\quad p_5=0.103-\mathrm{j}0.665 \]

设闭环根为 \(s\),根据根轨迹的幅值条件

\[ K=1=|s-p_1|\cdot|s-p_2|\cdot|s-p_3|\cdot|s-p_4|\cdot|s-p_5| \]

应用MATLAB方法可解得

\[ s_1=-2.38,\quad s_2=-1.69,\quad s_3=-0.486,\quad s_4=0.274+\mathrm{j}0.662,\quad s_5=0.274-\mathrm{j}0.662 \]

实际上,应用MATLAB求根命令roots,可直接求出本题要求的结果。

仿真曲线如图4-93、图4-94所示。

MATLAB程序:exe429.m

num1=[1]; den1=[1 2.1 6.2 0]; k1=4.4; [p1,z1]=pzmap(num1,den1);
figure, rlocus(num1,den1); hold on; rlocus(num1,den1,k1);
num2=[1]; den2=[1 4 4 1 2 0]; k2=1; [p2,z2]=pzmap(num2,den2);
figure, rlocus(num2,den2); hold on; rlocus(num2,den2,k2);

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