考研851 自动控制原理
题海 · 题海 · p.257

图5-23 续:MATLAB代码

figure(2);semilogx(x2,y2),grid;
figure(3);semilogx(x3,y3),grid;
figure(4);semilogx(x4,y4),grid;
figure(5);semilogx(x5,y5),grid;
figure(6);semilogx(x6,y6),grid;

图:自控原理题海_p257_fig1

图5-23 \(G(s)=\dfrac{8(s+0.1)}{s(s^2+s+1)(s^2+4s+25)}\)对数幅频渐近特性(MATLAB)

5-16

系统的开环传递函数\(G(s)H(s)=\dfrac{K(T_as+1)(T_bs+1)}{s^2(T_1s+1)}\)\(H(s)=1\)。试分别作出下列三种情况下的概略幅相特性曲线:

(1) \(T_a>T_b+T_1>0,T_b>T_1>0\); (2) \(T_a>T_1>T_b>0\); (3) \(T_1>T_a+T_b,T_a>0,T_b>0\)

系统的开环频率特性为

\[G(\mathrm{j}\omega)H(\mathrm{j}\omega)=\frac{K(1+\mathrm{j}T_a\omega)(1+\mathrm{j}T_b\omega)}{-\omega^2(1+\mathrm{j}T_1\omega)}\]
\[=-\frac{K[1+(T_1T_a+T_1T_b-T_aT_b)\omega^2]}{\omega^2(1+T_1^2\omega^2)}-\mathrm{j}\frac{K(T_a+T_b-T_1+T_1T_aT_b\omega^2)}{\omega(1+T_1^2\omega^2)}\]

(1) \(T_a>T_b+T_1>0,T_b>T_1>0\)时,有

开环幅相特性曲线的起点为\(G(\mathrm{j}0_+)=-\infty-\mathrm{j}\infty\);终点为\(G(\mathrm{j}\infty)=0\)

开环幅相特性曲线与虚轴的交点:令\(\mathrm{Re}[G(\mathrm{j}\omega)]=0\),解得

\[ \begin{cases} \omega_y=\sqrt{\dfrac{1}{T_aT_b-T_1T_a-T_1T_b}}\\[2ex] G(\mathrm{j}\omega_y)=\mathrm{Im}[G(\mathrm{j}\omega_y)]\\[1ex] =-\dfrac{K\sqrt{T_aT_b-T_1T_a-T_1T_b}[T_aT_b(T_a+T_b)+T_1T_a(T_1-T_a)+T_1T_b(T_1-T_b)-3T_1T_aT_b]}{T_aT_b+T_1^2-T_1T_a-T_1T_b} \end{cases} \]