\[
\varphi(\omega)=
\begin{cases}
-90^\circ+4\arctan\omega-\arctan\dfrac{\omega}{3}-\arctan\dfrac{\omega}{1-\omega^2}, & 0<\omega\leqslant 1 \\[2mm]
-270^\circ+4\arctan\omega-\arctan\dfrac{\omega}{3}+\arctan\dfrac{\omega}{\omega^2-1}, & \omega>1
\end{cases}
\]
故当\(\omega=2\)和\(\omega=20\)时,有
\[
\begin{cases}
A(2)=1.44\\
L(2)=3.18\text{dB},\\
\varphi(2)=-16.26^\circ
\end{cases}
\qquad
\begin{cases}
A(20)=2.96\\
L(20)=9.41\text{dB}\\
\varphi(20)=-0.049^\circ
\end{cases}
\]
MATLAB验证:系统开环对数幅频及相频曲线如图5-13所示。
图5-13 \(G(s)H(s)=\dfrac{3(s^2+3s^2+3s+1)(s^3+s^2)}{(s^2+s-6)(s^2+2s)(s^2+s+1)}\)对数频率特性(MATLAB)
MATLAB文本:exe509.m
G1=tf(50*conv([1,-3],[4,4,1]),conv([1,4,25,0],conv([1,-1],[1,5,6])));
G2=tf(3*conv([1,3,3,1],[1,1,0,0]),conv([1,1,-6],conv([1,2,0],[1,1,1])));
figure(1);bode(G1);grid
figure(2);bode(G2);grid
5-10 求下述系统的穿越频率\(\omega_x\):
(1) \(G(s)=\dfrac{K}{s(T_1s+1)(T_2s+1)}\), \(K,T_1,T_2>0\);
(2) \(G(s)=\dfrac{K}{s(T_1s+1)(T_2s+1)(T_3s+1)}\), \(K,T_1,T_2,T_3>0\);
(3) \(G(s)=\dfrac{K}{s(T_1s+1)(T_2s+1)(T_3s+1)(T_4s+1)}\), \(K,T_1,T_2,T_3,T_4>0\)。
解 已知系统的穿越频率定义为\(\varphi(\omega_x)=-180^\circ\)。
(1) 系统(1)的相频特性为
\[
\varphi(\omega)=-90^\circ-\arctan T_1\omega-\arctan T_2\omega
\]