\[
= \frac{1+4\omega^2}{\omega\sqrt{1+\omega^2}\sqrt{1+\left(\frac{\omega}{2}\right)^2}\sqrt{\left(1-\frac{\omega^2}{25}\right)^2+\left(\frac{4\omega}{25}\right)^2}}
\]
\[
L(\omega) = 20\lg(1+4\omega^2) - 20\lg\omega - 10\lg(1+\omega^2) - 10\lg\left[1+\left(\frac{\omega}{2}\right)^2\right]
\]
\[
- 10\lg\left[\left(1-\frac{\omega^2}{25}\right)^2+\left(\frac{4\omega}{25}\right)^2\right]
\]
\[
\varphi(\omega) =
\begin{cases}
-90° + 2\arctan 2\omega - 2\arctan\dfrac{\omega}{3} + \arctan\omega - \arctan\dfrac{\omega}{2} - \arctan\dfrac{4\omega}{25-\omega^2}, & 0<\omega\leqslant 5 \\[2mm]
-270° + 2\arctan 2\omega - 2\arctan\dfrac{\omega}{3} + \arctan\omega - \arctan\dfrac{\omega}{2} + \arctan\dfrac{4\omega}{\omega^2-25}, & \omega>5
\end{cases}
\]
故当 \(\omega=2\) 和 \(\omega=20\) 时,有
\[
\begin{cases}
A(2) = 2.99 \\
L(2) = 9.51\text{dB}, \\
\varphi(2) = -7.87°
\end{cases}
\qquad
\begin{cases}
A(20) = 0.026 \\
L(20) = -31.72\text{dB} \\
\varphi(20) = -240.91°
\end{cases}
\]
MATLAB 验证:系统对数幅频及相频曲线,如图 5-12 所示。
图 5-12 \(\displaystyle G(s)H(s)=\frac{50(s-3)(4s^2+4s+1)}{(s^3+4s^2+25s)(s-1)(s^2+5s+6)}\) 对数频率特性(MATLAB)
(2) \(\displaystyle G(s)H(s)=\frac{3(s^3+3s^2+3s+1)(s^3+s^2)}{(s^2+s-6)(s^2+2s)(s^2+s+1)}\)
由系统的开环传递函数可得,系统的开环幅频、对数幅频和相频特性如下:
\[
A(\omega) = \frac{\dfrac{\omega}{4}(1+\omega^2)^2}{\left[1+\left(\dfrac{\omega}{2}\right)^2\right]\sqrt{1+\left(\dfrac{\omega}{3}\right)^2}\sqrt{(1-\omega^2)^2+\omega^2}}
\]
\[
L(\omega) = 20\lg\frac{\omega}{4} + 40\lg(1+\omega^2) - 20\lg\left[1+\left(\frac{\omega}{2}\right)^2\right] - 10\lg\left[1+\left(\frac{\omega}{3}\right)^2\right] - 10\lg[(1-\omega^2)^2+\omega^2]
\]