Question

The asymptotic Bode phase plot of $G\left(s\right)=\frac{k}{(s+0.1)(s+10)(s+p_1)}$, with $k$ and $p_1$ both positive , is shown below.

The value of $p_1$ is

• A. 1

Answer 1

The correct option is A 1

$G\left(s\right)=\frac{K}{(s+0.1)(s+10)(s+p_1)}$

$\angle G\left(s\right)=-{\tan }^{-1}\frac{\omega }{0.1}-{\tan }^{-1}\frac{\omega }{10}-{\tan }^{-1}\frac{\omega }{p_1}$

$\angle G\left(s\right){|}_{\omega =1}=-{\tan }^{-1}\frac{1}{0.1}-{\tan }^{-1}\frac{1}{10}-{\tan }^{-1}\frac{1}{p_1}$

$-{\tan }^{-1}-{\tan }^{-1}0.1-{\tan }^{-1}\frac{1}{p_1}=-{135}^{\circ }$

$-{84.28}^{\circ }-{5.71}^{\circ }-{\tan }^{-1}\frac{1}{p_1}=-{135}^{\circ }$

$-{\tan }^{-1}\frac{1}{p_1}=-{135}^{\circ }+{90}^{\circ }$ ${\tan }^{-1}\frac{1}{p_1}={45}^{\circ }$

$\frac{1}{p_1}=1\Rightarrow p_1=1$