Question

The polar plot of
$G\left(s\right)=\frac{10}{s(s+1{)}^2}$
intercepts real axis at $\omega ={\omega }_0$. Then, the real part and ${\omega }_0$ are respectively given by:

• A. $-2.5,1$
• B. $-5,0.5$
• C. $-5,1$
• D. $-5,2$

Answer 1

The correct option is C $-5,1$

$G\left(s\right)=\frac{10}{s(s+1{)}^2}$
$\angle G=-{180}^{\circ }=-{90}^{\circ }-2{\tan }^{-1}\left(\omega \right)$ $2{\tan }^{-1}\left(\omega \right)={45}^{\circ }$
$\omega =1$
${\omega }_{pc}=1rad/sec$
$|G{|}_{\omega ={\omega }_{pc}}=\frac{10}{\omega (1+{\omega }^2)}=\frac{10}{1(1+1)}=5$
At $\omega =\omega$ the polar plot crosses the negative real axis at$-5.$