Question

The Bode asymptotic magnitude plot of a minimum phase system is shown in the figure.

If the system is connected in a unity negative feedback configuration, the steady state error of the closed loop system, to a unit ramp input is

• A. 0.5

Answer 1

The correct option is A 0.5

The open loop transfer function of the system is
$G\left(s\right)H\left(s\right)=\frac{K(\frac{s}{2}+1)}{s(\frac{s}{10}+1)}....\left(i\right)$
where
$26.02{|}_{\omega =0.1}=20logK-20log\omega$
or, $K=1.99\simeq 2....\left(ii\right)$
From equation (i) and (ii), we get
$G\left(s\right)=\frac{20(s+2)}{2s(s+10)}.....\left(iii\right)$
Steady state error for ramp input is
$e_{ss}=\frac{A}{K_V}\text{where}{K}_V=\underset{s\to 0}{lim}sG\left(s\right)$
From equation (iii),
$K_V=\underset{s\to 0}{lim}s\times \frac{10(s+2)}{s(s+10)}=2$
$\therefore e_{ss}=\frac{1}{2}=0.5$