Question

The closed-loop transfer function of a system is $T\left(s\right)=\frac{4}{(s^2+0.4s+4)}$. The steady state error due to unit step input is _____.

• A. 0

Answer 1

The correct option is A 0
Closed-loop transfer function.
$T\left(s\right)=\frac{4}{(s^2+0.4s+4)}$

$or,\frac{G\left(s\right)}{1+G\left(s\right)}=\left(\frac{4}{(s^2+0.4s+4)}\right)\left[H\right(s)=t]$

$or,(s^2+0.4s+4)=4+4.G\left(s\right)$

$or,G\left(s\right)=\frac{4}{s(s+0.4)}$

= Open loop transfer funciton (for unity feedback system)

Given, Input $=u\left(t\right)=r\left(t\right)$

$\therefore R\left(s\right)=1/s$

Steady state error,

$e_{ss}={lim}_{s\to 0}\left[\frac{sR\left(s\right)}{1+G\left(s\right)H\left(s\right)}\right]={lim}_{s\to 0}\left[\frac{sR\left(s\right)}{1+G\left(s\right)}\right]$ (For $H\left(s\right)=1$)

$={lim}_{s\to 0}\left[\frac{s\times \frac{1}{s}}{1+\frac{4}{s(s+0.4)}}\right]={lim}_{s\to 0}\left[\frac{s(s+0.4)}{s^2+0.4s+4}\right]$

$={lim}_{s\to 0}\left[\frac{s(s+0.4)}{s^2+0.4s+4}\right]=0$

$\therefore$ Steady-state error for step input, $e_{ss}=0$