Question

The steady state error of a unity feedback linear system for a unit step input is 0.1. The steady state error of the same system, for a pulse input r(t) having a magnitude of 10 and aduration of one second, as shown in the figure is

• A. 0
• B. 0.1
• C. 1
• D. 10

Answer 1

The correct option is A 0

Let the system is represented as


$\frac{y\left(s\right)}{x\left(s\right)}=\frac{G\left(s\right)}{1+G\left(s\right)H\left(s\right)}$

H(s) = 1 (unity feedback)

$Error=E\left(s\right)=\frac{x\left(s\right)}{1+G\left(s\right)}$

$Steady-stateerrorforx\left(s\right)=1/s.e_{ss}=0.1$

$e_{ss}={lim}_{s\to 0}sE\left(s\right)={lim}_{s\to 0}\frac{sx\left(s\right)}{1+G\left(s\right)}$

$\Rightarrow 0.1={lim}_{s\to 0}\frac{s\times \frac{1}{s}}{1+G\left(s\right)}$

$\Rightarrow {lim}_{s\to 0}\frac{1}{1+G\left(s\right)}=0.1$

When input,


$x\left(t\right)=10\left[u\right(t)-u(t-1\left)\right]$

$x\left(s\right)=10[\frac{1}{s}-\frac{e^{-s}}{s}]=\frac{10}{s}[1-e^{-s}]$

$e_{ss}={lim}_{s\to 0}sE\left(s\right)={lim}_{s\to 0}\frac{Sx\left(s\right)}{1+G\left(s\right)}$

$={lim}_{s\to 0}\frac{s\times \frac{10}{s}[1-e^{-s}]}{1+G\left(s\right)}$

$\Rightarrow e_{ss}={lim}_{s\to 0}\frac{10(1-e^{-s})}{1+G\left(s\right)}=0$