Question

What is the steady-state value of the output of a system with transfer function G(s)= 6/(12s+3), subject to a unit-step input?

Answer 1

Answer:

At steady state output will be 2

Explanation:

We have given transfer function $G(S)=\frac{6}{12S+3}$

Input is unit step so $X(S)=\frac{1}{S}$

We know that $G(S)=\frac{Y(S)}{X(S)}$, here $Y(S)$, is output

So output $Y(S)=G(S)\times X(S)$

$Y(S)=\frac{1}{S}\times \frac{6}{12S+3}$

Taking 12 common from denominator

$Y(S)=\frac{1}{2S(S+\frac{1}{4})}$

Now using partial fraction

$\frac{1}{2S(S+\frac{1}{4})}=\frac{A}{2S}+\frac{B}{(S+\frac{1}{4})}$

$\frac{1}{2S(S+\frac{1}{4})}=\frac{A(S+\frac{1}{4}+2BS)}{2S(S+\frac{1}{4})}$

$AS+\frac{A}{4}+2BS=1$

On comparing coefficient A=4 and B = -2

Putting the values of A and B in Y(S)

$Y(S)=\frac{4}{2S}-\frac{2}{S+\frac{1}{4}}$

Now taking inverse la place

$y(t)=2-2e^{\frac{-t}{4}}$

Steady state means t tends to infinite

So output at steady state = $y(t)=2-2e^{\frac{-\infty}{4}}$

$y(t)=2-0=2$