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?

Question

asked Aug 13, 2020 109k views

1 Answer

Jovana · Answer 1 · 2020-08-18T17:02:54+0000

Answer:

At steady state output will be 2

Step-by-step explanation:

We have given transfer function
G(S)=(6)/(12S+3)

Input is unit step so
X(S)=(1)/(S)

We know that
G(S)=(Y(S))/(X(S)) , here
Y(S) , is output

So output
Y(S)=G(S)* X(S)

Y(S)=(1)/(S)* (6)/(12S+3)

Taking 12 common from denominator

Y(S)=(1)/(2S(S+(1)/(4)))

Now using partial fraction

(1)/(2S(S+(1)/(4)))=(A)/(2S)+(B)/((S+(1)/(4)))

(1)/(2S(S+(1)/(4)))=(A(S+(1)/(4)+2BS))/(2S(S+(1)/(4)))

AS+(A)/(4)+2BS=1

On comparing coefficient A=4 and B = -2

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

Y(S)=(4)/(2S)-(2)/(S+(1)/(4))

Now taking inverse la place

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

Steady state means t tends to infinite

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

y(t)=2-0=2