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

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`