Consider the open-loop transfer function:

\(G\left( s \right)H\left( s \right) = \frac{{5\left( {s + 1} \right)}}{{{s^2}\left( {s + 5} \right)\left( {s + 12} \right)}}\)

The steady-state error due to a ramp input is

Concept:

The deviation of the output of the control system from the desired response during steady-state is known as steady-state error.

Steady-state error, \({e_{ss}} = \mathop {\lim }\limits_{s \to 0} \frac{{sR\left( s \right)}}{{1 + G\left( s \right)}}\)

Kp is the positional error coefficient, \({K_p} = \mathop {\lim }\limits_{s \to 0} G\left( s \right)\)

Kv­ is the velocity error coefficient, \({K_v} = \mathop {\lim }\limits_{s \to 0} sG\left( s \right)\)

Ka is the acceleration error coefficient, \({K_a} = \mathop {\lim }\limits_{s \to 0} {s^2}G\left( s \right)\)

Calculation:

Velocity error coefficient, \({K_v} = \mathop {\lim }\limits_{s \to 0} s\frac{{5\left( {s + 1} \right)}}{{{s^2}\left( {s + 5} \right)\left( {s + 12} \right)}} = \infty \)

Steady-state error = 0