Correct Answer  Option 2 : ∞
Concept:
KP = position error constant = \(\mathop {\lim }\limits_{s \to 0} G\left( s \right)H\left( s \right)\)
Kv = velocity error constant = \(\mathop {\lim }\limits_{s \to 0} sG\left( s \right)H\left( s \right)\)
Ka = acceleration error constant = \(\mathop {\lim }\limits_{s \to 0} {s^2}G\left( s \right)H\left( s \right)\)
Steady state error for different inputs is given by
Input

Type 0

Type  1

Type 2

Unit step

\(\frac{1}{{1 + {K_p}}}\)

0

0

Unit ramp

∞

\(\frac{1}{{{K_v}}}\)

0

Unit parabolic

∞

∞

\(\frac{1}{{{K_a}}}\)

From the above table, it is clear that for type – 1 system, a system shows zero steadystate error for stepinput, finite steadystate error for Rampinput and \(\infty \) steadystate error for parabolicinput.
Calculation:
Input = \(\frac{{5{t^2}}}{2}\)
It is a parabolic input.
The error coefficient for parabolic input is acceleration error constant.
\({K_a} = \mathop {\lim }\limits_{s \to 0} {s^2}G\left( s \right) = \mathop {\lim }\limits_{s \to 0} {s^2}\frac{{40}}{{s\left( {s + 2} \right)\left( {{s^2} + 2s + 30} \right)}} = 0\)
Steady state error, \({e_{ss}} = \frac{5}{{{K_a}}} = \infty \)