Question

For a unity feedback control system, the forward path transfer function is given by \(G\left( s \right) = \frac{{40}}{{s\left( {s + 2} \right)\left( {{s^2} + 2s + 30} \right)}}\)

The steady-state error of the system for the input \(\frac{{5{t^2}}}{2}\) is

Answer 1

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 steady-state error for step-input, finite steady-state error for Ramp-input and \(\infty \) steady-state error for parabolic-input.

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, \({e_{ss}} = \frac{5}{{{K_a}}} = \infty \)