# 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( { 0 votes 222 views closed 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

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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 = $\mathop {\lim }\limits_{s \to 0} sG\left( s \right)H\left( s \right)$

K= 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, ${e_{ss}} = \frac{5}{{{K_a}}} = \infty$