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The closed loop transfer function of a system is \(T\left( s \right) = \frac{4}{{\left( {{s^2} + 0.4s + 4} \right)}}\). The steady state error due to unit step input is ________.

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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:

The given closed-loop transfer function is \(T\left( s \right) = \frac{4}{{\left( {{s^2} + 0.4s + 4} \right)}}\)

To find a steady state error, we require an open loop transfer function.

The open-loop transfer function \( = \frac{{CLTF}}{{CLTF - 1}}\)

\( = \frac{{\frac{4}{{\left( {{s^2} + 0.4s + 4} \right)}}}}{{\frac{4}{{\left( {{s^2} + 0.4s + 4} \right)}} - 1}}\)

\( = \frac{4}{{s\left( {s + 0.4} \right)}}\)

Error constant \({K_P} = \mathop {\lim }\limits_{s \to 0} \frac{4}{{s\left( {s + 0.4} \right)}} = \infty \)

Steady-state error \( = \frac{1}{{1 + \infty }} = 0\)

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