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:
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 \( = \frac{1}{{1 + \infty }} = 0\)