Correct Answer - Option 2 : ∞, 0
Concept:
Steady-state error is defined as the difference between the input and the output of a system in the limit as time goes to infinity (i.e. when the response has reached steady state).
The steady-state error will depend on the type of input (step, ramp, etc.) as well as the system type (0, I, or II).
The steady-state error of a unity feedback stable system
\({e_{ss}} = \mathop {\lim }\limits_{s \to 0} \frac{{s.R\left( s \right)}}{{1 + G\left( s \right)}}\)
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
|
∞
|
∞
|
\(\frac{1}{{{K_a}}}\)
|
Explanation:
For type 2 system with input as unit ramp, velocity error constant and steady state error are respectively ∞, 0.
