By increasing the gain K of type 1 system, for a step input, the
A. increases
B. decreases
C. remains unaltered at a positive non zero value
D. remains zero
Please scroll down to see the correct answer and solution guide.
Right Answer is: D
SOLUTION
Concept:
Ka = 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.
Application:
For a type 1 system, for step input, the steady state error is independent of systems gain K and it is always zero. Therefore, by increasing the gain K, the steady state error remains zero.
Important Points:
As the type of the system increases, the steady-state error decreases.
The steady-state error is inversely proportional to the gain. Therefore, it can be reduced by increasing the system gain.