Question
Download Solution PDFThe steady-state error due to the ramp input for a type two system is equal to-
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
KP (position error constant) is given as:
\(K_p=\mathop {\lim }\limits_{s \to 0} G\left( s \right)H\left( s \right)\)
Kv (velocity error constant) is given as:
\(K_v=\mathop {\lim }\limits_{s \to 0} sG\left( s \right)H\left( s \right)\)
Ka (acceleration error constant) is given as:
\(K_a=\mathop {\lim }\limits_{s \to 0} {s^2}G\left( s \right)H\left( s \right)\)
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 |
∞ |
∞ |
\(\frac{1}{{{K_a}}}\) |
Observation:
From the above table, it is clear that for the type-2 system, a system shows zero steady-state error for step-input and ramp-input; and finite steady-state error for a parabolic input.
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