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Mechanical Engineering QUESTION #2387
Question 1
For a type-one system (one integrator in the open-loop transfer function) with a ramp input \(X_i(s)=1/s^2\), the steady-state error is:
  • \(E_s = 0\)
  • \(E_s = \frac{1}{K_v}\) where \(K_v = \lim_{s\to 0} sY_1(s)\) (velocity error constant)✔️
  • \(E_s = \infty\)
  • \(E_s = 1\)
Correct Answer Explanation
For a ramp input with a type-one system: \(E_s=\lim_{s\to 0}\frac{s}{1+Y_1(s)}\cdot\frac{1}{s^2}=\frac{1}{\lim_{s\to 0}sY_1(s)}=\frac{1}{K_v}\). The velocity error constant \(K_v=\lim_{s\to0}sY_1(s)\).