Discussion
For a type one system. the steady-state error due to step input is equal to :
Answer: B
For a type 1 system, the steady-state error due to a step input is zero.
A type 1 system has one pole at the origin in its open-loop transfer function, \(G(s)H(s) = K(\ldots)/s(\ldots)\). The position error constant is then
\[ K_{p} = \lim_{s \to 0} G(s)H(s) = \infty \]
and the step error is
\[ e_{ss} = \frac{1}{1 + K_{p}} = 0 \]
The integrator keeps changing the output until the error is exactly zero.
An infinite error occurs for a type 1 system only with a parabolic input; with a ramp input the error is finite, \(1/K_{v}\). Values such as 0.25 or 0.5 would be the finite step error of a type 0 system with \(K_{p} = 3\) or 1.
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