# Steady State Error Calculation to Input Step and Ramp Input

by spggodd
Tags: calculation, error, input, ramp, state, steady, step
 P: 526 Since G(s) is unstable in open loop (it has poles in the right half-plane), I'm going to assume we're talking about using it in a feedback configuration. Let the system error, e(t), be given as: e(t) = r(t) - c(t) where r(t) and c(t) are the system input and output, respectively. For a unity feedback system, the Laplace transform of e(t), E(s), is then given as: $$E(s) = \frac{1}{1 + G(s)} R(s)$$ The system steady-state error, e_ss, is then given by the final value theorem as: $$e_{ss} = \lim_{s \rightarrow 0} s \frac{1}{1 + G(s)} R(s)$$ For a step input, R(s) = 1/s, we have: $$e_{ss} = \lim_{s \rightarrow 0} s \frac{1}{1 + \frac{4.992}{s^2 + 3s - 1}} \frac{1}{s} = \frac{1}{1 - 4.992} \approx -0.2505$$ For a ramp input, R(s) = 1/s^2, we have: $$e_{ss} = \lim_{s \rightarrow 0} s \frac{1}{1 + \frac{4.992}{s^2 + 3s - 1}} \frac{1}{s^2} = \lim_{s \rightarrow 0} \frac{1}{s + \frac{4.992s}{s^2 + 3s - 1}} = \infty$$ The steady-state error for a step input will thus be a constant and for a ramp input it will be unbounded. This is what you would expect for a type 0 system (no free integrators), if that makes sense to you.