We have a system with output given by
$\frac{Y(s)}{R(s)} = \frac{F(s)G(s)}{1+F(s)G(s)}$
where $F(s)G(s) = K\frac{s+1}{s^2+s+1}$.
Let $K=4$ and $R(s) = 10/s$. Using the final value theorem, calculate the steady state error of the system.
Now, I thought the error would be given by $\lim_{s \rightarrow 0} \frac{10\cdot F(s)G(s)}{1+F(s)G(s)}$, but the solution says to calculate $\frac{10}{1+F(s)G(s)}$. Can anyone explain why?
$\endgroup$ 11 Answer
$\begingroup$We have to find the steady state error $$\lim_{t\rightarrow\infty}(r(t)-y(t))$$ According to the final value theorem $$\lim_{t\rightarrow\infty}(r(t)-y(t))=\lim_{s\rightarrow 0}\left[s(R(s)-Y(s))\right]$$ The Laplace transform of the error is given by $$R(s)-Y(s)=R(s)-\frac{F(s)G(s)}{1+F(s)G(s)}R(s)=\frac{1}{1+F(s)G(s)}R(s)$$ Hence, for a step reference with Laplace transform $R(s)=10/s$ the steady state error value can be derived by $$\lim_{t\rightarrow\infty}(r(t)-y(t))=\lim_{s\rightarrow 0}\left[s\frac{1}{1+F(s)G(s)}\frac{10}{s}\right]=\lim_{s\rightarrow 0}\left[\frac{10}{1+F(s)G(s)}\right]$$
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