(Note: I'm currently learning about this, but I'm having a hard time understanding why this system I am modelling is giving unexpected results when finding the steady state error)
I have a system with a transfer function $G(s)=\frac{48}{s^3 + 7s^2 + 6s}$ (in the Laplace domain). This system has unity negative feedback, meaning the overall transfer function is $\frac{O(s)}{I(s)}=\frac{48}{s^3 + 7s^2 + 6s + 48}$. This transfer function is unstable because the system has poles in the right half of the s-plane. For a step input, the output increases without bound to infinity.
My issue is this equation for the steady state error:
$$e(\infty)=\lim_{s\to0}\frac{sR(s)}{1+G(s)}$$
Plugging the values $R(s) = \frac{1}{s}$ (for a step input) and $G(s)=\frac{48}{s^3 + 7s^2 + 6s}$ gives:
$$\begin{align*}e(\infty) &= \lim_{s\to0}\frac{s\frac{1}{s}}{1+\frac{48}{s^3 + 7s^2 + 6s}}\\
&= \lim_{s\to0}\frac{1}{1+\frac{48}{s^3 + 7s^2 + 6s}}\\
&= \lim_{s\to0}\frac{s^3 + 7s^2 + 6s}{s^3 + 7s^2 + 6s+48}\\
&= \frac{0^3+7\cdot0^2+6\cdot0}{0^3+7\cdot0^2+6\cdot0+48}\\
&= \frac{0}{48}\\
&= 0
\end{align*}$$
This means the steady state error is 0, and since the input was a unit step then (to my understanding) this means the system has a steady state at an amplitude of 1. However, as stated earlier, the output never actually settles, so surely the steady state error should be infinity. Where have I gone wrong?
Best Answer
The finite value theorem (FVT) of the Laplace transform does only apply if the considered transfer function $F(s)$ is stable. In your case, you have poles with positive real part, so you can't deduce the steady state error in this case.
As stated on Wikipedia: "If $f$ is bounded on $(0,\infty)$ and
$$ \lim_{t \rightarrow \infty} f(t) $$
has a finite value then
$$ \lim_{t \rightarrow \infty} f(t) = \lim_{s \rightarrow 0} s F(s) $$
where $F(s)$ is the (unilateral) Laplace transform of $f(t)$." A link to a proof of this statement is also given there.
In your case, the limit has obviously no finite value and $f(t)$ is also not bounded, as the system is unstable.
Actually, Wikipedia has an example very similar to your case: Example where FVT does not hold.