I was reading about the Final Value Theorem for Laplace Transforms. It states that for if a function $f(t)$ in continuous time has Laplace Tranform $F(s)$, then:
$$\lim_{t \to \infty}{f(t)} = \lim_{s \to 0}{sF(s)}$$
If we let $f(t) = \sin(t)$, then we get:
$$ \lim_{t \to \infty}{\sin(t)} = \lim_{s \to 0}{\frac{s}{s^2+1}}$$
By L'hopital's rule, the RHS evaluates to:
$$\lim_{s \to 0}{\frac{s}{s^2+1}} = \lim_{s \to 0}{\frac{1}{2s + 1}} = 1$$
which results in:
$$\lim_{t \to \infty}{\sin(t)} = 1$$
Is this a valid conclusion? If not, where is the flaw in my logic? Thanks in advance
Best Answer
One version of the final value theorem states that if $f(t)$ and $f'(t)$ have Laplace transforms that exist for $s>0$ AND if $\lim_{t\to\infty}f(t)$ exists and $\lim_{s\to0}sF(s)$ exists, then $\lim_{t\to\infty}f(t)=\lim_{s\to0}sF(s)$.
Clearly $\lim_{t\to\infty}\sin(t)$ fails to exist (This).
Another version states if every pole of $F(s)$ is either in the open left-half plane or at the origin, and that $F(s)$ has at most a single pole at the origin, then $\lim_{s\to0}sF(s)=\lim_{t\to\infty}f(t)$.
Clearly, the poles of $F(s)=\frac1{s^2+1}$ are at $s=\pm i$ and are not in the left half plane. Hence, the theorem does not apply and we cannot infer equality of the limits.
Aside, $\lim_{s\to0}\frac{s}{s^2+1}=0$. L'Hospital's Rule does not apply since the limit is not of indeterminate form (i.e., $0/0$).