So I had this question:
Find the Laplace transforms of the function $ x(t)=[t \sin(2t) +e^{-2t}]u(t)$. The function $u(t)$ is the unit step function. Check if the initial value theorem and the final value theorem are applicable in each case.
To answer this question I found the Laplace transform using partial fraction which resulted in $\displaystyle X(S)=\frac{4s}{(4+s^2)^2}+\frac{1}{s+2}$
There's three poles $s=-2, 2i,-2i$. Since its a causal signal ROC must be $\operatorname{Re}{s}>0$.
For the initial value theorem i used the formula and got 1.
So my question is about the final Value theorem.
When I used the formula of the final value theorem it gives me a value of 0 for $x(\infty)$. However, when I sketched the initial function unto graphing tool (DESMOS), the values is increasingly oscillating in negative and positive values as it reaches infinity. So I'm not sure if it even applicable.
According to my textbook, to apply FVT, when $t<0$ then $f(t)=0$.
Final value theorem: A signal for which $sX (s)$ is analytic for all $s$ on the imaginary axis and the right half plane
Correct me if I'm wrong anywhere.
As I see it, they seem to be respecting both those conditions, I don't understand why it doesn't work.
Any help would be helpful.
Given that $$ X(s) = {4 s \over (s^2 + 4)^2} + {1 \over s+2} $$
The Final Value Theorem (FVT) states that $$ x(\infty) = \lim\limits_{t \rightarrow \infty} x(t) = \lim\limits_{s \rightarrow 0} \ [ s X(s) ] \tag{1} $$ when $s X(s)$ has no poles on the Closed Right-Half Plane, i.e. in the region, where $\mbox{Re}[s] \geq 0$.
For the given problem, $$ s X(s) = {4 s^2 \over (s^2 + 4)^2} + {s \over s+2} = { 4 s^2 (s + 2) + s (s^2 + 4)^2 \over (s + 2) (s^2 + 4)^2} $$ which has the poles $s = -2, \pm 2 j, \pm 2 j$.
Since $s X(s)$ has four poles on the imaginary axis, we cannot apply Final Value Theorem to deduce the value of $f(\infty)$.