How to use Final Value Theorem for Inverse Laplace transform?

Initial Value and Final Value Theorems
Initial Value and Final Value Theorems

This problem is incorrectly stated. The Final Value Theorem is stated as follows (or check out a signals and systems text):

If $f$ is bounded on $(0,\infty)$ and $\lim\limits_{t\rightarrow\infty}f(t) < M$ for some $M$, then
\begin{align*}
\lim\limits_{t\rightarrow\infty}f(t) = \lim\limits_{s \searrow 0} sF(s)
\end{align*}
where $F(s)$ is the unilateral Laplace Transform of $f$.

With this, the final value (DC Gain) is $0$. The OP has this in their solution, since $\lim\limits_{t\rightarrow \infty} e^{-at} \rightarrow 0$, assuming $a>0$. Otherwise, the function is unbounded, and the final value theorem does not apply since it has a pole outside of the open left half plane.

Here’s what I think the problem meant to say: Let $a>0$ and
$Y(s) = G(s)U(s) = \frac{1}{s}\frac{A}{s+a}$. Using the FVT, find the DC gain, and the time domain response.

Then, applying FVT gives,
\begin{align*}
\lim\limits_{s \searrow 0} sY(s) = \lim\limits_{s \searrow 0} s\frac{A}{s+a}\frac{1}{s} = \lim\limits_{s \searrow 0} \frac{A}{s+a} = \frac{A}{a}
\end{align*}

As the solution above suggests, partial fractions can be used to get the total response, or, if you have been introduced to some complex analysis, you can also use Cauchy’s Residue Theorem.

Applying partial fractions should give $\frac{A/a}{s} + \frac{-A/a}{s+a}$, which when factored gives the final answer. Note that if you take $t\rightarrow \infty$, you find the final value (DC gain) as expected.

You are watching: How to use Final Value Theorem for Inverse Laplace transform?. Info created by THVinhTuy selection and synthesis along with other related topics.

Rate this post