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I do not understand how the Inverse Laplace transform is taken of G(s).

Please see my working. I understand how the final value theorem is used, but I do not know why it is relevant to find the output, and I do not understand how the time response output is found.

I know the second line of my working is probably not right, where I take the constant A out, into it's own inverse Laplace transform. If I had

A + 1/(s+a)

then I could do it, and the two inverse laplace terms would be added together. But I am not sure how to tackle this example.

Thanks

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2 Answers

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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.

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I think the second line you have is correct. Just, since A is a constant you can say that L^-1{A} = A. And you probably don't even need to take that step.

Then, note that G(s) = Y(s)/U(s) = Y(s)/(1/s) = sY(s)

Therefore Y(s) = G(s)/s

Going forward in the problem you have to simplify and use partial fractions, then take the inverse Laplace transform. Hope this gets you going on the problem.

I'm also not exactly clear on how that fact about the FVT helps, though.

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