# Inverse Laplace Transform

1. Dec 7, 2014

### redundant6939

1. The problem statement, all variables and given/known data
Find $$H(s) = \frac{Y(s)}{X(s)}$$
$$\frac {d^2y(t)}{dt^2} + a\frac {dy(t)}{dt} = x(t) + by(t)$$

2. Relevant equations

3. The attempt at a solution
$$[s^2 + as - b] Y(s) = X(s)$$
$$H(s) = \frac{1}{s^2+as-b}$$

I assume the inverse is a sign or a cosine but unsure which one.

2. Dec 7, 2014

### Ray Vickson

If $Y(s)$ is the LT of $y(t)$, what are the LTs of $y'(t)$ and $y''(t)$?

Hint: not what you wrote.

3. Dec 7, 2014

### Dustinsfl

You can use the Bromwich integral to find the inverse Laplace transform.
The poles of $s$ are $s = \frac{-a\pm\sqrt{a^2 + 4b}}{2}$
\begin{align}
\mathcal{L}^{-1}\Bigl\{\frac{1}{s^2 + as - b}\Bigr\} &= \frac{1}{2\pi i}\int_{\gamma -i\infty}^{\gamma +i\infty}\frac{e^{st}}{s^2 + as - b}ds\\
&= \sum\text{Res}\\
&= \lim_{s\to s_+}(s - s_+)\frac{e^{st}}{s^2 + as - b} + \lim_{s\to s_-}(s - s_-)\frac{e^{st}}{s^2 + as - b}
\end{align}
where $s_+=\frac{-a + \sqrt{a^2 + 4b}}{2}$ and $s_-=\frac{-a - \sqrt{a^2 + 4b}}{2}$

4. Dec 7, 2014

### redundant6939

The LT of $y'(t)$ and $y''(t)$ are sY[s} and $s^2Y$ respectively. Im not sure what you mean exactly.

5. Dec 7, 2014

### redundant6939

The region of convergence then, would it be btw the two values of s?

6. Dec 7, 2014

### Dustinsfl

You need to find y(t) for the ROC.

7. Dec 7, 2014

### Ray Vickson

What I mean is that your statements above are false, in general. Consult a table of Laplace transforms to see what you should have written.