How do I take inverse Laplace transform of $\frac{-2s+3}{s^2-2s+2}$?

Question

How do I take inverse Laplace transform of $\frac{-2s+3}{s^2-2s+2}$? I have checked my transform table and there is not a suitable case for this expression.

Answer 1

You may want to try this (slighlty) different approach:

Let $F(s)$ be the function to be inverse-Laplace transformed. Then, $F(s)$ admits the following partial fraction decomposition:

$$F(s) = \frac{A_1}{s-s_1} + \frac{A_2}{s-s_2},$$ where $s_1 = 1-i$, $s_2 = \overline{s}_1 = 1+i$, are the complex roots of the denominator, $A_1 = \lim_{s\to s_1} (s-s_1) F(s) = -1+i/2$ and $A_2 = \overline{A}_1 = \lim_{s\to s_2}(s-s_2) F(s) = -1-i/2$. Now take inverse Laplace's transform on each term to get:

$$f(t) = \mathcal{L}^{-1}[F(s)] = A_1 e^{s_1 t} + A_2 e^{s_2 t},$$ since $A_1$ and $A_2$ are complex conjugates as well as $s_1$ and $s_2$, the result is real, which you can easily prove with the help of Euler's identity: $e^{ix} = \cos{x} + i \sin{x}$.

I hope this helps.

Cheers!

Answer 2

To perform the inverse Laplace transform you need to complete the square at the denominator, $$ s^2-2s+2=(s-1)^2+1$$ so you rewrite your expression as

$$\frac{-2s+3}{(s-1)^2+1}= -2 \frac{(s-1)}{(s-1)^2+1}+\frac{(3-2)}{(s-1)^2+1}$$

now these expressions are standard on tables "exponentially decaying sine/cosine wave. The inverse transform is then $$e^t(-2\cos t+\sin t).$$