Laplace Transformation question

Question

$$\mathcal{L^{-1}}\left\{e^{-2s}\frac{s}{s^2 + 9}\right\}$$

Hello, I am new to Laplace transformations. Can anybody help me with this equation? Is this convolution or multiplication? Quite unclear on this.

Answer 1

The ${\mathcal{L}^{-1}}$ is the Inverse Laplace Operator. If you do the Laplace Transform of a function, and then apply ${\mathcal{L}^{-1}}$ to it - you should get back the same function.

To invert ${e^{-2s}\frac{s}{s^2 + 9}}$, we will need to introduce a special function: the Heaviside function. It's defined as

$${H(x) = \left\{\begin{array}{cc}0,&x<0\\1,&x\geq 0\end{array}\right.}$$

Let's try to find the Laplace Transform of ${H(t-a)f(t-a)}$, where ${f(t)}$ is just some function and $a$ is just some number ${\geq 0}$:

$${\mathcal{L}\{H(t-a)f(t-a)\}=\int_{0}^{\infty}H(t-a)f(t-a)e^{-st}dt}$$

Well, whenever ${t-a<0\Leftrightarrow t<a}$, we know ${H(t-a)=0}$. Notice also that

$${\int_{0}^{\infty}H(t-a)f(t-a)e^{-st}dt=\int_{0}^{a}H(t-a)f(t-a)e^{-st}dt+\int_{a}^{\infty}H(t-a)f(t-a)e^{-st}dt}$$

The ${\int_{0}^{a}H(t-a)f(t-a)e^{-st}dt}$ will thus clearly ${=0}$, hence we are just left with

$${\mathcal{L}\{H(t-a)f(t-a)\}=\int_{a}^{\infty}H(t-a)f(t-a)e^{-st}dt=\int_{a}^{\infty}f(t-a)e^{-st}dt}$$

(the ${H(t-a)}$ is just $1$ on the interval in the last integral, and so we can just leave it out since anything multiplied by $1$ is just that thing). Substituting ${u=t-a}$ gives us

$${\Rightarrow \int_{0}^{\infty}f(u)e^{-s(u+a)}du=e^{-as}\int_{0}^{\infty}f(u)e^{-su}du=e^{-as}\mathcal{L}\{f(u)\}}$$

We have now derived an identity involving the Heaviside function:

$${\mathcal{L}\{H(t-a)f(t-a)\}=e^{-as}\mathcal{L}\{f(t)\}}$$

Now, since

$${\mathcal{L}\{\cos(3x)\}=\frac{s}{s^2 + 9}}$$

the function you are trying to invert is exactly of the form of the right hand side of the identity with ${f(t)=\cos(3t)}$ and ${a = 2}$:

$${e^{-2s}\mathcal{L}\{\cos(3t)\}=e^{-2s}\frac{s}{s^2 + 9}}$$

and so using the Heaviside identity we derived, you can see the Inverse Laplace Transform should clearly be

$${\mathcal{L}^{-1}\left\{e^{-2s}\frac{s}{s^2 + 9}\right\}=\mathcal{L}^{-1}\{\mathcal{L}\{H(t-2)\cos(3(t-2))\}\}=H(t-2)\cos(3(t-2))}$$

And so there we have it:

$${\mathcal{L^{-1}}\left\{e^{-2s}\frac{s}{s^2 + 9}\right\}=H(t-2)\cos(3(t-2))}$$

Where ${H(t)}$ represents the Heaviside function.