Question

Find the convolution$\mathbf{\text{f×g}}$of the given functions. After integrating find the Laplace transform of$\mathbf{\text{f×g}}$. Where$\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{e}}^{\mathbf{t}}$and $\mathbf{g}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{e}}^{\mathbf{t}}$.

Short answer

The Laplace transform is$\mathrm{L}\left\{{\mathrm{e}}^{-\mathrm{t}}*{\mathrm{e}}^{\mathrm{t}}\right\}=\left[\frac{1}{{\mathrm{s}}^2-1}\right]$

Step-by-step solution

Finding the convolution for the two function given;

$\begin{array}{l}\mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{-\mathrm{\tau }}\\ \mathrm{g}(\mathrm{t}-\mathrm{\tau })={\mathrm{e}}^{\left(\mathrm{t}-\mathrm{\tau }\right)}\end{array}$

If function f and g are piecewise continuous on the interval $[0,\infty )$then

$f*g=\underset{0}{\overset{t}{\int }}f\left(\tau \right)g(t-\tau )d\tau$

Using the formula to solve further;

As we clearly see that;

$\begin{array}{l}{e}^{-t}*e^t=\underset{0}{\overset{t}{\int }}{e}^{-\tau }{e}^{\left(t-\tau \right)}\\ e^{-t}*e^t=\underset{0}{\overset{t}{\int }}{e}^{\left(t-2\tau \right)}d\tau \\ e^{-t}*e^t={\left.\left[\frac{-e^{t-2\tau }}{2}\right]\right|}_0^t\\ e^{-t}*e^t=\left[-\frac{e^{-t}}{2}+\frac{e^t}{2}\right]\\ e^{-t}*e^t=\frac{1}{2}\left[e^t-e^{-t}\right]\end{array}$

$\begin{array}{l}L\left\{e^{-t}*e^t\right\}=\frac{1}{2}L\left\{e^t-e^{-t}\right\}\\ L\left\{e^{-t}*e^t\right\}=\frac{1}{2}L\left\{e^t\right\}-\frac{1}{2}L\left\{e^{-t}\right\}\\ L\left\{e^{-t}*e^t\right\}=\frac{1}{2}\left[\frac{1}{s-1}-\frac{1}{s+1}\right]\\ L\left\{e^{-t}*e^t\right\}=\frac{1}{2}\left[\frac{2}{s^2-1}\right]\\ L\left\{e^{-t}*e^t\right\}=\left[\frac{1}{s^2-1}\right]\end{array}$

Hence the final answer is, $\mathrm{L}\left\{{\mathrm{e}}^{-\mathrm{t}}*{\mathrm{e}}^{\mathrm{t}}\right\}=\left[\frac{1}{{\mathrm{s}}^2-1}\right]$