Question

In Problems 19-22 proceed as in Example 3 and find the convolution$\mathbf{f}\mathbf{*}\mathbf{g}$of the given functions. After integrating find the Laplace transform of$\mathbf{f}\mathbf{*}\mathbf{g}$.

$\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{t}\mathbf{,}\mathbf{}\mathbf{g}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}$

Short answer

The convolution of the given function is$\mathrm{t}-1+{\mathrm{e}}^{-\mathrm{t}}$ and the Laplace transform of the convolution is $\frac{1}{s^2(s+1)}$.

Step-by-step solution

Convolution of two functions

"If f and g are piecewise continuous on the interval $[0,\infty )$, then the convolution off and g, denoted by the symbol $f*g$, is a function defined by the integral ."

$\mathbf{f}\mathbf{*}\mathbf{g}\mathbf{=}{\int }_{\mathbf{0}}^{\mathbf{t}}\mathbf{f}\mathbf{\left(}\mathbf{\tau }\mathbf{\right)}\mathbf{g}\mathbf{(}\mathbf{t}\mathbf{-}\mathbf{\tau }\mathbf{)}\mathbf{d\tau }$

Apply the convolution of two functions

The given functions are, f (t) = t and $g\left(t\right)=e^{-t}$.

By the definition of f and g, the values are, $\mathrm{f}\left(\mathrm{\tau }\right)=\mathrm{\tau }$and $g(t-\tau )=e^{-(t-\tau )}$.

The convolution of the given functions is computed as follows.

$\begin{array}{l}f*g={\int }_0^{t}f\left(\tau \right)g(t-\tau )d\tau \\ ={\int }_0^t\left(\tau \right)\left[e^{-(t-\tau )}\right]d\tau \\ =e^{-t}{\int }_0^t\tau e^{\tau }d\tau \\ =e^{-t}{\left[\tau e^{\tau }\right]}_0^t-e^{-t}{\int }_0^t{e}^{\tau }d\tau \end{array}$

On further simplification,

$\begin{array}{l}f*g=e^{-t}\left[t{e}^t-0\right]-e^{-t}{\left[e^{\tau }\right]}_o^t\\ =t-e^{-t}\left[e^t-1\right]\\ =t-1+e^{-t}\end{array}$

So, the convolution of the given functions is

$t-1+e^{-t}$

$\begin{array}{l}f*g=e^{-t}\left[t{e}^t-0\right]-e^{-t}{\left[e^{\tau }\right]}_o^t\\ =t-e^{-t}\left[e^t-1\right]\\ =t-1+e^{-t}\end{array}$

Apply Laplace transform

$\frac{1}{s^2(s+1)}$Now, the given function's Laplace transform is obtained as follows.

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So, the convolution's Laplace transform is $\frac{1}{s^2(s+1)}$

Hence, the convolution of the given function is $t-1+e^{-t}$and the Laplace transform of the convolution is $\frac{1}{s^2(s+1)}$