Question

consider the following.
f(t) = t, g(t) = e^{-t}
proceed as in example 4 and find the convolution f * g of the given functions.
f * g =
after integrating find the laplace transform of f * g. (write your answer as a function of s.)
ℒ{f * g} =

Explanation:

Step1: Recall the formula for convolution

The convolution of two functions \(f(t)\) and \(g(t)\) is \((fg)(t)=\int_{0}^{t}f(\tau)g(t - \tau)d\tau\). Given \(f(t)=t\) and \(g(t)=e^{-t}\), then \((fg)(t)=\int_{0}^{t}\tau e^{-(t-\tau)}d\tau\).

Step2: Simplify the integral

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LATEXBLOCK0
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Use integration by parts. Let \(u = \tau\), \(dv=e^{\tau}d\tau\), then \(du = d\tau\), \(v = e^{\tau}\).
By the integration - by - parts formula \(\int_{a}^{b}u\;dv=uv|_{a}^{b}-\int_{a}^{b}v\;du\), we have \(\int_{0}^{t}\tau e^{\tau}d\tau=\tau e^{\tau}|_{0}^{t}-\int_{0}^{t}e^{\tau}d\tau\).

$$
LATEXBLOCK1
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So \((f*g)(t)=e^{-t}(t e^{t}-e^{t}+1)=t - 1+e^{-t}\)

Step3: Recall the Laplace transform formula

The Laplace transform of a function \(y(t)\) is \(L\{y(t)\}=\int_{0}^{\infty}y(t)e^{-st}dt\). Also, we know the property \(L\{f*g\}=L\{f\}\cdot L\{g\}\).
The Laplace transform of \(f(t) = t\) is \(L\{t\}=\frac{1}{s^{2}}\) (using the formula \(L\{t^{n}\}=\frac{n!}{s^{n + 1}},n = 1\)).
The Laplace transform of \(g(t)=e^{-t}\) is \(L\{e^{-t}\}=\frac{1}{s + 1}\) (using the formula \(L\{e^{at}\}=\frac{1}{s - a},a=-1\)).

Step4: Calculate \(L\{f*g\}\)

$$
LATEXBLOCK2
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Answer:

\(f*g=t - 1+e^{-t}\), \(L\{f*g\}=\frac{1}{s^{2}(s + 1)}\)