I am given that the functions $x(t)$ and $h(t)$ are defined by
$$ x(t) = t\, e^{-2t}\, u(t) \qquad \text{and} \qquad h(t) = e^{-4t}\, u(t)$$
where $u(t)$ denotes the unit step function
$$ u(t) = \begin{cases}
1 & \text{if $t \geq 0$}\\
0 & \text{otherwise}.
\end{cases}$$
The question is the following:
Compute the convolution $(x * h)(t)$ by finding the corresponding Fourier transform $X(\omega)$ and $H(\omega)$ using the convolution property, and then inverse transforming.
By the convolution property, we mean $(x*h)(t) \stackrel{\text{FT}}{\longleftrightarrow} X(\omega) H(\omega)$.
Now I found the Fourier transforms $X(\omega)$ and $H(\omega)$ using the following reasoning.
\begin{align*}
x(t) &= t\, e^{-2t} \,u(t)\\
& = t\; y(t) \qquad \text{where $y(t) := e^{-2t} \,u(t)$}\\
\implies X(\omega)&= i\, \frac{\partial Y}{\partial \omega}\qquad \text{where $Y$ is the F.T. of $y(t)$}\\
&= \frac{1}{(2+ i \omega)^2}.
\end{align*}
$H(\omega)$ is simply $1/(4+i\omega)$. Now
\begin{align*}
(x*h)(t) \stackrel{\text{FT}}{\longleftrightarrow} X(\omega) H(\omega) = \frac{1}{(2+i\omega)^2(4+i\omega)}.
\end{align*}
If at this point I were to employ the inverse Fourier transform, I would obtain $(x*h)(t) = \frac{1}{2\pi} \int_{-\infty}^\infty \frac{e^{i \omega t}}{(2+i\omega)^2(4+i\omega)} \, \mathrm d\omega$. But this seems quite a complicated integral (considering this was a 4 mark question in an exam). Is there a simpler way to go about obtaining the inverse Fourier transform? Am I going about the question correctly?
I appreciate any help.
Best Answer
I think you're expected to apply the partial fraction decomposition: $$\frac 1 {(4+i \omega)(2+i \omega)^2} = \frac 1 {4(4+i \omega)} - \frac 1 {4(2+i \omega)} + \frac 1 {2(2+i \omega)^2},$$ and you already have the functions whose transforms produce those terms.