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.

  • $\begingroup$ My impulse would be to evaluate that last integral by turning it into a complex contour integral. Since the integral only has two poles above the real line, this shouldn’t be hard. $\endgroup$ – Semiclassical Jan 19 '18 at 14:14
  • $\begingroup$ Also, how are you defining the Fourier transform? In the version I know, the Fourier transform of $(f*g)(t)$ would be $F(\omega)G(\omega)$ not $F(i\omega)G(i\omega)$. $\endgroup$ – Semiclassical Jan 19 '18 at 16:14
  • $\begingroup$ @Semiclassical sorry about that, it's a notation purely to show that the argument may be complex also, equivalently it's $X(\omega)$. $\endgroup$ – Luke Collins Jan 19 '18 at 18:17

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.


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