A convoluted convolution

Consider the signal $x(t) = 1/t$ . What is the convolution of $x$ with itself?

__________

My solution:

From the definition of convolution [1], we have

$(x * x) (t) = \displaystyle \int_{-\infty}^{+\infty}\frac{d \tau}{\tau (t - \tau)} = \displaystyle \frac{1}{t} \int_{-\infty}^{+\infty} \left(\frac{1}{\tau} + \frac{1}{t - \tau}\right) d \tau$ .

Unfortunately, the latter integral is plagued by convergence problems. To make matters worse, $x(t)$ is not even defined at $t = 0$ . Fortunately, the former integral is a familiar one: it’s a scaled Hilbert transform of $x$ . It is well-known [1] that the Fourier transform of $1 / \pi t$ (the impulse response of a Hilbert transformer) is the following

$\mathcal{F} \left(\displaystyle\frac{1}{\pi t}\right) = - j \,\text{sgn} (w) =\begin{cases} -j, & \omega > 0\\ +j, & \omega < 0\end{cases}$

where $\text{sgn}(\cdot)$ is the signum function. Thus, we have that the Fourier transform of $x$ is

$\mathcal{F} (x) = -j \pi \, \text{sgn} (w) = \begin{cases} -j \pi, & \omega > 0\\ +j \pi, & \omega < 0\end{cases}$

Convolution in the time domain corresponds to multiplication in the frequency domain. Hence, the Fourier transform of the convolution is

$\mathcal{F} (x * x) = \left(\mathcal{F} (x)\right)^2 = \left(-j \pi \, \text{sgn} (w)\right)^2 = - \pi^2$

and, taking the inverse Fourier transform, we finally obtain

$(x * x) (t) = \mathcal{F}^{-1} (- \pi^2) = -\pi^2 \delta (t)$

where $\delta (t)$ is the Dirac delta. This is a prime example of a problem that Signal Processing professors in engineering departments tend to love, but that drives (some) mathematicians up the wall.

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References

[1] Alan V. Oppenheim, Alan S. Willsky, S. Hamid Nawab, Signals & Systems, 2nd edition, Prentice-Hall, 1996.

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Tags: Hilbert Transform, Puzzles

This entry was posted on April 2, 2011 at 16:36 and is filed under Puzzles, Signal Processing. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

2 Responses to “A convoluted convolution”

Taole Zhu Says:
April 6, 2011 at 11:45 | Reply
I think this Hilbert transform is a little brilliant trick, but don’t $f$ and $g$ have to be in $L_1(\mathbb{R})$ (Lebesgue Integral function) so that the convolution of $f$ and $g$ exists? And $x(t)=1/t$ is not.
- Rod Carvalho Says:
  April 7, 2011 at 15:40 | Reply
  Your objection is valid. Like I wrote, this post is bound to upset mathematicians…
  
  From an engineering perspective, this post shows that the cascade of two scaled Hilbert transformers produces an output which is the input multiplied by $-\pi^2$ . The problem is that Hilbert transformers are non-causal systems. In other words, this post is neither mathematically rigorous nor of any practical interest.

Rod Carvalho

A convoluted convolution

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