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I was watching this video on signal processing:

, and the author was trying to derive a function such that

$$ y(t) = \begin{cases} x(\tau) & \mbox{if } t = \tau, \\ 0 & \mbox{otherwise} \end{cases} $$

The answer is $\int_{-\infty}^{\tau+} x(\tau) \delta(t-\tau) dt$, where $\delta$ is the unit impulse. I understand how this works, but I am wondering, why all the trouble for such a simple thing? Why not just use the first definition above?

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    Cf Dirac delta function (functional) – 2017-01-01
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    "Why not just use the first definition above" How? BTW, It is already "used"! – 2017-01-01
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    As long as $x(t)$ is defined, $y(t)$ is easily defined (in two cases, as shown), hence the integration approach doesn't seem necessary. – 2017-01-01
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    OK. I will give you an answer. But I think the condition should be $t=\tau$ (not $x=\tau$). Right? – 2017-01-01
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    Ah, yes, that's a typo. Will correct it. – 2017-01-01

3 Answers 3

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The dirac delta $\delta$ is not a function; it is a distribution. As an example application, the convolution property that you hint at is used to create fundamental solutions of linear operators. You may encounter such differential equations in signal processing.

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Even though the second form, with the integral, seems unnecessarily complex, it is more compact in the sense that it gets rid of the if/else condition and becomes a linear operation. This can be more readily plugged into other equations and derivations.

For example, this can be used when computing the Fourier transform of the dirac-delta function itself.

In my experience, it keeps popping up as a simplifying step in other derivations, where you notice the term $\int_{-\infty}^{\infty} x(\tau) \delta(t-\tau) dt$, as a part of something bigger, and simply replace with $x(\tau)$.

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The answer is simple, in signal processing we use filters to obtain the desired output. So for any desired output there exists a filter, and an LTI filter is identified by its impulse response.

LTI filtering is defined as a convolution integral. So we look for a function $h(t)$ that satisfies $$y(t)=\int_{-\infty}^{+\infty} x(\tau) h(t-\tau) d\tau$$ and $h(t)$ is simply the impulse response of the filter. If we put $h(t)=\delta(t)$, it means the output to the Dirac delta impulse input is the same Dirac delta impulse. Hence $h(t)=\delta(t)$ is the identity system. If you are familiar with Fourier (or $z$) transform, you can see that $$y(t)=x(t)\star\delta(t)=\int_{-\infty}^{+\infty} x(\tau) \delta(t-\tau) d\tau$$ is written in the Fourier (or $z$) domain as $$Y(\omega)=X(\omega)$$