Definition of a function in terms of shifted deltas

Question

This is how a function $x(t)$ is usually defined in terms of the dirac delta function $\delta (t)$:

\begin{equation} x(t)=\int_{-\infty}^{\infty}x(\tau)\delta(t-\tau)d\tau \end{equation}

Since $\delta(t)$ is even, this is equivalent to: $$\int_{-\infty}^{\infty}x(\tau)\delta(\tau-t)d\tau$$

The second expression is directly consistent with the sampling property of $\delta(t)$ $$\int_{-\infty}^{\infty}x(\tau)\delta(\tau-t)d\tau=\int_{-\infty}^{\infty}x(t)\delta(\tau-t)d\tau=x(t)\int_{-\infty}^{\infty}\delta(\tau-t)d\tau=x(t)$$

This cannot be immediately seen in the first definition.

My question is, why is this usually presented in the first form? What is the intuition behind that form?

Answer 1

I figured this out and am posting it in case anyone is interested.

The definition

\begin{equation} x(t)=\int_{-\infty}^{\infty}x(\tau)\delta(t-\tau)d\tau \end{equation}

occurs in the context of linear time-invariant (LTI) systems. Expressing $x(t)$ in this form where the deltas are shifted and scaled leads directly to the convolution definition of the output $y(t)$ for LTI systems:

$$y(t)=\int_{-\infty}^{\infty}x(\tau)h(t-\tau)d\tau $$

where $h(t)$ is the impulse response of the LTI system and is likewise shifted and scaled.

Although the second form

$$x(t)=\int_{-\infty}^{\infty}x(\tau)\delta(\tau-t)d\tau$$

is equivalent, it also involves a reflection of the deltas about the y-axis. This does not translate directly into a reflection of their impulse responses in an LTI system. The first form is preferred for its direct relation to the output of the system.

So the logical steps seem to be that $x(t)$ was first obtained in the second form. It was then found that it could be written in the first form to facilitate obtaining a general expression for the output of an LTI system.

Answer 2

The first form, $\int_{-\infty}^{\infty} x(\tau) \, \delta(t-\tau) \, d\tau$ has the form of a convolution: $$f*g(t) = \int_{-\infty}^{\infty} f(\tau) \, g(t-\tau) \, d\tau$$ with $f=x$ and $g=\delta$.

Convolutions are very useful, e.g. for smoothing functions or expressing and calculating solutions to differential equations.

The given expression, $x(t) = \int_{-\infty}^{\infty} x(\tau) \, \delta(t-\tau) \, d\tau = (x*\delta)(t)$ shows that $\delta$ acts as an identity for convolution (as does $1$ for multiplication and $0$ for addition).