Dirac delta convolution

Question

How can this identity convolution be shown?

$$\int^\infty_{-\infty} f(\tau)\delta(t-\tau)d\tau=f(t)$$

I keep getting stuck in traps when trying to show this and need a bit of assistance

Answer 1

$$ \int_{-\infty}^{\infty} f(\tau) \, \delta(t-\tau) \, d\tau = \{ \tau = t-\sigma \} = \int_{\infty}^{-\infty} f(t-\sigma) \, \delta(\sigma) \, (-d\sigma) \\ = \int_{-\infty}^{\infty} f(t-\sigma) \, \delta(\sigma) \, d\sigma = f(t-0) = f(t). $$

Answer 2

There is no proof of the formula you're asking, this is a definition. However, here is the way why the expression is used in so many contexts.

Consider a sequence of function $$ \delta_\epsilon(x)=\begin{cases}\frac{1}{2\epsilon},&-\epsilon<x<\epsilon,\\ 0,&\mbox{otherwise}. \end{cases} $$

It is reasonable to expect that $$ \delta(x)=\lim_{\epsilon\to 0}\delta_\epsilon(x). $$

Now consider the integral and assume that we are free to change the order of operations: $$ \int_{\mathbb R}\delta(x)f(x)dx=\lim_{\epsilon\to 0}\int_{\mathbb R}\delta_{\epsilon}(x)f(x)dx=\lim_{\epsilon\to 0}\int_{-\epsilon}^{\epsilon}\frac{1}{2\epsilon} f(x)dx=\lim_{\epsilon\to0}2\epsilon\cdot \frac{1}{2\epsilon} f(\xi), $$ where due to the mean value theorem $\xi\in(-\epsilon,\epsilon)$.

Hence we can conclude that $$ \lim_{\epsilon\to 0}f(\xi)=f(0), $$ which gives you a "proof" of the original formula.

Now just repeat the same for $\delta (x-a)$, or $\delta(t-\tau)$ if you prefer.

Answer 3

First of all, one needs to understand that the Dirac delta is not a function, but a distribution, ie an element of the dual space of some space of test functions. Thus, one defines the Dirac delta by its action on smooth, compactly supported function $$\delta(f):=f(0)\ ,$$ and one writes (this is just notation!) $$\delta(f) = \int_{\mathbb{R}}\delta(t)f(t)dt\ .$$ One then defines $\delta(t-\tau)$ by a formal change of variable $u = t-\tau$: $$\int_{\mathbb{R}}\delta(t-\tau)f(t)dt = \int_{\mathbb{R}}\delta(u)f(u+\tau)dt = f(\tau)\ .$$

Note: Sorry, I switched $t$ and $\tau$ with respect to the OP.

Answer 4

This answer does not provide the proof but aims to provide signal-theoretic insight.

I am myself figuring this, and the answer, I believe lies in distribution theory; for dirac delta has been defined so, for its purported usage in signal theory as singularity "signal". $$\int \delta(t) dt = 1 $$

All the answers above are justified and would satisfy any undergrad, such as me. To add one more observation. $$(f* \delta) (t) = \int f(\tau)\delta (t - \tau) d \tau \implies f(t)$$

Wouldn't that make $\delta(t)$ a "convolution kernel" for any function (signal).

If so, in a system (a transformation $f : x \to y$ ) having an impulse response $h(t) = f(\delta(t))$, it would be immensely helpful to realise that the output... $$ y(t) = x(t)*h(t) $$ how? with a criminal lack of mathematical insight, I believe, ($\delta(t)$ being the kernel) h(t) upon convoluting, purely (zero state?) communicates just specification of the system.