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I am confused on the interpretation of $$ f(x)\delta(x-t_0) $$ where $\delta(x)$ is the Dirac delta function: $$ \delta(x-t_0)= \left\{ \begin{array}{ll} 0 & x< t_0 \\ \text{undefined} & x=t_0 \\ 0 & x>t_0\\ \end{array} \right. $$

The delta function is used in the discrete probability density functions, and more importantly in signal processing where the function is used as multiplier to represent discrete stuff.

Given,

$$ g(x)= \sin(x)\sum^n_{i=0}\delta(x-t_i) $$ What does $g(x)$ actually represent? If the delta function is undefined at $t_i$, what does it mean for it to be multiplied to $\sin(x)$?

I've seen that the graph of the given looks like a periodic graph of $\delta(x)$ at $t_i$ which seems to be enveloped by $\sin(x)$. How can this happen since the said function is undefined at all $t_i$?

Rupam Sarkar
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wd violet
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  • $\delta$ is not a function. It's a distribution in mathematics. But in practice we treat it as a function. – Trebor Sep 18 '21 at 06:58
  • @Trebor then is it treated like a unit step function (in a very short span of time) when multiplied to $\sin(x)$? That is my reasoning when I saw the said graph – wd violet Sep 18 '21 at 07:01
  • Given a nonempty set $X$ and a $\sigma$-algebra $\mathcal{A}$ of subsets of $X$, and $x\in X$, $\delta_x$ is a measure on $\mathcal{A}$ such that $\delta_x(A)=1$ of $x\in A$ and $\delta_x(A)=0$ is $x\notin A$. As any other measure, $\delta_x$ acts on measurable functions ny integration. For any $f:(X,\mathcal{A})\rightarrow\mathbb{R}$, $\delta_xf :=\int_X f(s),\delta_x(ds)=f(x)$. When $X=\mathbb{R}^d$ and $\mathcal{A}$ is the Borel $\sigma$-algebra $\delta_0$ is known as the Dirac $\delta$-function, but in reality is not a function but, as as said before, a measure: – Mittens Sep 18 '21 at 07:03
  • @Oliver Diaz then it means that there is no actual interpretation for $\sin(x) \delta(x)$ but simply an equation, waiting to be processed later with integration or transforms? – wd violet Sep 18 '21 at 07:08
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    In the discrete world, we use Kronecker delta instead of Dirac Delta. The former has an amplitude of 1, instead of infinite. In your question, g(x) (Or usually, we would use g[n] to represent the discrete nature of it), is the sampled sine wave. – user12758604 Sep 18 '21 at 07:12
  • @user12758604 I think that the Dirac delta is used for its integral property rather than the Kronecker? Since $g(x)$ above will undergo Fourier transform later and the property built for Dirac delta will be invoked – wd violet Sep 18 '21 at 07:16
  • @mkcpz maybe the concept of 'Dirac Comb' or the sifting property of Dirac Delta could shed some light for you :) Just curious, you mentioned that you observed what 'looks like a periodic graph of $\delta(x)$ at $t_i$ which seems to be enveloped by $sin(x)$.', Does it have the same amplitude of $sin(x)$? – user12758604 Sep 18 '21 at 07:51
  • @user12758604 yes, it has the same amplitude with $\sin(x)$ – wd violet Sep 18 '21 at 10:26

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The Dirac delta $\delta_{t_0}$ (where I write $\delta_{t_0}(x) = \delta(x-t_0)$) is defined not as you say but by its action on continuous test functions $\varphi\in C^0$. $$ \langle\delta_{t_0},\varphi\rangle = ∫_{\Bbb R} \varphi\, \delta_{t_0} = \varphi(t_0). $$ If you multiply it by a function $f$ then you have $$ \langle f\,\delta_{t_0},\varphi\rangle = ∫_{\Bbb R} \varphi\, f\,\delta_{t_0} = \varphi(t_0) \,f(t_0) = f(t_0) ∫_{\Bbb R} \varphi\,\delta_{t_0} = f(t_0)\,\langle \delta_{t_0},\varphi\rangle. $$ so you just have $f\,\delta_{t_0} = f(t_0)\,\delta_{t_0}$, i.e. you are just multiplying the delta by a constant. This is a Distribution similar to the delta but with integral $$ \int_{\Bbb R} f\,\delta_{t_0} = f(t_0). $$ (instead of $∫ \delta_{t_0} = 1$). Notice that the dirac Delta is not a function but a measure or a distribution of order $0$.

In your particular case, $$ f = \sin(x)\sum_i\delta_{t_i} = \sum_i \sin(x)\,\delta_{t_i} = \sum_i \sin(t_i)\,\delta_{t_i} $$ which indicates indeed that there are "masses" of size $\sin(t_i)$ at each $t_i$ and no mass outside of these points (which is sometimes represented as you say). I say masses because as you say, the Dirac delta is not a function so one can not talk about the value of the Dirac at one point. But around these points one has $\int_{t_i-\varepsilon}^{t_i+\varepsilon} f = \sin(t_i)$ if $\varepsilon$ is sufficenlty small

LL 3.14
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