2
$\begingroup$

If a impulse function is mulplied with a function $f(x)$ then the formula will be apply $$f(x)\delta(x-a)=f(a)\delta(x-a)$$ so from this formula mulplication of two impulse function will be

$$\delta(t)\delta(t)=\delta(0)\delta(t)$$

But,what if i take two impulse function as the approximation of standard signal (rectangular or gaussian) and mulplied them directy then i will get again an impulse function of same property.

enter image description here

What is wrong with this two?

Improve this question
$\endgroup$
  • 2
    $\begingroup$ Strictly speaking the function multiplying the generalized function $\delta(t)$ should be a sufficiently smooth continuous one. Otherwise the multiplication is undefined. You can have a look at the generalized function theory and distribution theory books. $\endgroup$ – Fat32 Oct 1 '17 at 17:33
  • $\begingroup$ See the mathoverflow question Is square of Delta function defined somewhere? $\endgroup$ – Olli Niemitalo Oct 2 '17 at 6:58
5
$\begingroup$

The property

$$f(t)\delta(t-t_0)=f(t_0)\delta(t-t_0)\tag{1}$$

is only valid for a function $f(t)$ that is continuous at $t=t_0$. Since the Dirac delta impulse $\delta(t)$ is not a function (it is a distribution) and since $\delta(t)$ is not continuous, property $(1)$ does not hold (and does not make sense) for $f(t)=\delta(t)$.

The product $\delta(t)\cdot \delta(t)$ is undefined. Also note that the quantity $\delta(0)$ is meaningless. Since $\delta(t)$ is not a function you cannot determine its value for any value of its argument.

Improve this answer
$\endgroup$
  • $\begingroup$ have a look at the original question and consistent usage of $\delta(0)$ in its solution. (where $\delta(0)$ is used in both time domain and Fourier domain computation of an improper (divergent) convolution) $\endgroup$ – Fat32 Oct 2 '17 at 9:12
  • 2
    $\begingroup$ @Fat32: OK, but I still don't think that $\delta(0)$ makes any sense. $\delta(0)=\infty$ may be an intuitive way to think about it, but it is not correct. $\endgroup$ – Matt L. Oct 2 '17 at 16:05
1
$\begingroup$

You can use the following (weak) argumentation to deduce that the result is interpreted as the (not very meaningful statement of) $$ \delta(t) \cdot \delta(t) = \delta(0) \cdot \delta(t) $$

Define the impulse as a limit of the following pulse function

$$ \delta(t) = \lim_{\Delta \to 0} \delta_{\Delta}(t) $$ where $\delta_{\Delta}(t)$ is a rectangular function with the definition that

$$ \delta_{\Delta}(t) = \begin{cases} 0 &,\ t < 0 \\ \frac{1}{\Delta} &,\ 0 \le t < \Delta \\ 0 &,\ \Delta \le t \end{cases} $$

Then you can define the multiplication as: $$\begin{align} \delta(t)\delta(t) &= \left( \lim_{\Delta \to 0}\delta_{\Delta}(t) \right) \left( \lim_{\sigma \to 0}\delta_{\sigma}(t) \right)\\ &= \lim_{\Delta \to 0} \lim_{\sigma \to 0} \left[ \delta_{\Delta}(t) \delta_{\sigma}(t) \right]\\ &= \lim_{\Delta \to 0} \left( \delta_{\Delta}(t) \lim_{\sigma \to 0} \delta_{\sigma}(t) \right)\\ \text{now observe that}\\ \delta_{\Delta}(t) \lim_{\sigma \to 0} \delta_{\sigma}(t) &\approx \delta_{\Delta}(0) \lim_{\sigma \to 0} \delta_{\sigma}(t) = \delta_{\Delta}(0) \delta(t) \end{align} $$ where the equality holds in the limit. Then proceed with

$$ \begin{align} \delta(t)\delta(t) &= \lim_{\Delta \to 0} \left( \delta_{\Delta}(0) \delta(t) \right) \\ \delta(t)\delta(t) &= \left( \lim_{\Delta \to 0} \delta_{\Delta}(0) \right) \delta(t) \\ \delta(t)\delta(t) &= \left( \delta(t)|_{t=0} \right) \delta(t) \\ \delta(t)\delta(t) &= \delta(0) \delta(t) \\ \end{align} $$

Improve this answer
$\endgroup$
  • 1
    $\begingroup$ So why is it "not very meaningful"? All the math looks perfectly OK to me. $\endgroup$ – Dilip Sarwate Oct 1 '17 at 20:42
  • $\begingroup$ @DilipSarwate, does "$\delta(0)$" look perfectly OK? $\endgroup$ – robert bristow-johnson Oct 2 '17 at 4:52
  • $\begingroup$ and i would like to see anyone successfully convolve these two functions: $$ x(t) = 1 \qquad \forall t \in \mathbb{R} $$ and $$ h(t) = 1 \qquad \forall t \in \mathbb{R} $$ $\endgroup$ – robert bristow-johnson Oct 2 '17 at 4:59
  • $\begingroup$ @DilipSarwate not meaningful in the classical (19th century) mathematical sense... Classical point set topology is fundamentally based on smoothness and density. So that the concept of limit, integral and derivate makes sense. It cannot be used to model a blackhole nor a quark however... More modern mathematics deals with distributions in a more consistent way but I'm not sure of its justifications, which however are no less meaningful than the justifications of smoothness! $\endgroup$ – Fat32 Oct 2 '17 at 9:23
  • 1
    $\begingroup$ @robertbristow-johnson The point is that the limit does not exist at $t=0$ (the value of $\delta_{\Delta}(t)$ at $t=0$ is $\frac{1}{\Delta}$ which diverges to $\infty$ as $\Delta \to 0$) and even if the limit exists, we cannot claim that the "value" of $\delta(t)$ at $t=0$ equals this limit without assuming that $\delta(t)$ is continuous at $t=0$. In short, Fat32's "argument" perpetuates the myth that $\delta(t)$ is an ordinary function. That he knows better is shown in his comment above as well as in his comment on the OP's question. $\endgroup$ – Dilip Sarwate Oct 2 '17 at 15:15
-2
$\begingroup$

Well, since

$$\pi^2\delta(0)^2=2\int_0^\infty t dt-\frac1{12}$$ we have $$\delta(0)^2=\frac2{\pi^2}\int_0^\infty t dt-\frac1{12\pi^2}$$

So, we can write $\delta(x)^2$ as

$$\delta(x)^2=\frac1{\pi^2}\int_{-\infty}^\infty |t| \cos(xt) dt-\frac{0^{|x|}}{12\pi^2}$$

Improve this answer
$\endgroup$
  • 2
    $\begingroup$ This is complete nonsense $\endgroup$ – Jazzmaniac Dec 1 '20 at 10:04
  • 1
    $\begingroup$ @Jazzmaniac what exactly? This does not work for you? $\endgroup$ – Anixx Dec 1 '20 at 10:08
  • $\begingroup$ Do you actually believe that what you wrote here is proper math? If you do, then there's no point in argueing. $\endgroup$ – Jazzmaniac Dec 1 '20 at 10:10
  • 2
    $\begingroup$ No, that's not an explanation. It's just more incoherent, hand-waving and inconsistent impressionistic "math". This is very far from a rigorous argument. My comment above still stands. $\endgroup$ – Jazzmaniac Dec 1 '20 at 10:55
  • 1
    $\begingroup$ @Jazzmaniac if you see some inconsistencies, you are welcome to point them. There are definitely which I know, but the do not touch this result. $\endgroup$ – Anixx Dec 1 '20 at 11:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.