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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?

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    $\begingroup$ 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
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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.

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  • $\begingroup$ 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
  • $\begingroup$ @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
  • $\begingroup$ Instead of an endless theoretical discussion whose metaphysical basis is on the realness vs logicality of the so called entities of purely mathematical (imaginative!) nature, what I applied in these posts is that in certain mathematical statements, the expression $\delta(t)\delta(t)$ can be replaced by $\delta(0)\delta(t)$ without affecting the result. The meaning of $\delta(0)$ is another concern of course. Also we all know that this is in contrast to the basic statement that I posted as the comment to the question. My concern is on the result. You can upvote my answer too :-) ! $\endgroup$ – Fat32 Oct 2 '17 at 17:28
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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

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

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    $\begingroup$ 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
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    $\begingroup$ @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

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