# Time scaling of a shifted function

When we have say Delta at time $$t_0$$ on continuous time: $$\delta(t-t_0)$$.

• If we want to move it $$t_0/2$$, can we scale the time and squeeze by 2 instead of shifting?

• How does scaling in time of a shifted function look like? Is it $$f(a(t-t_0))$$ or $$f(at-t_0)$$?

• @DilipSarwate: If people seriously started reading books, we might as well shut down this site ... Jul 14, 2020 at 15:01
• I would suggest to keep it polite and civilized around here. Jul 14, 2020 at 15:11
• @VitaliPom to be very clear here: Dilip pointed out that your question wasn't well-researched, which is one of the measures we apply to questions here. His feedback is actually constructive – go read a signals textbook, it's not like any of this requires any skill that wasn't absolutely basic in your field of work. Attacking him definitely means I'll ignore your future questions for constructive feedback, and just downvote if I find a question doesn't fulfill the requirements for own research. Jul 14, 2020 at 15:18
• VitaliPom might have an exam tomorrow but instead of studying for it, he seems to have taken a break to go to my user page and downvote the top answers there (just as he has threatened to do to @MarcusMuller). Way to go! Jul 14, 2020 at 15:51
• @DilipSarwate I deleted the question, but it does not deletet immediately so I googled you. I found out you're from India. So I apologies about that, I have special attitude towards people from India. Hope you're doing well. Jul 14, 2020 at 16:25

Properties or effects of mathematical objects (operators, systems, functions, etc.) generally depend on hypotheses they are supposed to possess (or axioms they obey).

Here, one of the tag the OP used is very important: , and a mention in the question as well: "shifted function". Words are important. The Dirac $$\delta$$ is not a function. And from the realm of function, it cannot be considered continuous (with standard continuity), in the sense of topological continuity (note to self: ask to rename in ).

The Dirac $$\delta$$ is a kind of continuous analogue of the discrete Kronecker delta. Continuous here means that it is defined on continuous time. Its properties are odd, compared to classical functions. For the formula the OP considers, a useful property is: if $$g$$ is a continuously differentiable function with a real root at $$t_0$$, and its derivative (well-defined) does not vanish, one could write:

$$\delta(g(t)) = \frac{\delta(t-t_0)}{g'(t_0)}\,.$$

which applies well here. For details, see Dirac delta, Composition with a function.

• Thank you Laurent <3 have a great special day and I’m glad someone responded (even though I understood it. Sorry I didn’t mark as answered, but you leveraged my self esteem, which gives you credit points. I’ll mark you your answer as the answer, it’d be also stackoverflow’s way - people who dislocate like I did here will be able to read my answer.) Mercy Boku :) Jul 14, 2020 at 18:42
• And Matt was super Okay :) Jul 14, 2020 at 18:44
• Happy Bastille Day, Laurent! No brickbats from me. Jul 14, 2020 at 20:51
• And I learn a word today Jul 14, 2020 at 20:54

Despite all the uncensured content here in the comments, I'd like to answer my own question.

I understand that some have difficulties understanding the intuition behind the dilema here. I'll explain this first:

When we have deltas and we do Fourier to them, intuition says since we do scaling to a shifted delta function's time we might maybe somehow exchange a shifting operation on a scaling operation.

$$\delta(t-t_0)$$ can be scalled in time and since we have y=infinity at $$t_0$$ and the width of delta is infinitely small, then maybe we could make a little trick in future use?

Well then, wrong.

The thing is that when we do time scalling, the shift moves, that's right. But the time scalles as well. When we have $$\delta(a(t-t_0))$$ function, it might yield "the same" in time domain, but it's different.

The thing is that while doing fourier to it, since we have an integral during Fourier process, time scalling to delta function will have an impact on the integral and so that:

$$\int\delta(a(t-t_0))= {1 \over a}\int\delta(t-t_0)$$

Means that we have no way doing time scalling instead of shifting to delta functions despite appealing as much as it can seem to be.

• I believe that the answer to your last integral should be something like $1/a$ Jul 14, 2020 at 20:26
• I apologies, forgot the integral there, yes indeed I fixed tnx. Jul 14, 2020 at 21:34
• The $\mathrm{d}t$ is important as well Jul 15, 2020 at 21:34