Talk:Laplace transform

article states often L is for managing imaginary

No. It was investigated as to if it impacts imaginary and how: but was and is not used with imaginary numbers particularly. I could say D or Dx are for managing imaginary numbers. But neither are they to any extent except in afterthought.

Readability

The occurrence of many Heaviside stepfunctions $u(t)$ in the table "selected Laplace transforms" is redundant and only makes sense if one considers the two-sided Laplace transform. — Preceding unsigned comment added by Hart15 (talk • contribs) 09:46, 30 January 2022 (UTC)

Please put new talk page messages at the bottom of talk pages and sign your messages with four tildes (~~~~) — See Help:Using talk pages. Thanks. - DVdm (talk) 10:22, 31 January 2022 (UTC)

Integral Function is Potentially Confusing

The transform of the integral function is shown as:

$\int _{0}^{t}f(\tau )d\tau =(u*f)(t)$

While correct mathematically, why have the convolution on the right side? I don't think it adds much and potentially confuses new readers.

https://proofwiki.org/wiki/Laplace_Transform_of_Integral

Tables of Laplace transforms don't generally write it like this. Rhodydog (talk) 20:27, 24 February 2023 (UTC)

Initial/final value theorems

Can someone explain what it means for a complex number to go to infinity? I could understand |s| -> inf or Re(s) -> but just s? 138.232.68.221 (talk) 08:53, 31 March 2023 (UTC)

Probability Theory

It would be welcome to have $E$ defined. Can someone please edit to provide a proper definition for those who are trying to read/understand/use as reference? — Preceding unsigned comment added by DrKC MD (talk • contribs) 18:05, 15 February 2024 (UTC)

• Done

Constant314 (talk) 18:17, 15 February 2024 (UTC)

• • Thanks, but did you mean to say $E[r]$ for some random value $r$? — Preceding unsigned comment added by DrKC MD (talk • contribs) 22:29, 15 February 2024 (UTC)

    No. $r$ is a random variable. Constant314 (talk) 23:39, 15 February 2024 (UTC)

Fourier transform

Essentially expanding the Laplace transform into real and imaginary exponent parts. The real bit is just $e^{-\sigma t}$, imaginary $e^{-j\omega t}$. FT of $f(t)e^{-\sigma t}$ is $\int _{0}^{\infty }f(t)e^{-\sigma t}e^{-j\omega t}dt$ when $f(t)=0,t<0$. XabqEfdg (talk) 02:24, 23 May 2024 (UTC)

The term "transform" is a bit ambiguous. It can either refer to the operation of transforming a function, or to the result of this operation on a particular function. Using the second meaning, the claim would be false. The term "function" has the same ambiguity, when we say "x is a function of y." It would be more verbose, but maybe "the Laplace transform operator can be defined in terms of the Fourier transform operator" would be clearer? Albie's relation of misfortune (talk) 03:08, 23 May 2024 (UTC)

I don't think that, as put, this should go in the lead. Per MOS:LEAD, the lead should summarize the content in the article. There is a section on the Laplace transform's relationship to the FT, but it does not talk about real arguments. Before putting the interpretation of the Laplace transform in the lead, it should first go in the Fourier transform section. Its placement in the lead may be undue because the Fourier transform relationship is not very important in the rest of the article. I get that there may be some pedagogical value, but see WP:NOTTEXTBOOK. People can also simply scroll down to the Fourier transform section as well. I do think that the Fourier transform section should include a bit about equivalence to the bilateral Laplace transform (which is the context that Signals and Systems and Bracewell put it in), but not the unilateral one which is, after all, just a special case. XabqEfdg (talk) 04:06, 23 May 2024 (UTC)

I'm not too worried about where this equation ends up, but there is a reason that the Laplace transform always comes along with the other elements of Fourier analysis, and that needs to be mentioned somewhere in the lead. Albie's relation of misfortune (talk) 04:19, 23 May 2024 (UTC)

Demonstration that the Laplace transform can be expressed in terms of the Fourier transform.

$$LHS={\mathcal {L}}\{f(t)\}(\sigma \ +\ i\omega )$$

$$=\int _{0}^{\infty }f(t)e^{-(\sigma \ +\ i\omega )t}\,dt$$

$$RHS={\mathcal {F}}\{f(t)e^{-\sigma t}\}(\omega )$$

$$=\int _{-\infty }^{\infty }[f(t)e^{-\sigma t}]e^{-i\omega t}\,dt$$

$$=\int _{-\infty }^{\infty }f(t)e^{-(\sigma \ +\ i\omega )t}\,dt$$

So for suitable functions,

$$LHS=RHS.$$

Albie's relation of misfortune (talk) 02:30, 23 May 2024 (UTC)

Except the limits of integration are not the same. Constant314 (talk) 05:22, 23 May 2024 (UTC)