1
$\begingroup$

The unilateral Laplace transform of $f(t)$ is $\int_0^\infty e^{st} f(t) \mathrm{d}t$.

If we define the transform as $\int_{a}^\infty e^{st} f(t) \mathrm{d}t$, would it conserve all the nice properties of the true Laplace transform (e.g., the convolution theorem)?

How would its inverse be?

$\endgroup$
  • $\begingroup$ forget the uni-lateral Laplace transform $F(s) = \int_0^\infty f(t) e^{-st} dt$ and learn only the bilateral Laplace tranform $$\mathcal{L}[h(t)](s) = \int_{-\infty}^\infty h(t) e^{-st} dt$$ hence $$ \int_0^\infty f(t) e^{-st} dt = \mathcal{L}[f(t)1_{t >0}](s)$$ while $$ \int_a^\infty f(t) e^{-st} dt = \mathcal{L}[f(t)1_{t >a}](s)$$ $\endgroup$ – reuns May 23 '16 at 18:10
  • $\begingroup$ (and your teacher is bad) $\endgroup$ – reuns May 23 '16 at 18:10
  • 1
    $\begingroup$ I'm self-taught :,-( ! $\endgroup$ – altroware May 29 '16 at 7:53
0
$\begingroup$

This is what you get when you have a unit step function. (Heaviside function).

If $\int_0^{\infty} f(t) e^{-st} dt = F(s)$ then $\int_a^{\infty} f(t) e^{-st} dt = e^{-as} F(s)$

$\endgroup$
0
$\begingroup$

If you mean its inverse in time domain, then it is $$u(t - a) f(t)$$ where $$u = \begin{cases} 1, & t \geq a \\ 0, & \text{ otherwise} \end{cases} $$

And yes, the convolution theorem still applies.

$\endgroup$

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.