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Why is the Laplace transform commonly taught as the unilateral Laplace transform?

I mean, for the Fourier transform, we commonly have the bilateral transform... if the signal is 0 for $t<0$, then it turns into a unilateral Fourier transform. Why not have this same convention for Laplace transform? Why specifically introduce the unilateral version?

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The widespread use of the unilateral Laplace transform reflects the fact that in practice we often deal with causal systems and signals that have a defined starting time (usually chosen as $t_0=0$).

The Fourier transform is mainly used for analyzing ideal signals and systems, such as ideal filters (e.g., low pass, high pass, etc.) and ideal signals such as perfect sinusoids. In these cases we have to deal with non-causal systems with impulse responses that extend from $-\infty$ to $\infty$. The same is of course true for sinusoidal signals or complex exponentials. Note that neither ideal filters nor the signals mentioned above can be treated by the Laplace transform.

One of the most important features of the unilateral Laplace transform is that it can be used to elegantly solve differential equations with initial conditions. The initial conditions are taken into account by the well-known differentiation property of the unilateral Laplace transform:

$$\mathcal{L}\{f'(t)\}=sF(s)-f(0^-)\tag{1}$$

where $f(t)$ is a differentiable function, and $F(s)$ is its (unilateral) Laplace transform. The Fourier transform doesn't have an equivalent to $(1)$ for taking initial conditions into account. If the Fourier transform is to be used for solving a differential equation with non-zero initial conditions, then the initial conditions need to be modeled as additional sources.

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  • $\begingroup$ Can't we use the bilateral laplace transform for ideal filters and ideal signals? In these cases is it always preferable to use Fourier? $\endgroup$ – Ameet Sharma Nov 6 '19 at 14:37
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    $\begingroup$ @AmeetSharma: The bilateral Laplace transform of a sinusoid or of a sinc function doesn't exist, the Fourier transform does. $\endgroup$ – Matt L. Nov 6 '19 at 14:58
  • $\begingroup$ Thanks. Just to clarify, do you mean the bilateral transform only converges for these signals when s is purely imaginary? In that case, the bilateral laplace would be the same as the fourier transform right? $\endgroup$ – Ameet Sharma Nov 6 '19 at 16:29
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    $\begingroup$ @AmeetSharma: The bilateral Laplace transform always converges in a strip of the complex plane, if it converges at all. If that strip "degenerates" to a line, especially the imaginary axis, then it's actually no Laplace transform anymore, but the Fourier transform. In the cases mentioned in my answer (sinusoids, sinc functions, etc.), the meaning of convergence of the Fourier integral is different than for the standard Fourier or Laplace transforms. $\endgroup$ – Matt L. Nov 6 '19 at 16:36
  • $\begingroup$ So is this just some kind of a convention... ie: we allow delta functions in fourier transforms but not in laplace transforms? $\endgroup$ – Ameet Sharma Nov 6 '19 at 17:58

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