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I've read in numerous places that the unilateral laplace transform is extermely useful in solving differential equations with initial conditions based on the differentiation property of the unilateral transform:

$\mathscr{L}{f′(t)}=sF(s)−f(0_−)$

What i don't understand is why this is possible to do in Laplace but not in the Fourier Transform? Is this related to the dacaying exponential added by the Laplace transform?

I would be grateful for an in-depth answer involving intuition.

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  • $\begingroup$ If you look at "transient response" with Laplace xform you will find a lot :) The first real electronics book I read was Transient Response in linear systems. You should realize that these initial terms f(0), f'(0) are the boundary conditions and really are required to analyze a circuit. They can be folded into the Fourier transform but the interpretation/characterization is much harder; and not needed. Be aware that things like Power supplies, motors, and power semiconductors are most stressed during startup; i.e. the worst conditions can occur due to "transients". $\endgroup$
    – rrogers
    Mar 30 at 19:40
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You need to look at the derivation of that property. Integration by parts gives

$$\begin{align}\mathcal{L}\{f'(t)\}&=\int_{0^-}^{\infty}f'(t)e^{-st}dt\\&=f(t)e^{-st}\Big|_{0^-}^{\infty}+s\underbrace{\int_{0^-}^{\infty}f(t)e^{-st}dt}_{F(s)}\\&=\lim_{t\to\infty}f(t)e^{-st}-f(0^-)+sF(s)\tag{1}\end{align}$$

The first term in $(1)$ is only guaranteed to vanish for $t\to\infty$ if $\text{Re}\{s\}>\alpha$ for some value $\alpha$. So even if you define a unilateral Fourier transform, that term may not vanish in general.

However, if $\lim_{t\to\infty}f(t)=0$ holds, we could use the same property with a unilateral Fourier transform. It's just much more common to use the well-established (unilateral) Laplace transform in cases where non-zero initial conditions need to be taken into account.

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  • $\begingroup$ Ok thank you very much this is exactly what i was looking for!, and now if you could please complete my analysis: the above method with $0_-$ is meant to deal with transients e.g. a step function - say i've a a basic rc circuit and at time t=0 the switch closes introducing a step function into the system - its derivative is the delta function which also introduces a singularity at t=0, is all this just a method to deal with this singularity of transients? - by modeling the above as if it has initial conditions? thus enabling us to perdict the system's response in full? $\endgroup$ Mar 27 at 15:14
  • $\begingroup$ @meirfranco: The unilateral transform is used to take initial conditions into account. I'm not sure I understand your question though ... $\endgroup$
    – Matt L.
    Mar 28 at 11:10
  • $\begingroup$ They allow you to eliminate the delta function there. That's not their purpose but it's true. Switching a charged capacitor into a circuit requires careful analysis till the two systems come into adjustment; of course you can ignore it and watch :) You can also cut up things in the time domain and restart the Laplace to avoid nonlinearity (i.e. switches) or input delta functions $\endgroup$
    – rrogers
    Mar 30 at 20:01

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