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:


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.

  • $\begingroup$ 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, 2021 at 19:40

1 Answer 1


You need to look at the derivation of that property. Integration by parts gives


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.

  • $\begingroup$ 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, 2021 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, 2021 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, 2021 at 20:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.