# Is the Laplace transform redundant?

The Laplace transform is a generalization of the Fourier transform since the Fourier transform is the Laplace transform for $s = j\omega$ (i.e. $s$ is a pure imaginary number = zero real part of $s$).

Reminder:

Fourier transform: $X(\omega) = \int x(t) e^{-j\omega t} dt$

Laplace transform: $X(s) = \int x(t) e^{-s t} dt$

Besides, a signal can be exactly reconstructed from its Fourier transform as well as its Laplace transform.

Since only a part of the Laplace transform is needed for the reconstruction (the part for which $\Re(s) = 0$), the rest of the Laplace transform ($\Re(s) \neq 0$) seems to be unuseful for the reconstruction...

Is it true?

Also, can the signal be reconstructed for another part of the Laplace transform (e.g. for $\Re(s)=5$ or $\Im(s)=9$)?

And what happens if we compute a Laplace transform of a signal, then changing only one point of the Laplace transform, and compute the inverse transform: do we come back to the original signal?

• Why the downvote? Even if the question may contain false conclusions that is something you can very well deal with in a comment or an answer. Silently downvoting a question that someone apparently put some effort into is not very constructive. – Jazzmaniac Oct 1 '15 at 10:09
• i upvoted the question. if i am thinking in terms of angular frequency $\omega$, then i like to say Fourier Transform: $$X(j\omega) = \int\limits_{-\infty}^{\infty} x(t) e^{-j \omega t} \ dt$$ and Laplace Transform: $$X(s) = \int\limits_{-\infty}^{\infty} x(t) e^{-st} \ dt$$ . then it's pretty clear they're the same thing (sorta). – robert bristow-johnson Oct 2 '15 at 3:57

The Fourier and the Laplace transform obviously have many things in common. However, there are cases where only one of them can be used, or where it's more convenient to use one or the other.

First of all, even though in the definitions you simply replace $s$ by $j\omega$ or vice versa to go from one transform to the other, this cannot generally be done when given the Laplace transform $X_L(s)$ or the Fourier transform $X_F(j\omega)$ of a function. (I use different indices because the two functions can be different for the same time domain function). There are functions for which only the Laplace transform exists, e.g., $f(t)=e^{at}u(t)$, $a>0$, where $u(t)$ is the Heaviside step function. The reason is that the integral in the definition of the Laplace transform only converges for $\Re\{s\}>a$, which implies that the corresponding integral in the definition of the Fourier transform does not converge, i.e. the Fourier transform doesn't exist in this case.

There are functions for which both transforms exist, but $X_F(j\omega)\neq X_L(j\omega)$. One example is the function $f(t)=\sin(\omega_0t)u(t)$, for which the Fourier transform contains Dirac delta impulses.

Finally, there are also functions for which only the Fourier transform exists, but not the Laplace transform. This means that the integral in the definition of the Laplace transform only converges (in a specific sense) for $s=j\omega$, but for no other values of $s$. The Laplace transform is only said to exist if the integral converges in a half-plane or in a vertical strip of finite size of the complex $s$-plane. Such functions for which only the Fourier transform exists include complex exponentials and sinusoids ($-\infty<t<\infty$), and impulse responses of ideal brick-wall filters, which are related to the sinc function. So, e.g., the functions $f(t)=\sin(\omega_0 t)$ or $f(t)=\sin(\omega_ct)/\pi t$ do not have a Laplace transform but they do have a Fourier transform.

The Laplace transform can be a convenient tool for analyzing the behavior of linear time-invariant (LTI) systems by considering their transfer function, which is the Laplace transform of their impulse response. The poles and zeros of the transfer function in the complex $s$-plane conveniently characterize many system properties and are useful for an intuitive understanding of the system's behavior. Furthermore, the unilateral Laplace transform is very useful for analyzing LTI systems with non-zero initial conditions. The Fourier transform is a useful tool for analyzing ideal (non-causal, unstable) systems, such as ideal low pass or band pass filters.

Also have a look at this answer to a related question.

• The Fourier transform is a useful tool for analyzing ideal (non-causal, unstable) systems: you would say causal and stable? – Vinz Oct 5 '15 at 10:09
• @user17604: I meant what I wrote. Of course you can also use it for causal and stable (and non-ideal) systems. But one important use is the analysis of ideal system (such as ideal frequency-selective filters), where the Laplace transform can't be used. – Matt L. Oct 5 '15 at 10:13
• @MattL. Great answer, but I found “analyzing LTI systems with non-zero initial conditions” confusing, how can an LTI system have non-zero initial conditions? – 0MW Feb 23 '18 at 4:43
• @0MW: Yes, I probably should have said "systems that are otherwise LTI (if initially at rest)". – Matt L. Feb 23 '18 at 9:25