# Why Fourier transform is not sufficient and we have to use Laplace transform? [duplicate]

• Is there an easy way to explain the motivation behind the use of Laplace transform instead of Fourier transform?

• Isn't that any periodic function can be represented by sines and cosines? - Why to introduce exponential idea?

• Why not using differential equations with Fourier transform? An example would help.

*Asked the same question a while ago at math.stackexchange but no answers given.

• This is a nice but a problematic question. Like many inventions of mathematics (as applied to Physical problems of either pure or applied-engineering genre), the Laplace transform has pros and cons. It was modified a lot before it became as its standard form. To sum up I can say is 1-Like other transforms, it converts LCCDE (core of engineering analysis) into algebraic ones, easier to solve. 2-It enables mathematical analysis of complicated feedback-control systems in a simpler way. 3- It gained widespread acceptance after the first quarter of the 20th century. – Fat32 Jul 28 '16 at 13:46

The Laplace Transform is more representative of real systems that have a starting point, which is why the integral starts at 0, and also why the unit step function is generally talked about alongside the Laplace Transform. With the Laplace Transform, we can examine the transient and steady-state behavior of a system.

Using $e^{st}$ instead of $e^{iwt}$ allows us to examine different aspects of a physical system. The variable $s$ is complex, and if the real part was set to 0, it would reduce to a truncated Fourier Transform. The real part of $s$ is related to the amount of damping in the system. Also, with the Laplace Transform, a system's stability can be considered.

In short, Laplace is used to consider damping, stability, transient and steady-state behavior of a physical system (represented by a differential equation).

• why is Laplace Transform is preferred when dealing with differential equations? – user16307 Jul 28 '16 at 13:29
• The definition of the Laplace Transform fits better with modeling physical systems as differential equations: time starts at 0, a switch flips, and the system starts. And again, using a complex variable instead of a real variable in the transform kernel allows for more analysis like steady state, transient, stability, etc. – soultrane Jul 28 '16 at 13:33

The behavior of many systems in the "real" world is closer to that of decaying exponentials rather than to that of infinitely periodic sinusoids that extend into the past. Thus, playing with Laplace transforms gets one closer to applied engineering solutions to models of these systems while using less chalkboard.

Also, the limits of integration and the regions of convergence are different between Laplace and Fourier transforms.

Laplace transforms are often used in converting a differential equation into an algebraic equation that are easy to solve. Fourier transform is often more useful in de-constructing continuous signals.

Laplace is much better at understanding the stability of a system. You can't really find the fourier transform of exponentially growing functions.

• you are writing the conclusion but it doesn't make me understand why it is so. – user16307 Jul 28 '16 at 13:30
• Well, for example, you can't find the fourier transform of e^t but you can find the laplace transform of e^t. – iyop45 Jul 28 '16 at 13:34
• Oh really? Is that because it is non periodic? – user16307 Jul 28 '16 at 13:39
• @Florian Castellane That is true, but the same can be said about the Laplace Transform...it does not converge for every function. – soultrane Jul 28 '16 at 13:51
• @soultrane And in that case you wouldn't be using the laplace transform – iyop45 Jul 28 '16 at 13:53