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In solving a linear system, when would I use a Fourier transform versus a Laplace transform? I am not a mathematician, so the little intuition I have tells me that it could be related to the boundary conditions imposed on the solution I am trying to find, but I am unable to state this rigorously or find a reference that discusses this. Any help would be appreciated. Thanks.

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If you have an initial value problem, say an ODE for a function $x(t)$ with initial conditions at $t=0$, then the Laplace transform is the way to go. The Fourier transform is useful, among other things, to solve for steady-state response. –  José Figueroa-O'Farrill Feb 24 '10 at 17:08
    
FT uses periodic functions, LT does not: that's quite a striking difference when it comes to modelling boundary and other "time" related conditions, I think. –  Suvrit Sep 29 '12 at 8:40
    
simply use laplace transform to solve initial value ordinary differential equations problems and use fourier trnsform when facing partial differencial equation problem –  user38618 Aug 15 '13 at 0:22
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6 Answers

Here is a heuristic point of view from engineering considerations. I must confess I do not fully know the mathematical reasons.

Suppose you want to consider $f(t)$, a function of time, $t$. Imagine that as we look at the direction of positive $t$-axis, the graph of $f(t)$ s like looking behind to the trail $f$ left in time. If you do not care about the future, ie the case $t < 0$, then it makes sense to use Laplace transform, because the transform integral goes from $0$ to $\infty$. On the other hand, if you care about the future also, it makes more sense to consider the Fourier transform. The transformation integral here goes from $-\infty$ to $\infty$.

So if you want to include future in your analysis, then Fourier transform is the way. This makes sense in electrical engineering applications for example, where you consider sinusoidal signals and you have an idea of what is going to come.

However for some physical systems, you only have the data of what happened until then. And you want all your analysis to be based on this, without predicting the future. Then Laplace transforms is the way.

If you do not care about the future, ie if you can declare $f(t) = 0$ for $t < 0$, then the Laplace and Fourier transforms coincide: The Fourier transform is nothing but the Laplace transform evaluated on the imaginary axis. Such systems are called causal systems: the response depends only on what happened so far. This a terminology from control systems or signal processing.

For control systems engineering, stability of electrical networks, etc., Laplace transformation defines a more natural transfer function, and is easier to deal with, and the poles and zeros would immediately tell you about the stability of the network under consideration. Here we use Laplace transforms rather than Fourier, since its integral is simpler.

For instances where you look at the "frequency components", "spectrum", etc., Fourier analysis is always the best. The Fourier transform is simply the frequency spectrum of a signal. If you know that the sin/cos/complex exponentials would behave nicely, you might as well want to express a function in terms of these and observe how it behaves then.

Another example is solving the wave equation. Fourier himself used Fourier series/transforms for heat conduction problems.

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If you wanted to take an integral over R, you could still just use the two-sided Laplace transform instead of the Fourier transform. –  Qiaochu Yuan Feb 24 '10 at 19:05
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@Qiaochu: There is little difference between two-variable Laplace transform and the Fourier transform. Each can be got from the other looking at the imaginary axis. –  Anweshi Feb 24 '10 at 19:14
    
That was useful Anweshi, thanks. –  pirata Feb 25 '10 at 17:36
    
Again a practical suggestion, seeing below that you use RC filters: I have no idea how neurons are modelled. But, say, if you want to do control engineering, do Laplace transforms, and if you want to do signal analysis, processing, etc., then use Fourier transforms. If both looks usable, use Laplace, because Laplace is somewhat simpler. –  Anweshi Feb 25 '10 at 20:15
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$t<0$ is future? –  timur Nov 26 '10 at 20:46
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Laplace transforms appear in physics because of causality: a response function $R(t-t')$ which gives the response at time $t$ to a force at time $t'$ should vanish for $t\lt t'$, in order not to violate the temporal relation between cause and effect. Because $R(t)=0$ for $t<0$ its integral transform is the Laplace rather than the Fourier transform.

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For typical practical usage it is essential to know what properties has the system you are trying to describe. Usually you are able even get some insights about possible shape of solution before you really solve equations, only by means of symmetry considerations. So then you have to consider one or another transform not because of formal effectiveness, but because your solution has to have practical interpretation! What do you do with solution if you do not have any interpretation for example for coefficients of equations you get?

Fourier transformation sometimes has physical interpretation, for example for some mechanical models where we have quasi-periodic solutions ( usually because of symmetry of the system) Fourier transformations gives you normal modes of oscillations. Sometimes even for nonlinear system, couplings between such oscillations are weak so nonlinearity may be approximated by power series in Fourier space. Many systems has discrete spatial symmetry ( crystals) then solutions of equations has to be periodic so FT is quite natural ( for example in Quantum mechanics). With any of normal modes you may tie finite energy, sometimes momentum etc. invariants of motion. So during evolution, for linear system, such modes do not couple each other, and system in one of this state leaves in it forever. Every linear physical system has its spectrum of normal modes, and if coupled with some external random source of energy ( white noise), its evolution runs through such states from the lowest possible energy to the greatest.

It depends on initial conditions and boundary values and restrictions but for finite systems and linear equations Fourier Transform gives you transformation from linear differential equation to matrix one ( which is nearly always soluble and has clear theory and meaning) whilst Laplace Transform from DE to algebraic one with all advantages and disadvantages of it.

Laplace transform gives you solution in terms of decaying exponents so it is quite useful in relaxation processes, but it has no physical interpretation, usually no invariants are connected to any "vectors" of such representation, there is no discrete version of such transform with physical meaning. It is used in various engineering problems such that electrical circuits, queue theory etc. many equations in diffusion processes has easy Laplace transform solutions.

Definitely it would be easier to advice you what method of solution to use if you would describe what is the process you are trying to describe.

References: try to Google such words: energy spectrum, normal modes, eigenstates, eigenvectors in context of linear differential equations - solving DE by means of integral transforms in practical way is usually described in books on Mathematical Methods in Physics, and is connected to response functions, distribution theory, Hilbert and Banach functional spaces etc. It is very very broad area. What is more, if you asking in specific context ( for example in context of stochastic processes, or quantum mechanics) , then probably you are looking for some certain interpretation of such transforms and not for formal theory. This differences sometimes tricky because mount of mathematical books focus on existence theorems etc. which for many applications are obvious ( providing we have working, well formulated model, which ususaly we have!)

It is very difficult to get one useful reference without knowledge of area of application, because its are is such frequently used method! In analogue on mathematics level is like ask for application of metric spaces, or Stokes theorem and its meaning: it so broad area that probably you may just put in in every other area ad it fit! Nearly every Quantum Mechanics book will have explanation and interpretation of Fourier method. Laplace transform will be used in every books regarding signal processing! Many of them have very well and practical introduction to such methods. I prefer physical books, for example Byron Fuller "Mathematical Methods of Physics" for intermediate level.

Here you have many, many references: http://mathworld.wolfram.com/FourierTransform.html Here you have another one list: http://www.ericweisstein.com/encyclopedias/books/FourierTransforms.html

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In my experience, googling random keywords is a very inefficient way of learning. Maybe a pointer to a single coherent source would be more helpful? –  j.c. Feb 24 '10 at 22:58
    
OF course You are right, but unfortunately solving Differential Equations by means of integral transforms in context practical approach is very broad area. Probably I will be able to point to certain book if Pirata give us a name for area of the problem she/he is asking for. But I will try to put some more detailed references. –  kakaz Feb 25 '10 at 7:31
    
I am actually a biologist and I wanted to understand a simple RC circuit as a low pass filter as a first step to modeling some neurons. This is a simple linear system and one can easily obtain the transfer function etc... But in doing so, some texts use a FT and some use a LT. This difference may not matter to study the properties of the system at this level, but that got me thinking about this question. I understand the difference between an initial value problem and a boundary condition problem and was just looking for a statement of when to use one as opposed to the other. –  pirata Feb 25 '10 at 17:14
    
I also believe that Fourier modes are eigenfunctions of certain operators but it seems to me that the Laplace modes will be too (maybe with real/imaginary eigenvalues switched) so that led me to believe that the differential operator in question should not play a decisive role in choosing a FT or a LT (I am correct in saying this?). Maybe the point is, as mentioned in the other answer, that if the system is causal, by which I understand that the impulse-response function vanishes for t<0, there really is no difference. Google and wikipedia etc provided no satisfactory answers. –  pirata Feb 25 '10 at 17:23
    
In this context, Laplace transform gives You ability of analyzing properties of such system when changing its parameters, or parameters of initial values. This is typical analysis of stability of "black box" with some response function etc. From the other side Fourier transform gives You direct insight in properties of such system in this meaning that typical signals which drive such systems in practice are periodic and sinusoidal or simple superpositions of such signals. –  kakaz Feb 25 '10 at 20:13
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For many years I have tried to obtain a good answer for the Laplace and Fourier transforms relationship. Many of the explanations just mention that the relationship is that s=a+jw, so the Fourier transform becomes a special case of the laplace transform. Sad explanation. Better explanations deals that Laplace is used for stability studies and Fourier is used for sinusoidal responses of systems. Using that information, I conclude that as systems are stable if the real part of s is negative, that is to say there is a transient that will vanish in time, in those cases, it is enaugh to use Fourier. Of course you will lose the insight of the transient part. Laplace should be able to determine the full response of a system, be it stable or unstable, including transient parts.

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What exactly are you trying to teach us here? I think it would be better to think of the Fourier transform in terms of tempered distributions, and of the Laplace transform in term of causal systems and the corresponding properties of holomorphic functions, which are related to a variant of the Paley-Wiener theorem. In my own experience, different window functions (and low pass filters) are appropriate in the context of (numerical) Laplace transform (Hann-window related constructions well in this context) than in the context (numerical) Fourier transform (convolutions of rectangular functions) –  Thomas Klimpel Sep 28 '12 at 22:19
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Fourier transform (FT) - (roughly) a tool to visualize ANY signal as a sum of sinusoids.

Laplace transform (LT)- a tool to analyze the stability of systems.

Why can't we use FT to analyze systems? because it cannot handle exponentially growing signals. That in turn is because a signal needs to be absolutely integrable(necessary, but not sufficient condition) for it to have a FT.

On the other hand, LT is specially designed (by Laplace himself after he rejected the thesis of Fourier and extended FT) to handle exponentially growing signals so that now you could if a system has an exponentially growing output ( a > 0 where s = a + jw) or exponentially decaying output (a < 0)

Therefore if you have a square wave type signal and you wanted to break it up into sinusoids of various frequencies, you'd use FT and on the other hand, if you have a system and you want to understand how stable it is, then you'd try to get LT of it's impulse response to further analyze the systems relative stability.

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Exponentials have well-defined FTs in generalized functions. See mathoverflow.net/a/42017/121 –  S. Carnahan Jan 10 at 10:51
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The solution pattern behind is like this:

Using a transform is like changing your point of view. In some cases the problem might get such easy under the new point of view, that you are able to solve the problem there and then you take the obtained solution and transform back to your original point of view.

Here we might try to solve a differential equation, thus looking for some function (or distribution) which fullfills the differential equation and additional conditions.

The Laplace transform has this nice property:

$f'(t) = s F(s) - f(0)$

This means a transformed differential equation will have the derivative $f'(t)$ replaced by the above term, so a simple equation with just occurences of $F(s)$ and the start condition $f(0)$ are left. Such an equation we can solve for the unknown $F(s)$ by simple algebra.

The change of point of view from $t$-domain to $s$-domain made the problem of solving the differential equation easier by replacing it with an algebraic equation.

The trickier part is transforming back the found solution $F(s)$ to $f(t)$ by applying an inverse Laplace transformation (or looking it up in some table).

The Fourier transform has a similar property:

$f'(x) = 2\pi i \,t\, F(t)$

which in a sense transforms the differential operator $\partial_x = \partial / \partial x$ into a multiplication with the variable $t$.

This is used to solve differential equations like the diffusion equation, wave equation, Schrödinger equation, Klein-Gordon equation etc. by turning them into algebraic equations, as $\Delta = \sum \partial_i \partial_i$ gets transformed into $\sum x_i x_i$.

Again you solve it and then transform back.

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