I was wondering

  1. why we have both Laplace transform and Fourier transform, instead of just one?
  2. why we have both generating function and Z transform, instead of just one?

In other words, in each group,

  • in what cases one transform is better than the other?
  • in what cases the two transforms can be equally useful?

Since both transforms seem able to be converted to each other by change of variable, I am curious how people pick which tool/transform over the other to solve their problems。

Thanks and regards!

  • 2
    $\begingroup$ "Use the right tool for the job" and such. One constructs a transform simply because there are things that are easily done in the transform's domain as opposed to the original domain, much like one would transform coordinates to exploit symmetries in a system. $\endgroup$ – J. M. isn't a mathematician Apr 25 '11 at 0:23
  • $\begingroup$ @J.M.: Thanks! Are there some examples to show one transform is better than the other, or even using one is possible and using the other is impossible? $\endgroup$ – Tim Apr 25 '11 at 0:25

Yes, as the questioner notes, Fourier transform and Laplace transform are basically "the same", except in different coordinates. There is at least one further point, that the Laplace transform set-up puts "the point at minus infinity" at 0, rather than at infinity (as does Fourier), giving a handle on behavior there, boundary values and such. In effect, Laplace tolerates functions that don't decay much at minus-infinity, by throwing in exponentials to make integrals converge. That is, it's not just the object itself (in this case, the real line, or, by exponentiating, the positive reals), but a compactification of it, to be able to talk sensibly about behavior at the boundary, e.g., at "infinity".

In fact similarly, I've always interpreted Z-transform stuff as being "basically the same as" Fourier series: the discrete double-sided list of numbers I'd interpret as Fourier coefficients of a function on a circle, that is, a periodic function. This viewpoint does explain a certain number of things about Z-transforms, but it does also neglect the refinements (which I don't know much about, except their existence) that are parallel to the distinction mentioned above between Fourier and Laplace transforms.

Even apart from boundary-value subtleties, I think the point is that there is a widely-applicable "spirit" about integral transforms (that often convert differential operators into multiplication operators), but the details "matter", in the first place, and, second, can be exploited by people who know what they're doing. Hard to give a broad prescription of the latter, I fear.

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Let me give a very high brow example, partially inspired by Terry Tao's recent blog post.

In Euclidean spaces, the Fourier transform has the nice property that it interchanges derivation with multiplication by coordinate weights. As such it is very well suited to the analysis of differential equations. Formally speaking, by taking the Fourier transform, a differential equation can be converted to an algebraic equation. And similarly integral/algebraic inequalities can be then used to establish differential inequalities for solutions.

Another way to look at this is through the lens of spectral theory of linear operators. In many cases the Fourier transform allows one to vastly simplify the spectral analysis by reducing it to algebraic statements.

Now what if we want to do something similar on a manifold? Suppose we have a compact, closed Riemannian manifold, technically speaking we can decompose any square-integrable function $f$ using the eigenfunction decomposition relative to the Laplace-Beltrami operator corresponding to the Riemannian metric. In practice, this is not very ideal because this requires us knowing what the eigenvalues of the Laplacian are (something not easy to compute) and also what the eigenfunctions of the Laplacian are (something even harder to compute explicitly). (Basically, to decompose an arbitrary vector in an inner product space using some orthonormal basis, the usual thing to do is to describe the coefficients by studying the inner product of the arbitrary vector against the basis vectors; this is, in some sense, an informal description of the Fourier transform/series.) In Euclidean space, we know what the "eigenfunctions" are: they are just $e^{i\xi\cdot x}$, so the Fourier transform is effective. On manifolds, we don't usually know what the eigenfunctions are, and so it is a bit harder to use the Fourier transform "as it is".

However, we can massage the Laplace transform to get something more computable, yet sufficiently similar to a frequency decomposition to be useful.

A very rough description goes something like this: given $(M,g)$ a closed Riemannian manifold, with $\triangle$ its associated Laplacian. Given a smooth function $f:M\to\mathbb{R}$ and a non-negative smooth function $h:[0,\infty) \to [0,\infty)$, we define the $h$-frequency-weight of $f$ to be

$$ \mathcal{F}[f;h] = \int_0^\infty h(t) e^{t\triangle}f ~dt $$

Notice that it can be formally written as the operator $\int_0^\infty h(t) e^{t\triangle}dt = \mathcal{L}h(-\triangle)$ acting on $f$, where $\mathcal{L}h$ denotes the Laplace transform of $h$. $e^{t\triangle}$ should, of course, be interpreted in the sense of the heat-kernel. By appropriate choices of $h$, we can make it such that $\mathcal{F}[f;h]$ behaves like a frequency localised version of $f$, kind of like the plane wave $\hat{f}(k)e^{ikx}$ in the Fourier transform.

(If you want to read more about this, you will want to consult Stein, Topics in harmonic analysis related to Littlewood-Paley theory.)

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  • $\begingroup$ Thanks! I will try to understand what you have written. Do you happen to know about the differences when applying generating function and Z transform? $\endgroup$ – Tim Apr 25 '11 at 1:22
  • $\begingroup$ Nope. That's beyond my expertise. $\endgroup$ – Willie Wong Apr 25 '11 at 1:25

Unilateral Z transform and ordinary generating functions are equivalent (just take $x=1/z$). I guess the reason that some people (especially in engineering) prefer the Z transform is that the formulas look more like what they already know about the Laplace transform.

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