So I am getting to grasps with Fourier transforms. Intuitively now I definately understand what it does and will soon follow some classes on the mathematics (so the actual subject). But then I go on reading about the laplace transform and there I kind of lose it. What is the moment of a signal? Why is the fourier transform a special case of the laplace transform? How can I come to grips with the Laplace transform?

Ive looked at these sources before I asked this question:

What is meant by a system's "impulse response" and "frequency response?"

How to distinguish between the different frequency domains?

Amplitude vs Frequency Response

Why is the Fourier transform so important?


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    $\begingroup$ I think this is a good question because it is not an especially intuitive concept $\endgroup$ – PAK-9 Oct 8 '13 at 17:02

If you have an understanding of Fourier transforms then you probably already have a conceptual model of transforming signals into the frequency domain. The Laplace transform provides an alternative frequency domain representation of the signal - usually referred to as the "S domain" to differentiate it from other frequency domain transforms (such as the Z transform - which is essentially a descretised equivalent of the Laplace transform).

What is the moment of a signal?

As you are no doubt aware the Laplace transform gives us a description of a signal from it's moments, similar to how the Fourier transform gives us a description from phase and amplitudes.

Broadly speaking a moment can be considered how a sample diverges from the mean value of a signal - the first moment is actually the mean, the second is the variance etc... (these are known collectively as "moments of a distribution")

Given our function F(t) we can calculate the n'th derivative at t=0 to give our n'th moment. Just as a signal can be described completely using phase and amplitude, it can be described completely by all of its derivatives.

Why is the fourier transform a special case of the laplace transform?

If we look at the bilateral laplace transform:


It should be quite apparent that a substitution $s=i\omega$ will yield the familiar Fourier transform equation:

$${\int_{-\infty}^\infty}e^{-i\omega t}f(t)dt$$

There are some notes about this relationship (http://en.wikipedia.org/wiki/Laplace_transform#Fourier_transform) but the mathematics should be quite transparent.

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    $\begingroup$ I don't see how the Laplace Transform is the "Description of a signal from its moments". I'd be happy to learn this outlook of the things. $\endgroup$ – Royi Oct 8 '13 at 18:07
  • $\begingroup$ Interesting, thank you for your answer! Especially the explanation about what a moment is was much more clarifying than what Ive read so far. How the integrals result in the S and the frequency domain is still opaque to me, but how the fourier is a subset of the laplace is more obvious now. Thanks $\endgroup$ – Leo Oct 8 '13 at 19:35

Why is the fourier transform a special case of the laplace transform?

The Laplace transform produces a 2D surface of complex values, while the Fourier transform produces a 1D line of complex values. The Fourier transform is what you get when you slice the Laplace transform along the jω axis. For instance, a simple lowpass filter $H(s)=\frac{1}{s+1}$ has a single pole in the S plane to the left of the origin:

S plane and other plots

Viewed from the side, the magnitude of this Laplace transform forms a surface, with the pole acting like a tent pole that raises the amplitude to infinity at that point (and an implied zero at infinity that drops the amplitude to zero the farther away from the origin you get in any direction):

tent pole

If you now take the value of the surface along the jω axis only, that's the Fourier transform. It's the red curve in the image above, which you can see forms a lowpass filter. If you moved the pole farther away from the origin, the tent would move in the same direction, and the slice along the jω axis would drop, both reducing the gain (which we compensate for by adding overall gain) and increasing the cutoff frequency. I've been meaning to make some animations of stuff like this...




The best intuitive description of Laplace transform I've ever seen:

At first glance, it would appear that the strategy of the Laplace transform is the same as the Fourier transform: correlate the time domain signal with a set of basis functions to decompose the waveform. Not true! Even though the mathematics is much the same, the rationale behind the two techniques is very different.

The Laplace transform can be viewed as probing the system's impulse response with various exponentially decaying sinusoids. Probing waveforms that produce a cancellation are called poles and zeros.

This allows us instead of describing frequency response for every $\omega$ use a small set of feature points that determine the behavior of a system in all other points (including the part of $s$-plane $s=j\omega$ which is a frequency response).

There's a nice analogy for this in a book:

Now, think about how you understand the relationship between elevation and distance along the train route, compared to that of the conductor. Since you have directly measured the elevation along the way, you can rightly claim that you know everything about the relationship. In comparison, the conductor knows this same complete information, but in a simpler and more intuitive form: the location of the hills and valleys that cause the dips and humps along the path. While your description of the signal might consist of thousands of individual measurements, the conductor's description of the signal will contain only a few parameters.

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    $\begingroup$ This is a useful link, but it would be great if you added some details about what it exactly is that you find intuitive in that document. Link-only answers are usually discouraged here. $\endgroup$ – Matt L. Mar 19 '15 at 10:21
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    $\begingroup$ Welcome to DSP.SE! The system has flagged this as a low quality answer. Please do as Matt L. suggests and summarize what the description is on the link. $\endgroup$ – Peter K. Mar 19 '15 at 11:37

In terms of linear algebra, functions can be represented as vector with infinite number of components (in the Hilbert's space as they say in physics or $L2$ space in mathematics). The set of function $e^{-st}$ with complex $s$ forms a vector space, and Fourier and Laplace transform X(s) = $\int_{a}^{ \infty} x(t) e^{-st} $ is a projection of $x(t)$ to $e^{-st}$, i.e. an inner product with basis function $e^{-st}$. Here $a= -\infty$ for Fourier and $0$ for Laplace.

Now in the case of Fourier, your basis function is $e^{-iwt}$ with real $w$ (sine and cosine of frequency $w$). Thus Fourier transform is inner product of $x(t)$ with the basis function of a single frequency $w$.

In the case of Laplace, basis functions are $e^{-(\sigma + iw)t}$, with real $\sigma$ and $w$, i.e. $e^{-\sigma t} (cos(wt) + i sin(wt))$, Laplace transform is like an inner product of $x(t)$ with the basis function of frequency $w$ and decays at the rate specified by $\sigma$.

The transform gives you a weight at $s$, and the inverse transform is a linear combination of all the basis function $e^{st}$ in which forward transforms are weights, gives you back $x(t)$.


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