With Fourier you can analyze oscillatory characteristics of function (frequency and phase). With Laplace you can also analyze amplification/attenuation.

The Fourier transform, well, transforms a time-based phenomenon such as an alternating current sine wave into a frequency spectrum where you can observe the frequency spectrum components of the signal. A pure nice sine becomes a spike (delta function) located at a specific frequency. Music, as we observe it through our ears and can view it on an oscilloscope becomes moving spikes (lots of them :) in the frequency spectrum where the "amplitude" at a given frequency relates to the "amount" of that frequency in the music.

The behaviour of a filter is much easier to describe in the frequency spectral domain than it would be in the time domain.

Now to the direct current (DC) view. This cannot be handled by the Fourier transform -- at least the DC-part of the signal cannot be transformed to the frequency domain. As shown in the article, there were "steps", "ramps" and such. A typical scenario would be to describe what happens in your amplifier during startup, to describe how electrical circuits are behaving during startup before reaching the "running" state.

The Laplace transform will handle these types of scenarios, and can thus be used to study (or describe) systems during other types of transitions than the "steady state" when you are up and running.

Regarding filters, the Fourier transform describes things going on at the unit circle, while the Laplace transform can be used to study both the interior and exterior of the plane. In this sense, creating filters relates to locate "poles" and "zeros" in the plane (amplification and attenuation) which can be observed on the unit circle as the behaviour on periodic signals.

Why can DC not get fouriertransformed? Usually the f(p=0) is the DC component.

I recall from some undergraduate classes I took (in mechanical engineering) that you can "convert" between a Laplace transform and a Fourier transform by saying that s = i * omega. This has some appeal from a purely algebraic standpoint but doesn't seem rigorous to me. For one, the limits of the integral aren't the same. How valid is this? I always assumed it was an approximation.

Is the Laplace transform in some sense similar to a one-sided/semi-infinite Fourier transform, provided that change of variables is made?

Years ago in a complex analysis class I worked out the contour integration for a few Fourier transforms as I recall, but I've had no similar training for the Laplace transform and have forgotten many details.

Yeah, they're only equal when your f(t) = 0 for all t < 0. Otherwise they can be quite different, because the integral limits differ. In an undergrad engineering context you're often evaluating the response of a system to some input and it's common to have "at t=0 the switch is closed/mass is released/etc." where the function is assumed to have been 0 previously.

Seems obvious in retrospect but wasn't obvious to me. Thanks for the insight!

For a little more intuition about why you can do that replacement: laplace is a representation of a system as a sum of sinusoids * exponentials, which are your two axis in the laplace plane. Frequency on the iw axis and exponential on the a axis. If you think of that replacement as s = iw + a | a = 0, you'll see the exponential terms go away and you're left with just the sinusoidal parts:

  f(t) * e^(iw + 0)t 
  = f(t) * e^(iwt) * e(at) 
  = f(t) * e(iwt) * e^(0t) 
  = f(t) * e(iwt) 

integrated over time, which is your fourier transform subject to the condition above. It's just the laplace transform along the Y axis, or, the frequency response at steady state when not growing/decaying exponentially.

I believe you're confusing discrete and continuous time, Fourier in continuous time is Laplace evaluated along s = jw or the vertical axis and not e^jw, the unit circle

You are correct. I mixed how I tend to visualize and think of them with what actually happens....