Comment by btrettel | Hacker News Reader

Comment by btrettel | Hacker News Reader

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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.

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