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I'm sure this is a question that many other people have, so here's the answer.

I asked Mimi this some time ago, and essentially, the Fourier transform is a special case of the Laplace transform. If you recall, the Laplace transform uses the variable 's', which equals a + jb (for a,b real), and thus has both a real and imaginary part. The key difference is that the Fourier transform has no real part, it is purely imaginary, so the 'a' would be zero.

As for the mathematical definitions of each, the limits on the Fourier transform go from negative infinity to positive infinity, while for the (one-sided) Laplace transform, they go from zero to positive infinity.

Laplace (One-Sided)

$\mathfrak {L}(s)=\int_{0}^{\infty} f(t)e^{-st} dt$

Fourier

$\mathfrak {F}(\omega)=\int_{-\infty}^{\infty} f(t)e^{-j{\omega}t} dt$

Thanks --heather.r.barrett.1, Fri, 26 Oct 2007 17:01:45

I was wondering about that, too. Now I can sleep at night again. =P

• The easiest way to think about this is that the Fourier transform is the Laplace transform evaluated at $s = j \omega\$. Any signal that has a Fourier transform has a Laplace transform, but not conversely. When evaluating the Region of Convergence (ROC) of the Laplace transform, the signal has a Fourier transform if:
• CT: The ROC includes the imaginary axis, x = 0, in the complex plane.
• DT: The ROC includes the unit circle, z = 1, in the complex plane.

## Alumni Liaison

Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.