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Fourier transform spectroscopy

  1. Jun 9, 2009 #1
    Hey guys,

    There is something I have known and applied for a long time, that the greater the length of an interferogram the greater the resolution of the resulting frequency-domain spectrum. But I've never fully understood why, I've always waved it off as something to do with the uncertainty principle because this is what I was told many years ago in secondary school. I've read around and I can't really find a good explanation.

    It'd be great if somebody here knew. Thanks :)

    Regards,

    Chris.
     
  2. jcsd
  3. Jun 9, 2009 #2
    This is easy but it is the other way around. The uncertainty principle derives from the property of the Fourier transform that you describe. It basically works like this: If you have an interferogram of a certain length, then you can describe this as an infinite interferogram multiplied with a function that is 1 on the finite domain, and 0 outside. But the convolution theorem states that the Fourier transform of a product of two functions is the convolution of both functions Fourier transforms. The Fourier transform of the rectangular function can be calculated, and it gets sharper and sharper the wider the rectangle gets. This makes the resolution get sharper and sharper. Maybe look up Fourier transform, convolution theorem and rectangular aperture.

    It is also a general property of a function's Fourier transform to get thinner the wider the function gets even when they are not cut of at the ends, but this requires higher mathematics, I think it is linked to [tex]\ell^p[/tex]-spaces
     
  4. Jun 9, 2009 #3

    DrGreg

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    A less technical way of saying the same thing as 0xDEADBEEF it that it's quite easy to tell the difference between 324 and 325 wavelengths (in a long interval) but it's a lot harder to tell the difference between 3.24 and 3.25 wavelengths (in a 100 times shorter interval), especially if they are corrupted by noise.
     
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