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I Why is the Fourier transform of a sinusoid assumed as this?

  1. Jun 28, 2016 #1
    Hello everyone.

    I'm trying to better understand structured illumination microscopy and in the literature, I keep coming across bits of text like this.
    Source: http://www.optics.rochester.edu/workgroups/fienup/PUBLICATIONS/SAS_JOSAA09_PhShiftEstSupRes.pdf

    From Fourier analysis, if I take the Fourier transform ##X(f)## of a time-varying function ##x(t)## that is a cosine, I get a pair of delta functions (quick derivation below).
    X(f) & = \int_{-\infty}^{\infty} x(t)\exp(-i2\pi f t)\; dt\\
    x(t) &= \cos(2 \pi f_0 t) = \frac{1}{2}\left[\exp(i2\pi f_0 t) + \exp(-i2\pi f_0 t)\right]\\
    X(f) &= \int_{-\infty}^{\infty} \frac{1}{2}\left[\exp(i2\pi f_0 t) + \exp(-i2\pi f_0 t)\right]\exp(-i2\pi f t)\; dt\\
    &= \frac{1}{2}[\int_{-\infty}^{\infty} \exp(i2\pi f_0 t)\exp(-i2\pi f t)\; dt + \int_{-\infty}^{\infty} \exp(-i2\pi f_0 t)\exp(-i2\pi f t)\; dt\\
    &= \frac{1}{2}[\mathcal{F}\left\{\exp(i2\pi f_0 t)\right\} + \mathcal{F}\left\{\exp(-i2\pi f_0 t)\right\}]\\
    \mathcal{F}\left\{\exp(i2\pi f_0 t)\right\} &= \delta(f-f_0)\\
    \mathcal{F}\left\{\exp(-i2\pi f_0 t)\right\} &= \delta(f+f_0)\\
    \therefore X(f) &= \frac{1}{2}[\delta(f-f_0) + \delta(f+f_0)]
    But the paper says that I should be getting three impulses. One at the origin, and the two I have detailed above. Where does the one at the origin come from?

    My only hunch so far is it might have something to do with the fact that this derivation I just did was in 1D, and what they are describing is in 2D (a surface). Of course, the other minor difference is that they are describing spatial frequency and I am describing temporal frequency, but replacing ##t## with ##x## and ##f## with a scaled ##k## (spatial frequency) isn't a big deal.

    Any tips?
    Last edited: Jun 28, 2016
  2. jcsd
  3. Jun 28, 2016 #2


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    An aside: the argument for the exp function should have "i" in it. [itex]2\pi ift[/itex] etc.
  4. Jun 28, 2016 #3
    You're right. I will fix that.
  5. Jun 28, 2016 #4


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    Gold Member

    The word "intensity" usually means the square of the signal - from a quick glance at the paper they are indeed considering the square. Look at equation (1) in the paper, and note
    \cos^2(x) = \frac{1}{2}(1 + \cos(2x))
  6. Jun 28, 2016 #5
    That's exactly it. The "DC" term comes from the square-to-double-argument trigonometric identity. Got it. Thank you very much! The rest of it is straightforward. :)
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