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

  • #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.
Structured illumination is one such method where the object is illuminated with a sinusoidal pattern instead of the conventional uniform illumination. The Fourier transform of the intensity of a sinusoid is three impulses—one at the origin and the other two at the positive and negative spatial frequency of the sinusoid. Therefore, when a sinusoidal illumination is incident on an object, the Fourier transform of the image consists of three replicas of the object Fourier transform, each centered at one of the three impulses.
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).
\begin{align*}
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)]
\end{align*}
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?
 
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Answers and Replies

  • #2
mathman
Science Advisor
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An aside: the argument for the exp function should have "i" in it. [itex]2\pi ift[/itex] etc.
 
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  • #3
An aside: the argument for the exp function should have "i" in it. [itex]2\pi ift[/itex] etc.
You're right. I will fix that.
 
  • #4
jasonRF
Science Advisor
Gold Member
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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
[tex]
\cos^2(x) = \frac{1}{2}(1 + \cos(2x))
[/tex]
 
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  • #5
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
[tex]
\cos^2(x) = \frac{1}{2}(1 + \cos(2x))
[/tex]
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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