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Fourier transform of a triangle function

  1. Mar 20, 2013 #1
    1. The problem statement, all variables and given/known data

    Hello

    I'm learning Fourier transforms via the Stanford lecture series on Youtube. In the 6th lecture, the professor claims that the FT of a triangle function is the square of the sinc function. I'm trying to derive this, but I can't get my math to work out. Could someone please help me figure out where I went wrong?

    2. Relevant equations

    FT(f(t)) = ∫exp(i2πst)*f(t) dt

    sin(t) = [exp(it) - exp(-it)] / 2i

    3. The attempt at a solution

    edit: Please let me know if the images below aren't legible on your monitor. If so, I will increase the resolution.

    http://imageshack.us/a/img716/9334/fourierpage1.png [Broken]

    http://imageshack.us/a/img46/6585/fourierpage2.png [Broken]
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Mar 20, 2013 #2
    This can be done by the convolution theorem.

    [itex]\mathcal{F} \left[ f * g\right] = \mathcal{F} \left[f \right] \mathcal{F} \left[g \right][/itex]

    So, if sinc^2(ω) corresponds to a triangle function, then a triangle function would be the convolution of the inverse Fourier transform of sinc with itself. The inverse Fourier transform of a sinc is a rectangle function. So, all you need to do is show a triangle function is the convolution of a rectangle function with itself.
     
  4. Mar 20, 2013 #3
    Thanks for your help. Yes, I can see that it can be done via the convolution theorem, and the convolution of a rectangle with itself being a triangle is intuitively clear, but shouldn't this be a relatively simple transformation to perform directly? It's making me very frustrated that I can't get it right.
     
  5. Mar 21, 2013 #4

    jbunniii

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    The triangle function is even, so the Fourier transform takes a simpler form:
    $$\begin{align}
    \int_{-1}^{1}f(t) e^{i\omega t} dt
    &= \int_{-1}^{1} f(t) \cos(\omega t) dt + i \int_{-1}^{1} f(t) \sin(\omega t) dt\\
    &= 2 \int_{0}^{1} f(t) \cos(\omega t) dt + 0\\
    &= 2 \int_{0}^{1} f(t) \cos(\omega t) dt
    \end{align}$$
    This is nice because we only have to deal with one integral. Now ##f(t) = 1-t## in the interval ##[0,1]##, so the Fourier transform becomes
    $$2\int_{0}^{1} \cos(\omega t) dt - 2\int_{0}^{1} t \cos(\omega t) dt$$
    Try using that as a starting point. It should make your life easier.
     
  6. Apr 3, 2013 #5
    Thanks to everyone for the help. I had to take a hiatus due to exams, but now I'm trying to figure it out again. I'm using jbunnii's tip to set it up, but I still can't get the correct answer. I've checked over my math three times and can't see what I'm doing wrong. I'm getting the ω^2 in the denominator, but I don't see how the integration can produce a sin^2 in the numerator. Could someone please help me out one more time?

    http://img600.imageshack.us/img600/2717/spage001.jpg [Broken]

    http://imageshack.us/a/img683/5251/spage002.jpg [Broken]
     
    Last edited by a moderator: May 6, 2017
  7. Apr 4, 2013 #6

    jbunniii

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    I think all your work is fine. To get the result into the desired form, try using this trig identity:
    $$\sin^2(x) = \frac{1 - \cos(2x)}{2}$$
     
  8. Apr 4, 2013 #7
    jbunnii, thank you very much for your help! It's a relief that I haven't forgotten my basic calculus. I've always been horrible at trig identities; maybe this lecture series will finally bring me up to par. I have one quick question remaining:

    In order to apply the identity above, I had to make the substitution x = ω/2. I ended up with FT = sinc^2(x), or equivalently FT = sinc^2(ω/2). I'm trying to understand the implications of this substitution.

    ω = 2 π s, where s = k/λ. k is the integer frequency multiplier of the wave, and λ is the wavelength. For this non-periodic function, λ = ∞; therefore, s is infinitesimal. Is it valid to make the argument that since ω is 2π * an infinitesimal, x = (2π)/2 * an infinitesimal, so x = ω and therefore FT = sinc^2(ω)?
     
  9. Apr 4, 2013 #8

    jbunniii

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    I'm not sure what you are trying to argue here.

    It is true that (reasonably well behaved) periodic functions can be represented by a Fourier series, which involves terms with ##\omega = 2\pi k /\lambda## where ##k## is any integer.

    For a non-periodic function, we need a "continuum" of ##\omega## values. But outside of nonstandard analysis, we don't talk about infinitesimals as they are not well defined.

    If you are trying to work out whether the answer should be ##\text{sinc}^2(\omega/2)## or ##\text{sinc}^2(\omega)##, that depends on how ##\text{sinc}## is defined. This doesn't appear to be completely standardized. According to Wikipedia, there are at least two definitions in common use:
    $$\text{sinc}(x) = \frac{\sin(x)}{x}$$
    and
    $$\text{sinc}(x) = \frac{\sin(\pi x)}{\pi x}$$
    It would not surprise me if there are other conventions as well. Similarly, there is no universal consensus regarding the definition of the Fourier transform. Some authors move the various ##2\pi## factors around, or invert the signs on the complex exponential, etc. Best to check your textbook or instructor's notes to see what definition you should be using. Or, to be on the safe side, just leave the answer in the form ##\sin^2(\omega/2)/(\omega/2)^2##.
     
  10. Apr 4, 2013 #9
    Thank you again. I was trying to reconcile my result with the professor's assertion that the Fourier transform will be "sinc squared". Thank you very much for your help; I really appreciate it.
     
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