Shape of fourier transform

  • Thread starter Nikitin
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Hi! I am taking a second look on fourier transforms. While I am specifically asking about the shape of the fourier transform, I'd appreciate if you guys could also proof-read the question below as well, as I've written down allot of assumptions that I've gained, which might be wrong.

OK.

As far as I am aware, the nastier/"less smooth" (ie sharp, large and discontinious derivative and so on) a function ##f(x)## is (for ex. ##f(x) = \delta (x)##, the more "spread out" its fourier transform is. That is, the sines and cosines summed by the fourier-inverse integral get weighed increasingly equally irregardless of their frequencies, as ##f(x)## becomes less periodic.

However if you got a very spread out and slowly changing ##f(x)## (for ex. ##f(x)=1##), the fourier transform will be narrow around 0, meaning only the low-frequency sines and cosines in the fourier-inverse integral will dominate.

Why is this so? This applies to periodic functions as well, so let me rephrase the question in case you don't get me: Why can a "smooth and slow" function be described adequately with less terms in a fourier series, than a nasty one? Is it perhaps because the further you go out in a fourier series, the bigger the derivatives will be and thus these violent sines and cosines can adequately describe a swiftly changing function?
 

Answers and Replies

  • #2
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And can somebody tell me how to evaluate the infinite integral of the complex exponential function so I can get something representing the dirac delta? (I didn't want to open a new thread for this question alone as it's related to the OP, so don't murder me moderators).

I read that you can multiply it a converging term as a trick, like this, ##\lim_{\epsilon \to 0} \int_{-\infty}^{\infty} e^{-\epsilon x^2} \cdot e^{isx} dx##, but I don't remember how to evaluate a gaussian integral multiplied by another function.
 
  • #3
mathman
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And can somebody tell me how to evaluate the infinite integral of the complex exponential function so I can get something representing the dirac delta? (I didn't want to open a new thread for this question alone as it's related to the OP, so don't murder me moderators).

I read that you can multiply it a converging term as a trick, like this, ##\lim_{\epsilon \to 0} \int_{-\infty}^{\infty} e^{-\epsilon x^2} \cdot e^{isx} dx##, but I don't remember how to evaluate a gaussian integral multiplied by another function.
##\int_{-\infty}^{\infty} e^{-\epsilon x^2} \cdot cos(sx) dx=\sqrt{{\pi}/{\epsilon}}\cdot e^{{-s^2}/{4\epsilon}}##, from Gradshteyn and Ryzhik. Note that the imaginary part = 0.
 
  • #4
Why can a "smooth and slow" function be described adequately with less terms in a fourier series, than a nasty one? Is it perhaps because the further you go out in a fourier series, the bigger the derivatives will be and thus these violent sines and cosines can adequately describe a swiftly changing function?
You confuse two things: number (amount) of non-zero terms, and their frequencies (i.e. this [itex]n[/itex] in your [itex]\cos nx[/itex] and [itex]\sin nx[/itex] or, in the standard presentation, [itex]\exp(inx)[/itex]).

Also,
  • Do not use such terms as “nasty function”. Not clear.
  • Make distinction between Fourier transform of functions on ℝ and Fourier series on the circle (a.k.a. for periodic functions).
  • Learn to think in exponents, not ugly real trigonometry.
 

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