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Hi : I'm reading an introductory book on Fourier transforms. After explaining the forward and inverse transformation clearly, the author then states:

" We realize the dual character of the forward and inverse transformations:

a very slowly varying function will have a very high spectral density for very small frequencies; the spectral density will go down quickly and rapidly approaches zero. Conversely, a quickly varying function f(t) will show spectral density over a very wide frequency range. "

Then some figures are shown will illustrate the statement above. What I can't understand is why this is so mathematically. It seems obvious to the author but not to me. If someone understands it and can explain it, the expressions the author uses for the forward and inverse transformations are the following:

$F(\omega) = \int_{-\infty}^{\infty} f(t) e^{-i\omega t} dt $

$f(t) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} F(\omega) e^{i\omega t}$

Thank you very much for any intuition regarding the mathematics behind the statement.

mark leeds
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  • A guess that is less precise than hcabral's answer, but might be informative: Informally, think of a slowly varying function as one in which long wavelengths dominate, and short wavelengths have little effect. This is what the text describes. When a function varies significantly "quickly", that means that short wavelength patterns must be having a large effect, as well as or instead of long wavelengths. The "as well as" variant is what the text describes. – Mars Feb 08 '19 at 04:35
  • @Mars: Thanks. That helps also. – mark leeds Feb 09 '19 at 15:58

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You have to understand that the term "very slowly varying function" is not very meaningful unless compared to another function, as in "this functions varies slower than that one".

If you restrict yourself to comparing two functions $f(t)$ and $g(t)$ where the second is just a time-scaled version of the first (i.e., $g(t) = f(\alpha t)$), it is quite easy to see the effect in the spectral density. You can quickly show that

$\alpha G(\omega) = \alpha \int_{-\infty}^\infty g(t)e^{-i\omega t}dt = \int_{-\infty}^\infty f(\alpha t)e^{-i\omega t}d(\alpha t) = F(\omega /\alpha),$

that is, if $g(t)$ varies faster that $f(t)$, i.e., $\alpha > 1$, then its spectral density $G(\omega)$ spreads out more in frequency than $F(\omega)$.

If $f(t)$ and $g(t)$ are not related, you can probably use an approximation by stepwise functions and apply the definition of Fourier Transform to convince yourself of the validity of the statement. Specifically, assuming $f(t)$ varies slower than $g(t)$, a stepwise approximation to $f(t)$ will have longer intervals than a similar approximation (one with the max error similar to the first approximation) to $g(t)$. Using these approximations and the fact that

$\int_A^B e^{i\omega t}dt \approx 0\quad\textrm{if}\ \ \omega\gg 2\pi/(B-A)$

we can see that $F(\omega)$ goes down quicker than $G(\omega)$.

hcabral
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