Consider the Fourier transform $F(\omega)$ of the function $f(t)$. The magnitude of $F(\omega)$ depends on $\omega$ and thus also depends on the scale of the $t$-axis. For example, when $f_1(t)$ is a box function which is 1 between $[0,1]$ s and $f_2(t)$ is a box function which is 1 between $[0,1]$ ns = $[0,10^{-9}]$ s, then the amplitude of $F_1(\omega)$ is $10^9$ times larger than the amplitude of $F_2(\omega)$. This is because the time scale is reduced by a factor $10^9$ in the second case.

However, no difference would be obtained when you do a numerical Fourier transform. If you have 100 points of the function $f_1(t)$ equally spaced between $[0,10]$ and also 100 points of the function $f_2(t)$ equally spaced between $[0,10\times10^{-9}]$, then numerically the amplitude of the Fourier transform of both functions is the same (which is not correct). Numerically, you only consider the Fourier transform of the "y-values" $f(t_i)$, so the information about the time scale is lost...

You can solve this problem by multiplying the result with $10^{-9}$ in the second case because the Fourier transform of the rectangular function is proportional with $1/\omega$. However, in general, this proportionality is not present... How to solve this problem for a general function $F(\omega)$? For example, when the Fourier transform is proportional to $e^\omega$ or proportional to $1/\omega^2$ or ...? Is there a numerical approach to take the scale of the time axis into account?

  • $\begingroup$ You can approximate the continuous-time Fourier transform (CTFT) by a discrete Fourier transform (DFT) including scaling: cf. this answer. You just need to scale the DFT by the sampling interval. $\endgroup$
    – Matt L.
    Jul 24, 2020 at 12:36
  • $\begingroup$ Maybe this will help: dsp.stackexchange.com/questions/69186/… $\endgroup$ Jul 24, 2020 at 12:51
  • $\begingroup$ Is this scaling by the sampling interval valid for any kind of function f(t)? What if $f(t)=e^{i\omega_0 t}$? $\endgroup$
    – Frederic
    Jul 24, 2020 at 12:51
  • 1
    $\begingroup$ @Frederic: Yes, it's valid for any kind of function, but the general problem with using the DFT to approximate the CTFT remains: apart from the aliasing error, you'll have a truncation error because you can only process a finite time window of a theoretically infinitely long function. $\endgroup$
    – Matt L.
    Jul 24, 2020 at 16:06

1 Answer 1


Most of the classical linear transformations (even filtering) may have three main types of scalings:

  1. natural or none: sum or integral does not have an explicit scaling factor (but the inverse may need one)
  2. in amplitude: because of linearity, at least one reference "unit" signal should have a unit amplitude in the transformed domain, for an easy read of graphs. Typically a unit amplitude sine (which is not $L_1$-integrable) should have a one amplitude at its frequency in the Fourier domain
  3. in energy: to have constant energy in the original or the transformed domain.

Each of these options (sometimes, two of them are equal) depends on the purpose. Fourier has a specific relation to scale (Where in “Discrete Time Signal Processing” (Oppenheim et al.) can I find the scale-change theorem?):

$$ s(\alpha t) \mapsto \frac{1}{|\alpha|} S \left( \frac{f}{|\alpha|}\right) $$

Obtaining scale-, shift- or orientation-invariant features is a long-lasting issue in DSP, and is still a concern in machine intelligence or deep learing. Along the Fourier line, you can look at the Fourier-Mellin transform or scale represensation (by L. Cohen), and the many feature representations like SIFT, SURF, ORB, BRIEF: Image Matching Using SIFT, SURF, BRIEF and ORB: Performance Comparison for Distorted


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.