# Time scale and Fourier transform

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?

• 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. – Matt L. Jul 24 at 12:36
• Maybe this will help: dsp.stackexchange.com/questions/69186/… – Cedron Dawg Jul 24 at 12:51
• Is this scaling by the sampling interval valid for any kind of function f(t)? What if $f(t)=e^{i\omega_0 t}$? – Frederic Jul 24 at 12:51
• @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. – Matt L. Jul 24 at 16:06

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