# Scale of x-axis for Fourier transform

Consider a function $$f(t)$$ and its Fourier transform $$F(\omega)$$. The amplitude of the Fourier transform $$F(\omega)$$ depends on the frequency $$\omega$$ and thus also depends on the scale of the $$t$$-axis. For example, if $$f_1(t)$$ is a box function which is 1 between $$[0,10^9]$$ and $$f_2(t)$$ is a box function which is 1 between $$[0,1]$$, then the amplitude of $$F_1(\omega)$$ is $$10^9$$ times larger than the amplitude of $$F_2(\omega)$$.

However, numerically, this is not the case. If you have 100 points of the function $$f_1(t)$$ equally spaces between $$[0,10\times10^9]$$ and also 100 points of the function $$f_2(t)$$ equally spaced between $$[0,10]$$, then numerically the amplitude of the Fourier transform of both functions will be 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 $$t$$-axis scale is lost...

In this example of the Fourier transform of the rectangular function, you can solve this problem by multiplying the result with $$10^9$$ in the first case because the Fourier transform of the rectangular function is proportional with $$1/\omega$$. Now, 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 "x-axis" into account?

• I think I don't understand the question. Are you asking what (i) the magnitude of the Fourier transform should be theoretically, or (ii) in practice? I believe that you're saying that you know what it should be in theory, but that the practice deviates. In both cases, it would probably help if you provided details on how exactly you define the discrete Fourier transform you seem to be computing. – Wolfgang Bangerth Jul 25 '20 at 0:40