# Inverse Fourier transform of a step-function

If the Fourier transform of a function h(t) is the step-function

$H(k) = \left\{\begin{matrix} 1, \;\;\; 0 \le k \le 6 \\ 0, \;\;\;\;\;\;\;\;\;\;\;\;\; \text{else}\end{matrix}\right.$

How can I find the original function h(t)?

The function $h(t)$ will be the inverse Fourier trasform of $H(k)$, i.e.
$h(t)=\int_{\infty}^{\infty} H(k)e^{2\pi i t k}dk=\int_0^6 e^{2 \pi i tk} dk$
$=\frac{e^{2 \pi itk}}{2\pi i t}\big |_{0}^{6}=\frac{1}{2\pi i t} (e^{12\pi i t}-e^{0})$
$-\frac{i}{2\pi t} (e^{12\pi i t}-1)=-\frac{i}{2\pi t} (\cos (12 \pi t)+i \sin (12 \pi t)-1)$

I am using the definiton of the Fourier transform in https://en.wikipedia.org/wiki/Fourier_transform

Erdos
4.7K
• I see, thank you :)

You take the inverse fourier transformation, which is roughly the same up to the sign in the exponential and possibly a factor of $1/2\pi$ in fromt of the integral ; this depends on the conventions used. The integral will be easy to compute, it's just an exponential integrated fr om 0 to 6.

• Thank you :)

Join Matchmaticians Affiliate Marketing Program to earn up to a 50% commission on every question that your affiliated users ask or answer.