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)?

2 Answers

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

Erdos Erdos

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.

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