Solve Fourier Transform: f(t)=sin(2πt)/t

[lucad93]

I don't know if it is the right section to post in. I have a problem with a "simple" Fourier transform. This is the function to transform: \(\displaystyle f(t)=\frac{\sin\left({2\pi t}\right)}{t}\). My first idea was to write that as \(\displaystyle \sin\left({2\pi t}\right)\cdot\frac{1}{t}\) but then my fantasy crashed against a wall not finding the right transform for \(\displaystyle \frac{1}{t}\). Can you help me please?

 

[Joppy]

I don't know if it is the right section to post in. I have a problem with a "simple" Fourier transform. This is the function to transform: \(\displaystyle f(t)=\frac{\sin\left({2\pi t}\right)}{t}\). My first idea was to write that as \(\displaystyle \sin\left({2\pi t}\right)\cdot\frac{1}{t}\) but then my fantasy crashed against a wall not finding the right transform for \(\displaystyle \frac{1}{t}\). Can you help me please?

The following may be useful,

\(\displaystyle \frac{sin(2\pi t)}{t} = 2\pi sinc(2\pi t)\).

and,

\(\displaystyle \int_{-\infty}^{\infty} \,2\pi sinc(2\pi t) = \pi\).

Also, if \(\displaystyle f(t) = f(-t)\) and \(\displaystyle f: \Bbb{R}\rightarrow \Bbb{R}\) then \(\displaystyle \mathbb{I}\mathbb{m}({{F}_{n}}) = 0\)

EDIT: I forgot to add that this function is not periodic (\(\displaystyle f(t) \ne f(t+T)\)). We can deal with this by taking the limit as \(\displaystyle T \rightarrow \infty\) or equivalently, the limit as \(\displaystyle f \rightarrow 0\). This is because, very loosely speaking, the function is periodic with \(\displaystyle \infty\).

Did you manage to solve this?

 

[Opalg]

I don't know if it is the right section to post in. I have a problem with a "simple" Fourier transform. This is the function to transform: \(\displaystyle f(t)=\frac{\sin\left({2\pi t}\right)}{t}\). My first idea was to write that as \(\displaystyle \sin\left({2\pi t}\right)\cdot\frac{1}{t}\) but then my fantasy crashed against a wall not finding the right transform for \(\displaystyle \frac{1}{t}\). Can you help me please?

This is not an easy problem, but there is a "back door" way to sneak up on the answer.

Start with the "rectangular" function \(\displaystyle g(x) = \begin{cases}1&\text{if }|x|\leqslant 1, \\ 0&\text{otherwise.} \end{cases}\) Its Fourier transform is given by $$\hat{g}(t) = \int_{\mathbb R}g(x)e^{-itx}dx = \int_{-1}^1e^{-itx}dx = \Bigl[\frac{e^{-itx}}{-it}\Bigr]_{-1}^1 = \frac{e^{-it} - e^{it}}{-it} = \frac{2\sin t}t.$$ So the Fourier transform of the rectangular function is your function (give or take a couple of constants). You can now apply the Fourier inversion theorem to deduce that the Fourier transform of your function is a rectiangular function.