Fourier transform - application, integral calculation

**liquidFuzz** · November 15th 2011, 05:05 AM

I'm trying to brush the dust of my transform tinkering. (Have to use it in some electronic stuff.) I just did an exercise where I calculate the F-transform of an even function.
Let f(x) be:
$\displaystyle f(t)=\left\{\begin{matrix} 0 & x > 2\\ 2-x & 0 < x < 2 \end{matrix}\right.$
Find the even Fourier transform.

I calculated the transform with this formula
$\displaystyle f(\omega) = \int_{-\infty}^{\infty} f(t) e^{-i \omega t} dt = \int_{-\infty}^{\infty} f(t) (cos(\omega t) + i sin(\omega t))dt$

Knowing that the function is even I write:
$\displaystyle f(\omega) = 2 \int_{0}^{\infty} (2-t) cos(\omega t) dt$

A bit integration gives, don't hesitate to call out errors.
$\displaystyle f(\omega) = \frac{4sin^{2}w}{w^2}$

Now as an application to this calculation I'm supposed to determine this integral.
$\displaystyle I = \int_0^{\infty} \frac{sin^{2}w}{w^2}$

This is where I like to get some help... I simply don't know how to do it. Calculating the integral, in an other way, I get $\displaystyle \frac{\pi}{2}$

**Opalg** · November 15th 2011, 08:06 AM

Originally Posted by liquidFuzz

I'm trying to brush the dust of my transform tinkering. (Have to use it in some electronic stuff.) I just did an exercise where I calculate the F-transform of an even function.
Let f(x) be:
$\displaystyle f(t)=\left\{\begin{matrix} 0 & x > 2\\ 2-x & 0 < x < 2 \end{matrix}\right.$
Find the even Fourier transform.

I calculated the transform with this formula
$\displaystyle f(\omega) = \int_{-\infty}^{\infty} f(t) e^{-i \omega t} dt = \int_{-\infty}^{\infty} f(t) (cos(\omega t) + i sin(\omega t))dt$

Knowing that the function is even I write:
$\displaystyle f(\omega) = 2 \int_{0}^{\infty} (2-t) cos(\omega t) dt$

A bit integration gives, don't hesitate to call out errors.
$\displaystyle f(\omega) = \frac{4sin^{2}w}{w^2}$

Now as an application to this calculation I'm supposed to determine this integral.
$\displaystyle I = \int_0^{\infty} \frac{sin^{2}w}{w^2}$

This is where I like to get some help... I simply don't know how to do it. Calculating the integral, in an other way, I get $\displaystyle \frac{\pi}{2}$

The notation for the Fourier transform of a function should be different from the notation for the function itself. So you should write $\hat{f}(\omega)$ rather than $f(\omega).$

The Fourier inversion theorem says that you can retrieve $f(t)$ from $\hat{f}(\omega)$ by the formula $f(t) = \frac1{2\pi}\int_{-\infty}^\infty \hat{f}(\omega)e^{it\omega}d\omega.$ Put t=0 in that formula and you will get the result you are looking for.

**liquidFuzz** · November 15th 2011, 09:12 AM

Thanks! For slapping my fingers, and explaining how to do this.

**liquidFuzz** · November 16th 2011, 11:40 AM

How about this..?

Assume that $\hat{f}(w)$ converges to f(x), $\hat{f}(w) = f(x)$ , and that the Fourier transform is correct, $\hat{f}(w) = \frac{4 sin^2 w}{w^2}$

Taking a arbitrary point on f(x)
$f(0) = 1 = \lim_{a \to \infty} \frac{1}{2 \pi} \int_0^a \frac{sin^2 x}{x^2} dx$

Gives
$1 = \frac{1}{2 \pi} \int_0^\infty \frac{4sin^2 x}{x^2} dx$
$\drarrow \frac{\pi}{2} = \int_0^\infty \frac{sin^2 x}{x^2} dx$

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