0
$\begingroup$

Fourier transform $f(x) = \cos{x}$ , $x \in [-\pi,\pi]$

$f(x) = \begin{cases} \cos{x}, & \text{$x \in [-\pi,\pi]$} \\[2ex] 0, & \text{else} \end{cases}$

Well, after some calculations I got to

$\frac{1}{2\pi}(\frac{\sin(\pi-\pi w)}{1-w} + \frac{\sin(\pi+\pi w)}{1+w})$,

I don't know how to play with the numerator so I get to this:

$\frac{1}{2\pi}\cdot \frac{2w\sin(\pi w)}{1-w^2}$

Any help would be much appreciated.

  • 1
    Note that $\sin(\pi - \pi w) = \sin (\pi w)$ and $\sin(\pi + \pi w) = -\sin(\pi w)$ – 2017-01-13
  • 0
    @OpenBall Still didn't get to the answer. – 2017-01-13
  • 1
    Really?! ${}{}{}$ – 2017-01-13
  • 0
    @OpenBall Just kidding. I guess it's the same trick with $cos(\pi - \pi w) = cos(\pi w)$.. And thanks! :) – 2017-01-13

1 Answers 1

2

We can represent $f(x)$ by

$$f(x)=H(\pi-|x|)\cos(x)$$

where $H(x)$ is the Heaviside step function.

The Fourier Transform of $\cos(x)$ is

$$\mathcal{F}[\cos(x)](\omega)=\sqrt{\frac{\pi}{2}}\left(\delta(\omega-1)+\delta(\omega+1)\right)$$

and the Fourier Transform of $H(\pi-|x|)$ is easily calculated:

\begin{align} \mathcal{F}[H(\pi-|x|)](\omega)&=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty H(\pi-|x|)e^{-i\omega x}dx\\ &=\frac{1}{\sqrt{2\pi}}\int_{-\pi}^\pi e^{-i\omega x}dx\\ &=-\frac{2}{2i\omega\sqrt{2\pi}}\left[e^{-i\omega\pi}-e^{i\omega\pi}\right]\\ &=\sqrt{\frac{2}{\pi}}\frac{\sin(\pi\omega)}{\omega} \end{align}

Now, by the Convolution Theorem, the Fourier Transform of the product of two functions is given by the convolution of their Fourier Transforms:

\begin{align} \mathcal{F}[H(\pi-|x|)\cos(x)](\omega)&=\int_{-\infty}^\infty\left(\delta(s-\omega-1)+\delta(s-\omega+1)\right)\frac{\sin(\pi s)}{s}ds\\ &=\frac{\sin(\pi(\omega-1))}{\omega-1}+\frac{\sin(\pi(\omega+1))}{\omega+1} \end{align}