Fourier Series/ fourier transform

Question

What is the Fourier series of the following piece-wise function?

$$ f(x) = \begin{cases} 0 & -1 \leq x < -0.5 \\ \cos (3 \pi x) & -0.5 < x < 0.5 \\ 0 & 0.5 \leq x < 1 \end{cases} $$

Answer 1

Define problem

Piecewise function: Resolve $f(x)$ into a left, center, and right piece

$$ \begin{align} % l(x) &= 0, \qquad \qquad \, -1 \le x < -\frac{1}{2} \\ % c(x) &= \cos \left( 3\pi x \right) \quad\ -\frac{1}{2} \le x \le \frac{1}{2} \\ % r(x) &= 0, \qquad \qquad \ \ \ \frac{1}{2} \le x \le 1 % \end{align} $$ The length of the domain is $2$.

Find the Fourier expansion $$ f(x) = \frac{1}{2}a_{0} + \sum_{k=1}^{\infty} \left( a_{k} \cos \left( k \pi x \right) + \color{gray}{b_{k} \sin \left( k \pi x \right)} \right) $$ where the amplitudes are given by

$$ \begin{align} % a_{0} &= \int_{-1}^{1} f(x) dx \\ % a_{k} &= \int_{-1}^{1} f(x) \cos \left( k \pi x \right) dx \\ % \color{gray}{b_{k}} &= \color{gray}{\int_{-1}^{1} f(x) \sin \left( k \pi x \right) dx} \\ % \end{align} $$

Basic integrals

Center piece

$$ \begin{align} % \int_{-\frac{1}{2}}^{\frac{1}{2}} c(x) dx &= -\frac{2}{3 \pi } \\ % \int_{-\frac{1}{2}}^{\frac{1}{2}} c(x) \sin \left( k \pi x \right) dx &= 0 \\ \end{align} $$

For $m\ne n$, $$ \int \cos (m \pi x) \cos (n \pi x) \, dx = \left( 2\pi \right)^{-1} \left( \frac{\sin (\pi (m-n) x )}{m-n}+\frac{\sin (\pi (m+n) x )}{m+n} \right) $$

$$ \int \cos (3 \pi x) \cos (3 \pi x) \, dx = \frac{x}{2}+\frac{\sin (6 \pi x)}{12 \pi } $$

Amplitudes

$$ \begin{align} a_{0} &= -\frac{2}{3 \pi }, \\ a_{3} &= \frac{1}{2}. \end{align} $$

For $k=1,2,\dots..$ $$ a_{2k} = \frac{6 \cos \left( \pi k \right)}{\pi \left((2k)^2-9\right)} $$

Convergence sequence

The parameter $n$ represents the highest frequency term in the series, $$ f(x) = \frac{1}{2}a_{0} + \sum_{k=1}^{n} a_{k} \cos \left( \frac{k \pi x}{2} \right) $$