Fourier series - by Ray Betz

Fourier Series

If

1. $x(t)=x(t+T)$
2. Dirichlet conditions satisfied

then we can write

${\mathbf {x}}(t)=\sum _{k=-\infty }^{\infty }\alpha _{k}e^{\frac {j2\pi kt}{T}}$

The above equation is called the complex fourier series. Given $x(t)$, we may determine $\alpha _{k}$ by taking the inner product of $\alpha _{k}$ with $x(t)$. Let us assume a solution for $\alpha _{k}$ of the form $e^{\frac {j2\pi nt}{T}}$. Now we take the inner product of $\alpha _{k}$ with $x(t)$. $<\alpha _{k}|x(t)>=<e^{\frac {j2\pi nt}{T}}|\sum _{k=-\infty }^{\infty }\alpha _{k}e^{\frac {j2\pi kt}{T}}>$ $=\int _{-{\frac {T}{2}}}^{\frac {T}{2}}x(t)e^{\frac {-j2\pi nt}{T}}dt$ $=\int _{-{\frac {T}{2}}}^{\frac {T}{2}}\sum _{k=-\infty }^{\infty }\alpha _{k}e^{\frac {j2\pi kt}{T}}e^{\frac {-j2\pi nt}{T}}dt$ $=\sum _{k=-\infty }^{\infty }\alpha _{k}\int _{-{\frac {T}{2}}}^{\frac {T}{2}}e^{\frac {j2\pi (k-n)t}{T}}dt$

If $k=n$ then,

$\sum _{k=-\infty }^{\infty }\alpha _{k}\int _{-{\frac {T}{2}}}^{\frac {T}{2}}e^{\frac {j2\pi (k-n)t}{T}}dt=\int _{-{\frac {T}{2}}}^{\frac {T}{2}}1dt=T$

If $k\neq ;n$ then,

$\sum _{k=-\infty }^{\infty }\alpha _{k}\int _{-{\frac {T}{2}}}^{\frac {T}{2}}e^{\frac {j2\pi (k-n)t}{T}}dt=0$

We can simplify the above two conclusion into one equation.

$\sum _{k=-\infty }^{\infty }\alpha _{k}\int _{-{\frac {T}{2}}}^{\frac {T}{2}}e^{\frac {j2\pi (k-n)t}{T}}dt=\sum _{k=-\infty }^{\infty }T\delta _{k,n}\alpha _{k}=T\alpha _{n}$

So, we may conclude $\alpha _{n}={\frac {1}{T}}\int _{-{\frac {T}{2}}}^{\frac {T}{2}}x(t)e^{\frac {-j2\pi nt}{T}}dt$