10/02 - Fourier Series

Fourier Series (as compared to vectors)

If a function is periodic, $x(t)=x(t+T)\,\!$, and it meets the Dirichlet conditions, then we can write it as $x(t)=\sum _{n=-\infty }^{\infty }\alpha _{n}e^{j2\pi nt/T}$

• Dirichlet conditions
  • x(t) must have a finite number of extrema in any given interval
  • x(t) must have a finite number of discontinuities in any given interval
  • x(t) must be absolutely integrable over a period
  • x(t) must be bounded

Like vectors we can change to a new basis function by taking the inner product of $x(t)\,\!$ with the mth basis function. Don't forget that the inner product of two vectors requires a complex conjugate. ${\vec {a}}\cdot {\vec {b}}=\sum _{i=1}^{n}a_{i}\cdot b_{i}^{*}$

$x(t)\cdot e^{j2\pi mt/T}$	$=\int _{-T/2}^{T/2}x(t)e^{-j2\pi mt/T}\,dt$
	$=\int _{-T/2}^{T/2}\left(\sum _{n=-\infty }^{\infty }\alpha _{n}e^{-j2\pi nt/T}\right)e^{j2\pi mt/T}\,dt$
	$=\sum _{n=-\infty }^{\infty }\alpha _{n}\int _{-T/2}^{T/2}e^{j2\pi nt/T}e^{-j2\pi mt/T}\,dt$
	$=\sum _{n=-\infty }^{\infty }\alpha _{n}\int _{-T/2}^{T/2}e^{j2\pi (n-m)t/T}\,dt$
	$=\sum _{n=-\infty }^{\infty }\alpha _{n}{\begin{Bmatrix}T,&n=m\\0,&n\neq m\end{Bmatrix}}$
	$=\sum _{n=-\infty }^{\infty }\alpha _{n}T\delta (n-m)$
	$=T\sum _{n=-\infty }^{\infty }\alpha _{n}\delta (n-m)$
	$=T\alpha _{m}\,\!$

Sticking this back into the top equation gives

$\alpha _{m}={\frac {1}{T}}\int _{-T/2}^{T/2}x(t)e^{-j2\pi mt/T}\,dt$

Notes

• The integral range reflects the period T as defined at the top of the page. The current range carries over to the Fourier series better than going from 0 to T.

• When switching the order of integrals and summations, you can "cavalierly" switch the order as long as there aren't summations/integrals to infinity.

• To see ${\begin{Bmatrix}T,&n=m\\0,&n\neq m\end{Bmatrix}}$ remember Euler's identities for sine and cosine. Since we are integrating the sine and cosine waves over a single period (assuming $m\neq n$), the value integrates to 0.

$\cos(x)={\frac {e^{ix}+e^{-ix}}{2}}$

$\sin(x)={\frac {e^{ix}-e^{-ix}}{2i}}$