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 $x(t)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{\alpha }_n{e}^{j2\pi nt/T}$

• 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. $\overrightarrow{a}\cdot \overrightarrow{b}=\sum _{i=1}^n{a}_i\cdot b_i^{*}$

$x(t)\cdot e^{j2\pi mt/T}$	$=\underset{-T/2}{\overset{T/2}{\int }}x(t)e^{-j2\pi mt/T}dt$
	$=\underset{-T/2}{\overset{T/2}{\int }}\left(\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{\alpha }_n{e}^{-j2\pi nt/T}\right)e^{j2\pi mt/T}dt$
	$=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{\alpha }_n\underset{-T/2}{\overset{T/2}{\int }}{e}^{j2\pi nt/T}{e}^{-j2\pi mt/T}dt$
	$=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{\alpha }_n\underset{-T/2}{\overset{T/2}{\int }}{e}^{j2\pi (n-m)t/T}dt$
	$=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{\alpha }_n\left\{\begin{array}{cc}T,& n=m\\ 0,& n\ne m\end{array}\right\}$
	$=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{\alpha }_{n}T\delta (n-m)$
	$=T\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{\alpha }_n\delta (n-m)$
	$=T{\alpha }_m$

Sticking this back into the top equation gives

${\alpha }_m=\frac{1}{T}\underset{-T/2}{\overset{T/2}{\int }}x(t)e^{-j2\pi mt/T}dt$

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 $\left\{\begin{array}{cc}T,& n=m\\ 0,& n\ne m\end{array}\right\}$ remember Euler's identities for sine and cosine. Since we are integrating the sine and cosine waves over a single period (assuming $m\ne n$), the value integrates to 0.

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

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