View source for 10/02 - Fourier Series

==Fourier Series (as compared to vectors)==
If a function is periodic, <math> x(t) = x(t+T) \,\!</math>, and it meets the Dirichlet conditions, then we can write it as <math> x(t) = \sum_{n=-\infty}^\infty \alpha_n e^{j2\pi nt/T}</math>
*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 <math>x(t)\,\!</math> with the mth basis function. Don't forget that the inner product of two vectors requires a complex conjugate. <math> \vec a \cdot \vec b = \sum_{i=1}^n a_i \cdot b_i^* </math>

{| border="0" cellpadding="0" cellspacing="0"
|-
|<math>x(t) \cdot e^{j2\pi mt/T}</math>
|<math>=\int_{-T/2}^{T/2} x(t) e^{-j2\pi mt/T}\, dt</math>
|-
|
|<math>=\int_{-T/2}^{T/2}\left (\sum_{n=-\infty}^\infty \alpha_n e^{-j2\pi nt/T} \right ) e^{j2\pi mt/T}\, dt</math>
|-
|
|-
|
|<math>=\sum_{n=-\infty}^\infty \alpha_n \int_{-T/2}^{T/2}e^{j2\pi nt/T}  e^{-j2\pi mt/T}\, dt</math>
|-
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|<math>=\sum_{n=-\infty}^\infty \alpha_n \int_{-T/2}^{T/2}e^{j2\pi (n-m)t/T}\, dt</math>
|-
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|<math>=\sum_{n=-\infty}^\infty \alpha_n \begin{Bmatrix} T, & n=m \\ 0, & n \ne m \end{Bmatrix} </math>
|-
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|<math>=\sum_{n=-\infty}^\infty \alpha_n T \delta(n-m) </math>
|-
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|<math>=T\sum_{n=-\infty}^\infty \alpha_n \delta(n-m) </math>
|-
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|<math>=T \alpha_m \,\!</math>
|}

Sticking this back into the top equation gives

<math> \alpha_m = \frac{1}{T}\int_{-T/2}^{T/2} x(t) e^{-j2\pi mt/T}\, dt</math>

==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 <math> \begin{Bmatrix} T, & n=m \\ 0, & n \ne m \end{Bmatrix} </math> remember Euler's identities for sine and cosine. Since we are integrating the sine and cosine waves over a single period (assuming <math> m \ne n </math>), the value integrates to 0.
:<math>\cos(x) = \frac{e^{ix} + e^{-ix}}{2} </math>
:<math>\sin(x) = \frac{e^{ix} - e^{-ix}}{2i}</math>