10/02 - Fourier Series: Difference between revisions
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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> |
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> |
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<math> 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 |
:<math> 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 </math> |
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:<math> = \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</math> |
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*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. |
*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. |
Revision as of 16:40, 10 November 2008
Fourier Series (as compared to vectors)
If a function is periodic, , and it meets the Dirichlet conditions, then we can write it as
- 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 with the mth basis function. Don't forget that the inner product of two vectors requires a complex conjugate.
- 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.