Signals and systems/GF Fourier: Difference between revisions

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==Fourier series==
The Fourier series is used to analyze arbitrary periodic functions by showing them as a composite of sines and cosines.

A function is considered periodic if <math> x(t) = x(t+T)\, </math> for <math> T \neq 0 </math>.

The exponential form of the Fourier series is defined as <math> x(t) = \sum_{n=-\infty}^\infty \alpha_n e^{{j2\pi nt}/T} \, </math>

==Determining the coefficient <math> \alpha_n \,</math> ==

<math> x(t) = \sum_{n=-\infty}^\infty \alpha_n e^{{j2\pi nt}/T} \, </math>

*The definition of the Fourier series

<math> \int_{-T/2}^{T/2} x(t)\, dt = \sum_{n=-\infty}^\infty \alpha_n \int_{-T/2}^{T/2} e^{{j2\pi nt}/T} dt</math>

*Integrating both sides for one period. The range of integration is arbitrary, but using <math> \int_{-T/2}^{T/2} </math> scales nicely when extending the Fourier series to a non-periodic function

<math> \int_{-T/2}^{T/2} x(t) 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</math>

*Multiply by the complex conjugate

<math> \int_{-T/2}^{T/2} x(t) e^{{-j2\pi mt}/T} dt = \sum_{n=-\infty}^\infty \alpha_n \frac{Te^{{j2\pi (n-m)t}/T}}{{j2\pi (n-m)}} \bigg|_{-T/2}^{T/2} = \sum_{n=-\infty}^\infty \alpha_n T\delta_{n,m} = T\alpha_m</math>

*<math> \frac{Te^{{j2\pi (n-m)t}/T}}{{j2\pi (n-m)}} \bigg|_{-T/2}^{T/2} = T\frac{e^{j\pi(n-m)}-e^{-j\pi(n-m)}}{j2\pi(n-m)} = T \frac{\sin\pi(n-m)}{\pi(n-m)} = \begin{Bmatrix} T, n=m \\ 0, n\ne m \end{Bmatrix} = T\delta_{n,m}</math>

** Using L'Hopitals to evaluate the <math>\frac{T\cdot 0}{0}</math> case. Note that n & m are integers

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

== <math> \left \langle Bra \mid Ket \right \rangle </math> Notation ==

<math> \left \langle \ n \mid m \ \right \rangle = m \cdot n </math>

==Linear Time Invariant Systems==

==Changing Basis Functions==

==Identities==
<math>e^{j \theta} = \cos \theta + j \sin \theta \, </math>

<math>\sin x = \frac{e^{jx}-e^{-jx}}{2j} \,</math>

<math>\cos x = \frac{e^{jx}+e^{-jx}}{2} \,</math>

Latest revision as of 08:29, 4 November 2006