Signals and systems/GF Fourier: Difference between revisions

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


<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
*The definition of the Fourier series


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


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

Revision as of 22:12, 29 October 2006

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 x(t)=x(t+T) for T0.

The exponential form of the Fourier series is defined as x(t)=n=αnej2πnt/T

Determining the coefficient αn

x(t)=n=αnej2πnt/T

  • The definition of the Fourier series

T/2T/2x(t)dt=n=αnT/2T/2ej2πnt/Tdt

  • Integrating both sides for one period. The range of integration is arbitrary, but using T/2T/2 scales nicely when extending the Fourier series to a non-periodic function

T/2T/2x(t)ej2πmt/Tdt=n=αnT/2T/2ej2πnt/Tej2πmt/Tdt=n=αnT/2T/2ej2π(nm)t/Tdt

  • Multiply by the complex conjugate

T/2T/2x(t)ej2πmt/Tdt=n=αnTej2π(nm)t/Tj2π(nm)|T/2T/2=Tejπ(nm)ejπ(nm)j2π(nm)=Tsinπ(nm)π(nm)={T,n=m0,nm}=Tδn,m

  • Using L'Hopitals to evaluate the T00 case. Note that n & m are integers

BraKet Notation

Linear Time Invariant Systems

Changing Basis Functions

Identities

ejθ=cosθ+jsinθ

sinx=ejxejx2j

cosx=ejx+ejx2

nm=Tδn,m