Fourier Series: Difference between revisions

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


<math> \left \langle \ Bra \mid Ket \ \right \rangle = Ket \cdot Bra </math>
<math> <u|v> = \int_{-\infty}^\infty u^*(x) v(x) dx </math>


<math> \alpha_{-m} = \alpha^* \,</math>
<math> \alpha_{-m} = \alpha^* \,</math>


The dirac delta has an infinite height and an area of 1
The dirac delta has an infinite height and an area of 1

Revision as of 00:04, 21 November 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=n=αnTδn,m=Tαm

  • Tej2π(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

αm=1TT/2T/2x(t)ej2πmt/Tdt


Linear Time Invariant Systems

Must meet the following criteria

  • Time independance
  • Linearity
    • Superposition (additivity)
    • Scaling (homogeneity)

The Dot Product, Complex Conjugates, and Orthogonality

File:300px-Scalarproduct.gif

Geometrically, the dot product is a scalar projection of a onto b

  • ab=|a||b|cosθ

Arthimetically, multiply like terms and add

  • (3,2,1)(5,6,7)=35*+26*+17*

Lets imagine that we are only have one dimension

  • (a+jb)i^(a+jb)i^a2+b2

In order to get the real parts and imaginary parts to multiply as like terms, we need to take the complex conjugate of one of the terms

  • (a+jb)i^(ajb)i^=a2+b2

To test for orthogonality, take the complex conjugate of one of the vectors and multiply.

  • ϕn(t)ϕm*(t)dt=0

Changing Basis Functions

We'd like to change from n=αnej2πnt/T to m=0cmcos(2πmtT+Θm)

x(t)=n=αnej2πnt/T=n=1αnej2πnt/Tn=n+α0+n=1αnej2πnt/T=n=1αnej2πnt/Tm=n+α0+n=1αnej2πnt/Tm=n

=α0+m=1(αmej2πmt/T+αmej2πmt/T)

If we assume x(t)m, then to make the imaginary parts cancel out

  • αm=αm*
  • u+u*=2[u]
  • αm=|αm|ejϕm

=α0+m=12[αmej2πmt/T]=α0+m=12[|αm|ejϕmej2πmt/T]=α0+m=1|αm|2[ej(2πmt/T+ϕm)]=α0+m=1|αm|2cos(2πmtT+ϕm)

Changing variables

  • c0=α0
  • cm=2|αm|
  • Θm=ϕm

=m=0cmcos(2πmtT+Θm)

Identities

ejθ=cosθ+jsinθ Euler's identity linking rectangular and polar coordinates

sinx=ejxejx2j

cosx=ejx+ejx2

<u|v>=u*(x)v(x)dx

αm=α*

The dirac delta has an infinite height and an area of 1