Fourier Series: Difference between revisions

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<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)\, 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
*Integrating both sides for one period. The range of integration is arbitrary, but using <math> \int_{-T/2}^{T/2} </math> instead of <math> \int_{0}^{T} </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>  
<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>  
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<math> \alpha_m = \frac{1}{T}\int_{-T/2}^{T/2} x(t) e^{{-j2\pi mt}/T} dt </math>
<math> \alpha_m = \frac{1}{T}\int_{-T/2}^{T/2} x(t) e^{{-j2\pi mt}/T} dt </math>


==Linear Time Invariant Systems==
==Linear Time Invariant Systems==

Latest revision as of 15:54, 4 October 2007

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 T≠0.

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=−∞∞αn∫−T/2T/2ej2πnt/Tdt

  • Integrating both sides for one period. The range of integration is arbitrary, but using ∫−T/2T/2 instead of ∫0Tscales nicely when extending the Fourier series to a non-periodic function

∫−T/2T/2x(t)e−j2πmt/Tdt=∑n=−∞∞αn∫−T/2T/2ej2πnt/Te−j2πmt/Tdt=∑n=−∞∞αn∫−T/2T/2ej2π(n−m)t/Tdt

  • Multiply by the complex conjugate

∫−T/2T/2x(t)e−j2πmt/Tdt=∑n=−∞∞αnTej2π(n−m)t/Tj2π(n−m)|−T/2T/2=∑n=−∞∞αnTδn,m=Tαm

  • Tej2π(n−m)t/Tj2π(n−m)|−T/2T/2=Tejπ(n−m)−e−jπ(n−m)j2π(n−m)=Tsin⁡π(n−m)π(n−m)={T,n=m0,n≠m}=Tδn,m
    • Using L'Hopitals to evaluate the T⋅00 case. Note that n & m are integers

αm=1T∫−T/2T/2x(t)e−j2πmt/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

  • a→⋅b→=|a||b|cos⁡θ

Arthimetically, multiply like terms and add

  • (3,2,1)⋅(5,6,7)=3⋅5*+2⋅6*+1⋅7*

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^⋅(a−jb)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=0∞cmcos⁡(2πmtT+Θm)

x(t)=∑n=−∞∞αnej2πnt/T=∑n=−∞−1αnej2πnt/T⏟n′=−n+α0+∑n=1∞αnej2πnt/T=∑n′=1∞αnej2πnt/T⏟m=n′+α0+∑n=1∞αnej2πnt/T⏟m=n

=α0+∑m=1∞(αmej2πmt/T+α−me−j2πmt/T)

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

  • α−m=αm*
  • u+u*=2ℜ[u]
  • αm=|αm|ejϕm

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

Changing variables

  • c0=α0
  • cm=2|αm|
  • Θm=ϕm

=∑m=0∞cmcos⁡(2πmtT+Θm)

Identities

ejθ=cos⁡θ+jsin⁡θ Euler's identity linking rectangular and polar coordinates

sin⁡x=ejx−e−jx2j

cos⁡x=ejx+e−jx2

<u|v>=∫−∞∞u*(x)v(x)dx

α−m=α*

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