Afdeling Toegepaste Wiskunde/ Division of Applied Mathematics Fourier analysis 1(Alternative to Gonzalez& Woods: ) SLIDE 1/13 L 2 L 2

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1 Fourier analysis (Alternative to Gonzalez& Woods: ) SLIDE /3 FOURIER SERIES Considerafunction,f(x),definedontheinterval[ L,L ]: f ( x) L L We now want to approximate f(x) on the interval [ L,L ] with a Fourier series, f(x). AFourierseriesisalinearcombinationofasetofbasis functions,{φ n (x)=e πinx/l, n Z},thatis f(x)= n= c n e πinx/l wherec n arethefouriercoefficients.

2 Fourier analysis (Alternative to Gonzalez& Woods: ) SLIDE /3 These basis functions have two important properties. They are ()PERIODICwithperiodL,thatisφ n (x+l)=φ n (x),and ()ORTHOGONALontheinterval[ L,L ],thatis (verify) L e πinx/l e πikx/l dx= {, if n k L, if n=k (verify) Property () implies that f(x+l) = f(x). Therefore f(x) does not only approximatef(x),butalsotheperiodiccontinuationoff(x)ontheinterval [ L,L ],thatisp(x): p( x) 3 L L L 3L x

3 Fourier analysis (Alternative to Gonzalez& Woods: ) SLIDE 3/3 Property()enablesonetocalculatetheFouriercoefficients,c n,easilyand thereforealsothefourierseries f(x): f(x)e πikx/l dx = c n e πinx/l e πikx/l dx L n= L = c k L Therefore... c n = L f(x)e πinx/l dx L Example: Find f(x)whenf(x)= x ontheinterval[,] c n = f(x)e πinx dx= f(x)[cos(nπx) isin(nπx)]dx = ( x ) x cos(nπx)dx = nπ sin(nπx) sin(nπx)dx nπ = ( ) { nπ nπ cos(nπx) = ( )n, if n even(n ) = n π n π, if n odd

4 Fourier analysis (Alternative to Gonzalez& Woods: ) SLIDE 4/3 Therefore Speed of convergence c = f(x) = π n= (n odd) = 4 π n=,3,5,... e inπx n cos(nπx) n (verify) (verify) Howfastdoesthecoefficients c n decreasewhenn? ForExamplewehave: c n = n π, nodd c = c n =, neven(n ) Thereforethecoefficients c n decreaselike n

5 Fourier analysis (Alternative to Gonzalez& Woods: ) SLIDE 5/3 abs ( c n ) Coefficients of LOW frequency modes. 3 n Example Coefficients of HIGH frequency modes Find f(x)whenf(x)=xontheinterval[,] c n = i( )n nπ, c = c n = nπ, c = Thereforethecoefficients c n decreaselike n and f(x)= i π n= ( ) n e inπx n = π n= ( ) n sin(nπx) n (verify)

6 Fourier analysis (Alternative to Gonzalez& Woods: ) SLIDE 6/3 abs ( c n ). 3 Coefficients of LOW frequency modes 6.. n Coefficients of HIGH frequency modes Ingeneralwehavethefollowing... Whenp(x)istheperiodiccontinuationoff(x)ontheinterval[ L,L ],then Periodic continuation Fourier series p(x)discontinuous Convergeslike n etc. p(x) continuous p (x)discontinuous p(x) continuous p (x)continuous p (x)discontinuous Convergeslike n Convergeslike n 3

7 Fourier analysis (Alternative to Gonzalez& Woods: ) SLIDE 7/3 Therefore, the smoother the periodic continuation of f(x) is, the faster its Fourier series will converge This property is important when it comes to the compression of images (LATER) Example 3 Find f(x)whenf(x)=x, x [,] Option () Option () Why is option() better?

8 Fourier analysis (Alternative to Gonzalez& Woods: ) SLIDE 8/3 The difference in the speed of convergence of the Fourier series in Examples and is clear when the truncated Fourier series are plotted for differentvaluesofn... till n = Example till n = till n = 3 till n = 5 till n = 7

9 Fourier analysis (Alternative to Gonzalez& Woods: ) SLIDE 9/3.5 till n = Example till n = till n = till n = till n = 7 3 3

10 Fourier analysis (Alternative to Gonzalez& Woods: ) SLIDE /3 till n = 9 Example (continued...) 3 3 till n = 3 3 till n = till n = till n = 3 3 The Gibbs phenomenon occurs in Example, since the periodic continuation has step discontinuities

11 Fourier analysis (Alternative to Gonzalez& Woods: ) SLIDE /3 We now set out to develop the Fourier transform (Fourier integral) of a function f(x), since Fourier series have two important deficiencies: ()Theyutilizeonlyperiodicfunctionsφ(x)=e πinx/l withfrequenciesof n L, where n Z, and no periodic functions with frequencies between these discrete values ()Theycanonlyapproximateperiodicfunctions,thatisLhastobefinite TogofromaFourierseriestoaFouriertransform,weneedtoletL THE FOURIER TRANSFORM [ f(x) = ] f(t)e πint/l dt L = L = L n= n= n= L L L f(t)e πin(x t)/l dt e πinx/l f(t)cos(πn(x t)/l)dt+ L L f(t)dt (Derivation) Let u n = n L then u n+= n+ L =u n+ u,oru n+ u n = u= L n,

12 Fourier analysis (Alternative to Gonzalez& Woods: ) SLIDE /3 then: f(x)= u f(t)cos(πu n (x t))dt+ L f(t)dt n= L L abs( c n ) L L u L L n L LetL u theabovebecomescontinuous deficiencies() and() are addressed Ifweassumethat f(x)=f(x) = lim u = f(t)dt< (f(t) when t ± ),then... n= = () H(u n ) { }}{ u f(t)cos(πu n (x t))dt f(t)cos(πu(x t))dtdu(continuous!) f(t)cos(πu(x t))dtdu...

13 Fourier analysis (Alternative to Gonzalez& Woods: ) SLIDE 3/3 Also +i : f(x) = = f(t)sin(πu(x t))dtdu=... f(t)e πui(x t) dtdu e πiux f(t)e πiut dtdu } {{} F(u) In summary, we have the following for the one dimensional continuous case: FT{f(x)}=F(u)= IFT{F(u)}=f(x)= f(x)e πiux dx F(u)e πiux du Physical space FT f ( x) F( u) IFT Fourier space or frequency space