Fourier transforms (Chapter 15) Fourier integrals are generalizations of Fourier series. The series representation

Transcription

1 Pof. D. I. Nass Phys57 (T-3) Sptmb 8, 03 Foui_Tansf_phys57_T3 Foui tansfoms (Chapt 5) Foui intgals a gnalizations of Foui sis. Th sis psntation a0 nπx nπx f ( x) = + [ an cos + bn sin ] n = of a function f ( x ) is a piodic fom on < x < obtaind by gnating th cofficints fom th function s dfinition on th last piod [, ]. If a function f ( x ) dfind on th st of all al numbs has no piod, thn an analogy to Foui intgals can b nvisiond as ltting and placing th intg valud indx, n, by a al valud function k. Th cofficints a n and b n thn tak th fom ak ( ) and bk ( ). This mod of thought lads to th following dfinition. W will assum th following conditions on f ( x ) - f ( x ) satisfis th Diichlt conditions in vy finit intval[, ]. + - f ( x) dx finit, convgs, i.. f ( x ) is absolutly intgabl in[,. ] Foui's fomula fo -piodic functions using sins and cosins Fo a -piodic function ƒ(x) that is intgabl on[, ], th numbs a0 nπx nπx f ( x) = + [ an cos + bn sin ] () n = is calld th Foui sis of f ( x ). Using th intgals: mπx nπx mπx nπx cos( )cos( ) dx = δmn, sin( )sin( ) dx mn = δ, on finds: a0 = f ( x) dx, nπx nπx sin( )cos( ) dx = 0 nπ x an = f ( x)cos( ) dx and nπ x bn = f ( x)sin( ) dx ; n > 0 a calld th Foui cofficints of ƒ. Fo continuous ang, i.. ±, Equation () ducs to (not that f ( x ) 0 as ±): nπx nπx ' f ( x) = f ( x ') dx ' cos f ( x ')cos( ) dx ' + n = nπx nπx ' + sin f ( x ')sin( ) dx ' n = nπ x ' = f ( x ') dx ' + f ( x ')cos ( x x ') dx ' n = nπ π π t, k = ; k = n =. If f ( x ) is finit, thn ( using = dk ) n = π 0 ()

2 Pof. D. I. Nass Phys57 (T-3) Sptmb 8, 03 Foui_Tansf_phys57_T3 + f ( x) = dk f ( x ')cos{ k( x x ')} dx ' π 0 + = dk f ( x ')cos{ k( x x ')} dx ' Sinc: + dk f ( x ')sin{ k ( x x ')} dx ' 0 = thn, w can hav: ( ') ik x x ikx ikx ' f ( x) = dk f ( x ') dx ' dk f ( x ') dx ' = Wh g( k ) is known as th Foui tansfom of f ( x ). Applications: Exampl: Calculat th Foui tansfomation of th Gaussian function Answ: g( k ) k ik k x x ikx + g( k) = dx = dx = /4 /4 Not that: Th Foui tansfom of a Gaussian functions is a Gaussian function. H.W. Plot both functions. Pov that if f ( x) N αx =, thn g( k ) N k /4 ( α ) =, and vic vsa. α π f ( x) = x. (3) Th Foui tansfom of th Gaussian function is anoth Gaussian:

3 Pof. D. I. Nass Phys57 (T-3) Sptmb 8, 03 Foui_Tansf_phys57_T3 Not that th width sigma is oppositly positiond in th agumnts of th xponntials. This mans th naow a Gaussian is in on domain, th boad it is in th oth domain. IMAGE QUAITY Th Gaussian function can appoximat th bhavio of an imaging systm. In paticula, if w think of a vy naow slit of x-ays as bing a lin of dlta functions, an x-ay scn will blu this dlta lin into a boad "idg". It should b obvious that w want this idg to b as naow as possibl. Th imaging systm's spons to a dlta function lin input is calld th INE SPREAD FUNCTION, o SF in th spatial domain. Th magnitud of th complx function which is th Foui tansfom of th SF is th fquncy-dpndnt function known as th MODUATION TRANSFER FUNCTION, o MTF. Using what w hav just land about Gaussian functions, w conclud that th naow th SF, th boad th MTF in fquncy spac. Sinc w want naow SF's to poduc shap imags, w want MTF's to stay high until a high spatial fquncy is achd bfo it falls to zo. High fquncis a associatd with shap fatus in th imag, and th MTF is th systm's ability to cod infomation as a function of fquncy. HEISENBERG UNCERTAINTY PRINCIPE In Quantum Mchanics, th Hisnbg unctainty pincipl stats that w cannot simultanously know a paticl's position and mommtum (o diction of motion). This is bcaus th position wav function and th momntum wav function a Foui tansfom pais. Th naow on function bcoms, th wid th pai bcoms. Th btt w know position, th wos w know momntum. 3

4 Pof. D. I. Nass Phys57 (T-3) Sptmb 8, 03 Foui_Tansf_phys57_T3 Fo 4

5 Pof. D. I. Nass Phys57 (T-3) Sptmb 8, 03 Foui_Tansf_phys57_T3 Exampl (5.5a): Calculat th Foui tansfomation of th function, x a f ( x) =, ( a> 0). 0, x > a Answ: Its Foui tansfom ads a ika ika ikx ' sin( ak) F( k) = dx ' = = a ( ik ) π k Th functions f ( k ) and F ( k ) a shown in th following figu. Commnt: This is th singl-slit diffaction poblm of physical optics. Th slit is dscibd by f ( x ). Th diffaction pattn amplitud is givn by th Foui tansfom F( k ), wh F(0) = a. π Exampl: Diac Dlta function ikx ikx ' Stat with f ( x ) = g ( k) dk, and using g ( k) = f ( x ') dx ' thn ikx ' ikx ik ( x x ') f ( x ) = f ( x ') dx ' dk = f ( x ') dk dx ' = f ( x ) Fo a continuous function in th dimnsions, on finds: i k. ( ' ) 3 δ ( ' ) = d k 3 ( ) δ ( x x ') - Find th Foui tansfomation of th tiangula puls. x x < ; f ( x ) = 0 Othwis Answ: Fo th givn function (it is vn function), th Foui tansfom is: (5.d) 5

6 Pof. D. I. Nass Phys57 (T-3) Sptmb 8, 03 Foui_Tansf_phys57_T3 - Find th Foui tansfomation of th function f () =.{Hint: us th sphical coodinats wh dτ = sinθdθdϕd ] Answ: + i q f () = f () q = dτ, dτ = sinθdθdϕd ( ) 3/ b + i qi b iq cosθ I = d = d d d cos φ θ 0 0 iq iq b ( iq b ) ( iq + b ) = d d = iq iq 0 0 ( iq b ) ( iq + b ) 0 0 4π = iq + = + = iq b iq + b iq iq b iq + b b + q + i q 4π f ( q ) = 3/ dτ = π 3/ q = + π + q ( ) ( ) Exampl: Find th Foui tansfomation of th function f () = b. [Hint: b ± i qi 4π I = dτ = ], dτ = sinθdθdϕd b + q b+ i q f () = f () q = d 3/ ( ) τ b ± i qi 4π 8πb = 3/ dτ = = 3/ 3/ ( ) b ( ) b b + q ( ) ( b + q ) H.W. Calculat th following intgals: i k ir I = dk = k R b ± i qi I 8π b I = dτ = = ; b ( b + q ) ± i q i ± i q i( -' ) 4π I3 = d = d = lim I = 0 -' -' b q ± iqi' ± iqi' ± iqi' 6

7 Pof. D. I. Nass Phys57 (T-3) Sptmb 8, 03 Foui_Tansf_phys57_T3 7