FOURIER ANALYSIS: LECTURE 6

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1 FOURIER ANALYSIS: LECTURE Convergence of Fourier series Fourier series (rel or complex) re very good wys of pproximting functions in finite rnge, by which we men tht we cn get good pproximtion to the function by using only the first few modes (i.e. truncting the sum over n fter some low vlue n = N). This is how music compression works in MP3 plyers, or how digitl imges re compressed in JPEG form: we cn get good pproximtion to the true wveform by using only limited number of modes, nd so ll the modes below certin mplitude re simply ignored. We sw relted exmple of this in our pproximtion to using Eqn. (2.79) nd Tble. Not exminble: Mthemticlly, this trnsltes s the Fourier components converging to zero i.e. n,b n! 0s n!,providedf(x) is bounded (i.e. hs no divergences). But how quickly do the high order coe cients vnish? There re two common cses:. The function nd its first p derivtives(f(x),f 0 (x),...f (p ) (x)) re continuous, but the p th derivtive f (p) (x) hsdiscontinuities: n,b n /n p+ for lrge n. (2.82) An exmple of this ws our expnsion of f(x) =x 2. When we periodiclly extend the function, there is discontinuity in the grdient (p =derivtive)ttheboundriesx = ±L. Wehve lredy seen n /n 2 s expected (with b n =0). 2. f(x) is periodic nd piecewise continuous (i.e. it hs jump discontinuities, but only finite number within one period): ) n,b n /n for lrge n. (2.83) An exmple of this is the expnsion of the odd function f(x) = x, which jumps t the boundry. The Fourier components turn out to be b n /n (with n =0). End of non-exminble section How close does it get? Convergence of Fourier expnsions We hve seen tht the Fourier components generlly get smller s the mode number n increses. If we truncte the Fourier series fter N terms, we cn define n error D N tht mesures how much the truncted Fourier series di ers from the originl function: i.e. if we define the error s f N (x) = N X D N = n= Z L L h n cos n x + b n sin L n x i. (2.84) L dx f(x) f N (x) 2 0. (2.85) 9

2 Figure 2.6: The Gibbs phenomenon for truncted Fourier pproximtions to the signum function Eqn Note the di erent x-rnge in the lower two pnels. Tht is, we squre the di erence between the originl function nd the truncted Fourier series t ech point x, thenintegrtecrossthefullrngeofvlidityofthefourierseries. Techniclly,this is wht is known s n L 2 norm. Some things you should know, but which we will not prove: if f is resonbly well-behved (no nonintegrble singulrities, nd only finite number of discontinuities), the Fourier series is optiml in the lest-squres sense i.e. if we sk wht Fourier coe cients will minimise D N for some given N, they re exctly the coe cients tht we obtin by solving the full Fourier problem. Furthermore, s N!, D N! 0. This sounds like we re gurnteed tht the Fourier series will represent the function exctly in the limit of infinitely mny terms. But looking t the eqution for D N,itcnbeseenthtthisisnotso:it slwyspossibletohve(sy)f N =2f over some rnge x, ndthebestwecnsyistht x must tend to zero s N increses. EXAMPLE: As n exmple of how Fourier series converge (or not), consider the signum function which picks out the sign of vrible: ( if x<0, f(x) =signumx = (2.86) + if x 0, 20

3 N Tble 2: Error D N on the N-term truncted Fourier series pproximtion to the signum function Eqn D N which we will expnd in the rnge pple x pple (i.e.wesetl =). Thefunctionisodd,so n =0 nd we find Z b n =2 dx sin(n x) = 2 n [ ( )n ]. (2.87) 0 f(x) hsdiscontinuitiestx =0ndx = ±L = ± (due to the periodic extension), so from Sec. 2.. we expected n /n. In Tble 2 we show the error D N for the signum function for incresing vlues of D N. As expected the error decreses s N gets lrger, but reltively slowly. We ll see why this is in the next section Ringing rtefcts nd the Gibbs phenomenon We sw bove tht we cn define n error ssocited with the use of truncted Fourier series of N terms to describe function. Note tht D N mesures the totl error by integrting the devition t ech vlue of x over the full rnge. It does not tell us whether the devitions between f N (x) nd f(x) werelrgendconcentrtedtcertinvluesofx, orsmllerndmoreevenlydistributed over ll the full rnge. An interesting cse is when we try to describe function with finite discontinuity (i.e. jump) using truncted Fourier series, such s our discussion of the signum function bove. In Fig. 2.6 we plot the originl function f(x) ndthetrunctedfourierseriesforvriousn. We find tht the truncted sum works well, except ner the discontinuity. Here the function overshoots the true vlue nd then hs dmped oscilltion. As we increse N the oscillting region gets smller, but the overshoot remins roughly the sme size (bout 8%). This overshoot is known s the Gibbs phenomenon. Lookingttheplot,wecnseethtittendsto be ssocited with extended oscilltions either side of the step, known s ringing rtefcts. Such rtefcts will tend to exist whenever we try to describe shrp trnsitions with Fourier methods, nd re one of the resons tht MP3s cn sound bd when the compression uses too few modes. We cn reduce the e ect by using smoother method of Fourier series summtion, but this is well beyond this course. For the interested, there re some more detils t Gibbs_phenomenon. 2.2 Prsevl s theorem There is useful reltionship between the men squre vlue of the function f(x) ndthefourier coe cients. Prsevl s formul is 2L Z L L f(x) 2 dx = 0 / X n= n 2 + b n 2, (2.88)

4 or, for complex Fourier Series, Z L f(x) 2 dx = 2L L X n= c n 2. (2.89) The simplicity of the expression in the complex cse is n exmple of the dvntge of doing things this wy. The quntity c n 2 is known s the power spectrum. This is by nlogy with electricl circuits, where power is I 2 R.Sothemenoff 2 is like the verge power, nd c n 2 shows how this is contributed by the di erent Fourier modes. Proving Prsevl is esier in the complex cse, so we will stick to this. The equivlent for the sin+cos series is included for interest, but is not exminble. First, note tht f(x) 2 = f(x)f (x) nd expnd f nd f s complex Fourier Series: f(x) 2 = f(x)f (x) = X n= c n n (x) X m c m m(x) (2.90) (recll tht n(x) =e iknx ). Then we integrte over L pple x pple L, notingtheorthogonlityof n nd m: = X m,n= Z L L f(x) 2 dx = c n c m(2l mn ) = 2L X m,n= X n= c n c m Z L L c n c n =2L n(x) m(x) dx (2.9) X n= where we hve used the orthogonlity reltion R LL n (x) m(x) dx =2L if m = n, ndzerootherwise. c n Summing series vi Prsevl Consider once gin the cse of f = x 2.ThelhsofPrsevl stheoremis(/2l) R LL x4 dx =(/5)L 4. The complex coe cients were derived erlier, so the sum on the rhs of Prsevl s theorem is X c n 2 = c X L c n X 2L 2 ( ) n = +2 3 n 2 2 n6=0 n= n= Equting the two sides of the theorem, we therefore get X n= 2 = L4 9 + X n= 8L 4 n 4 4. (2.92) m 4 =( 4 /8)(/5 /9) = 4 /90. (2.93) This is series tht converges fster thn the ones we obtined directly from the series t specil vlues of x 22

5 FOURIER ANALYSIS: LECTURE 7 3 Fourier Trnsforms Lerning outcomes In this section you will lern bout Fourier trnsforms: their definition nd reltion to Fourier series; exmples for simple functions; physicl exmples of their use including the di rction nd the solution of di erentil equtions. You will lern bout the Dirc delt function nd the convolution of functions. 3. Fourier trnsforms s limit of Fourier series We hve seen tht Fourier series uses complete set of modes to describe functions on finite intervl e.g. the shpe of string of length `. Inthenottionwehveusedsofr,` =2L. Insome wys, it is esier to work with `, whichwedobelow;butmosttextbookstrditionllycoverfourier series over the rnge 2L, ndthesenotesfollowthistrend. Fourier trnsforms (FTs) re n extension of Fourier series tht cn be used to describe nonperiodic functions on n infinite intervl. The key ide is to see tht non-periodic function cn be viewed s periodic one, but tking the limit of `!.Thisisreltedtoourerlierideofbeingbleto construct number of di erent periodic extensions of given function. This is illustrted in Fig. 3. for the cse of squre pulse tht is only non-zero between <x<+. When ` becomes lrge compred to, theperiodicreplicsofthepulserewidelyseprted,ndinthelimitof `!we hve single isolted pulse. Figure 3.: Di erent periodic extensions of squre pulse tht is only non-zero between < x<+. As the period of the extension, `, increses,thecopiesofthepulsebecomemorewidely seprted. In the limit of `!,wehvesingleisoltedpulsendthefourierseriesgoesover to the Fourier trnsform. 23

6 Fourier series only include modes with wvenumbers k n = 2n with djcent modes seprted by ` k =. Wht hppens to our Fourier series if we let `!?Considerginthecomplexseries 2 ` for f(x): X f(x) = C n e iknx, (3.) where the coe cients re given by C n = ` Z `/2 n= `/2 dx f(x) e iknx. (3.2) nd the llowed wvenumbers re k n =2n /`. The seprtion of djcent wvenumbers (i.e. for n! n +)is k =2 /`; sos`!,themodesbecomemorendmorefinelyseprtedink. In the limit, we re then interested in the vrition of C s function of the continuous vrible k. The fctor /` outside the integrl looks problemtic for tlking the limit `!, but this cn be evded by defining new quntity: f(k) ` C(k) = Z dx f(x) e ikx. (3.3) The function f(k) (o cillyclled ftilde,butmorecommonly ftwiddle ;f k is nother common nottion) is the Fourier trnsform of the non-periodic function f. To complete the story, we need the inverse Fourier trnsform: thisgivesusbckthefunctionf(x) if we know f. Here, we just need to rewrite the Fourier series, remembering the mode spcing k =2 /`: f(x) = X C(k)e ikx = X (`/2 ) C(k)e ikx k = X f(k) e ikx k. (3.4) 2 k n k n k n In this limit, the finl form of the sum becomes n integrl over k: X g(k) k! Z g(k) dk s k! 0; (3.5) this is how integrtion gets defined in the first plce. We cn now write n eqution for f(x) in which ` does not pper: f(x) = Z dk 2 f(k) e ikx. (3.6) Note the infinite rnge of integrtion in k: thiswslredypresentinthefourierseries,wherethe mode number n hd no limit. EXAM TIP: You my be sked to explin how the FT is the limit of Fourier Series (for perhps 6or7mrks),somkesureyoucnreproducethestu inthissection. The density of sttes In the bove, our sum ws over individul Fourier modes. But if C(k) is continuousfunctionofk, we my s well dd modes in bunches over some bin in k, ofsize k: f(x) = X C(k)e ikx N bin, k bin (3.7) 24

7 where N bin is the number of modes in the bin. Wht is this? It is just k divided by the mode spcing, 2 /`, sowehve f(x) = ` X C(k)e ikx k (3.8) 2 k bin The term `/2 is the density of sttes: ittellsushowmnymodesexistinunitrngeofk. This is widelyusedconceptinmnyresofphysics,especillyinthermodynmics.oncegin,wecn tke the limit of k! 0ndobtintheintegrlfortheinverseFouriertrnsform. Summry Afunctionf(x) nditsfouriertrnsform f(k) rethereforereltedby: f(x) = 2 f(k) = Z Z dk f(k) e ikx ; (3.9) dx f(x) e ikx. (3.0) We sy tht f(k) is the FT of f(x), nd tht f(x) is the inverse FT of f(k). EXAM TIP: If you re sked to stte the reltion between function nd its Fourier trnsform (for mybe 3 or 4 mrks), it is su cient to quote these two equtions. If the full derivtion is required, the question will sk explicitly for it. Note tht, since the Fourier Trnsform is liner opertion, FT[f(x)+g(x)] = f(k)+ g(k). (3.) For rel function f(x), its FT stisfies the sme Hermitin reltion tht we sw in the cse of Fourier series: f( k) = f (k) (3.2) Exercise: prove this. FT conventions Eqns. (3.0) nd (3.9) re the definitions we will use for FTs throughout this course. Unfortuntely, there re mny di erent conventions in ctive use for FTs. Aside from using di erent symbols, these cn di er in: The sign in the exponent The plcing of the 2 prefctor(s) (nd sometimes it is p 2 ) Whether there is fctor of 2 in the exponent The bd news is tht you will probbly come cross ll of these di erent conventions. The good news is tht tht it is reltively esy to convert between them if you need to. The best news is tht you will lmost never need to do this conversion. 25

8 k spce nd momentum spce The Fourier convention presented here is the nturl one tht emerges s the limit of the Fourier series. But it hs the disdvntge tht it trets the Fourier trnsform nd the inverse Fourier trnsform di erently by fctor of 2, wheresinphysicsweneed to lern to tret the functions f(x)nd f(k)sequllyvlidformsofthesmething: the rel-spce nd k-spce forms. This is most obvious in quntum mechnics, where wve function exp(ikx) represents prticle with well-defined momentum, p = hk ccording to de Broglie s hypothesis. Thus the description of function in terms of f(k) isoftenclledthe momentum-spce version. The result tht illustrtes this even-hnded pproch most clerly is to relise tht the Fourier trnsform of f(x) cnitselfbetrnsformed: g f(k)(k) = Z dk f(k) e ikk. (3.3) We will show below tht g f(k)(k) =2 f( K) : (3.4) so in essence, repeting the Fourier trnsform gets you bck the function you strted with. f nd f re relly just two sides of the sme coin. FOURIER ANALYSIS: LECTURE Some simple exmples of FTs In this section we ll find the FTs of some simple functions. EXAM TIP: sketch f(k). You my be sked to define nd sketch f(x) inechcse,ndlsotoclcultend 3.2. The top-ht A top-ht function (x) ofheighth nd width 2 ( ssumed positive), centred t x = d is defined by: ( h, if d <x<d+, (x) = (3.5) 0, otherwise. The function is sketched in Fig Its FT is: f(k) = Z dx (x) e ikx = h Z d+ d dx e ikx =2h e ikd sinc(k) (3.6) The derivtion is given below. The function sinc x sin x is sketched in Fig. 3.3 (with notes on x how to do this lso given below). f(k) willlookthesme(ford =0),butthenodeswillnowbet k = ± n nd the intercept will be 2h rther thn. You re very unlikely to hve to sketch f(k) for d 6= 0. 26

9 EXAM TIPS: x to y. If the question sets d =0,clerlythereisnoneedtodovriblechngefrom Sometimes the question specifies tht the top-ht should hve unit re i.e. h (2) =,soyou cn replce h. The width of the top-ht won t necessrily be 2... Deriving the FT: f(k) = Z dx (x) e ikx = h Z d+ d dx e ikx (3.7) Now we mke substitution u = x d (which now centres the top-ht t u =0). Theintegrnd e ikx becomes e ik(u+d) = e iku e ikd. We cn pull the fctor e ikd outside the integrl becuse it does not depend on u. The integrtion limits become u = ±. There is no scle fctor, i.e. du = dx. This gives Z pple e f(k) =he ikd du e iku = he ikd iku ik = he ikd 2 k eik e ik 2i e = he ikd ik e ik ik =2he ikd sin(k) k Note tht we conveniently multiplied top nd bottom by 2 midwy through. =2h e ikd sinc(k) (3.8) Sketching sinc x: You should think of sinc x sin x s sin x oscilltion (with nodes t x = ±n x for integer n), but with the mplitude of the oscilltions dying o s /x. Note tht sinc x is n even function, so it is symmetric when we reflect bout the y-xis. The only compliction is t x =0,whensinc0= 0 which ppers undefined. To del with this, 0 expnd sin x = x x 3 /3! + x 5 /5! +...,soitisobviousthtsinx/x! sx! 0. EXAM TIP: intercepts. Mke sure you cn sketch this, nd tht you lbel ll the zeros ( nodes ) nd Figure 3.2: Sketch of top-ht function defined in Eqn. (3.5) 27

10 3.2.2 The Gussin The Gussin curve is lso known s the bell-shped or norml curve. centred t x = d is defined by: (x d) 2 f(x) =N exp 2 2 A Gussin of width (3.9) where N is normliztion constnt, which is often set to. We cn insted define the normlized Gussin, where we choose N so tht the re under the curve to be unity i.e. N =/ p 2 2. This normliztion cn be proved by net trick, which is to extend to two-dimensionl Gussin for two independent (zero-men) vribles x nd y, bymultiplyingthetwoindependentgussin functions: p(x, y) = 2 e (x2 +y2 )/2 2. (3.20) 2 The integrl over both vribles cn now be rewritten using polr coordintes: ZZ Z p(x, y) dx dy = nd the finl expression clerly integrtes to so the distribution is indeed correctly normlized. p(x, y) 2 rdr= 2 2 Z 2 re r2 /2 2 dr (3.2) P (r >R)=exp R 2 /2 2, (3.22) The Gussin is sketched for d =0ndtwodi erentvluesofthewidthprmeter. Fig. 3.4 hs N = in ech cse, wheres Fig. 3.5 shows normlized curves. Note the di erence, prticulrly in the intercepts. For d = 0, the FT of the Gussin is Z x 2 f(k) = dx N exp e ikx = p 2 N i.e. the FT of Gussin is nother Gussin (this time s function of k). 2 2 k 2 2 exp, (3.23) 2 Figure 3.3: Sketch of sinc x sin x x 28

11 Deriving the FT For nottionl convenience, let s write =,so 2 2 f(k) =N Z dx exp x 2 + ikx (3.24) Now we cn complete the squre inside [...]: x 2 ikx = x + ik 2 k (3.25) giving Z f(k) =Ne k2 /4 We then mke chnge of vribles: u = p dx exp x + ik 2 pple x + ik 2 2!. (3.26). (3.27) This does not chnge the limits on the integrl, nd the scle fctor is dx = du/ p,giving f(k) = p N Z r e k2 /4 du e u2 = N e k2 /4 = e k2 /4. (3.28) where we chnged bck from to. To get this result, we hve used the stndrd result Z du e u2 = p. (3.29) 3.3 Reciprocl reltions between function nd its FT These exmples illustrte generl nd very importnt property of FTs: there is reciprocl (i.e. inverse) reltionship between the width of function nd the width of its Fourier trnsform. Tht is, nrrow functions hve wide FTs nd wide functions hve nrrow FTs. This importnt property goes by vrious nmes in vrious physicl contexts, e.g.: Figure 3.4: Sketch of Gussins with N = 29

12 Heisenberg Uncertinty Principle: the rms uncertinty in position spce ( x) ndtherms uncertinty in momentum spce ( p) reinverselyrelted: ( x)( p) h/2. The equlity holds for the Gussin cse (see below). Bndwidth theorem: to crete very short-lived pulse (smll wide rnge of frequencies (lrge!). t), you need to include very In optics, this mens tht big objects (big reltive to wvelength of light) cst shrp shdows (nrrow FT implies closely spced mxim nd minim in the interference fringes). We discuss two explicit exmples in the following subsections: 3.3. The top-ht The width of the top-ht s defined in Eqn. (3.5) is obviously 2. For the FT, whilst the sinc k function extends cross ll k, itdieswyinmplitude,soitdoes hve width. Exctly how we define the width does not mtter; let s sy it is the distnce between the first nodes k = ± / in ech direction, giving width of 2 /. Thus the width of the function is proportionl to, nd the width of the FT is proportionl to /. Note tht this will be true for ny resonble definition of the width of the FT The Gussin Agin, the Gussin extends infinitely but dies wy, so we cn define width. For Gussin, it is esy to do this rigorously in terms of the stndrd devition (squre root of the verge of (x d) 2 ), which is just (check you cn prove this). Compring the form of FT in Eqn. (3.23) to the originl definition of the Gussin in Eqn. (3.9), if the width of f(x) is,thewidthof f(k) is/ by the sme definition. Agin, we hve reciprocl reltionship between the width of the function nd tht of its FT. Since p = hk, the width in momentum spce is h times tht in k spce. Figure 3.5: Sketch of normlized Gussins. The intercepts re f(0) = p

13 Figure 3.6: The Fourier expnsion of the function f(x) =/(4 x /2 ), x < isshowninthelh pnel ( cosine series, up to n =5). TheRHpnelcompresdf /dx with the sum of the derivtive of the Fourier series. The mild divergence in f mens tht the expnsion converges; but for df /dx it does not. The only subtlety in relting this to the uncertinty principle is tht the probbility distributions use 2, not. Ifthewidthof (x) is,thenthewidthof 2 is / p 2. Similrly, the uncertinty in momentum is (/ )/ p 2, which gives the extr fctor /2 in ( x)( p) = h/ Di erentiting nd integrting Fourier series Once we hve function expressed s Fourier series, this cn be useful lterntive wy of crrying out clculus-relted opertions. This is becuse di erentition nd integrtion re liner opertions tht re distributive over ddition: this mens tht we cn crry out di erentition or integrtion term-by-term in the series: X f(x) = C n e iknx (3.30) ) df dx = Z ) fdx= n= X n= X n= C n (ik n ) e iknx (3.3) C n (ik n ) e iknx +const. (3.32) The only compliction rises in the cse of integrtion, if C 0 6=0: thentheconstnttermintegrtes to be / x, ndthisneedstobehndledseprtely. From these reltions, we cn see immeditely tht the Fourier coe cients of function nd its derivtive re very simply relted by powers of k: if the n th Fourier coe cient of f(x) isc n,the n th Fourier coe cient of df (x)/dx is (ik n )C n. Tking the limit of non-periodic function, the sme reltion clerly pplies to Fourier Trnsforms. Thus, in generl, multiple derivtives trnsform s: FT f (p) (x) pple d p f = FT =(ik) p f(k) (3.33) dx p 3

14 The min cvet with ll this is tht we still require tht ll the quntities being considered must be suitble for Fourier representtion, nd this my not be so. For exmple, f(x) =/ p x for 0 <x<isncceptblefunction:ithssingulritytx = 0, but this is integrble, so ll the Fourier coe cients converge. But f 0 (x) = x 3/2 /2, which hs divergent integrl over 0 <x<. Attempts to use Fourier representtion for f 0 (x) would come drift in this cse, s is illustrted in Fig FOURIER ANALYSIS: LECTURE 9 4 The Dirc delt function The Dirc delt function is very useful tool in physics, s it cn be used to represent very loclised or instntneous impulse whose e ect then propgtes. Informlly, it is to be thought of s n infinitely nrrow (nd infinitely tll) spike. Mthemticins think it s not proper function, since function is mchine, f(x), tht tkes ny number x nd replces it with well-defined number f(x). Dirc didn t cre, nd used it nywy. Eventully, the theory of distributions ws invented to sy he ws right to follow his intuition. 4. Definition nd bsic properties The Dirc delt function (x d) isdefinedbytwo expressions. First, it is zero everywhere except t the point x = d where it is infinite: ( 0 for x 6= d, (x d) = (4.34)! for x = d. Secondly, it tends to infinity t x = d in such wy tht the re under the Dirc delt function is unity: Z dx (x d) =. (4.35) 4.. The delt function s limiting cse To see how spike of zero width cn hve well-defined re, it is helpful (lthough not strictly necessry) to think of the delt function s the limit of more fmilir function. The exct shpe of this function doesn t mtter, except tht it should look more nd more like (normlized) spike s we mke it nrrower. Two possibilities re the top-ht s the width! 0(normlizedsothth =/(2)), or the normlized Gussin s! 0. We ll use the top-ht here, just becuse the integrls re esier. Let (x) benormlizedtop-htofwidth2centred t x =0sinEqn.(3.5) we vemde the width prmeter obvious by putting it s subscript here. The Dirc delt function cn then be defined s (x) =lim (x). (4.36)!0 32

15 EXAM TIP: When sked to define the Dirc delt function, mke sure you write both Eqns. (4.34) nd (4.35) Sifting property The sifting property of the Dirc delt function is tht, given some function f(x): Z dx (x d) f(x) =f(d) (4.37) i.e. the delt function picks out the vlue of the function t the position of the spike (so long s it is within the integrtion rnge). This is just like the sifting property of the Kronecker delt inside discretesum. EXAM TIP: If you re sked to stte the sifting property, it is su cient to write Eqn. (4.37). You do not need to prove the result s in Sec unless specificlly sked to. Technicl side: The integrtion limits don t techniclly need to be infinite in the bove formulæ. If we integrte over finite rnge <x<bthe expressions become: Z ( b for <d<b, dx (x d) = (4.38) 0 otherwise. Z ( b f(d) for <d<b, dx (x d) f(x) = (4.39) 0 otherwise. Tht is, we get the bove results if the position of the spike is inside the integrtion rnge, nd zero otherwise Compre with the Kronecker delt The Kronecker delt mn = m = n 0 m 6= n (4.40) plys similr sifting role for discrete modes, s the Dirc delt does for continuous modes. For exmple: X A n mn = A m (4.4) which is obvious when you look t it. Kronecker delt in it. n= Be prepred to do this whenever you see sum with 4..4 Delt function of more complicted rgument Sometimes you my come cross the Dirc delt function of more complicted rgument, [f(x)], e.g. (x 2 4). How do we del with these? Essentilly we use the definition tht the delt function integrtes to unity when it is integrted with respect to its rgument. i.e. Z [f(x)]df = (4.42) 33

16 Chnging vribles from f to x, Z [f(x)] df dx = (4.43) dx where we hve not put the limits on x, s they depend on f(x). Compring with one of the properties of (x), we find tht [f(x)] = (x x 0) (4.44) df /dx x=x0 where the derivtive is evluted t the point x = x 0 where f(x 0 ) = 0. Note tht if there is more thn one solution (x i ; i =,...)tof =0,then (f) issum [f(x)] = X i (x x i ) df /dx x=xi (4.45) 4..5 Proving the sifting property We cn use the limit definition of the Dirc delt function [Eqn. (4.36)] to prove the sifting property given in Eqn. (4.37): Z dx f(x) (x) = Z Z dx f(x)lim (x) =lim dx f(x) (x). (4.46)!0!0 We re free to pull the limit outside the integrl becuse nothing else depends on. Substituting for (x), the integrl is only non-zero between nd. Similrly,wecnpullthenormliztion fctor out to the front: Z Z dx f(x) (x) =lim dx f(x). (4.47)!0 2 Wht this is doing is verging f over nrrow rnge of width 2 round x = 0. Provided the function is continuous, this will converge to well-defined vlue f(0) s! 0(thisisprettywell the definition of continuity). Alterntively, suppose the function ws di erentible t x = 0(whichnotllcontinuousfunctions will be: e.g. f(x) = x ). Then we cn Tylor expnd the function round x =0(i.e. theposition of the centre of the spike): Z dx f(x) (x) =lim!0 2 Z dx pplef(0) + xf 0 (0) + x2 2! f 00 (0) (4.48) The dvntge of this is tht ll the f (n) (0) re constnts, which mkes the integrl esy: Z pple x dx f(x) (x) =lim f(0) [x]!0 2 + f 0 2 (0) + f pple 00 (0) x ! 3 =lim f(0) + 2!0 6 f 00 (0) +... which gives the sifting result. (4.49) (4.50) = f(0). (4.5) EXAM TIP: An exm question my sk you to derive the sifting property in this wy. Mke sure you cn do it. Note tht the odd terms vnished fter integrtion. This is specil to the cse of the spike being centred t x = 0. It is useful exercise to see wht hppens if the spike is centred t x = d insted. 34

17 4..6 Some other properties of the delt function These include: ( x) = (x) x (x) =0 (x) = (x 2 2 )= (x) (x )+ (x + ) 2 (4.52) The proofs re left s exercises Clculus with the delt function The -function is esily integrted: Z x dy (y d) = (x d), (4.53) where (x d) = ( 0 x<d x d which is clled the Heviside function, orjustthe stepfunction. The derivtive is lso esy to write down, just by pplying the product rule to x (x) =0: (4.54) (x)+x d dx (x) =0) d dx In the workshops, we will look t how the derivtive of the (x) = (x)/x. (4.55) -function cn be used More thn one dimension Finlly, in some situtions (e.g. point chrge t r = r 0 ), we might need 3D Dirc delt function, which we cn write s product of three D delt functions: (r r 0 )= (x x 0 ) (y y 0 ) (z z 0 ) (4.56) where r 0 =(x 0,y 0,z 0 ). Note tht (r ) isnotthesmes (r ): the former picks out point t position, buttheltterpicksoutnnnulusofrdius. Supposewehdsphericlly symmetric function f(r). The sifting property of the 3D function is Z f(r) (r ) d 3 x = f() =f(), (4.57) wheres Z Z f(r) (r ) d 3 x = f(r) (r ) 4 r 2 dr =4 2 f(). (4.58) 35

18 4..9 Physicl importnce of the delt function The -function is tool tht rises gret del in physics. There re number of resons for this. One is tht the clssicl world is mde up out of discrete prticles, even though we often tret mtter s hving continuously vrying properties such s density. Individul prticles of zero size hve infinite density, nd so re perfectly suited to be described by -functions. We cn therefore write the density field produced by set of prticles t positions x i s (x) = X i M i (x x i ). (4.59) This expression mens we cn tret ll mtter in terms of just the density s function of position, whether the mtter is continuous or mde of prticles. This decomposition mkes us look in new wy t the sifting theorem: Z f(x) = f(q) (x q) dq. (4.60) The integrl is the limit of sum, so this ctully sys tht the function f(x) cnbethoughtofs mde up by dding together infinitely mny -function spikes. This turns out to be n incredibly useful viewpoint when solving liner di erentil equtions: the response of given system to n pplied force f cn be clculted if we know how the system responds to single spike. This response is clled Green s function, nd will be mjor topic lter in the course. 4.2 FT nd integrl representtion of (x) The Dirc delt function is very useful when we re doing FTs. The FT of the delt function follows esily from the sifting property: f(k) = Z In the specil cse d = 0, we get simply f(k) =. dx (x d) e ikx = e ikd. (4.6) The inverse FT gives us the integrl representtion of the delt function: (x d) = 2 = 2 Z Z dk f(k)e ikx = 2 Z dk e ikd e ikx (4.62) dk e ik(x d). (4.63) You ought to worry tht it s entirely unobvious whether this integrl converges, since the integrnd doesn t die o t. A sfer pproch is to define the -function (sy) in terms of Gussin of width, where we know tht the FT nd inverse FT re well defined. Then we cn tke the limit of! 0. In the sme wy tht we hve defined delt function in x, wecnlsodefinedeltfunction in k. This would, for instnce, represent signl composed of oscilltions of single frequency or wvenumber K. Agin, we cn write it in integrl form if we wish: (k K) = 2 Z 36 e i(k K)x dx. (4.64)

19 This k-spce delt function hs exctly the sme sifting properties when we integrte over k s the originl version did when integrting over x. Note tht the sign of the exponent is irrelevnt: (x) = 2 Z e ±ikx dk (4.65) which is esy to show by chnging vrible from k to chnge dk! dk). k (the limits swp, which cncels the sign FOURIER ANALYSIS: LECTURE 0 5 Ordinry Di erentil Equtions We sw erlier tht if we hve liner di erentil eqution with driving term on the right-hnd side which is periodic function, then Fourier Series my be useful method to solve it. If the problem is not periodic, then Fourier Trnsforms my be ble to solve it. 5. Solving Ordinry Di erentil Equtions with Fourier trnsforms The dvntge of pplying FT to di erentil eqution is tht we replce the di erentil eqution with n lgebric eqution, which my be esier to solve. Let us illustrte the method with couple of exmples. 5.. Simple exmple The eqution to be solved is Tke the FT, which for z is: nd noting tht d/dt! i!, sod 2 /dt 2! d 2 z dt 2! 2 0z = f(t). (5.66) z(!) = Z z(t)e i!t dt (5.67)! 2,theequtionbecomes! 2 z(!)! 2 0 z(!) = f(!). (5.68) Rerrnging, with solution z(!) = z(t) = Z 2 f(!)! 2 0 +! 2 (5.69) f(!)! 2 0 +! 2 ei!t d!. (5.70) 37

20 Wht this sys is tht single oscillting f(t), with mplitude, willgenerteresponseinphse with the pplied oscilltion, but of mplitude /(! 2 0 +! 2 ). For the generl cse, we superimpose oscilltions of di erent frequency, which is wht the inverse Fourier trnsform does for us. Note tht this gives prticulr solution of the eqution. Normlly, we would rgue tht we cn lso dd solution of the homogeneous eqution (where the rhs is set to zero), which in this cse is Ae! 0t + Be! 0t.BoundryconditionswoulddeterminewhtvluestheconstntsA nd B tke. But when deling with Fourier trnsforms, this step my not be pproprite. This is becuse the Fourier trnsform describes non-periodic functions tht stretch over n infinite rnge of time so the mnipultions needed to impose prticulr boundry condition mount to prticulr imposed force. If we believe tht we hve n expression for f(t) thtisvlidforlltimes,then boundry conditions cn only be set t t =. Physiclly,wewouldnormllylcknyresonfor displcementinthislimit,sothehomogeneoussolutionwouldtendtobeignored eventhough it should be included s mtter of mthemticl principle. We will come bck to this problem lter, when we cn go further with the clcultion (see Convolution, section6). 5.2 LCR circuits Let us look t more complicted problem with n electricl circuit. LCR circuits consist of n inductor of inductnce L, cpcitorofcpcitncec nd resistor of resistnce R. If they re in series, then in the simplest cse of one of ech in the circuit, the voltge cross ll three is the sum of the voltges cross ech component. The voltge cross R is IR, wherei is the current; cross the inductor it is LdI/dt, ndcrossthecpcitoritisq/c, whereq is the chrge on the cpcitor: V (t) =L di dt + RI + Q C. (5.7) Now, since the rte of chnge of chrge on the cpcitor is simply the current, dq/dt = I, wecn di erentite this eqution, to get second-order ODE for I: L d2 I dt 2 + RdI dt + I C = dv dt. (5.72) If we know how the pplied voltge V (t) vrieswithtime,wecnusefouriermethodstodetermine I(t). With Ĩ(!) =R I(t)e i!t dt, nd noting tht the FT of di/dt is i!ĩ(!), nd of d2 I/dt 2 it is! 2 Ĩ(!). Hence where Ṽ (!) is the FT of V (t). Solving for Ĩ(!): nd hence! 2 LĨ(!)+i!RĨ(!)+ Ĩ(!) =i!ṽ (!), (5.73) C Ĩ(!) = I(t) = Z 2 i!ṽ (!) C + i!r! 2 L, (5.74) i!ṽ (!) C + i!r! 2 L ei!t d!. (5.75) So we see tht the individul Fourier components obey form of Ohm s lw, but involving complex impednce, Z: Ṽ (!) =Z(!)Ĩ(!); Z = R + i!l i!c. (5.76) 38

21 Figure 5.7: A simple series LCR circuit. This is very useful concept, s it immeditely llows more complex circuits to be nlysed, using the stndrd rules for dding resistnces in series or in prllel. The frequency dependence of the impednce mens tht di erent kinds of LCR circuit hve functions s filters of the time-dependent current pssing through them: di erent Fourier components (i.e. di erent frequencies) cn be enhnced or suppressed. For exmple, consider resistor nd inductor in series: Ĩ(!) = Ṽ (!) R + i!l. (5.77) For high frequencies, the current tends to zero; for! R/L, theoutputofthecircuit(current over voltge) tends to the constnt vlue Ĩ(!)/Ṽ (!) =R. Sothiswouldbeclledlow-pss filter: it only trnsmits low-frequency vibrtions. Similrly, resistor nd cpcitor in series gives Ĩ(!) = Ṽ (!) R +(i!c). (5.78) This cts s high-pss filter, removingfrequenciesbelowbout(rc). Note tht the LR circuit cn lso ct in this wy if we mesure the voltge cross the inductor, V L,rtherthnthecurrent pssing through it: Ṽ (!) Ṽ L (!) =i!lĩ(!) =i!l R + i!l = Ṽ (!) +R(i!L). (5.79) Finlly, full series LCR circuit is bnd-pss filter, whichremovesfrequenciesbelow(rc) nd bove R/L from the current. 5.3 The forced dmped Simple Hrmonic Oscilltor The sme mthemtics rises in completely di erent physicl context: imgine we hve mss m ttched to spring with spring constnt k, ndwhichislsoimmersedinviscousfluidtht exerts resistive force proportionl to the speed of the mss, with constnt of proportionlity D. Imgine further tht the mss is driven by n externl force f(t). The eqution of motion for the displcement z(t) is m z = kz Dż + f(t). (5.80) 39

22 This is the sme eqution s the LCR cse, with (z,f,m,k,d)! (I, V,L,C,R). (5.8) To solve the eqution for z(t), first define chrcteristic frequency by! 0 2 = k/m, ndlet = D/m. Then z + ż +! 0z 2 = (t) (5.82) where (t) =f(t)/m. Now tke the FT of the eqution with respect to time, nd note tht the FT of ż(t) isi! z(w), nd the FT of z(t) is! 2 z(!). Thus where ã(!) is the FT of (t). Hence! 2 z(!)+i! z(!)+! 2 0 z(!) =ã(!), (5.83) z(!) = ã(!). (5.84)! 2 + i! +! Explicit solution vi inverse FT This solution in Fourier spce is generl nd works for ny time-dependent force. Once we hve specificformfortheforce,wecninprincipleusethefourierexpressiontoobtintheexct solution for z(t). How useful this is in prctice depends on how esy it is to do the integrls, first to trnsform (t) intoã(!), nd then the inverse trnsform to turn z(!) intoz(t). For now, we shll just illustrte the pproch with simple cse. Consider therefore driving force tht cn be written s single complex exponentil: Fourier trnsforming, we get (t) =A exp( t). (5.85) ã(!) = Z Ae i t e i!t dt =2 A (!) =2 A (! ). (5.86) Unsurprisingly, the result is -function spike t the driving frequency. Since we know tht z(!) = ã(!)/(! 2 + i! +! 0), 2 we cn now use the inverse FT to compute z(t): z(t) = 2 = A = A Z Z ã(!) e i!t d! (5.87)! 2 + i +! 0 2 (! ) e i!t d!! 2 + i! +! 0 2 e i t 2 + i +! 2 0 This is just the nswer we would hve obtined if we hd tken the usul route of trying solution proportionl to exp(i t) but the nice thing is tht the inverse FT hs produced this for us utomticlly, without needing to guess. 40

23 5.3.2 Resonnce The result cn be mde bit more intuitive by splitting the vrious fctors into mplitudes nd phses. Let A = A exp(i )nd( 2 + i +! 0)= 2 exp(i ), where q = (! ) (5.88) nd Then we hve simply tn = /(! ). (5.89) z(t) = A exp[i( t + )], (5.90) so the dynmicl system returns the input oscilltion, modified in mplitude by the fctor / nd lgging in phse by. For smll frequencies, this phse lg is very smll; it becomes /2 when =! 0 ;forlrger,thephselgtendsto. The mplitude of the oscilltion is mximized when is minimum. Di erentiting, this is when we rech the resonnt frequency: q = res =! /2, (5.9) i.e. close to the nturl frequency of the oscilltor when is smll. At this point, =( 2! /4) /2,whichis! 0 to leding order. From the structure of, wecnseethtitchngesbylrge fctor when the frequency moves from resonnce by n mount of order (i.e. if is smll then the width of the response is very nrrow). To show this, rgue tht we wnt the term (! ) 2, which is negligible t resonnce, to be equl to 2! 0.Solvingthisgives 2 =(! 2 0! 0 ) /2 =! 0 ( /! 0 ) /2 '! 0 /2. (5.92) Thus we re mostly interested in frequencies tht re close to! 0,ndwecnpproximte by Thus, if we set = res +, then ' [( 2! 2 0) 2 + 2! 2 0] /2 ' [4! 2 0(! 0 ) 2 + 2! 2 0] /2. (5.93) ' (! 0 ), (5.94) ( / 2 ) /2 which is Lorentzin dependence of the squre of the mplitude on frequency devition from resonnce Tking the rel prt? The introduction of complex ccelertion my cuse some concern. A common trick t n elementry level is to use complex exponentils to represent rel oscilltions; the rgument being tht (s long s we del with liner equtions) the rel nd imginry prts process seprtely nd so we cn just tke the rel prt t the end. Here, we hve escped the need to do this by sying tht rel functions require the Hermitin symmetry c m = c n. If (t) istoberel,wemusttherefore lso hve the negtive-frequency prt: ã(!) =2 A (! ) + 2 A (! + ). (5.95) 4

24 The Fourier trnsform of this is (t) =A exp(i t)+a exp( i t) =2 A cos( t + ), (5.96) where A = A exp(i ). Aprt from fctor 2, this is indeed wht we would hve obtined vi the trditionl pproch of just tking the rel prt of the complex oscilltion. Similrly, therefore, the time-dependent solution when we insist on this rel driving force of given frequency comes simply from dding the previous solution to its complex conjugte: z(t) = A exp[i( t + A )] + exp[ i( t + )] = 2 A cos( t + ). (5.97) 42