Royal Holloway Uiversity of odo Departet of Physics Orthogoal Fuctios Motivatio Aalogy with vectors You are probably failiar with the cocept of orthogoality fro vectors; two vectors are orthogoal whe they ake a agle of 90º with each other We will explore the utility of this property of (basis) vectors, before applyig the ideas by aalogy to fuctios A vector v i 3d space ca be expressed i ters of its copoets v = v i+ v x yj+ vxk Here i, j, kare uit vectors (basis vectors) i the x, y, ad z directios, ad v x, v y, ad v z are the copoets of the vector i these directios The physical vector v is specified whe we have a) the basis set of uit vectors,, i j kad b) the coefficiets v x, v y, ad v z You ca chage the basis set (chagig the orietatio of the coordiate frae) The the coefficiets v x, v y, ad v z will chage, but they describe the sae physical vector The orthogoality of i, j, k is expressed i ters of the dot product relatios ij = jk = ki = 0 ii = jj = kk = t is a iportat property of the dot product that it takes two vectors ad gives a scalar fro the Deteriig the coefficiets Give a vector v ad a basis set = i, j, k B, the orthogoality of the basis set allows us to deterie the coefficiets = vx, vy, vzb = B we take the dot product of v with the,, To fid the vx, vy, vz i j k Recall that the dot product of v with a vector i gives the copoet of v alog i: vi = vi cosθ Takig the dot product we fid vi = 3v i j k x + vy + vx 8 i = v ii + v ji + v ki x y x Orthogoality causes the last two ters to vaish, ad ii =, so that vi = v x Applyig the sae procedure to the y ad the z copoets we obtai PH30 Matheatical Methods
Royal Holloway Uiversity of odo Departet of Physics v v v x y z = vi = vj = vk More copact otatio We ca siplify the equatios by usig the sybol α (or ay other Greek character) to deote the directios x, y, z Thus if we deote vx, vy, vz by v α where α = x, y, z i, j, k by i α where α = x, y, z the we ay express v i ters of its copoets as v = v αiα Here α = xyz,, v is the vector {v α } represet the coefficiets { i α } represet the basis vectors The orthogoality of the basis vectors is expressed by i α iβ = 0 if α β = if α = β Ad this ay be writte i a eve ore copact for by the use of the Kroeker delta sybol δ αβ which has the defiig property δ = αβ 0 if α β = if α = β ters of the Kroeker delta sybol the orthogoality of the basis vectors is siply i i = δ α β αβ Deteriig the coefficiets (agai) Usig the ore copact otatio together with the Kroeker delta sybol, we shall see how to obtai the vector s coefficiets i a ore straightforward aer The vector is writte, i ters of its coefficiets ad basis vectors as v = v αiα α = xyz,, We take the dot product of this with a basis vector i β (Sice α was used as the duy variable i the su, we use a differet variable β here) Takig the dot product gives vi i β = vα α iβ But α = xyz,, i i = δ α β αβ PH30 Matheatical Methods
Royal Holloway Uiversity of odo Departet of Physics so that vi β = vαδ αβ α = xyz,, Now α rages over x, y, z But δ αβ akes each ter zero except whe α = β e the effect of the Kroeker delta is to pick out just the β ter fro the su over α Thus we coclude that vi β = v β So this has deteried the coefficiets v α = vi α Reeber that α is a duy variable, = x, y, z The iportat cocepts of this sectio are: Orthogoality Basis set of uit vectors Coefficiets/coordiates v α, i α otatio, where α rages over x, y, z Kroeker delta sybol Deterie coefficiets, usig orthogoality of basis vectors, by takig dot product with each basis vector PH30 Matheatical Methods 3
Royal Holloway Uiversity of odo Departet of Physics Orthogoal fuctios Fourier series Reeber the Fourier series (o the iterval x ): a0 % f x a x b 6= + x & + ' ( cos si ) = * Outside the iterval x, f(x) repeats itself a ifiite, spatially periodic fuctio We ight exploit this, or we ight just be cosiderig x withi the specified iterval The a 0 / ter is really the = 0 ter of the cos series The factor ½ is coveiet as the the sae forula (whe we fid it) holds for all the a The Fourier series is a bit like writig a vector v as a liear su of basis vectors with appropriate coefficiets Here we have a fuctio f(x) writte as a liear su of basis fuctios cos(x/), si(x/) (iteger ) with appropriate coefficiets a, b The questio is: How to fid the coefficiets a, b? a is the aout of cos(x/) i f(x) b is the aout of si(x/) i f(x) We could do with soethig like a dot product ad the cocept of orthogoality We will itroduce the idea of a ier product as the geeralizatio of the dot product This will take two fuctios ad give a scalar fro the Cosider the followig itegrals for positive itegers, cos cos x x x d = 0 uless = si si x x x d = 0 uless = cos si x x x d = 0 eve if = We ca regard the itegrals as givig the ier products of the basis fuctios cos(x/) ad si(x/) Whe = the itegrals are cos si This gives us the orthogoality relatios = = x d x x d x PH30 Matheatical Methods 4
Royal Holloway Uiversity of odo Departet of Physics cos cos x x x δ d = si si x x x δ d = cos si x x x d = 0 Note that this aspect of orthogoality has othig to do with agles; it relates solely to the Kroeker delta properties of the ier products of the basis fuctios just as the orthogoality of vectors relates to the Kroeker delta properties of the basis uit vectors We ca use the orthogoality properties of the basis fuctios cos(x/) ad si(x/) to fid the Fourier copoets a ad b We use the basic rule: RUE Take the ier product of the fuctio f(x) ad oe of the basis fuctios other words, ultiply f(x) by oe of the basis fuctios, say cos(x/), ad itegrate over the iterval x Sice a0 % f x a x b 6= + x & + ' ( cos si ) = * we the have f x x x a 6cos cos x x 0 = d d a x + x x cos cos d = b x + x x si cos d = So log as 0, the first lie is zero The si cos itegral esures the third lie vaishes The secod lie is a su over iteger ad the orthogoality relatio picks out the = ter oly Thus the expressio reduces to f x x x a 6cos x x cos = d d = a This gives the expressio for the cos coefficiets of the Fourier series We ca perfor the sae process usig the si basis fuctios That is, ultiplyig f(x) by si(x/) ad itegratig over the iterval x This gives, i exactly the sae way, the si coefficiets of the Fourier series Thus we fid a f x x x = 6 cos d b f x x x = 6 si d PH30 Matheatical Methods 5
Royal Holloway Uiversity of odo Departet of Physics These are kow as the Euler forulae for the Fourier coefficiets You ca see that the forula for a gives the correct result for the = 0 ter of the series: a 0 = f6d x x This is the ea value of f(x) over the iterval Exaple sawtooth wave 05-05 Sawtooth wave x This fuctio is specified by f6= x, x The Euler forulae for the Fourier coefficiets are a f x x x = 6 cos d b f x x x = 6 si d So i this case we have a x x x = cos d b x x x = si d The itegral for a vaishes This is because f(x) is a odd fuctio The expressio for a idicates that a odd fuctio has o cos ters f we evaluate the b itegral usig Matheatica the it gives the result b = si cos 6 This is ot i its siplest for sice Matheatica does ot kow that we are oly iterested i itegral - PH30 Matheatical Methods 6
Royal Holloway Uiversity of odo Departet of Physics Treatig quatities like si The easiest way of workig with quatities like si ad cos whe is restricted to iteger values is to write these quatities as the iagiary part ad the real part of e i, ad to use de Moivre s theore i cos + isi = e i =e 7 Now e i =, so that cos + isi = 6 ad upo takig the real ad iagiary parts of this we obtai, for iteger cos = 6 si = 0 The si Fourier copoet of the sawtooth wave b = si cos 6 is the siply b = 6 The Fourier series for the sawtooth fuctio f 6= x b x b + x + b 3 x si si 3si + is thus give by f 6= x x x + 3 x 4 si si si si x 3 + Recall that there are o cos ters because f(x) is a odd fuctio of x Covergece of the Fourier series Usig just the first ter of the series gives the fudaetal copoet of the series 06 04 0-0 -04-06 First ter of the Fourier series PH30 Matheatical Methods 7
Royal Holloway Uiversity of odo Departet of Physics This reflects the periodicity of the sawtooth fuctio Addig the secod ter of the series gives a slight iproveet (does it?) 075 05 05-05 -05-075 First two ters of the Fourier series We ca see how the ters gradually build up to the required fuctio by lookig at the partial sus of the first oe, two, three ad four ters 05-05 - Sus of oe, two, three ad four ters For twety ters the sawtooth is lookig pretty realistic Observe the wiggles i the viciity of the sharp corers This is kow as the Gibbs pheoeo ad it is a feature of Fourier series whe there are discotiuities i the fuctio 05-05 - The first 0 ters of the Fourier series PH30 Matheatical Methods 8
Royal Holloway Uiversity of odo Departet of Physics Addig ore ters gives a iproveet to the probleatic parts: 05-05 - The first 50 ters of the Fourier series The Gibbs pheoeo is still there, but o a fier scale Exaple triagular wave 05-05 Triagular wave This fuctio ust be specified i a piecewise fashio The fuctio is defied o the iterval x Ad the triagular profile ay be expressed as f x x 6= + x for 0 = x x for 0 We wat to fid the Fourier series a0 f x a x b 6 = + x + cos si %&' = ()* for this fuctio, ad we have the Euler expressio for the coefficiets: a f x x x = 6 cos d b f x x x = 6 si d - PH30 Matheatical Methods 9
Royal Holloway Uiversity of odo Departet of Physics The itegrals ust be doe i two parts sice the expressio for f(x) is differet for positive ad for egative x a f x x x = 6 cos d f x x x = f x x x 0 6 + cos d 6cos d 0 x x x = + x x x 0 + cos d cos d 0 with a siilar expressio for the b : a x x x x x x = + 0 + cos d cos d 0 b x x x x x x = + 0 + si d si d 0 These itegrals ay be evaluated (usig Matheatica or by had), givig 4 a = cos 6 b = 0 this case there are o si ters, because this fuctio is eve i x You ca see this geeral property fro the Euler expressio for the b coefficiets We ca siplify the cos ter as we did i the previous case There we saw that cos = 6, so that 4 a = 4 69 Observe that this is tellig us that all the eve ters vaish (icludig he costat = 0 ter) The values of a are 8 a % = &' 0 (,,,,,,,,, 0,,, )* 0 0 3 5 7 ad these ay be represeted i the bar chart: 08 0 9 0 06 04 0 3 4 5 6 7 8 9 0 Fourier cos coefficiets of triagular wave PH30 Matheatical Methods 0
Royal Holloway Uiversity of odo Departet of Physics Covergece of the Fourier series Usig just the first ter of the series gives the fudaetal copoet of the series 075 05 05-05 -05-075 First ter of the Fourier series This reflects the periodicity of the triagular fuctio Addig the ext ter of the series gives a sigificat iproveet 075 05 05-05 -05-075 First two ozero ters of the Fourier series We ca see how the ters gradually build up to the required fuctio by lookig at the partial sus of the first, third, fifth ad seveth 075 05 05-05 -05-075 Sus of the first, third, fifth ad seveth ters For 0 ters the triagle wave looks very good PH30 Matheatical Methods
Royal Holloway Uiversity of odo Departet of Physics 05-05 Su of the first twety ters There is o Gibbs pheoeo sice the fuctio is ot discotiuous - Exaple 3 square wave 05-05 - Square wave The square wave fuctio show is specified by f6= x for x 0 =+ for 0 x We wat to fid the Fourier series for this fuctio We ote that as represeted here, this is a odd fuctio Thus there are o cos ters i the expasio The coefficiets if the si ters, the b, are give by b f x x x = 6si d x x = x x + 0 si d si d 0 We ca siplify this expressio a little by substitutig x xi the first ter This gives b x x = si d 0 which, upo itegratio is PH30 Matheatical Methods
Royal Holloway Uiversity of odo Departet of Physics 6 b = cos = 4 69 As i the case of the triagular wave, the eve ters vaish The values of b are b % = 4 &' ( 0,,, 3 0,, 5 0,, 7 0,, 9 0,, 0, )* ad these ay be represeted i the bar chart: 08 06 04 0 3 4 5 6 7 8 9 0 Fourier si coefficiets of square wave Covergece of the Fourier series Usig just the first ter of the series gives the fudaetal copoet of the series 05-05 - First ter of the Fourier series This reflects the periodicity of the square wave Addig the ext ter looks a little better PH30 Matheatical Methods 3
Royal Holloway Uiversity of odo Departet of Physics 05-05 Fudaetal ad ext (3 rd ) haroic of square wave We ca see how the ters gradually build up to the required fuctio by lookig at the partial sus of the first, third, fifth ad seveth - 05-05 - Gradual build-up of square wave The su up to the 0 th haroic looks quite good 05-05 Fourier series for square wave up to 0 th haroic The Gibbs pheoeo is apparet i the viciity of the discotiuities This ay be see ost clearly i the expaded regio - PH30 Matheatical Methods 4
Royal Holloway Uiversity of odo Departet of Physics 005 0 05 0 08 06 04 Viciity of x = 0 showig the Gibbs pheoeo The su to the 50 th haroic looks uch iproved, but the Gibbs pheoeo is still preset 05-05 Su to 50 th haroic Exaple 4 Full wave rectified sie curve full wave rectificatio positive regios of a sigal reai positive, ad egative regios of a sigal are iverted to that they, also, appear positive We shall cosider a full wave rectifies si curve(actually a full wave rectified cos curve) - 08 06 04 0 Full wave rectified (co)sie wave PH30 Matheatical Methods 5
Royal Holloway Uiversity of odo Departet of Physics The fuctio is specified i the iterval x this iterval f(x) is give, siply, by f 6= x cos x This is a eve fuctio so we kow that there will oly be cos ters i the Fourier series The Euler forula gives the Fourier cos coefficiets a x x x = cos cos d 4 cos = 4 Usig the trick for cos we write this as 6 4 a = 4 The values of a are 4 a =,,,,,,,, %&' 3 5 35 63 99 43 95 55 ad these ay be represeted i the bar chart: ()* 04 03 0 0 3 4 5 6 7 8 9 0 Fourier coefficiets of full wave rectified cos wave this case there is a = 0 (costat) ter, sice the ea of the fuctio is ot zero This is give by a 0 = 4/ PH30 Matheatical Methods 6
Royal Holloway Uiversity of odo Departet of Physics Covergece of the Fourier series Usig the costat a 0 together with just the first ter of the series gives the fudaetal copoet of the series 08 06 04 0 Fudaetal of full rectified sie wave Addig the secod haroic flattes the top ad sharpes the botto 08 06 04 0 Ters up to the secod haroic The first few partial sus show how the Fourier series approaches the fuctio 08 06 04 0 Gradual build-up of full wave rectified sie wave With the su to 0 ters the fuctio is lookig pretty good PH30 Matheatical Methods 7
Royal Holloway Uiversity of odo Departet of Physics 08 06 04 0 Su to 0 ters Fially we plot the differece betwee the 0 ter series ad the origial fuctio 005 00 0005 - -05 05-0005 Error i 0 ter Fourier series The iportat cocepts of this sectio are: Expressio of a fuctio as a Fourier series Orthogoality itegrals for sies ad cosies Euler forulae for Fourier coefficiets For piecewise defied fuctios do Euler itegrals i separate bits Odd fuctios use oly sies; eve fuctios use oly cosies Costat a 0 ter, eeded whe ea of fuctio is ot zero Gibbs pheoeo whe fuctio is discotiuous Teriology: fudaetal, haroics Sall uber of ters eeded whe the coefficiets decrease rapidly PH30 Matheatical Methods 8