4 Fourier series Any LI sysem is compleely deermined by is impulse response h(). his is he oupu of he sysem when he inpu is a Dirac dela funcion a he origin. In linear sysems heory we are usually more ineresed in how a sysem responds o signals a differen frequencies. When we alk abou a signal of frequency ω, we mean he signal e jω. his is he only signal ha will conain boh a variable and a ω variable in is specificaion i defines he relaionship beween he ime and frequency domains. Noe ha j =, so he signal is complex valued. he perinen quesion is his: wha happens o a pure frequency when i passes hrough a paricular linear sysem? Assuming he inpu is x() = e jω, and assuming ha he frequency ω is fixed a some value of ineres, he oupu is simple o derive from convoluion: y() = h(τ)x( τ)dτ = h(τ)e jω( τ) dτ = ( ) = h(τ)e jωτ dτ e jω = H(ω)e jω. h(τ)e jωτ e jω dτ Since ω is fixed, H(ω) is a number, possibly complex valued, ha depends on he impulse response h(). We see ha he sysem herefore has he following inpu-oupu pair: e jω H(ω)e jω. Looking a i differenly, if he inpu-oupu relaion of he sysem is wrien as y() = {x()}, hen he complex exponenial inpu saisfies he propery { e jω} = H(ω)e jω. his is an eigen relaion: he ransformaion applied o he signal e jω resuls in he same signal e jω, bu muliplied by a consan H(ω) ha depends on he frequency ω and is deermined by he sysem. Complex exponenial funcions (or pure frequencies) are characerisic funcions for LI sysems: hey propagae hrough wihou change, excep for an overall (complex) scaling. For any given sysem he scaling depends on he frequency of he complex exponenial. Alernaively, LI sysems canno creae frequencies a he oupu ha were no already presen a he inpu: hey can only modify hem by a complex-valued scaling. he funcion H(ω) = h(τ)e jωτ dτ is called he frequency response of he sysem. For any value of ω i ells us how he sysem responds o an inpu frequency e jω : he oupu will be H(ω)e jω. If you know he impulse response h() of a sysem, hen you also know is frequency response H(ω) from he formula given. he quaniy H(ω) is also called he ransfer funcion of he sysem. he Fourier series decomposiion allows us o express any periodic signal x() wih period as a linear combinaion (or weighed sum) of a counable se of frequencies: x() = c k e jkω for some coefficiens c k. Puing his signal hrough an LI sysem is simple: he oupu is { } y() = {x()} = c k e jkω = c k { e jkω} = c k H(kω )e jkω = d k e jkω, 4-
wih d k = c k H(kω ). he ransformaion of coefficiens d k = c k H(kω ) ells us exacly how he differen frequency componens in he signal are affeced by he sysem. hus if we can ge used o hinking abou signals in he frequency domain, we have a very simple way of inerpreing he acion of LI sysems on signals. Complex exponenial are he eigenfuncions of LI sysems: hey are he only funcions ha propagae hrough linear sysems wihou change excep for muliplicaion by a complex scale facor. Coupled wih he fac ha any periodic signal can be expressed as a weighed sum of a se of complex exponenials, his gives a very inuiive descripion of a sysem: inpus are linear combinaions of complex exponenials, and oupus are linear combinaions of responses o complex exponenials. he weighs in he linear combinaions a he oupu are relaed o he weighs a he inpu by muliplicaion wih he sysem ransfer funcion H(ω). Mos signals in he world are real. I may seem srange o express a real-valued signal as a linear combinaion of complex signals. Noneheless, he ruh of he maer is ha i makes hings simpler, boh in erms of undersanding and in erms of algebra. Complex exponenials are he naural building blocks of signals, even hough we re usually only ineresed linear combinaions of complex exponenials ha end up being real valued. 4. Complex numbers: a review Leing j = we can wrie a complex number in wo ways: Recangular form: s = a+jb, a = Re(s) (real par) b = Im(s) (imaginary par) Polar form: s = ρe jθ, ρ = s (nonnegaive magniude) θ = s (phase). Euler s formula saes ha e jθ = cos(θ)+jsin(θ) and links hese wo represenaions algebraically. his can be used direcly o conver a complex number from polar o recangular form: s = ρe jθ = ρ(cos(θ)+jsin(θ)) = ρcos(θ)+jρsin(θ) = a+jb, wih a = ρcos(θ) and b = ρsin(θ). o cover from recangular o polar form we noe ha and a 2 +b 2 = ρ 2 cos 2 (θ)+ρ 2 sin 2 (θ) = ρ 2 (cos 2 (θ)+sin 2 (θ)) = ρ 2 b/a = sin(θ) cos(θ) = an(θ), so he magniude and phase componens of he complex number are given by ρ = a 2 +b 2 { arcan(b/a) a θ = 8 +arcan (b/a) a <. hese represenaions are linked very naurally via he correspondence of complex numbers wih he 2D plane, also called he Argand diagram: 4-2
Im b s = a + jb s s a Re Doing algebra on complex numbers is easy as long as hey are expressed in he appropriae form. Defining he wo complex numbers s and s 2 according o he following operaions are simple: s = a +jb = ρ e jθ and s 2 = a 2 +jb 2 = ρ 2 e jθ2, Addiion: s +s 2 = (a +jb )+(a 2 +jb 2 ) = (a +a 2 )+j(b +b 2 ) Subracion: s s 2 = (a +jb ) (a 2 +jb 2 ) = (a a 2 )+j(b b 2 ) Muliplicaion: s s 2 = (ρ e jθ )(ρ 2 e jθ2 ) = ρ ρ 2 e j(θ+θ2) Division: s /s 2 = (ρ e jθ )(ρ 2 e jθ2 ) = (ρ /ρ 2 )e j(θ θ2). o add or subrac wo complex numbers, express hem in recangular form and he operaion follows. o muliply or divide complex numbers, use polar forms. Noe ha as far as algebra is concerned you can hink of j as a normal variable i jus happens o have a value of. Consider a plo of he funcion x() = sin(2), expressed in polar form. he signal x() is shown below, along wih x() and x(): x() - - -.5.5.5 2 x().5 - -.5.5.5 2 x() - -.5.5.5 2 Noe ha, even hough he signal is real, he phase is nonzero. his becomes obvious if you plo he number on he Argand diagram: 4-3
Im Re Evidenly when expressed in polar form we have = e j, which has a magniude of and a phase of. he same is rue for any negaive real number: i has a phase of, because he magniude mus be posiive. However, noe also ha = e j we re jus measuring he angle in he negaive direcion insead of he posiive direcion. Phase is no deermined up o addiion by a muliple of 2: a phase of.3 radians is equivalen o a phase of.3+2 radians or.3+(2)2 radians or.3 2 radians we re jus going around he uni circle by muliples of a full circle, bu he poin in he Argand diagram (and hence he corresponding complex number) is unchanged. hus, in he plo of x() we could have given he porions of he signal indicaed wih he value he value +k(2) for any ineger k. Posiive real numbers have phases of,±2,±4,...; negaive real numbers have phases of ±,±3,... 4.2 Fourier series formulaion Suppose we are given a signal x s (), which can be a real or complex valued funcion of a (real) ime variable. Furhermore, suppose ha he signal is periodic wih period : for all we have x s () = x s (+). We define he (parameric) signal x() o be he weighed sum of an infinie se of complex exponenials x() = a k e j(2k/), for some given se of coefficien weighs {a k } = {...,a,a,a,...}. Noe ha we allow hese coefficiens o be complex numbers, and in general he funcion x() can be complex valued. We can describe he relaion above by saying ha x() can be wrien as a linear combinaion or weighed sum of complex exponenial signals e j(2k/). he weighs of each complex exponenial are he values a k in he summaion. We will no prove i in his course, bu if x s () is sufficienly well behaved (like all real-world signals are), hen we can always find a se of coefficiens a k for k Z such ha x() is almos exacly equal o x s (). heperiodicsignalx() canbe compleelydescribedbyhe seofvaluesha iakesfor <. Formally, o know x() we jus need o know all values of he (infinie) se {x() R, < } (he se of values of x() such ha is real and < ). From he previous asserion x() can alernaively be described by he se of coefficiens a k for k Z, or he elemens of he (infinie) Almos exacly in his conex means ha for any ǫ > we can find a se of coefficiens a k such ha x() x s() 2 d < ǫ. his means ha he difference in he energy of he wo signals is arbirarily small, bu does no necessarily mean ha x() = x s() for all. (he difference beween hese wo saemens is quie suble, and unimporan for mos pracical purposes.) 4-4
se {a k k Z} (he se of values of a k such ha k is any ineger). For our purposes hese wo descripions are exacly equivalen: if a k for k Z are specified hen x() is compleely deermined, and any x() corresponds o a se of coefficiens a k ha compleely describes i. he imporan (basis) funcions in he Fourier series represenaion are he complex exponenials wih frequencies ω k = 2k, for ineger k (posiive or negaive). here are an infinie number of hese complex exponenial funcions:...,e j 4,e j 2,e j,e j 2,e j 4,..., corresponding o frequencies..., 4, 2 2,,, 4,... Each of hese funcions can be regarded as a complex-valued signal. In he Fourier series represenaion each of hese complex exponenials is assigned a weigh, and he sum of all of he weighed complex exponenials yields he desired signal. People (especially engineers) will ofen alk abou he amoun of a frequency presen in a signal: by his hey mean some quaniy relaed direcly o he coefficien weighing he complex exponenial a ha frequency. he fac ha he Fourier series represenaion for a periodic signal wih period exiss leads o he ineresing observaion ha such a signal only conains frequencies a ineger muliples of ω = 2. his is he fundamenal frequency of he signal, expressed in radians per second. he Fourier series represenaion is consruced as he sum of an infinie se of producs of he form ae jω, where a is generally complex. I is really worh undersanding how muliplicaion by a affecs he complex exponenial e jω. Firs of all le s hink of a specific complex exponenial signal x() = e jω a some frequency ω. Noe ha x() = e jω = e jω is a complex-valued funcion, expressed in polar form. hus he magniude is x() =, a consan funcion, and he phase is x() = ω. Now consider he signal y() = ax(), where a is a consan. he righ hand side involves muliplicaion of complex quaniies, so we should wrie a in polar form: a = ρe jφ. hen y() = ax() = ρe jφ e jω = ρe j(ω+φ) which is also a complex valued funcion in polar form: y() = ρ and y() = ω + φ. Comparing he represenaions we see ha muliplicaion by a changes he magniude of he complex exponenial by a, and adds an overall phase of a. o really see he effec, we can noe ha y() can be wrien in erms of x() as ( ( )) ) y() = ρe j(ω+φ) = ρe j ω + φ ω = ρx (+ φω. (his is only rue if x() = e jω.) Muliplicaion by a has caused a change in he overall scale by a, and has ime-shifed he complex exponenial by some amoun proporional o a. In general muliplicaion of a complex exponenial by a complex consan causes a change in is magniude and a change in is posiion along he ime axis. [Exercise: do one of hese explicily] In summary, and modifying noaion slighly, he Fourier series represenaion claims he following: Any periodic signal x(), for which x() = x(+) for all, can be expressed in he form x() = a k e jkω, where ω = 2 is he fundamenal frequency. his is he Fourier series synhesis equaion: i ells us how o reconsruc (or synhesise) he signal x() from he coefficien values a k. 4-5
[Discuss harmonics here] he only quesion ha remains is his: How do we find he coefficiens a k for a given signal x()? 4.3 Finding coefficien values Exercise: he complex exponenials in he Fourier expansion for x() have an imporan propery: hey are all orhogonal o one anoher wih respec o a paricular scalar produc, namely inegraion over one period. Specifically, consider he inegral e j( 2 )k e j(2 )l d = e j( 2 )(k l) d. When k = l his inegral is simply e d = d =. When k l we have since e j2m = (e j2 ) m = m =. e j( 2 )(k l) d = Solve he inegral and convince yourself ha his is rue. j2(k l)/ (ej2(k l) ) =, hese wo resuls can be summarised ino he single expression e j(2 )k e j(2 )l d = δ kl, where δ kl is called he Kronecker dela: he subscrips k and l are ineger, and { k = l δ kl = k l. his resul can be used o find he coefficiens or weighs in he Fourier series expansion. Muliplying x() by e j(2k/) (for some fixed k) and inegraing over one period gives x()e j( [ ] 2 )k d = a m e j(2 )m e j( 2 )k d = m= m= a m e j( 2 )m e j( 2 )k d = m= a m δ mk = a k. In he above we used he dummy variable m in he Fourier series expansion of x(). hus for each ineger k we have he Fourier series analysis equaion a k = x()e j( 2 )k d. Given a signal x() in he ime domain, his expression can be used o find he coefficiens or weighs associaed wih each frequency componen in he Fourier series represenaion or in oher words, how o analyse he signal o find is frequency componens. In pracice, everyhing in his secion remains rue if he inegraion domain is changed from [,] o [b,b + ] for any value a: i is simple o show ha he complex exponenials are orhogonal as long as he inegraion is over any complee period. his modificaion leads o he analysis equaion b+ a k = x()e j( 2 )k d, b where b can be any number. I is ofen slighly simpler o use his more general form for he analysis sage. 4-6
4.4 Example: recangular pulse rain Suppose we are given he signal x() below, and wan o find he Fourier series represenaion: x().5-8 -6-4 -2 2 4 6 8 (seconds) he signal is periodic. By inspecion he fundamenal period is = 8, so he fundamenal frequency is ω = 2 8 = 4 rad/s. he signal can herefore be expressed as a Fourier series x() = a k e jkω = a k e jk( 4), and he coefficiens can be found using he analysis equaion a k = /2 /2x()e jkω d = 8 4 4 x()e jk( 4) d. he inegraion inerval [ 4, 4] corresponds o one complee period and is convenien for his problem. o evaluae a k we noe ha over he inegraion limis he signal is only nonzero from 2 o 2, and wihin hese limis i has a value of exacly one. he coefficiens are herefore a k = 8 I usually helps o calculae a separaely: a = 8 2 2 2 e jk( 4) d. 2e j(4) d = 8 For k, he remaining coefficiens are given by a k = [ 2 8 2e jk( 4) d = 8 jk ( 4 ( = jk ( ) e jk( 4)2 e jk( ) 4)2 = 4 8 k )e jk( 4) 2j 2 ] =2 2 d = /2. [ = jk ( ) e jk( ] =2 4) = 2 4 8 (e jk( 4)2 e jk( ) 4)2 = sin(( = 2 2) k) k Each of hese coefficien values is real, which is no ypically he case i only happens when he signal x() is real and even. As always i is useful o plo resuls in order o visualise hem. Consider he erms in he Fourier series represenaion x() = a k e jk( 4). Each value of k relaes o a frequency ω = ( 4) k, and here is a corresponding coefficien value a k. Ploing he value of a k as a funcion of he relaed frequency is meaningful. However, since a k is generally complex, wo plos are needed. I is easier o inerpre magniude and phase plos, so his is wha is commonly done. A plo of each coefficien verses he corresponding frequency is shown below:. 4-7
.4 ak.2 2 ω 2 ak 2 ω 2 he erm for k = corresponds o a frequency ω =, and he coefficien value is a = /2 = 2 ej. his is indicaed by a do of value.5 a frequency zero of he magniude plo, and a do of value a he same frequency in he phase plo. Similarly, k = corresponds o a frequency ω = 4, for which he coefficien a =. Again his is real and posiive, so he phase is zero. he coefficien a 2 =, which has zero magniude and any phase we have arbirarily chosen a zero value for he phase, and he poin is indicaed a ω = 2. he erm for k = 3 is differen. I corresponds o a frequency ω = 3 4, bu he value is a 3 = 3 = 3 ej. he nonzero phase is required o make he value negaive, since he magniude canno be negaive. he oher poins can be ploed similarly. Nonzero poins on he negaive ω axis of he phase plo have been indicaed wih values of. Since phase is undeermined up o addiion of a muliple of 2, hese poins could equally validly have been drawn going upwards raher han downwards. he reason hey are drawn as hey are is his: since i is rue ha a k = a k = x()e j( 2 )k d, x()e j( 2 )k d and a k = x ()e j( 2 )k d. If x() is real hen x() = x (), and we see from above ha i mus be rue ha a k = a k. Wriing a k = a k e j a k and a k = a k e j a k, his condiion saes ha for real signals we mus have a k e j a k = a k e j a k. Equaliy of complex numbers means ha heir magniudes and phases mus be equal, so For real-valued signals i is always he case ha a k = a k and a k = a k. he magniude of he Fourier coefficiens is an even funcion of k, and he phase is an odd funcion. For his reason he phase plo in he previous figure was drawn as i was in order o make he phase plo look more like an odd funcion. I seems eviden from he magniude plo of a k ha he coefficien values are ending o zero as k. his can indeed be shown o be he case. he naure of he Fourier series represenaion can herefore be explored by considering he parial reconsrucion N x N () = a k e jkω, k= N 4-8
where he runcaed sum means ha he reconsrucion only uses a subse of he erms in he Fourier series. We should have x() = lim N x N (), and indeed his is he case in an appropriaely-defined conex. Reconsruced signals for differen values of N are shown below: x2() x5() x() x2().5.5 -.5-8 -6-4 -2 2 4 6 8.5.5 -.5-8 -6-4 -2 2 4 6 8.5.5 -.5-8 -6-4 -2 2 4 6 8.5.5 -.5-8 -6-4 -2 2 4 6 8 Small values of N yield a poor reconsrucion, bu as N increases he reconsruced signal ends owards he original signal x(). In pracice one seldom acually reconsrucs a signal from is Fourier series coefficiens i is sufficien jus o know ha i can be done, and ha hey compleely describe he signal. 4.5 Example: impulse rain One of he mos imporan periodic signals is he impulse rain p () shown below: () () () () () p () One possible mahemaical expression for his signal is p () = δ( k). One reason for is imporance is ha if we convolve any signal wih i, hen he resul will be periodic wih a period of. For example, if we convolve p () wih x() below x() 4-9
hen he resul y() = x() p () will be y() Mahemaically, his follows from (bu is no proved by) he lineariy and shif invariance of convoluion: ( ) y() = x() δ( k) = (x() δ( k)) = x( k), Exercise: Show ha he signal y() as defined above is periodic wih a period. Having a Fourier series represenaion of he impulse rain is useful. Firsly, we noe ha i is clearly periodic wih a period of. I herefore has a represenaion of he form p () = c k e jkω for some se of coefficiens c k and ω = 2. he coefficien values are given by c k = /2 /2p ()e jkωk d = /2 /2δ()e jkω d = /2 /2 δ()d =, where he sifing propery was used and he inegraion inerval was chosen o ensure ha we didn ry o inegrae over half of a Dirac dela. Subsiuing back ino he Fourier series synhesis equaion gives he required represenaion: p () = 4.6 rigonomeric Fourier series 2 ejk( ). If he signal x() is real, hen i seems a lile srange ha he Fourier series represenaion requires a linear combinaion of complex exponenial componens. Shouldn a real signal have a real-valued decomposiion? he answer is yes: a real-valued signal can be wrien as a linear combinaion of real-valued componens a ineger muliples of a fundamenal frequency. In a sense, hough, he complex decomposiion is simpler. If x() is real, hen i has been shown ha he Fourier coefficiens saisfy a k = a k. In general a k is complex, so we can wrie i in magniude-phase form as a k = a k e j a k. Wih his definiion i is eviden ha a k = a k = ( a k e j a k ) = a k e j a k. Consider he Fourier series reconsrucion formula in his case: x() = a k e jkω = +a 2 e j2ω +a e jω +a +a e jω +a 2 e j2ω + Pulling ou he k = erm his can be wrien as x() = a + (a k e jkω +a k e jkω ). k= 4-
Noing ha cosθ = 2 (ejθ +e jθ ) we see ha x() = a + = a + = a + = a + ( a k e j a k e jkω + a k e j a k e jkω ) k= a k (e j(kω+ ak) +e j(kω+j a k ) k= 2 a k 2 (ej(kω+ a k) +e j(kω+j a k ) k= 2 a k cos(kω + a k ). k= he componen funcions in he weighed sum are now real-valued sinusoids. his form of decomposiion is called he rigonomeric Fourier series. One can develop direc equaions for calculaing he parameers of he rigonomeric Fourier series for real-valued signals. hey look similar o hose for he complex exponenial case, bu are slighly moreinricaeand hardero remember. heaboveanalysisshowsha in he casewhereasignalis real, he complex exponenial Fourier coefficiens have srucure ha essenially cases cancellaion of all he complex pars of he complex exponenial componens. 4.7 Parseval s heorem A curren I hrough a resisor R dissipaes a oal power of P = I 2 R. A volage V across a resisor R resuls in a power dissipaion of P = V 2 /R. he imporan hing o noe is ha he power dissipaed is proporional o he square of eiher he curren or he volage. he power of a signal x() is defined o be he average of is squared values, or is mean square value. Formally, his can be calculaed as P = lim x() 2 d. he limiing process is required because he signal can have infinie duraion. If x() is a curren, his calculaion finds he average power dissipaed when he signal is passed hrough a reference R = ohm resisor. If x() is a volage hen i finds he averagepower dissipaed when he volage signal is held across a reference R = ohm resisor. If he signal x() is periodic, hen we can find is mean-square value wihou using he limi. he average power of he signal will jus be he average power over one cycle: P = x() 2 d. Given he descripion of a signal x() in he ime domain, i is herefore quie simple o calculae he average power. Parseval s heorem relaes power in he ime domain o power in he frequency domain, in erms of he Fourier series coefficiens. 4-
Assuming ha x() has a Fourier series represenaion Parseval s heorem saes ha x() = x() 2 d = a k e jkω, a k 2. he heorem can be parially jusified as follows. Consider he kh erm in he Fourier series represenaion: i is a signal of he form x k () = a k e jω. In isolaion, his erm has an average power of P k = = a ka k x k () 2 d = x k ()x k ()d = e jkω e jkω d = a ka k a k e jkω a k e jkω d d = a k a k = a k 2, where we ve used he fac ha z 2 = zz for any complex number z. hus he average power in he kh componen of x() is a k 2. Parseval s heorem saes ha he oal power in he signal is he sum of he power in each of he individual componens in he Fourier series represenaion. (Noe ha his is only possible because he complex exponenials in he expansion are orhogonal over one period.) hus he oal power in x() can be found eiher from he ime-domain descripion, or from he Fourier series coefficiens. However, he Fourier coefficiens conain more informaion: hey addiionally le you deermine he frequencies a which he power conribuions arise. In he example of Secion 4.4, he averagesignal power can be found in he ime domain as follows: P = 8 4 4 x() 2 d = 8 2 2 d = 4 8 = 2. his would be measured in unis of power, probably was. he same resul can be obained from he magniude plo of a k : using Parseval s heorem we have P = = 4 + 2 2 a k 2 = + (3) 2 + () 2 + (2) 2 + () 2 + (3) 2 + k= (2(k +) ) 2 = 4 + 2 2 2 8 = 2. (his is far from obvious.) However, he heorem also ells us where he power is locaed in erms of frequency. he k = erm corresponds o he componen of he signal a e jω = a, which is consan and has frequency zero (i.e. he DC componen of he signal). he power in he signal a frequency is herefore a 2 = 4 was. Half of he signal power is herefore conained in he DC componen. he firs harmonic corresponds o he fundamenal frequency: he erms for k = ±. he power conained in a e jω is, and he power in he componen a 2 e jω is also. he oal power 2 conained in he firs harmonic is herefore 2 was. here is no power in he second harmonic, 2 since he Fourier coefficiens are zero, bu he power in he hird harmonic can similarly be shown 2 o be 9 was. 2 he pariion of power o differen frequencies in he signal allows us o inerpre he effec of modifying he signal. Suppose we were o pu x() hrough a sysem ha removes he componens 4-2
of he signal above he hird harmonic his is he operaion of a filer, which will be discussed a lengh laer on. Such a filer would null he componens of he signal corresponding o frequencies ω 4ω : he Fourier componens a hese frequencies would be se o zero by he acion of he sysem. he remaining nonzero coefficiens are hen jus {a 3,a,a,a,a 3 }, which have an average power of 9 + 2 pi + 2 4 + pi + 2 9 =.4752 was. he sysem herefore removes.248 2 was of signal power. Alernaively, we could say ha.4752/.5 = 95.32% of he oal signal power is conained in he firs hree harmonics. 4-3