About the HELM Project HELM (Helping Engineers Learn Mathematics) materials were the outcome of a three-year curriculum development project

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2 Abou he HELM Projec HELM (Helping Engineers Learn Mahemaics) maerials were he oucome of a hree-year curriculum developmen projec underaken by a consorium of five English universiies led by Loughborough Universiy, funded by he Higher Educaion Funding Council for England under he Fund for he Developmen of eaching and Learning for he period Ocober Sepember 5. HELM aims o enhance he mahemaical educaion of engineering undergraduaes hrough a range of flexible learning resources in he form of Workbooks and web-delivered ineracive segmens. HELM suppors wo CAA regimes: an inegraed web-delivered implemenaion and a CD-based version. HELM learning resources have been produced primarily by eams of wriers a six universiies: Hull, Loughborough, Mancheser, Newcasle, Reading, Sunderland. HELM graefully acknowledges he valuable suppor of colleagues a he following universiies and colleges involved in he criical reading, rialling, enhancemen and revision of he learning maerials: Ason, Bournemouh & Poole College, Cambridge, Ciy, Glamorgan, Glasgow, Glasgow Caledonian, Glenrohes Insiue of Applied echnology, Harper Adams Universiy College, Herfordshire, Leiceser, Liverpool, London Meropolian, Moray College, Norhumbria, Noingham, Noingham ren, Oxford Brookes, Plymouh, Porsmouh, Queens Belfas, Rober Gordon, Royal Fores of Dean College, Salford, Sligo Insiue of echnology, Souhampon, Souhampon Insiue, Surrey, eesside, Ulser, Universiy of Wales Insiue Cardiff, Wes Kingsway College (London), Wes Nos College. HELM Conacs: Pos: HELM, Mahemaics Educaion Cenre, Loughborough Universiy, Loughborough, LE 3U. helm@lboro.ac.uk Web: hp://helm.lboro.ac.uk HELM Workbooks Lis Basic Algebra 6 Funcions of a Complex Variable Basic Funcions 7 Muliple Inegraion 3 Equaions, Inequaliies & Parial Fracions 8 Differenial Vecor Calculus 4 rigonomery 9 Inegral Vecor Calculus 5 Funcions and Modelling 3 Inroducion o Numerical Mehods 6 Exponenial and Logarihmic Funcions 3 Numerical Mehods of Approximaion 7 Marices 3 Numerical Iniial Value Problems 8 Marix Soluion of Equaions 33 Numerical Boundary Value Problems 9 Vecors 34 Modelling Moion Complex Numbers 35 Ses and Probabiliy Differeniaion 36 Descripive Saisics Applicaions of Differeniaion 37 Discree Probabiliy Disribuions 3 Inegraion 38 Coninuous Probabiliy Disribuions 4 Applicaions of Inegraion 39 he Normal Disribuion 5 Applicaions of Inegraion 4 Sampling Disribuions and Esimaion 6 Sequences and Series 4 Hypohesis esing 7 Conics and Polar Coordinaes 4 Goodness of Fi and Coningency ables 8 Funcions of Several Variables 43 Regression and Correlaion 9 Differenial Equaions 44 Analysis of Variance Laplace ransforms 45 Non-parameric Saisics z-ransforms 46 Reliabiliy and Qualiy Conrol Eigenvalues and Eigenvecors 47 Mahemaics and Physics Miscellany 3 Fourier Series 48 Engineering Case Sudies 4 Fourier ransforms 49 Suden s Guide 5 Parial Differenial Equaions 5 uor s Guide Copyrigh Loughborough Universiy, 6

3 3 Conens Fourier Series 3. Periodic Funcions 3. Represening Periodic Funcions by Fourier Series Even and Odd Funcions Convergence Half-range Series he Complex Form An Applicaion of Fourier Series 68 Learning oucomes In his Workbook you will learn how o express a periodic signal f() in a series of sines and cosines. You will learn how o simplify he calculaions if he signal happens o be an even or an odd funcion. You will learn some brief facs relaing o he convergence of he Fourier series. You will learn how o approximae a non-periodic signal by a Fourier series. You will learn how o re-express a sandard Fourier series in complex form which paves he way for a laer examinaion of Fourier ransforms. Finally you will learn abou some simple applicaions of Fourier series.

4 Represening Periodic Funcions by Fourier Series 3. Inroducion In his Secion we show how a periodic funcion can be expressed as a series of sines and cosines. We begin by obaining some sandard inegrals involving sinusoids. We hen assume ha if f() is a periodic funcion, of period, hen he Fourier series expansion akes he form: f() = a + (a n cos n + b n sin n) Our main purpose here is o show how he consans in his expansion, a n (for n =,,, 3... and b n (for n =,, 3,...), may be deermined for any given funcion f(). Prerequisies Before saring his Secion you should... Learning Oucomes On compleion you should be able o... know wha a periodic funcion is be able o inegrae funcions involving sinusoids have knowledge of inegraion by pars calculae Fourier coefficiens of a funcion of period calculae Fourier coefficiens of a funcion of general period HELM (8): Secion 3.: Represening Periodic Funcions by Fourier Series 9

5 . Inroducion We recall firs a simple rigonomeric ideniy: cos = + cos or equivalenly cos = + cos () Equaion can be inerpreed as a simple finie Fourier series represenaion of he periodic funcion f() = cos which has period. We noe ha he Fourier series represenaion conains a consan erm and a period erm. A more complicaed rigonomeric ideniy is sin 4 = 3 8 cos + cos 4 () 8 which again can be considered as a finie Fourier series represenaion. (Do no worry if you are unfamiliar wih he resul ().) Noe ha he funcion f() = sin 4 (which has period ) is being wrien in erms of a consan funcion, a funcion of period or frequency (he firs harmonic ) and a funcion of period or frequency (he second harmonic ). he reason for he consan erm in boh () and () is ha each of he funcions cos and sin 4 is non-negaive and hence each mus have a posiive average value. Any sinusoid of he form cos n or sin n has, by symmery, zero average value. herefore, so would a Fourier series conaining only such erms. A consan erm can herefore be expeced o arise in he Fourier series of a funcion which has a non-zero average value.. Funcions of period We now discuss how o represen periodic non-sinusoidal funcions f() of period in erms of sinusoids, i.e. how o obain Fourier series represenaions. As already discussed we expec such Fourier series o conain harmonics of frequency n (n =,, 3,...) and, if he periodic funcion has a non-zero average value, a consan erm. hus we seek a Fourier series represenaion of he general form f() = a + a cos + a cos b sin + b sin +... he reason for labelling he consan erm as a will be discussed laer. he ampliudes a,a,... b,b,... of he sinusoids are called Fourier coefficiens. Obaining he Fourier coefficiens for a given periodic funcion f() is our main ask and is referred o as Fourier Analysis. Before embarking on such an analysis i is insrucive o esablish, a leas qualiaively, he plausibiliy of approximaing a funcion by a few erms of is Fourier series. HELM (8): Workbook 3: Fourier Series

6 Even and Odd Funcions 3.3 Inroducion In his Secion we examine how o obain Fourier series of periodic funcions which are eiher even or odd. We show ha he Fourier series for such funcions is considerably easier o obain as, if he signal is even only cosines are involved whereas if he signal is odd hen only sines are involved. We also show ha if a signal reverses afer half a period hen he Fourier series will only conain odd harmonics. Prerequisies Before saring his Secion you should... Learning Oucomes On compleion you should be able o... know how o obain a Fourier series be able o inegrae funcions involving sinusoids have knowledge of inegraion by pars deermine if a funcion is even or odd or neiher easily calculae Fourier coefficiens of even or odd funcions 3 HELM (8): Workbook 3: Fourier Series

7 Convergence 3.4 Inroducion In his Secion we examine, briefly, he convergence characerisics of a Fourier series. We have seen ha a Fourier series can be found for funcions which are no necessarily coninuous (here may be jumps in he curve) he only requiremen ha we have made is ha he funcion be periodic. We have seen ha he more erms we ake in he Fourier series he beer is he approximaion o he given signal. Bu an obvious quesion o ask is wha happens a he poins of disconinuiy? Wha does he Fourier series converge o a hese poins? I mus converge o somehing (finie) since a Fourier series is a sum of very smooh coninuous funcions. In his Secion we give he answer o his quesion. Prerequisies Before saring his Secion you should... Learning Oucomes On compleion you should be able o... know how o obain a Fourier series be familiar wih he limi process as applied o funcions deermine wha a Fourier series converges o a each poin, including a a poin of disconinuiy use he convergence propery of Fourier Series o obain series for he number 4 HELM (8): Workbook 3: Fourier series

8 . Convergence of a Fourier series We have now shown how o obain a Fourier series for periodic funcions. We have suggesed ha we would expec o be able o approximae such funcions by using a few erms of he Fourier series. he deailed quesion of he convergence or oherwise of Fourier series has no been discussed. he reason for his is ha he grea majoriy of funcions likely o be encounered in pracice have Fourier series ha do indeed converge and can herefore be safely used as approximaions. he precise condiions ha have o be fulfilled for a Fourier series o converge are known as Dirichle condiions afer he French mahemaician who invesigaed he maer. he hree condiions are lised in he following Key Poin. Key Poin 6 he Dirichle condiions for he convergence of a Fourier series of a periodic funcion f() are:. f() mus have only a finie number of finie disconinuiies, wihin one period. f() mus have a finie number of maxima and minima over one period 3. he inegral f() d mus be finie. I follows, for example, ha if f() is defined over (, ) as one of he following funcions 3 or /( 4) or 3 +and f( +) =f(), hen f() can indeed be represened as a Fourier series as each funcion saisfies he Dirichle condiions. On he oher hand, if, over (, ), f() is or or an hen f() canno be expanded in a Fourier series because each of hese funcions has an infinie disconinuiy wihin (, ). If he Dirichle condiions are saisfied a a poin = where f() is coninuous hen, as we would expec, he Fourier series a given by n n a + a n cos + b n sin A a poin, say =, a which f() has a disconinuiy hen he series a + n n a n cos + b n sin converges o converges o he funcion value f( ) {f( )+f( + )} where f( ) is he limi of f() as approaches from he lef and f( + ) is he limi as approaches from he righ (Figure 9). HELM (8): Secion 3.4: Convergence 4

9 Half-Range Series 3.5 Inroducion In his Secion we address he following problem: Can we find a Fourier series expansion of a funcion defined over a finie inerval? Of course we recognise ha such a funcion could no be periodic (as periodiciy demands an infinie inerval). he answer o his quesion is yes bu we mus firs conver he given non-periodic funcion ino a periodic funcion. here are many ways of doing his. We shall concenrae on he mos useful exension o produce a so-called half-range Fourier series. Prerequisies Before saring his Secion you should... Learning Oucomes On compleion you should be able o... know how o obain a Fourier series be familiar wih odd and even funcions and heir properies have knowledge of inegraion by pars choose o expand a non-periodic funcion eiher as a series of sines or as a series of cosines 46 HELM (8): Workbook 3: Fourier Series

10 he Complex Form 3.6 Inroducion In his Secion we show how a Fourier series can be expressed more concisely if we inroduce he complex number i where i =. By uilising he Euler relaion: e iθ cos θ + i sin θ we can replace he rigonomeric funcions by complex exponenial funcions. By also combining he Fourier coefficiens a n and b n ino a complex coefficien c n hrough c n = (a n ib n ) we find ha, for a given periodic signal, boh ses of consans can be found in one operaion. We also obain Parseval s heorem which has imporan applicaions in elecrical engineering. he complex formulaion of a Fourier series is an imporan precursor of he Fourier ransform which aemps o Fourier analyse non-periodic funcions. Prerequisies Before saring his Secion you should... Learning Oucomes On compleion you should be able o... know how o obain a Fourier series be compeen working wih he complex numbers be familiar wih he relaion beween he exponenial funcion and he rigonomeric funcions express a periodic funcion in erms of is Fourier series in complex form undersand Parseval s heorem HELM (8): Secion 3.6: he Complex Form 53

11 . Complex exponenial form of a Fourier series So far we have discussed he rigonomeric form of a Fourier series i.e. we have represened funcions of period in he erms of sinusoids, and possibly a consan erm, using f() = a + n n a n cos + b n sin. If we use he angular frequency ω = we obain he more concise form f() = a + (a n cos nω + b n sin nω ). We have seen ha he Fourier coefficiens are calculaed using he following inegrals: a n = f()cosnω d n =,,,... () b n = f() sin nω d n =,,... () An alernaive, more concise form, of a Fourier series is available using complex quaniies. his form is quie widely used by engineers, for example in Circui heory and Conrol heory, and leads naurally ino he Fourier ransform which is he subjec of 4.. Revision of he exponenial form of a complex number Recall ha a complex number in Caresian form which is wrien as z = a + ib, where a and b are real numbers and i =, can be wrien in polar form as z = r(cos θ + i sin θ) where r = z = a + b and θ, he argumen or phase of z, is such ha a = r cos θ b = r sin θ. A more concise version of he polar form of z can be obained by defining a complex exponenial quaniy e iθ by Euler s relaion e iθ cos θ + i sin θ he polar angle θ is normally expressed in radians. Replacing i by i we obain he alernaive form e iθ cos θ i sin θ 54 HELM (8): Workbook 3: Fourier Series

12 An Applicaion of Fourier Series 3.7 Inroducion In his Secion we look a a ypical applicaion of Fourier series. he problem we sudy is ha of a differenial equaion wih a periodic (bu non-sinusoidal) forcing funcion. he differenial equaion chosen models a lighly damped vibraing sysem. Prerequisies Before saring his Secion you should... Learning Oucomes On compleion you should be able o... know how o obain a Fourier series be compeen o use complex numbers be familiar wih he relaion beween he exponenial funcion and he rigonomeric funcions solve a linear differenial equaion wih a periodic forcing funcion using Fourier series 68 HELM (8): Workbook 3: Fourier Series

13 . Modelling vibraion by differenial equaion Vibraion problems are ofen modelled by ordinary differenial equaions wih consan coefficiens. For example he moion of a spring wih siffness k and damping consan c is modelled by m d y d + cdy + ky = d where y() is he displacemen of a mass m conneced o he spring. I is well-known ha if c < 4mk, usually referred o as he lighly damped case, hen y() =e α (A cos ω + B sin ω) () i.e. he moion is sinusoidal bu damped by he negaive exponenial erm. In () we have used he noaion α = c ω = 4km c o simplify he equaion. m m he values of A and B depend upon iniial condiions. he sysem represened by (), whose soluion is (), is referred o as an unforced damped harmonic oscillaor. A lighly damped oscillaor driven by a ime-dependen forcing funcion F () is modelled by he differenial equaion m d y d + cdy + ky = F () d he soluion or sysem response in (3) has wo pars: (a) A ransien soluion of he form (), (b) A forced or seady sae soluion whose form, of course, depends on F (). If F () is sinusoidal such ha F () =A sin(ω + φ) where Ω and φ are consans, hen he seady sae soluion is fairly readily obained by sandard echniques for solving differenial equaions. If F () is periodic bu non-sinusoidal hen Fourier series may be used o obain he seady sae soluion. he mehod is based on he principle of superposiion which is acually applicable o any linear (homogeneous) differenial equaion. (Anoher engineering applicaion is he series LCR circui wih an applied periodic volage.) he principle of superposiion is easily demonsraed:- Le y () and y () be he seady sae soluions of (3) when F () =F () and F () =F () respecively. hen m d y d + cdy d + ky = F () m d y d + cdy d + ky = F () Simply adding hese equaions we obain m d d (y + y )+c d d (y + y )+k(y + y )=F ()+F () () (3) HELM (8): Secion 3.7: An Applicaion of Fourier Series 69

14 from which i follows ha if F () =F ()+F () hen he sysem response is he sum y ()+y (). his, in is simples form, is he principle of superposiion. More generally if he forcing funcion is N F () = F n () hen he response is y() = N y n () where y n () is he response o he forcing funcion F n (). Reurning o he specific case where F () is periodic, he soluion procedure for he seady sae response is as follows: Sep : Obain he Fourier series of F (). Sep : Solve he differenial equaion (3) for he response y n () corresponding o he n h harmonic in he Fourier series. (he response y o o he consan erm, if any, in he Fourier series may have o be obained separaely.) Sep 3: Superpose he soluions obained o give he overall seady sae moion: y() =y ()+ N y n () he procedure can be lenghy bu he soluion is of grea engineering ineres because if he frequency k of one harmonic in he Fourier series is close o he naural frequency of he undamped sysem m hen he response o ha harmonic will dominae he soluion.. Applying Fourier series o solve a differenial equaion he following ask which is quie long will provide useful pracice in applying Fourier series o a pracical problem. Essenially you should follow Seps o 3 above carefully. ask he problem is o find he seady sae response y() of a spring/mass/damper sysem modelled by m d y d + cdy + ky = F () (4) d where F () is he periodic square wave funcion shown in he diagram. F F () F 7 HELM (8): Workbook 3: Fourier Series

15 Sep : Obain he Fourier series of F () noing ha i is an odd funcion: Your soluion he calculaion is similar o hose you have performed earlier in his Workbook. Since F () is an odd funcion and has period so ha ω = =, i has Fourier coefficiens: b n = o n F sin d n =,, 3,... = F cos n n = F ( cos n) = n 4F n odd n n even so F () = 4F sin nω n (where he sum is over odd n only). Sep (a): Since each erm in he Fourier series is a sine erm you mus now solve (4) o find he seady sae response y n o he n h harmonic inpu: F n () =b n sin nω n =, 3, 5,... From he basic heory of linear differenial equaions his response has he form y n = A n cos nω + B n sin nω (5) where A n and B n are coefficiens o be deermined by subsiuing (5) ino (4) wih F () =F n (). Do his o obain simulaneous equaions for A n and B n : HELM (8): Secion 3.7: An Applicaion of Fourier Series 7

16 Your soluion We have, differeniaing (5), y n = nω( A n sin nω + B n cos nω) y n = (nω) ( A n cos nω B n sin nω) from which, subsiuing ino (4) and collecing erms in cos nω and sin nω, ( m(nω) A n + cnωb n + ka n )cosnω +( m(nω) B n cnωa n + kb n ) sin nω = b n sin nω hen, by comparing coefficiens of cos nω and sin nω, we obain he simulaneous equaions: (k m(nω) )A n + c(nω)b n = (6) c(nω)a n +(k m(nω) )B n = b n (7) Sep (b): Now solve (6) and (7) o obain A n and B n : Your soluion 7 HELM (8): Workbook 3: Fourier Series

17 cω n b n A n = (k mωn) + ωnc B n = (k mω n)b n (k mω n) + ω nc where we have wrien ω n for nω as he frequency of he n h harmonic (8) (9) I follows ha he seady sae response y n o he n h harmonic of he Fourier series of he forcing funcion is given by (5). he ampliudes A n and B n are given by (8) and (9) respecively in erms of he sysems parameers k, c, m, he frequency ω n of he harmonic and is ampliude b n. In pracice i is more convenien o represen y n in he so-called ampliude/phase form: y n = C n sin(ω n + φ n ) where, from (5) and (), Hence A n cos ω n + B n sin ω n = C n (cos φ n sin ω n + sin φ n cos ω n ). () so C n sin φ n = A n C n cos φ n = B n an φ n = A n B n = cω n (mω n k) () Sep 3: C n = A n + B n = b n (mω n k) + ω nc () Finally, use he superposiion principle, o sae he complee seady sae response of he sysem o he periodic square wave forcing funcion: Your soluion y() = y n () = (n odd) C n (sin ω n + φ n ) where C n and φ n are given by () and (). In pracice, since b n = 4F n i follows ha he ampliude C n also decreases as. However, if one of n k he harmonic frequencies say ωn is close o he naural frequency of he undamped oscillaor m hen ha paricular frequency harmonic will dominae in he seady sae response. he paricular value ωn will, of course, depend on he values of he sysem parameers k and m. HELM (8): Secion 3.7: An Applicaion of Fourier Series 73

18 ask Wrie down in cos θ±i sin θ form and also in Caresian form (a) e i/6 (b) e i/6. Use Euler s relaion: Your soluion We have, by definiion, (a) e i/6 =cos + i sin = =cos i (b) e i/6 i sin = i ask Wrie down (a) cos 6 (b) sin 6 in erms of e i/6 and e i/6. Your soluion We have, adding he wo resuls from he previous ask e i/6 + e i/6 = cos or cos = e i/6 + e i/6 6 6 Similarly, subracing he wo resuls, e i/6 e i/6 =isin 6 (Don forge he facor i in his laer case.) or sin = e i/6 e i/6 6 i Clearly, similar calculaions could be carried ou for any angle θ. he general resuls are summarised in he following Key Poin. HELM (8): Secion 3.6: he Complex Form 55

19 Key Poin 8 Euler s Relaions e iθ cos θ + i sin θ, e iθ cos θ i sin θ cos θ e iθ + e iθ sin θ e iθ e iθ i Using hese resuls we can redraf an expression of he form a n cos nθ + b n sin nθ in erms of complex exponenials. (his expression, wih θ = ω, is of course he n h harmonic of a rigonomeric Fourier series.) ask Using he resuls from he Key Poin 8 (wih nθ insead of θ) rewrie a n cos nθ + b n sin nθ in complex exponenial form. Firs subsiue for cos nθ and sin nθ wih exponenial expressions using Key Poin 8: Your soluion We have so a n cos nθ = a n e inθ + e inθ a n cos nθ + b n sin nθ = a n b n sin nθ = b n e inθ e inθ i e inθ + e inθ + b n e inθ e inθ i 56 HELM (8): Workbook 3: Fourier Series

20 Now collec he erms in e inθ and in e inθ and use he fac ha i = i: Your soluion We ge a n + b n i or, since i = e inθ + a n b n e inθ i i i = i (a n ib n )e inθ + (a n + ib n )e inθ. Now wrie his expression in more concise form by defining c n = (a n ib n ) which has complex conjugae c n = (a n + ib n ). Wrie he concise complex exponenial expression for a n cos nθ + b n sin nθ: Your soluion a n cos nθ + b n sin nθ = c n e inθ + c ne inθ Clearly, we can now rewrie he rigonomeric Fourier series a + (a n cos nω + b n sin nω ) as a + cn e inω + c ne inω (3) A neaer, and paricularly concise, form of his expression can be obained as follows: Firsly wrie a = c (which is consisen wih he general definiion of c n since b =). he second erm in he summaion c ne inω = c e iω + c e iω +... can be wrien, if we define c n = c n = (a n + ib n ), as c e iω + c e iω + c 3 e 3 iω +...= Hence (3) can be wrien c + c n e inω + n= n= c n e inω c n e inω or in he very concise form n= c n e inω. HELM (8): Secion 3.6: he Complex Form 57

21 he complex Fourier coefficiens c n can be readily obained as follows using () and () for a n,b n. Firsly c = a = f() d For n =,, 3,... we have c n = (a n ib n )= Also for n =,, 3,... we have f()(cos nω i sin nω ) d i.e. c n = (4) f()e inω d (5) c n = c n = (a n + ib n )= f()e inω d his las expression is equivalen o saing ha for n =,, 3,... c n = f()e inω d he hree equaions (4), (5), (6) can hus all be conained in he one expression (6) c n = f()e inω d for n =, ±, ±, ±3,... he resuls of his discussion are summarised in he following Key Poin. Key Poin 9 Fourier Series in Complex Form A funcion f() of period has a complex Fourier series f() = c n e inω where c n = f()e inω d n= For he special case =, sohaω =, hese formulae become paricularly simple: f() = n= c n e in c n = f()e in d. 58 HELM (8): Workbook 3: Fourier Series

22 3. Properies of he complex Fourier coefficiens Using properies of he rigonomeric Fourier coefficiens a n, b n we can readily deduce he following resuls for he c n coefficiens:. c = a is always real.. Suppose he periodic funcion f() is even so ha all b n are zero. hen, since in he complex form he b n arise as he imaginary par of c n, i follows ha for f() even he coefficiens c n (n = ±, ±,...) are wholly real. ask If f() is odd, wha can you deduce abou he Fourier coefficiens c n? Your soluion Since, for an odd periodic funcion he Fourier coefficiens a n (which consiue he real par of c n ) are zero, hen in his case he complex coefficiens c n are wholly imaginary. 3. Since c n = f()e inω d hen if f() is even, c n will be real, and we have wo possible mehods for evaluaing c n : (a) Evaluae he inegral above as i sands i.e. over he full range,. Noe carefully ha he second erm in he inegrand is neiher an even nor an odd funcion so he inegrand iself is ( even funcion) ( neiher even nor odd funcion) = neiher even nor odd funcion. hus we canno wrie c n = / (b) Pu e inω =cosnω i sin nω so f()e inω d f()e inω = f()cosnω if() sin nω = ( even)( even) i( even)( odd) Hence c n = f()cosnω d = a n. = ( even) i( odd). 4. If f( + )= f() hen of course only odd harmonic coefficiens c n (n = ±, ±3, ±5,...) will arise in he complex Fourier series jus as wih rigonomeric series. HELM (8): Secion 3.6: he Complex Form 59

23 Example 4 Find he complex Fourier series of he saw-ooh wave shown in Figure 4: f() A Figure 4 Soluion We have f() = A << f( + )=f() he period is in his case so ω =. Looking a he graph of f() we can say immediaely (a) he Fourier series will conain a consan erm c (b) if we imagine shifing he horizonal axis up o A he signal can be wrien f() = A + g(), where g() is an odd funcion wih complex Fourier coefficiens ha are purely imaginary. Hence we expec he required complex Fourier series of f() o conain a consan erm A and complex exponenial erms wih purely imaginary coefficiens. We have, from he general heory, and using << as he basic period for inegraing, c n = A e inω d = A We can evaluae he inegral using pars: e inω d = e inω ( inω ) + inω = e inω ( inω ) ( inω ) e inω d e inω e inω d 6 HELM (8): Workbook 3: Fourier Series

24 Soluion (cond.) Bu ω = so e inω = e in = cos n i sin n Hence he inegral becomes Hence = i = e inω inω ( inω ) c n = A Noe ha c n = Also c = inω = ia n ia ( n) = ia n = c n A d = A as expeced. n = ±, ±,... as i mus Hence he required complex Fourier series is f() = A + ia e inω n n= n= which could be wrien, showing only he consan and he firs wo harmonics, as f() = A... i e iω ie iω + + ie iω + i e iω he corresponding rigonomeric Fourier series for he funcion can be readily obained from his complex series by combining he erms in ±n, n =,, 3,... For example his firs harmonic is A ie iω + ie iω = A { i(cos ω i sin ω )+ i(cos ω + i sin ω )} = A ( sin ω ) = A sin ω Performing similar calculaions on he oher harmonics we obain he rigonomeric form of he Fourier series f() = A A sin nω. n HELM (8): Secion 3.6: he Complex Form 6

25 ask Find he complex Fourier series of he periodic funcion: f() =e << f( +) =f() f() 3 Firsly wrie down an inegral expression for he Fourier coefficiens c n : Your soluion We have, since =, soω = c n = e e in d Now combine he real exponenial and he complex exponenial as one erm and carry ou he inegraion: Your soluion We have c n = e ( in) d = e ( in) = e ( in) ( e in) ( in) ( in) 6 HELM (8): Workbook 3: Fourier Series

26 Now simplify his as far as possible and wrie ou he Fourier series: Your soluion e ( in) = e e in = e (cos n i sin n) =e cos n e ( in) = e e in = e cos n Hence c n = ( in) (e e )cosn = sinh ( + in) cos n ( + n ) Noe ha he coefficiens c n n = ±, ±,... have boh real and imaginary pars in his case as he funcion being expanded is neiher even nor odd. Also c n = sinh ( in) ( in) cos( n) =sinh ( + ( n) ) ( + n ) cos n = c n as required. his includes he consan erm c = sinh. Hence he required Fourier series is f() = sinh n ( + in) ( ) ( + n ) e in since cos n =( ) n. n= HELM (8): Secion 3.6: he Complex Form 63

27 4. Parseval s heorem his is essenially a mahemaical heorem bu has, as we shall see, an imporan engineering inerpreaion paricularly in elecrical engineering. Parseval s heorem saes ha if f() is a periodic funcion wih period and if c n (n =, ±, ±,...) denoe he complex Fourier coefficiens of f(), hen f () d = n= c n. In words he heorem saes ha he mean square value of he signal f() over one period equals he sum of he squared magniudes of all he complex Fourier coefficiens. Proof of Parseval s heorem. Assume f() has a complex Fourier series of he usual form: f() = c n e inω ω = where hen Hence c n = n= f()e inω d f () =f()f() =f() c n e inω = c n f()e inω f () d = = cn f()e inω d cn f()e inω d = c n c n = n= c n which complees he proof. Parseval s heorem can also be wrien in erms of he Fourier coefficiens a n,b n of he rigonomeric Fourier series. Recall ha so c = a c n = a n ib n n =,, 3,... c n = a n + ib n n =,, 3,... so c n = a n + b n 4 n = ±, ±, ±3, HELM (8): Workbook 3: Fourier Series

28 n= c n = a 4 + a n + b n 4 and hence Parseval s heorem becomes f ()d = a 4 + (a n + b n) (7) he engineering inerpreaion of his heorem is as follows. Suppose f() denoes an elecrical signal (curren or volage), hen from elemenary circui heory f () is he insananeous power (in a ohm resisor) so ha f () d is he energy dissipaed in he resisor during one period. Now a sinusoid wave of he form A cos ω ( or A sin ω) has a mean square value A A so a purely sinusoidal signal would dissipae a power in a ohm resisor. Hence Parseval s heorem in he form (7) saes ha he average power dissipaed over period equals he sum of he powers of he consan (or d.c.) componens and of all he sinusoidal (or alernaing) componens. ask he riangular signal shown below has rigonomeric Fourier series f() = 4 cos n. n ( odd n) [his was deduced in he ask in Secion 3.3, page 39.] f() Use Parseval s heorem o show ha (n odd) n 4 = HELM (8): Secion 3.6: he Complex Form 65

29 Firs, idenify a, a n and b n for his siuaion and wrie down he definiion of f() for his case: Your soluion We have a = 4 n =, 3, 5,... a n = n n =, 4, 6,... Also b n = n =,, 3, 4,... f() = f( +) =f() << Now evaluae he inegral on he lef hand side of Parseval s heorem and hence complee he problem: Your soluion 66 HELM (8): Workbook 3: Fourier Series

30 We have f () = so f () d = d = 3 he righ-hand side of Parseval s heorem is a 4 + a n = n 4 Hence 3 = (n odd) n 4 (n odd) 3 8 = 3 (n odd) n 4 = (n odd) n 4 = Exercises Obain he complex Fourier series for each of he following funcions of period.. f() =. f() = 3. f() =e s. i ( ) n n ein (sum from o excluding n =).. + i n ein (sum from o excluding n =). 3. sinh ( ) n ( + in) ( + n ) ein (sum from o ). HELM (8): Secion 3.6: he Complex Form 67

31 . Half-range Fourier series So far we have shown how o represen given periodic funcions by Fourier series. We now consider a sligh variaion on his heme which will be useful in 5 on solving Parial Differenial Equaions. Suppose ha insead of specifying a periodic funcion we begin wih a funcion f() defined only over a limied range of values of, say <<. Suppose furher ha we wish o represen his funcion, over <<, by a Fourier series. (his siuaion may seem a lile arificial a his poin, bu his is precisely he siuaion ha will arise in solving differenial equaions.) o be specific, suppose we define f() = << f() Figure We shall consider he inerval <<o be half a period of a periodic funcion. We mus herefore define f() for < < o complee he specificaion. ask Complee he definiion of he above funcion f() =, << by defining i over << such ha he resuling funcions will have a Fourier series conaining (a) only cosine erms, (b) only sine erms, (c) boh cosine and sine erms. Your soluion HELM (8): Secion 3.5: Half-Range Series 47

32 (a) We mus complee he definiion so as o have an even periodic funcion: f() =, << f () (b) We mus complee he definiion so as o have an odd periodic funcion: f() =, << f () (c) We may define f() in any way we please (oher han (a) and (b) above). For example we migh define f() =over <<: f 3 () he poin is ha all hree periodic funcions f (),f (),f 3 () will give rise o a differen Fourier series bu all will represen he funcion f() = over <<. Fourier series obained by exending funcions in his sor of way are ofen referred o as half-range series. Normally, in applicaions, we require eiher a Fourier Cosine series (so we would complee a definiion as in (i) above o obain an even periodic funcion) or a Fourier Sine series (for which, as in (ii) above, we need an odd periodic funcion.) he above consideraions apply equally well for a funcion defined over any inerval. 48 HELM (8): Workbook 3: Fourier Series

33 Example 3 Obain he half range Fourier Sine series o represen f() = <<3. Soluion We firs exend f() as an odd periodic funcion F () of period 6: f() =, F () 3 << 3 Figure We now evaluae he Fourier series of F () by sandard echniques bu ake advanage of he symmery and pu a n =,n=,,,... Using he resuls for he Fourier Sine coefficiens for period from 3. subsecion 5, b n = n F () sin d, we pu =6and, since he inegrand is even (a produc of odd funcions), we can wrie b n = 3 n F () sin d = 3 n sin d (Noe ha we always inegrae over he originally defined range, in his case <<3.) We now have o inegrae by pars (wice!) b n = 3 3 n 3 3 n 3 n cos + cos d 3 n 3 = n n cos n + sin sin d 3 n n n 3 n n 3 = 7 8 cos n 3 3 n 3 n n n cos = 7 54 cos n + (cos n ) 3 3 n n n =, 4, 6,... n = 8 n 7 n =, 3, 5,... n 3 3 So he required Fourier Sine series is F () = 8 4 sin sin sin() HELM (8): Secion 3.5: Half-Range Series 49

34 ask Obain a half-range Fourier Cosine series o represen he funcion f() =4 <<4. 4 f() 4 Firs complee he definiion o obain an even periodic funcion F () of period 8. Skech F (): Your soluion 4 F() 4 4 Now formulae he inegral from which he Fourier coefficiens a n can be calculaed: Your soluion We have wih =8 a n = 4 n F ()cos d Uilising he fac ha he inegrand here is even we ge a n = 4 n (4 )cos d 4 5 HELM (8): Workbook 3: Fourier Series

35 Now inegrae by pars o obain a n and also obain a : Your soluion Using inegraion by pars we obain for n =,, 3,... a n = (4 ) 4 4 n n sin n = n cos n n 4 n =, 4, 6,... i.e. a n = 6 n =, 3, 5,... n Also a = 4 n sin d 4 = 8 [ cos(n)+] n (4 ) d =4. So he consan erm is a =. Now wrie down he required Fourier series: Your soluion We ge + 6 cos cos cos HELM (8): Secion 3.5: Half-Range Series 5

36 Noe ha he form of he Fourier series (a consan of ogeher wih odd harmonic cosine erms) could be prediced if, in he skech of F (), we imagine raising he -axis by unis i.e. wriing F () =+G() G() 4 4 Figure 3 Clearly G() possesses half-period symmery G( +4)= G() and hence is Fourier series mus conain only odd harmonics. Exercises Obain he half-range Fourier series specified for each of he following funcions:. f() = (sine series). f() = (sine series) 3. (a) f() =e (cosine series) (b) f() =e (sine series) 4. (a) f() = sin (cosine series) (b) f() = sin (sine series) s sin + sin 3 + sin 5 +. {sin sin + sin 3 } 3 3. (a) e n {e cos(n) } cos n (b) 4. (a) + n 4+n { e cos(n)} sin n n= (b) sin iself (!) ( cos( n))+ ( cos( + n)) cos n n +n 5 HELM (8): Workbook 3: Fourier Series

37 f() Figure 9 Key Poin 7 If Dirichle condiions are saisfied hen a a poin of coninuiy = o f( )= a + n n a n cos + b n sin whereas a a poin of disconinuiy = he Fourier series converges o he average of he wo limiing values {f( )+f( + )} = a + n n a n cos + b n sin Example Suppose we consider he square wave f() Figure Here f() has finie disconinuiies a, and. he Fourier series of f() is (see Secion 3.3, subsecion ) + sin + 3 sin sin HELM (8): Workbook 3: Fourier series

38 A =, for example, where f() is coninuous he square wave converges o f =. On he oher hand a = he Fourier series clearly has he value (since all he sine erms are zero here). his value agrees wih he average of he wo limiing values of f() a = : ( + ) =. If we acually pu = in he Fourier series we obain + sin sin sin +... = Since he series converges, as we have seen, o f =, we obain he ineresing resul = or = 4 ask he funcion f() = << << f( +) =f() f () has Fourier series (see Exercise 4 a he end of Secion 3.) cos cos 3 cos sin sin sin sin By using a suiable value of show ha = 6 HELM (8): Secion 3.4: Convergence 43

39 Firs decide on he appropriae value of o use: Your soluion Looking a he Fourier series, he numerical series we seek is presen in he cosine erms so we need o remove he sine erms. his we can do by selecing =or =. he choice =will make he cosine erms become: which is no wha we seek. Hence we pu =. Now pu = in he series and decide wha he Fourier series will converge o a his value. Hence complee he quesion: Your soluion A = he Fourier series is 6 cos cos 4 = 6 cos = A = he Fourier series will converge o So + = (he average of he lef and righ hand limis) = = 6 = 6 Noe ha in he las ask if we subsiue =in he Fourier series (which converges o f() = ) we obain anoher infinie series bu wih alernaing signs: = or = 6 44 HELM (8): Workbook 3: Fourier series

40 Exercises. Obain he Fourier series of f() = f( +) =f() By puing =show ha (n ) = 8. (a) Obain he Fourier series of he periodic funcion f() = 4 and use i o obain he following ideniies: (i) = 4 6 (b) Show ha = Obain he Fourier series of he periodic funcion (ii) = 4 f() = Use he series o show ha s = 4 + ( 4) cos[(n )] (n ). (a) + 3. cos(n) n cos n (i) Pu = (ii) Pu = (b) Add he wo series from (a). ( ) n sin n n HELM (8): Secion 3.4: Convergence 45

41 . Even and odd funcions We have shown in he previous Secion how o calculae, by inegraion, he coefficiens a n (n =,,, 3,...)andb n (n =,, 3,...) in a Fourier series. Clearly his is a somewha edious process and i is advanageous if we can obain as much informaion as possible wihou recourse o inegraion. In he previous Secion we showed ha he square wave (one period of which shown in Figure ) has a Fourier series conaining a consan erm and cosine erms only (i.e. all he Fourier coefficiens b n are zero) while he funcion shown in Figure 3 has a more complicaed Fourier series conaining boh cosine and sine erms as well as a consan. 4 Figure : Square wave f() Figure 3: Saw-ooh wave ask Conras he symmery or oherwise of he funcions in Figures and 3. Your soluion he square wave in Figure has a graph which is symmerical abou he y-axis and is called an even funcion. he saw-ooh wave shown in Figure 3 has no paricular symmery. In general a funcion is called even if is graph is unchanged under reflecion in he y-axis. his is equivalen o f( ) =f() for all Obvious examples of even funcions are, 4,, cos, cos, sin, cos n. A funcion is said o be odd if is graph is symmerical abou he origin (i.e. symmery abou he origin). his is equivalen o he condiion f( ) = f() i has roaional HELM (8): Secion 3.3: Even and Odd Funcions 3

42 Figure 4 shows an example of an odd funcion. f() Figure 4 Examples of odd funcions are, 3, sin, sin n. A periodic funcion which is odd is he saw-ooh wave in Figure 5. f() Figure 5 Some funcions are neiher even nor odd. he periodic saw-ooh wave of Figure 3 is an example; anoher is he exponenial funcion e. ask Sae he period of each of he following periodic funcions and say wheher i is even or odd or neiher. (a) f() (b) f() 4 4 Your soluion (a) is neiher even nor odd (wih period ) (b) is odd (wih period ). 3 HELM (8): Workbook 3: Fourier Series

43 A Fourier series conains a sum of erms while he inegral formulae for he Fourier coefficiens a n and b n conain producs of he ype f()cosn and f() sin n. We need herefore resuls for sums and producs of funcions. Suppose, for example, g() is an odd funcion and h() is an even funcion. Le F () = g() h() (produc of odd and even funcions) so F ( ) = g( )h( ) (replacing by ) = ( g())h() (since g is odd and h is even) = g()h() = F () So F () is odd. Now suppose F () = g()+h() (sum of odd and even funcions) F ( ) = g( )+h() = g()+h() We see ha F ( ) = F () and F ( ) = F () So F () is neiher even nor odd. ask Invesigae he odd/even naure of sums and producs of (a) wo odd funcions g (),g () (b) wo even funcions h (),h () Your soluion HELM (8): Secion 3.3: Even and Odd Funcions 33

44 G () = g ()g () G ( ) = ( g ())( g ()) = g ()g () = G () so he produc of wo odd funcions is even. G () = g ()+g () G ( ) = g ( )+g ( ) = g () g () = G () so he sum of wo odd funcions is odd. H () = h ()h () H () = h ()+h () A similar approach shows ha H ( ) = H () H ( ) = H () i.e. boh he sum and produc of wo even funcions are even. hese resuls are summarized in he following Key Poin. Key Poin 5 Producs of funcions (even) (even) = (even) (even) (odd) = (odd) (odd) (odd) = (even) Sums of funcions (even) + (even) = (even) (even) + (odd) = (neiher) (odd) + (odd) = (odd) 34 HELM (8): Workbook 3: Fourier Series

45 Useful properies of even and of odd funcions in connecion wih inegrals can be readily deduced if we recall ha a definie inegral has he significance of giving us he value of an area: y = f() a b b a Figure 6 f() d gives us he ne value of he shaded area, ha above he -axis being posiive, ha below being negaive. ask For he case of a symmerical inerval ( a, a) deduce wha you can abou a g() d and a a a h() d where g() is an odd funcion and h() is an even funcion. g() h() a a a a Your soluion We have a a g() d = for an odd funcion a h() d = a a h() d for an even funcion (Noe ha neiher resul holds for a funcion which is neiher even nor odd.) HELM (8): Secion 3.3: Even and Odd Funcions 35

46 . Fourier series implicaions Since a sum of even funcions is iself an even funcion i is no unreasonable o sugges ha a Fourier series conaining only cosine erms (and perhaps a consan erm which can also be considered as an even funcion) can only represen an even periodic funcion. Similarly a series of sine erms (and no consan) can only represen an odd funcion. hese resuls can readily be shown more formally using he expressions for he Fourier coefficiens a n and b n. ask Recall ha for a -periodic funcion b n = f() sin n d If f() is even, deduce wheher he inegrand is even or odd (or neiher) and hence evaluae b n. Repea for he Fourier coefficiens a n. Your soluion We have, if f() is even, f() sin n =(even) (odd) =odd hence b n = (odd funcion) d = hus an even funcion has no sine erms in is Fourier series. Also f()cosn =(even) (even) =even a n = (even funcion) d = I should be obvious ha, for an odd funcion f(), a n = b n = f()cosn d = f() sin n d f()cosn d. (odd funcion) d = Analogous resuls hold for funcions of any period, no necessarily. 36 HELM (8): Workbook 3: Fourier Series

47 For a periodic funcion which is neiher even nor odd we can expec a leas some of boh he a n and b n o be non-zero. For example consider he square wave funcion: f() Figure 7: Square wave his funcion is neiher even nor odd and we have already seen in Secion 3. ha is Fourier series conains a consan and sine erms. his resul could be expeced because we can wrie f() = + g() where g() is as shown: g() Figure 8 Clearly g() is odd and will conain only sine erms. he Fourier series are in fac f() = + sin + 3 sin sin and sin + 3 sin sin g() = HELM (8): Secion 3.3: Even and Odd Funcions 37

48 ask For each of he following funcions deduce wheher he corresponding Fourier series conains (a) sine erms only or cosine erms only or boh (b) a consan erm y y 3 y 4 y a a 5 y 6 y 7 y Your soluion. cosine erms only (plus consan). 5. sine erms only (no consan).. cosine erms only (no consan). 6. sine and cosine erms (plus consan). 3. sine erms only (no consan). 7. cosine erms only (plus consan). 4. cosine erms only (plus consan). 38 HELM (8): Workbook 3: Fourier Series

49 ask Confirm he resul obained for he riangular wave, funcion 7 in he las ask, by finding he Fourier series fully. he funcion involved is f() = << f( +) = f() Your soluion Since f() is even we can say immediaely b n = n =,, 3,... Also a n = Also a = cos n d = n even 4 n n odd d = so he Fourier series is cos + 9 cos cos (afer inegraion by pars) f() = 4 HELM (8): Secion 3.3: Even and Odd Funcions 39

50 ask Consider he square wave of period one period of which is shown in Figure. 4 (a) Wrie down he analyic descripion of his funcion, (b) Sae wheher you expec he Fourier series of his funcion o conain a consan erm, (c) Lis any oher possible feaures of he Fourier series ha you migh expec from he graph of he square-wave funcion. Your soluion (a) We have f() = f( +) = f() 4 << <<, << (b) he Fourier series will conain a consan erm since he square wave here is non-negaive and canno herefore have a zero average value. his consan erm is ofen referred o as he d.c. (direc curren) erm by engineers. (c) Since he square wave is an even funcion (i.e. he graph has symmery abou he y axis) hen is Fourier series will conain cosine erms bu no sine erms because only he cosines are even funcions. (Well done if you spoed his a his early sage!) HELM (8): Secion 3.: Represening Periodic Funcions by Fourier Series

51 I is possible o show, and we will do so laer, ha he Fourier series represenaion of his square wave is + 8 cos 3 cos cos 5 7 cos i.e. he Fourier coefficiens are a =, a = 8, a =, a 3 = 8 3, a 4 =, a 5 = 8 5,... Noe, as well as he presence of he consan erm and of he cosine (bu no sine) erms, ha only odd harmonics are presen i.e. sinusoids of period, 3, 5,,... or of frequency, 3, 5, 7,... 7 imes he fundamenal frequency. We now show in Figure 8 graphs of (i) he square wave (ii) he firs wo erms of he Fourier series represening he square wave (iii) he firs hree erms of he Fourier series represening he square wave (iv) he firs four erms of he Fourier series represening he square wave (v) he firs five erms of he Fourier series represening he square wave Noe: We show he graphs for <<only since he square wave and is Fourier series are even. (i) 4 (ii) (iii) + 8 cos + 8 (cos cos 3 ) (iv) + 8 (v) (cos 3 cos cos 5 ) (cos 3 cos cos 5 cos 7 ) Figure 8 HELM (8): Workbook 3: Fourier Series

52 We can clearly see from Figure 8 ha as he number of erms is increased he graph of he Fourier series gradually approaches ha of he original square wave - he ripples increase in number bu decrease in ampliude. (he behaviour near he disconinuiy, a =, is slighly more complicaed and i is possible o show ha however many erms are aken in he Fourier series, some overshoo will always occur. his effec, which we do no discuss furher, is known as he Gibbs Phenomenon.) Orhogonaliy properies of sinusoids As saed earlier, a periodic funcion f() wih period has a Fourier series represenaion f() = a + a cos + a cos b sin + b sin +..., = a + (a n cos n + b n sin n) (3) o deermine he Fourier coefficiens a n,b n and he consan erm a use has o be made of cerain inegrals involving sinusoids, he inegrals being over a range α, α+, where α is any number. (We will normally choose α =.) ask Find sin n d and cos n d where n is an ineger. Your soluion In fac boh inegrals are zero for sin n d = n cos n = ( cos n + cos n) = n n = (4) cos n d = sin n = n = (5) n As special cases, if n =he firs inegral is zero and he second inegral has value. N.B. Any inegraion range α, α +, would give hese same (zero) answers. hese inegrals enable us o calculae he consan erm in he Fourier series (3) as in he following ask. HELM (8): Secion 3.: Represening Periodic Funcions by Fourier Series 3

53 ask Inegrae boh sides of (3) from o and use he resuls from he previous ask. Hence obain an expression for a. Your soluion We ge for he lef-hand side f()d (whose value clearly depends on he funcion f()). Inegraing he righ-hand side erm by erm we ge a d + a n cos n d + b n sin n d = (using he inegrals (4) and (5) shown above). hus we ge f() d = (a ) or a = a + {+} f() d (6) Key Poin he consan erm in a rigonomeric Fourier series for a funcion of period is a = f() d = average value of f() over period. 4 HELM (8): Workbook 3: Fourier Series

54 his resul ies in wih our earlier discussion on he significance of he consan erm. Clearly a signal whose average value is zero will have no consan erm in is Fourier series. he following square wave (Figure 9) is an example. f() Figure 9 We now obain furher inegrals, known as orhogonaliy properies, which enable us o find he remaining Fourier coefficiens i.e. he ampliudes a n and b n (n =,, 3,...) of he sinusoids. ask Using he sandard rigonomeric ideniy ha sin n cos m {sin(n + m) + sin(n m)} evaluae sin n cos m d where n and m are any inegers. Your soluion We ge sin n cos m d = sin(n + m) d + using he resuls (4) and (5) since n + m and n m are also inegers. his resul holds for any inerval of. sin(n m) d = {+} = HELM (8): Secion 3.: Represening Periodic Funcions by Fourier Series 5

55 Key Poin Orhogonaliy Relaion For any inegers m, n, including he case m = n, sin n cos m d = We shall use his resul shorly bu need a few more inegrals firs. Consider nex cos n cos m d where m and n are inegers. Using anoher rigonomeric ideniy we have, for he case n = m, cos n cos m d = {cos(n + m) + cos(n m)}d = {+} = using he inegrals (4) and (5). For he case n = m we mus ge a non-zero answer since cos n is non-negaive. In his case: cos n d = = ( + cos n) d + sin n n = ( provided n = ) For he case n = m =we have cos n cos m d = ask Proceeding in a similar way o he above, evaluae sin n sin m d for inegers m and n. Again consider separaely he hree cases: (a) n = m, (b) n = m = and (c) n = m =. 6 HELM (8): Workbook 3: Fourier Series

56 Your soluion (a) Using he ideniy sin n sin m {cos(n m) cos(n + m)} and inegraing he righhand side erms, we ge, using (4) and (5) sin n sin m d = n, m inegers n = m (b) Using he ideniy cos θ = sin θ wih θ = n gives for n = m = (c) When n = m =, sin n d = sin n sin m d =. We summarise hese resuls in he following Key Poin: ( cos n)d = For inegers n, m Key Poin 3 sin n cos m d = cos n cos m d = sin n sin m d = n = m n = m = n = m = n = m, n = m = n = m All hese resuls hold for any inegraion range of widh. HELM (8): Secion 3.: Represening Periodic Funcions by Fourier Series 7

57 3. Calculaion of Fourier coefficiens Consider he Fourier series for a funcion f() of period : f() = a + (a n cos n + b n sin n) (7) o obain he coefficiens a n (n =,, 3,...), we muliply boh sides by cos m where m is some posiive ineger and inegrae boh sides from o. For he lef-hand side we obain f()cosm d For he righ-hand side we obain a cos m d + a n cos n cos m d + b n he firs inegral is zero using (5). sin n cos m d Using he orhogonaliy relaions all he inegrals in he summaion give zero excep for he case n = m when, from Key Poin 3 cos m d = Hence f()cosm d = a m from which he coefficien a m can be obained. Rewriing m as n we ge a n = f()cosn d for n =,, 3,... (8) Using (6), we see he formula also works for n =(bu we mus remember ha he consan erm is a.) From (8) a n = average value of f()cosn over one period. 8 HELM (8): Workbook 3: Fourier Series

58 ask By muliplying (7) by sin m obain an expression for he Fourier Sine coefficiens b n, n =,, 3,... Your soluion A similar calculaion o ha performed o find he a n gives f() sin m d = a sin m d + a n cos n sin m d + All erms on he righ-hand side inegrae o zero excep for he case n = m where b m sin m d = b m Relabelling m as n gives b n = b n sin n sin m d f() sin n d n =,, 3,... (9) (here is no Fourier coefficien b.) Clearly b n = average value of f() sin n over one period. HELM (8): Secion 3.: Represening Periodic Funcions by Fourier Series 9

59 Key Poin 4 A funcion f() wih period has a Fourier series f() = a + (a n cos n + b n sin n) he Fourier coefficiens are a n = b n = f()cosn d n =,,,... f() sin n d n =,,... In he inegrals any convenien inegraion range exending over an inerval of may be used. 4. Examples of Fourier series We shall obain he Fourier series of he half-recified square wave shown in Figure. f() period We have Figure f() = f( +) = f() << << he calculaion of he Fourier coefficiens is merely sraighforward inegraion using he resuls already obained: a n = f()cosn d in general. Hence, for our square wave a n = () cos n d = sin n n = provided n = HELM (8): Workbook 3: Fourier Series

60 Bu a = () d =so he consan erm is a =. (he square wave akes on values and over equal lengh inervals of so value.) Similarly b n = () sin n d = Some care is needed now! b n = ( cos n) n Bu cos n = + n =, 4, 6,..., cos n n is clearly he mean b n = n =, 4, 6,... However, cos n = n =, 3, 5,... b n = ( ( )) = n n i.e. b =,b 3 = 3,b 5 = 5,... Hence he required Fourier series is n =, 3, 5,... f() = a + (a n cos n + b n sin n) in general f() = + sin + 3 sin sin in his case Noe ha he Fourier series for his paricular form of he square wave conains a consan erm and odd harmonic sine erms. We already know why he consan erm arises (because of he non-zero mean value of he funcions) and will explain laer why he presence of any odd harmonic sine erms could have been prediced wihou inegraion. he Fourier series we have found can be wrien in summaion noaion in various ways: + (n odd) Fourier series as sin n or, since n is odd, we may wrie n =k k =,,... and wrie he n + k= sin(k ) (k ) HELM (8): Secion 3.: Represening Periodic Funcions by Fourier Series

61 ask Obain he Fourier series of he square wave one period of which is shown: 4 Your soluion HELM (8): Workbook 3: Fourier Series

62 We have, since he funcion is non-zero only for <<, a = 4 d =4 a =is he consan erm as we would expec. Also a n = = 4 n sin 4 cos n d = 4 n sin n n sin n = 8 n sin n I follows from a knowledge of he sine funcion ha n =, 4, 6,... n =,, 3,... Also a n = b n = 8 n 8 n n =, 5, 9,... n =3, 7,,... 4 sin n d = 4 cos n n = 4 n cos n Hence, he required Fourier series is f() =+ 8 cos 3 cos cos 5 7 cos cos n = which, like he previous square wave, conains a consan erm and odd harmonics, bu in his case odd harmonic cosine erms raher han sine. You may recall ha his paricular square wave was used earlier and we have already skeched he form of he Fourier series for, 3, 4 and 5 erms in Figure 8. Clearly, in finding he Fourier series of square waves, he inegraion is paricularly simple because f() akes on piecewise consan values. For oher funcions, such as saw-ooh waves his will no be he case. Before we ackle such funcions however we shall generalise our formulae for he Fourier coefficiens a n,b n o he case of a periodic funcion of arbirary period, raher han confining ourselves o period. HELM (8): Secion 3.: Represening Periodic Funcions by Fourier Series 3

63 5. Fourier series for funcions of general period his is a sraighforward exension of he period case ha we have already discussed. Using x (insead of ) emporarily as he variable. We have seen ha a periodic funcion f(x) has a Fourier series f(x) = a + (a n cos nx + b n sin nx) wih a n = f(x)cosnx dx n =,,,... b n = f(x) sin nx dx n =,,... Suppose we now change he variable o where x =. hus x = corresponds o = / and x = corresponds o = /. Hence regarded as a funcion of, we have a funcion wih period. Making he subsiuion x =, and hence dx = a n = b n = d, in he expressions for a n and b n we obain n f()cos d n =,,... n f() sin d n =,... hese inegrals give he Fourier coefficiens for a funcion of period whose Fourier series is f() = a + n n a n cos + b n sin Various oher noaions are commonly used in his case e.g. i is someimes convenien o wrie he period =. (his is paricularly useful when Fourier series arise in he soluion of parial differenial equaions.) Anoher alernaive is o use he angular frequency ω and pu =/ω. ask Wrie down he form of he Fourier series and expressions for he coefficiens if (a) = (b) =/ω. Your soluion 4 HELM (8): Workbook 3: Fourier Series

64 (a) f() = a + n n a n cos + b n sin wih a n = n f()cos d and similarly for b n. (b) f() = a + {a n cos(nω)+b n sin(nω)} wih a n = ω ω f() cos(nω) d ω and similarly for b n. You should noe ha, as usual, any convenien inegraion range of lengh (or or ω used in evaluaing a n and b n. ) can be Example Find he Fourier series of he funcion shown in Figure which is a saw-ooh wave wih alernae porions removed. f() Figure Soluion Here he period = =4so =. he Fourier series will have he form f() = a + n n a n cos + b n sin he coefficiens a n are given by a n = n f()cos d where f() = Hence a n = << << n cos d. f( +4)=f() HELM (8): Secion 3.: Represening Periodic Funcions by Fourier Series 5

65 Soluion (cond.) he inegraion is readily performed using inegraion by pars: n cos d = = = 4 n n sin cos n n 4 (cos n ). n Hence, since a n = cos( n )d n =, 4, 6,... a n = 4 n =, 3, 5,... n he consan erm is a where a = Similarly b n = n sin d where n sin d = n cos n d =. n + n he second inegral gives zero. Hence b n = n =, 4, 6,... n cos n = n + n =, 3, 5,... n n sin d n = n cos d. Hence, using all hese resuls for he Fourier coefficiens, he required Fourier series is f() = 4 cos cos cos sin sin sin... Noice ha because he Fourier coefficiens depend on n (raher han as was he case for n he square wave) he sinusoidal componens in he Fourier series have quie rapidly decreasing ampliudes. We would herefore expec o be able o approximae he original saw-ooh funcion using only a quie small number of erms in he series. 6 HELM (8): Workbook 3: Fourier Series

66 ask Obain he Fourier series of he funcion f() = << f( +) = f() f() Firs wrie ou he form of he Fourier series in his case: Your soluion Since = =and since he funcion has a non-zero average value, he form of he Fourier series is a + {a n (cos n)+b n sin(n)} Now wrie ou inegral expressions for a n and b n. Will here be a consan erm in he Fourier series? Your soluion Because he funcion is non-negaive here will be a consan erm. Since = =hen = and we have a n = b n = cos(n) d n =,,,... sin(n) d n =,,... he consan erm will be a where a = d. HELM (8): Secion 3.: Represening Periodic Funcions by Fourier Series 7

67 Now evaluae he inegrals. ry o spo he value of he inegral for b n so as o avoid inegraion. Noe ha he inegrand is an even funcions for a n and an odd funcon for b n. Your soluion he inegral for b n is zero for all n because he inegrand is an odd funcion of. Since he inegrand is even in he inegrals for a n we can wrie a n = cos n d n =,,,... he consan erm will be a o where a = d = 3. For n =,, 3,... we mus inegrae by pars (wice) a n = n sin(n) sin(n) d n = 4 n n cos(n) + cos(n) d. n he inegral in he second erm gives zero so a n = 4 cos n. n Now wriing ou he final form of he Fourier series we have f() = cos n cos(n) = n cos()+ 4 cos() 9 cos(3) HELM (8): Workbook 3: Fourier Series

68 Exercises For each of he following periodic signals skech he given funcion over a few periods find he rigonomeric Fourier coefficiens wrie ou he firs few erms of he Fourier series. <</. f() = f( +) =f() / << square wave. f() = << f( +)=f() / << 3. f() = f( + )=f() <</ << 4. f() = f( +) =f() << / << 5. f() = A sin <</ s cos 3 cos 3... f( + )=f() cos sin sin 3 sin 5 sin cos cos 3 cos 4 cos sin ω+ 3 sin 3 ω+ 5 sin 5ω+... where ω =/. 6 cos + A + A sin ω A square wave half-wave recifier sin cos cos sin sin sin sin cos ω cos 4ω + ()(3) (3)(5) +... HELM (8): Secion 3.: Represening Periodic Funcions by Fourier Series 9

69 Periodic Funcions 3. Inroducion You should already know how o ake a funcion of a single variable f(x) and represen i by a power series in x abou any poin x of ineres. Such a series is known as a aylor series or aylor expansion or, if x =, as a Maclaurin series. his opic was firs me in 6. Such an expansion is only possible if he funcion is sufficienly smooh (ha is, if i can be differeniaed as ofen as required). Geomerically his means ha here are no jumps or spikes in he curve y = f(x) near he poin of expansion. However, in many pracical siuaions he funcions we have o deal wih are no as well behaved as his and so no power series expansion in x is possible. Neverheless, if he funcion is periodic, so ha i repeas over and over again a regular inervals, hen, irrespecive of he funcion s behaviour (ha is, no maer how many jumps or spikes i has), he funcion may be expressed as a series of sines and cosines. Such a series is called a Fourier series. Fourier series have many applicaions in mahemaics, in physics and in engineering. For example hey are someimes essenial in solving problems (in hea conducion, wave propagaion ec) ha involve parial differenial equaions. Also, using Fourier series he analysis of many engineering sysems (such as elecric circuis or mechanical vibraing sysems) can be exended from he case where he inpu o he sysem is a sinusoidal funcion o he more general case where he inpu is periodic bu non-sinsusoidal. Prerequisies Before saring his Secion you should... Learning Oucomes On compleion you should be able o... be familiar wih rigonomeric funcions recognise periodic funcions deermine he frequency, he ampliude and he period of a sinusoid represen common periodic funcions by rigonomeric Fourier series HELM (8): Workbook 3: Fourier Series

70 . Inroducion You have me in earlier Mahemaics courses he concep of represening a funcion by an infinie series of simpler funcions such as polynomials. For example, he Maclaurin series represening e x has he form e x =+x + x! + x3 3! +... or, in he more concise sigma noaion, e x x n = n! n= (remembering ha! is defined as ). he basic idea is ha for hose values of x for which he series converges we may approximae he funcion by using only he firs few erms of he infinie series. Fourier series are also usually infinie series bu involve sine and cosine funcions (or heir complex exponenial equivalens) raher han polynomials. hey are widely used for approximaing periodic funcions. Such approximaions are of considerable use in science and engineering. For example, elemenary a.c. heory provides echniques for analyzing elecrical circuis when he currens and volages presen are assumed o be sinusoidal. Fourier series enable us o exend such echniques o he siuaion where he funcions (or signals) involved are periodic bu no acually sinusoidal. You may also see in 5 ha Fourier series someimes have o be used when solving parial differenial equaions.. Periodic funcions A funcion f() is periodic if he funcion values repea a regular inervals of he independen variable. he regular inerval is referred o as he period. See Figure. f() period If P denoes he period we have f( + P )=f() for any value of. Figure HELM (8): Secion 3.: Periodic Funcions 3

71 he mos obvious examples of periodic funcions are he rigonomeric funcions sin and cos, boh of which have period (using radian measure as we shall do hroughou his Workbook) (Figure ). his follows since sin( +) = sin and cos( +) = cos y = sin y = cos period period Figure he ampliude of hese sinusoidal funcions is he maximum displacemen from y =and is clearly. (Noe ha we use he erm sinusoidal o include cosine as well as sine funcions.) More generally we can consider a sinusoid y = A sin n which has maximum value, or ampliude, A and where n is usually a posiive ineger. For example y = sin is a sinusoid of ampliude and period because sin ( + ) = sin( +) = sin for any value of. = (Figure 3). he fac ha he period is follows y = sin period Figure 3 4 HELM (8): Workbook 3: Fourier Series

72 We see ha y = sin has half he period of sin, as opposed o (Figure 4). his can alernaively be phrased by saing ha sin oscillaes wice as rapidly (or has wice he frequency) of sin. y = sin y = sin Figure 4 In general y = A sin n has ampliude A, period and complees n oscillaions when changes n by. Formally, we define he frequency of a sinusoid as he reciprocal of he period: frequency = period and he angular frequency, ofen denoed he Greek Leer ω (omega) as hus angular frequency = frequency = period y = A sin n has frequency n and angular frequency n. ask Sae he ampliude, period, frequency and angular frequency of (a) y = 5 cos 4 (b) y = 6 sin 3. Your soluion (a) ampliude 5, period 4 =, frequency, angular frequency 4 Your soluion (b) ampliude 6, period 3, frequency 3, angular frequency 3 HELM (8): Secion 3.: Periodic Funcions 5

73 Harmonics In represening a non-sinusoidal funcion of period by a Fourier series we shall see shorly ha only cerain sinusoids will be required: (a) A cos (and B sin ) hese also have period and ogeher are referred o as he firs harmonic (or fundamenal harmonic). (b) A cos (and B sin ) hese have half he period, and double he frequency, of he firs harmonic and are referred o as he second harmonic. (c) A 3 cos 3 (and B 3 sin 3) hese have period 3 and consiue he hird harmonic. In general he Fourier series of a funcion of period will require harmonics of he ype A n cos n ( and B n sin n) where n =,, 3,... Non-sinusoidal periodic funcions he following are examples of non-sinusoidal periodic funcions (hey are ofen called waves ). Square wave f() Figure 5 Analyically we can describe his funcion as follows: << f() = (which gives he definiion over one period) + << f( +) =f() (which ells us ha he funcion has period ) Saw-ooh wave f() 4 4 Figure 6 In his case we can describe he funcion as follows: f() = << f( +)=f() Here he period is, he frequency is and he angular frequency is =. 6 HELM (8): Workbook 3: Fourier Series

74 riangular wave f() Figure 7 Here we can convenienly define he funcion using <<as he basic period : << f() = << or, more concisely, f() = << ogeher wih he usual saemen on periodiciy f( +) =f(). ask Wrie down an analyic definiion for he following periodic funcion: f() Your soluion f() = <<3 3 <<5 f( +5)=f() HELM (8): Secion 3.: Periodic Funcions 7

75 ask Skech he graphs of he following periodic funcions showing all relevan values: / <<4 (a) f() = 8 4 <<6 f( +8)=f() 6 <<8 (b) f() = << f( +)=f() Your soluion (a) 8 f() (b) period f() Figure 9 period 8 HELM (8): Workbook 3: Fourier Series

76 Index for Workbook 3 Ampliude 4 Angular frequency 5 Argumen 54 Complex form of Fourier series Convergence of Fourier series 4-45 Differenial equaion 69 Dirichle condiions 4 Euler s relaion 54, 56 Even funcion 3, 34 Fourier coefficiens, 4, 8, 4 Frequency 5 Funcion - odd 3 - even 3 Half-range Fourier series 46-5 Harmonic oscillaor 69 Harmonics 6 Modelling vibraing sysem 68 Odd funcion 3, 34 Orhogonaliy properies 3, 6 Parseval s heorem 64 Periodic funcions 3, 9-9 Phase 54 Pi Saw-ooh wave 6, 5, 3, 6 Square wave 6,,,, 9, 3 37, 4, 7 riangular wave 7, 38, 39, 65 Vibraion 69 Waves - saw-ooh 6, 5, 3, 6 - square 6,,,, 9 3, 37, 4, 7 - riangular 7, 38, 39, 65 EXERCISES 9, 45, 5, 67

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