# Representing Periodic Functions by Fourier Series. (a n cos nt + b n sin nt) n=1

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1 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,n=, 1,, 3... and b n,,, 3,... may be deermined for any given funcion f(). Prerequisies Before saring his Secion you should... Learning Oucomes Afer compleing his Secion you should be able o... 1 know wha a periodic funcion is be able o inegrae funcions involving sinusoids 3 have knowledge of inegraion by pars calculae Fourier coefficiens of a funcion of period calculae Fourier coefficiens of a funcion of general period

2 1. Inroducion We recall firs a simple rigonomeric ideniy: cos = 1+cos or equivalenly cos = cos (1) Equaion 1 can be inerpreed as a simple finie Fourier Series represenaion of he periodic funcion f() =cos which has period. We jus noe ha he Fourier Series represenaion conains a consan erm and a period erm. A more complicaed rigonomeric ideniy is sin 4 = cos + 1 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 1 (he firs harmonic ) and a funcion of period or frequency (he second harmonic ). The reason for he consan erm in boh (1) 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 as, herefore, 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() ofperiod 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 =1,, 3,...) and, if he periodic funcion has a non-zero average value, a consan erm. Thus we seek a Fourier Series represenaion of he general form f() = a + a 1 cos + a cos b 1 sin + b sin +... The reason for labelling he consan erm as a will be discussed laer. The ampliudes a 1,a,... b 1,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 (VERSION 1: March 18, 4): Workbook Level 3.: Represening Periodic Funcions by Fourier Series

3 Consider he square wave of period one period of which is shown in Figure 1. 4 Wrie down Figure 1 i. he analyic descripion of his funcion, ii. wheher you expec he Fourier Series of his funcion o conain a consan erm, iii. any oher possible feaures of he Fourier Series ha you migh expec from he graph of he square-wave funcion. Your soluion (iii) Since he square wave is an even funcion (i.e. he graph in Figure 1 has symmery abou he y axis) hen is Fourier Series will conain cosine erms bu no sine erms because only he former are even funcions. (Well done if you spoed his a his early sage!) (ii) The Fourier Series will conain a consan erm (ofen referred o as he d.c. (direc curren) erm by engineers) since he square wave here is non-negaive and canno herefore have a zero average value) <<, << 4 << f( +) = f() f() = (i) We have To be precise i is possible o show, and we will do so laer, ha he Fourier Series represenaion 3 HELM (VERSION 1: March 18, 4): Workbook Level 3.: Represening Periodic Funcions by Fourier Series

4 of his square wave is {cos 13 cos cos 5 17 cos i.e. he Fourier coefficiens are a =, a 1 = 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,,,,... or of frequency , 3, 5, 7,... imes he fundamenal frequency 1. We now show in Figure graphs (for <<only since he square wave and is Fourier Series are even) of (i) he square wave (ii) he firs wo erms of he Fourier Series (iii) he firs hree erms of he Fourier Series (iv) he firs four erms of he Fourier Series (v) he firs five erms of he Fourier Series (i) 4 (ii) (iii) + 8 cos + 8 (cos 1 cos 3 ) (iv) + 8 (v) (cos 1 3 cos cos 5 ) (cos 1 3 cos cos 5 1 cos 7 ) Figure We can clearly see from Figure 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. (The behaviour near he disconinuiy, a =,isslighly more complicaed and i is possible o show ha however many erms are aken in he Fourier Series, some overshoo will always occur. This effec, which we do no discuss furher, is known as Gibbs Phenomenon.) HELM (VERSION 1: March 18, 4): Workbook Level 3.: Represening Periodic Funcions by Fourier Series 4

5 Orhogonaliy properies of sinusoids As saed earlier, a periodic funcion f() wih period has a Fourier Series represenaion f() = a + a 1 cos + a cos b 1 sin + b sin +... = a + (a n cos n + b n sin n) (3) To 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 = ). Find sin n d and cos n d where n is an ineger Your soluion [ 1n cos n ] sin n d = [ 1 n = 1 n { cos n + cos n = n (4) cos n d = sin n ] In fac boh inegrals are zero for = n (5) 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. These inegrals enable us o calculae he consan erm in he Fourier Series (3) as in he following guided exercise. Inegrae boh sides of (3) from o and use he above resuls. Hence obain an expression for a. 5 HELM (VERSION 1: March 18, 4): Workbook Level 3.: Represening Periodic Funcions by Fourier Series

6 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 1 a d + { a n cos n d + b n sin n d = 1 [a ] + {+ (using he inegrals (4) and (5) shown above). Thus we ge f()d = 1 (a ) or a = 1 f()d (6) Key Poin The consan erm in a rigonomeric Fourier Series for a funcion of period is a = 1 f()d = average value of f() over 1period. This 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. The following square wave is an example. HELM (VERSION 1: March 18, 4): Workbook Level 3.: Represening Periodic Funcions by Fourier Series 6

7 1 f() 1 Figure 3 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 =1,, 3,...)ofhe sinusoids. Recall, using a sandard rigonomeric ideniy ha sin n cos m = 1 {sin(n + m) + sin(n m) Hence evaluae where n and m are any inegers. sin n cos m d Your soluion We ge { sin n cos m d = 1 sin(n + m) d + sin(n m) d = 1 {+ = using he resuls (4) and (5) since n + m and n m are also inegers. This resul holds for any inerval α, α +. 7 HELM (VERSION 1: March 18, 4): Workbook Level 3.: Represening Periodic Funcions by Fourier Series

8 Key Poin For any inegers m, n, including he case m = n, wehave he orhogonaliy relaion 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 = 1 {cos(n + m) + cos(n m)d = 1 {+ = 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 = 1 = 1 (1 + cos n)d [ + 1 sin n n For he case n = m = we have ] = (provided n ) cos n cos m d = Proceeding in a similar way o he above, obain sin n sin m d for inegers m and n. Again consider separaely he cases n m, n = m and n = m =. Your soluion HELM (VERSION 1: March 18, 4): Workbook Level 3.: Represening Periodic Funcions by Fourier Series 8

9 Using he ideniy sin n sin m = 1 {cos(n m) cos(n + m) and inegraing he righ hand side erms, we ge, using (4) and (5) sin n sin m d = n, m inegers n m using he ideniy cos θ =1 sin θ wih θ = n gives for n = m sin n d = 1 (1 cos n)d = Of course, when n = m =, sin n sin m d =. We summarise hese resuls in he following key poin: For inegers n, m Key Poin sin n cos m d = cos n cos m d = sin n sin m d = All hese resuls hold for any inegraion range α, α +. n m n = m n = m = { n m, n = m = n = m 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) To obain he coefficiens a n (n =1,, 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() cos m d 9 HELM (VERSION 1: March 18, 4): Workbook Level 3.: Represening Periodic Funcions by Fourier Series

10 for he righ hand side we obain a cos m d + {a n cos n cos m d + b n sin n cos m d The firs inegral is zero using (5). Using he orhogonaliy relaions all he inegrals in he summaion give zero excep for he case n = m when, from he las key poin cos m d = Hence f() cos m d = a m from which he coefficien a m can be obained. Rewriing m as n we ge a n = 1 f() cos n d for n =1,, 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() cos n over one period. By muliplying (7) by sin m obain an expression for he Fourier Sine coefficiens b n ; n =1,, 3,... Your soluion HELM (VERSION 1: March 18, 4): Workbook Level 3.: Represening Periodic Funcions by Fourier Series 1

11 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 + b n sin 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. b n = 1 (There is no Fourier coefficien b.) Clearly b n = average value of f() sin n over one period. f() sin n d n =1,, 3,... (9) Key Poin A funcion f() wih period has a Fourier Series f() = a + (a n cos n + b n sin n) The Fourier coefficiens are a n = 1 b n = 1 f() cos n d n =, 1,,... f() sin n d n =1,,... In he inegrals any convenien inegraion range α, α + may be used. 11 HELM (VERSION 1: March 18, 4): Workbook Level 3.: Represening Periodic Funcions by Fourier Series

12 4. Examples of Fourier Series We shall obain he Fourier Series of he half-recified square wave shown. 1 f() period We have Figure 4 { 1 << f() = << f( +) = f() The calculaion of he Fourier coefficiens is merely sraighforward inegraion using he resuls already obained: in general. Hence, for our square wave a n = 1 f() cos n d a n = 1 (1) cos n d = 1 [ sin n n ] = provided n Bu a = 1 (1) d =1so he consan erm is a = 1. (The square wave akes on values 1 and over equal lengh inervals of so 1 value.) Similarly b n = 1 (1) sin n d = 1 [ ] cos n n is clearly he mean Some care is needed now! b n = 1 (1 cos n) n Bu cos n =+1 n =, 4, 6,..., However, cos n = 1 n =1, 3, 5,... b n = n =, 4, 6,... b n = 1 (1 ( 1)) = n n n =1, 3, 5,... i.e. b 1 =,b 3 = 3,b 5 = 5,... HELM (VERSION 1: March 18, 4): Workbook Level 3.: Represening Periodic Funcions by Fourier Series 1

13 Hence he required Fourier Series is f() = a + (a n cos n + b n sin n) in general f() = 1 + (sin + 13 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. The Fourier series we have found can be wrien in summaion noaion in various ways: or, since n is odd, we may wrie 1 + (n odd) 1 sin n n n =k 1 k =1,,... and wrie he Fourier Series as 1 + k=1 1 sin(k 1) (k 1) Obain he Fourier Series of he square wave one period of which is shown: 4 Figure 5 13 HELM (VERSION 1: March 18, 4): Workbook Level 3.: Represening Periodic Funcions by Fourier Series

14 Your soluion a = 1 4d =4 a =is he consan erm as we would expec. Also [ sin n ] a n = 1 = 4 n ( sin 4 cos n d = 4 ( n ) sin ( n n )) = 8 ( n n sin ) n =1,, 3,... I follows from a knowledge of he sine funcion ha n =, 4, 6,... 8 a n = n =1, 5, 9,... n 8 n =3, 7, 11,... n Also b n = 1 4 sin n d = 4 [ cos n n ] = 4 n ( cos ( n We have, since he funcion is non-zero only for <<, ) cos ( n )) = Hence, he required Fourier Series is f() =+ 8 (cos 13 cos cos 5 17 ) cos 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 was used earlier and we have already skeched he form of he Fourier Series for, 3, 4 and 5 erms. Clearly in finding he Fourier Series of square waves he inegraion is paricularly simple because HELM (VERSION 1: March 18, 4): Workbook Level 3.: Represening Periodic Funcions by Fourier Series 14

15 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. 5. Fourier Series for funcions of general period This 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 = 1 b n = 1 Suppose we now change he variable o where f(x) cos nx dx n =, 1,,... f(x) sin nx dx n =1,,... x = P Thus x = corresponds o = P and x = corresponds o = P. Hence regarded as a funcion of, wehave a funcion wih period P. Making he subsiuion x =, and hence dx = d, inhe expressions for a P P n and b n we obain a n = P b n = P P P P P ( ) n f() cos d P n =, 1,... ( ) n f() sin d P n =1,... These inegrals give he Fourier coefficiens for a funcion of period P whose Fourier Series is f() = a + [ ( ) ( )] n n a n cos + b n sin P P Various oher noaions are commonly used in his case e.g. i is someimes convenien o wrie he period P = l. (This is paricularly useful when Fourier Series arise in he soluion of parial differenial equaions.) Anoher alernaive is o use he angualr frequency ω and pu P =/ω. 15 HELM (VERSION 1: March 18, 4): Workbook Level 3.: Represening Periodic Funcions by Fourier Series

16 Wrie down he form of he Fourier Series and expressions for he coefficiens if (i) P =l (ii) P = ω. Your soluion d ) l f() cos(nω) d ( n f() cos l l ω ω [a n cos(nω)+b n sin(nω)] wih a n = ω wih a n = 1 l )] l ( n + b n sin ) l ( n a n cos [ (ii) f() = a + and similarly for b n. (i) f() = a + and similarly for b n. You should noe ha, as usual, any convenien inegraion range of lengh P (or l or ω be used in evaluaing a n and b n. ) can Example Find he Fourier Series of he funcion shown in Figure 6, viz a saw ooh wave wih alernaive porions removed f() Figure 6 Here he period P =l =4sol =. The Fourier Series will have he form f() = a + ( ) ( ) n n a n cos + b n sin The coefficiens a n are given by where Hence a n = 1 cos ( ) n d a n = 1 ( ) n f() cos d { << f() = << HELM (VERSION 1: March 18, 4): Workbook Level 3.: Represening Periodic Funcions by Fourier Series 16

17 The inegraion is readily performed using inegraion by pars: ( ) [ ( )] n n cos d = n sin ( ) n sin d n [ ( )] 4 n = cos n n Hence, since a n = 1 n cos( )d a n = The consan erm is a where Similarly where = ( ) n sin d = The second inegral gives zero. Hence b n = 1 = 4 {cos n 1. n n =, 4, 6,... 4 n n =1, 3, 5,... a = 1 [ n cos d =1. ( ) n sin d ( )] n + n b n = cos n n n =, 4, 6,... n + n n =1, 3, 5,... ( ) n cos d. Hence, using all hese resuls for he Fourier coefficiens, he required Fourier Series is f() = 1 4 ( ) (cos + 1 ( ) 3 9 cos + 1 ( ) ) 5 5 cos ( ( ) sin 1 ( ) sin + 1 ( ) ) 3 3 sin... Noice ha because he Fourier coefficiens depend on 1 (raher han 1 as was he case for n 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 sawooh funcion using only a quie small number of erms in he series. 17 HELM (VERSION 1: March 18, 4): Workbook Level 3.: Represening Periodic Funcions by Fourier Series

18 Obain he Fourier Series of he funcion f() = 1 <<1 f( +) = f() f() 1 1 Figure 7 Firs wrie ou he form of he Fourier Series in his case Your soluion Since P =l =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 P =l =sol =1 we have a n = b n = cos(n)d n =, 1,,... sin(n)d n =1,,... The consan erm will be a where a = 1 1 d. HELM (VERSION 1: March 18, 4): Workbook Level 3.: Represening Periodic Funcions by Fourier Series 18

19 Now evaluae he inegrals. Try 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 The inegral for b n is zero for all n because he inegrand is an odd funcion of. (We shall cover his poin more fully in he nex uni.) Since he inegrand is even in he inegrals for a n we can wrie a n = The consan erm will be ao where 1 cos n d n =, 1,,... a = 1 d = 3. For n =1,, 3,... we mus inegrae by pars (wice) { [ ]1 a n = n sin(n) n = 4 n { [ n cos(n) ] n sin(n)d 1 cos(n)d. The inegral gives zero so a n = 4 cos n. n Now wrie ou he final form of he Fourier Series. We have f() = cos n n { 4 cos() 1 9 cos(3)+... cos(n) = cos() HELM (VERSION 1: March 18, 4): Workbook Level 3.: Represening Periodic Funcions by Fourier Series

20 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. 1 <</ 1. f() = f( +) =f() / << square wave. f() = 1 <<1 f( +)=f() 1 T/ << 3. f() = f( + T )=f() square wave 1 <<T/ << 4. f() = f( +) =f() <<, T/ << 5. f() = A sin half wave recifier T, <<T/ HELM (VERSION 1: March 18, 4): Workbook Level 3.: Represening Periodic Funcions by Fourier Series

21 Answers { cos cos 3 3 cos { sin + sin + sin sin sin cos 4 { sin ω+ 13 sin 3 ω+ 15 sin 5ω+... { cos cos 3 cos where ω =/T { cos + cos + {( 4 cos ) sin ( sin ) sin 3 4 sin A + A sin ω A { cos ω (1)(3) + cos 4ω (3)(5) HELM (VERSION 1: March 18, 4): Workbook Level 3.: Represening Periodic Funcions by Fourier Series

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