Production of this 2015 edition, containing corrections and minor revisions of the 2008 edition, was funded by the sigma Network.

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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 Teaching and Learning for he period Ocober 2002 Sepember 2005, wih addiional ransferabiliy funding Ocober 2005 Sepember HELM aims o enhance he mahemaical educaion of engineering undergraduaes hrough flexible learning resources, mainly hese Workbooks. HELM learning resources were 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 Technology, Harper Adams, Herfordshire, Leiceser, Liverpool, London Meropolian, Moray College, Norhumbria, Noingham, Noingham Tren, Oxford Brookes, Plymouh, Porsmouh, Queens Belfas, Rober Gordon, Royal Fores of Dean College, Salford, Sligo Insiue of Technology, Souhampon, Souhampon Insiue, Surrey, Teesside, Ulser, Universiy of Wales Insiue Cardiff, Wes Kingsway College (London), Wes Nos College. HELM Conacs: Pos: HELM, Mahemaics Educaion Cenre, Loughborough Universiy, Loughborough, LE 3TU. helm@lboro.ac.uk Web: hp://helm.lboro.ac.uk HELM Workbooks Lis Basic Algebra 26 Funcions of a Complex Variable 2 Basic Funcions 27 Muliple Inegraion 3 Equaions, Inequaliies & Parial Fracions 28 Differenial Vecor Calculus 4 Trigonomery 29 Inegral Vecor Calculus 5 Funcions and Modelling 30 Inroducion o Numerical Mehods 6 Exponenial and Logarihmic Funcions 3 Numerical Mehods of Approximaion 7 Marices 32 Numerical Iniial Value Problems 8 Marix Soluion of Equaions 33 Numerical Boundary Value Problems 9 Vecors 34 Modelling Moion 0 Complex Numbers 35 Ses and Probabiliy Differeniaion 36 Descripive Saisics 2 Applicaions of Differeniaion 37 Discree Probabiliy Disribuions 3 Inegraion 38 Coninuous Probabiliy Disribuions 4 Applicaions of Inegraion 39 The Normal Disribuion 5 Applicaions of Inegraion 2 40 Sampling Disribuions and Esimaion 6 Sequences and Series 4 Hypohesis Tesing 7 Conics and Polar Coordinaes 42 Goodness of Fi and Coningency Tables 8 Funcions of Several Variables 43 Regression and Correlaion 9 Differenial Equaions 44 Analysis of Variance 20 Laplace Transforms 45 Non- parameric Saisics 2 z- Transforms 46 Reliabiliy and Qualiy Conrol 22 Eigenvalues and Eigenvecors 47 Mahemaics and Physics Miscellany 23 Fourier Series 48 Engineering Case Sudy 24 Fourier Transforms 49 Suden s Guide 25 Parial Differenial Equaions 50 Tuor s Guide Copyrigh Loughborough Universiy, 205 Producion of his 205 ediion, conaining correcions and minor revisions of he 2008 ediion, was funded by he sigma Nework.

3 24 Conens Fourier Transforms 24. The Fourier Transform Properies of he Fourier Transform Some Special Fourier Transform Pairs 27 Learning oucomes In his Workbook you will learn abou he Fourier ransform which has many applicaions in science and engineering. You will learn how o find Fourier ransforms of some sandard funcions and some of he properies of he Fourier ransform. You will learn abou he inverse Fourier ransform and how o find inverse ransforms direcly and by using a able of ransforms. Finally, you will learn abou some special Fourier ransform pairs.

4 The Fourier Transform 24. Inroducion Fourier ransforms have for a long ime been a basic ool of applied mahemaics, paricularly for solving differenial equaions (especially parial differenial equaions) and also in conjuncion wih inegral equaions. There are really hree Fourier ransforms, he Fourier Sine and Fourier Cosine ransforms and a complex form which is usually referred o as he Fourier ransform. The las of hese ransforms in paricular has exensive applicaions in Science and Engineering, for example in physical opics, chemisry (e.g. in connecion wih Nuclear Magneic Resonance and Crysallography), Elecronic Communicaions Theory and more general Linear Sysems Theory. Prerequisies Before saring his Secion you should... Learning Oucomes On compleion you should be able o... be familiar wih basic Fourier series, paricularly in he complex form calculae simple Fourier ransforms from he definiion sae how he Fourier ransform of a funcion (signal) depends on wheher ha funcion is even or odd or neiher 2 HELM (205): Workbook 24: Fourier Transforms

5 . The Fourier ransform Unlike Fourier series, which are mainly useful for periodic funcions, he Fourier ransform permis alernaive represenaions of mosly non-periodic funcions. We shall firsly derive he Fourier ransform from he complex exponenial form of he Fourier series and hen sudy is various properies. 2. Informal derivaion of he Fourier ransform Recall ha if f() is a period T funcion, which we will emporarily re-wrie as f T () for emphasis, hen we can expand i in a complex Fourier series, f T () = c n e in 0 () n= where 0 = 2π T. In words, harmonics of frequency n 0 = n 2π T he series and hese frequencies are separaed by n = 0, ±, ±2,... are presen in n 0 (n ) 0 = 0 = 2π T. Hence, as T increases he frequency separaion becomes smaller and can be convenienly wrien as. This suggess ha as T, corresponding o a non-periodic funcion, hen 0 and he frequency represenaion conains all frequency harmonics. To see his in a lile more deail, we recall ( coefficiens c n are given by 23: Fourier series) ha he complex Fourier c n = T T 2 T 2 f T ()e in 0 d. (2) Puing T as 0 and hen subsiuing (2) in () we ge 2π { } T 0 2 f T () = f T ()e in0 d e in0. 2π T 2 n= In view of he discussion above, as T we can pu 0 as and replace he sum over he discree frequencies n 0 by an inegral over all frequencies. We replace n 0 by a general frequency variable. We hen obain he double inegral represenaion { } f() = f()e i d e i d. (3) 2π The inner inegral (over all ) will give a funcion dependen only on which we wrie as F (). Then (3) can be wrien where f() = 2π F ()e i d (4) HELM (205): Secion 24.: The Fourier Transform 3

6 F () = f()e i d. The represenaion (4) of f() which involves all frequencies can be considered as he equivalen for a non-periodic funcion of he complex Fourier series represenaion () of a periodic funcion. The expression (5) for F () is analogous o he relaion (2) for he Fourier coefficiens c n. The funcion F () is called he Fourier ransform of he funcion f(). Symbolically we can wrie F () = F{f()}. Equaion (4) enables us, in principle, o wrie f() in erms of F (). f() is ofen called he inverse Fourier ransform of F () and we denoe his by wriing f() = F {F ()}. Looking a he basic relaion (3) i is clear ha he posiion of he facor in (4) and (5). If insead of (5) we define F () = 2π hen (4) mus be wrien f() = f()e i d. F ()e i d. A hird, more symmeric, alernaive is o wrie F () = 2π and, consequenly: f() = 2π f()e i d F ()e i d. 2π (5) is somewha arbirary We shall use (4) and (5) hroughou his Secion bu you should be aware of hese oher possibiliies which migh be used in oher exs. Engineers ofen refer o F () (whichever precise definiion is used!) as he frequency domain represenaion of a funcion or signal and f() as he ime domain represenaion. In wha follows we shall use his language where appropriae. However, (5) is really a mahemaical ransformaion for obaining one funcion from anoher and (4) is hen he inverse ransformaion for recovering he iniial funcion. In some applicaions of Fourier ransforms (which we shall no sudy) he ime/frequency inerpreaions are no relevan. However, in engineering applicaions, such as communicaions heory, he frequency represenaion is ofen used very lierally. As can be seen above, noaionally we will use capial leers o denoe Fourier ransforms: hus a funcion f() has a Fourier ransform denoed by F (), g() has a Fourier ransform wrien G() and so on. The noaion F (i), G(i) is used in some exs because occurs in (5) only in he erm e i. 4 HELM (205): Workbook 24: Fourier Transforms

7 3. Exisence of he Fourier ransform We will discuss his quesion in a lile deail a a laer sage when we will also consider briefly he relaion beween he Fourier ransform and he Laplace Transform ( 20). For now we will use (5) o obain he Fourier ransforms of some imporan funcions. Example Find he Fourier ransform of he one-sided exponenial funcion { 0 < 0 f() = e α > 0 where α is a posiive consan, shown below: f() Figure Soluion Using (5) hen by sraighforward inegraion F () = = = = 0 0 e α e i d (since f() = 0 for < 0) e (α+i) d [ e (α+i) (α + i) α + i ] 0 since e α 0 as for α > 0. This imporan Fourier ransform is wrien in he following Key Poin: HELM (205): Secion 24.: The Fourier Transform 5

8 Key Poin F{e α u()} = α + i, α > 0. Noe ha his real funcion has a complex Fourier ransform. Noe ha if u() is used o denoe he Heaviside uni sep funcion: { 0 < 0 u() = > 0 hen we can wrie he funcion in Example as: concise noaion for one-sided funcions. f() = e α u(). We shall frequenly use his Task Wrie down he Fourier ransforms of (a) e u() (b) e 3 u() (c) e 2 u() Use Key Poin : Your soluion (a) (b) (c) Answer (a) α = so F{e u()} = + i (b) α = 3 so F{e 3 u()} = 3 + i (c) α = so F{e 2 u()} = 2 + i 2 6 HELM (205): Workbook 24: Fourier Transforms

9 Task Obain, using he inegral definiion (5), he Fourier ransform of he recangular pulse { a < < a p() =. 0 oherwise Noe ha he pulse widh is 2a as indicaed in he diagram below. p() a a Firs use (5) o wrie down he inegral from which he ransform will be calculaed: Your soluion Answer P () F{p()} = a a ()e i d using he definiion of p() Now evaluae his inegral and wrie down he final Fourier ransform in rigonomeric, raher han complex exponenial form: Your soluion Answer P () = = a a [ e ()e i i d = ( i) ] a a = e ia e +ia ( i) (cos a i sin a) (cos a + i sin a) ( i) = 2i sin a i i.e. 2 sin a P () = F{p()} = Noe ha in his case he Fourier ransform is wholly real. (6) HELM (205): Secion 24.: The Fourier Transform 7

10 Engineers ofen call he funcion sin x he sinc funcion. Consequenly if we wrie, he ransform x (6) of he recangular pulse as sin a P () = 2a a, we can say P () = 2a sinc(a). Using he resul (6) in (4) we have he Fourier inegral represenaion of he recangular pulse. p() = 2π 2 sin a ei d. As we have already menioned, his corresponds o a Fourier series represenaion for a periodic funcion. If p a () = { a < < a 0 oherwise Key Poin 2 The Fourier ransform of a Recangular Pulse hen: sin a F{p a ()} = 2a a = 2a sinc(a) Clearly, if he recangular pulse has widh 2, corresponding o a = we have: P () F{p ()} = 2 sin. As 0, hen 2 sin 2. Also, he funcion 2sin odd funcions 2 sin and. The graph of P () is as follows: P () is an even funcion being he produc of wo 2 π π Figure 2 8 HELM (205): Workbook 24: Fourier Transforms

11 Task Obain he Fourier ransform of he wo sided exponenial funcion { e α < 0 f() = e α > 0 where α is a posiive consan. f() Your soluion Answer We mus separae he range of he inegrand ino [, 0] and [0, ] since he funcion f() is defined separaely in hese wo regions: hen F () = = = 0 [ e (α i) e α e i d + (α i) ] 0 α i + α + i = 0 e α e i d = [ e (α+i) + (α + i) 2α α ] 0 0 e (α i) d + 0 e (α+i) d HELM (205): Secion 24.: The Fourier Transform 9

12 Noe ha, as in he case of he recangular pulse, we have here a real even funcion of giving a Fourier ransform which is wholly real. Also, in boh cases, he Fourier ransform is an even (as well as real) funcion of. Noe also ha i follows from he above calculaion ha F{e α u()} = (as we have already found) α + i and F{e α u( )} = { e α i where α < 0 eα u( ) = 0 > Basic properies of he Fourier ransform Real and imaginary pars of a Fourier ransform Using he definiion (5) we have, F () = f()e i d. If we wrie e i = cos i sin, hen F () = f() cos d i f() sin d where boh inegrals are real, assuming ha f() is real. Hence he real and imaginary pars of he Fourier ransform are: Re (F ()) = f() cos d Im (F ()) = f() sin d. Task Recalling ha if h() is even and g() is odd hen a a (a) (b) g() d = 0, deduce Re(F ()) and Im(F ()) if f() is a real even funcion f() is a real odd funcion. a h() d = 2 a a 0 h() d and Your soluion (a) 0 HELM (205): Workbook 24: Fourier Transforms

13 Answer If f() is real and even R() Re F () = 2 I() Im F () = 0 f() cos d (because he inegrand is even) f() sin d = 0 (because he inegrand is odd). Thus, any real even funcion f() has a wholly real Fourier ransform. Also since cos(( )) = cos( ) = cos he Fourier ransform in his case will be a real even funcion. Your soluion (b) Answer Now and Re F () = f() cos d = Im F () = f() sin d = 2 (odd) (even) d = (because he inegrand is (odd) (odd)=(even)). 0 f() sin d (odd) d = 0 Also since sin(( )) = sin, he Fourier ransform in his case is an odd funcion of. These resuls are summarised in he following Key Poin: HELM (205): Secion 24.: The Fourier Transform

14 Key Poin 3 f() real and even real and odd neiher even nor odd F () =F{f()} real and even purely imaginary and odd complex,f() =R()+iI() Polar form of a Fourier ransform Task The one-sided exponenial funcion f() = e α u() has Fourier ransform F () =. Find he real and imaginary pars of F (). α + i Your soluion Answer F () = α + i = α i α Hence R() = Re F () = α α I() = Im F () = α We can rewrie F (), like any oher complex quaniy, in polar form by calculaing he magniude and he argumen (or phase). For he Fourier ransform in he las Task F () = α R 2 () + I 2 () = (α ) = 2 α2 + 2 ( ) I() and arg F () = an R() = an. α 2 HELM (205): Workbook 24: Fourier Transforms

15 F () α argf () π/2 π/2 Figure 3 In general, a Fourier ransform whose Caresian form is F () = F () e iφ() where φ() arg F (). F () = R() + ii() has a polar form Graphs, such as hose shown in Figure 3, of F () and arg F () ploed agains, are ofen referred o as magniude specra and phase specra, respecively. Exercises. Obain he Fourier ransform of he recangular pulses { (a) f() = (b) f() = > 0 > 3 2. Find he Fourier ransform of f() = > 2 Answers.(a) F () = 2 sin sin 3 (b) F () = 2 cos HELM (205): Secion 24.: The Fourier Transform 3

16 Properies of he Fourier Transform 24.2 Inroducion In his Secion we shall learn abou some useful properies of he Fourier ransform which enable us o calculae easily furher ransforms of funcions and also in applicaions such as elecronic communicaion heory. Prerequisies Before saring his Secion you should... Learning Oucomes On compleion you should be able o... be aware of he basic definiions of he Fourier ransform and inverse Fourier ransform sae and use he lineariy propery and he ime and frequency shif properies of Fourier ransforms sae various oher properies of he Fourier ransform 4 HELM (205): Workbook 24: Fourier Transforms

17 . Lineariy properies of he Fourier ransform (i) If f(), g() are funcions wih ransforms F (), G() respecively, hen F{f() + g()} = F () + G() i.e. if we add 2 funcions hen he Fourier ransform of he resuling funcion is simply he sum of he individual Fourier ransforms. (ii) If k is any consan, F{kf()} = kf () i.e. if we muliply a funcion by any consan hen we mus muliply he Fourier ransform by he same consan. These properies follow from he definiion of he Fourier ransform and from he properies of inegrals. Examples. F{2e u() + 3e 2 u()} = F{2e u()} + F{3e 2 u()} = 2F{e u()} + 3F{e 2 u()} 2. = 2 + i i If f() = hen f() = 4p 3 () using he sandard resul for F{p a ()}. { oherwise so F () = 4P 3 () = 8 sin 3 Task If f() = { oherwise wrie down F (). Your soluion Answer We have f() = 6p 2 () so F () = 2 sin 2. HELM (205): Secion 24.2: Properies of he Fourier Transform 5

18 2. Shif properies of he Fourier ransform There are wo basic shif properies of he Fourier ransform: (i) Time shif propery: F{f( 0 )} = e i 0 F () (ii) Frequency shif propery F{e i 0 f()} = F ( 0 ). Here 0, 0 are consans. In words, shifing (or ranslaing) a funcion in one domain corresponds o a muliplicaion by a complex exponenial funcion in he oher domain. We omi he proofs of hese properies which follow from he definiion of he Fourier ransform. Example 2 Use he ime-shifing propery o find he Fourier ransform of he funcion { 3 5 g() = 0 oherwise g() 3 5 Figure 4 Soluion g() is a pulse of widh 2 and can be obained by shifing he symmerical recangular pulse { p () = 0 oherwise by 4 unis o he righ. Hence by puing 0 = 4 in he ime shif heorem G() = F{g()} = e 4i 2 sin. 6 HELM (205): Workbook 24: Fourier Transforms

19 Task Verify he resul of Example 2 by direc inegraion. Your soluion Answer G() = 5 3 e i d = [ e i i ] 5 as obained using he ime-shif propery. 3 = e 5i e 3i i ( ) e = e 4i i e i = e 4i 2 sin i, Task Use he frequency shif propery o obain he Fourier ransform of he modulaed wave g() = f() cos 0 where f() is an arbirary signal whose Fourier ransform is F (). Firs rewrie g() in erms of complex exponenials: Your soluion Answer ( e i 0 ) + e i 0 g() = f() = 2 2 f()ei f()e i 0 HELM (205): Secion 24.2: Properies of he Fourier Transform 7

20 Now use he lineariy propery and he frequency shif propery on each erm o obain G(): Your soluion Answer We have, by lineariy: F{g()} = 2 F{f()ei 0 } + 2 F{f()e i 0 } and by he frequency shif propery: G() = 2 F ( 0) + 2 F ( + 0). F () G() Inversion of he Fourier ransform Formal inversion of he Fourier ransform, i.e. finding f() for a given F (), is someimes possible using he inversion inegral (4). However, in elemenary cases, we can use a Table of sandard Fourier ransforms ogeher, if necessary, wih he appropriae properies of he Fourier ransform. The following Examples and Tasks involve such inversion. 8 HELM (205): Workbook 24: Fourier Transforms

21 Example 3 Find he inverse Fourier ransform of F () = 20 sin 5 5. Soluion The appearance of he sine funcion implies ha f() is a symmeric recangular pulse. sin a We know he sandard form F{p a ()} = 2a or F sin a {2a a a } = p a(). Puing a = 5 F sin 5 {0 5 } = p 5(). Thus, by he lineariy propery f() = F sin 5 {20 5 } = 2p 5() f() Figure 4 Example 4 Find he inverse Fourier ransform of G() = 20 sin 5 5 exp ( 3i). Soluion The occurrence of he complex exponenial facor in he Fourier ransform suggess he ime-shif propery wih he ime shif 0 = +3 (i.e. a righ shif). From Example 3 F sin 5 {20 5 } = 2p 5() so g() = F sin 5 {20 5 e 3i } = 2p 5 ( 3) 2 g() 2 8 Figure 5 HELM (205): Secion 24.2: Properies of he Fourier Transform 9

22 Task Find he inverse Fourier ransform of sin 2 H() = 6 e 4i. Firsly ignore he exponenial facor and find he inverse Fourier ransform of he remaining erms: Your soluion Answer We use he resul: F {2a Puing a = 2 gives sin a a } = p a() F sin 2 {2 } = p 2() F sin 2 {6 } = 3p 2() Now ake accoun of he exponenial facor: Your soluion Answer Using he ime-shif heorem for 0 = 4 h() = F sin 2 {6 e 4i } = 3p 2 ( 4) h() HELM (205): Workbook 24: Fourier Transforms

23 Example 5 Find he inverse Fourier ransform of 2 K() = + 2( )i Soluion The presence of he erm ( ) insead of suggess he frequency shif propery. Hence, we consider firs ˆK() = 2 + 2i. The relevan sandard form is F{e α u()} = α + i or F { α + i } = e α u(). Hence, wriing ˆK() = + i ˆk() = e 2 u(). 2 Then, by he frequency shif propery wih 0 = k() = F 2 { + 2( )i } = e 2 e i u(). Here k() is a complex ime-domain signal. Task Find he inverse Fourier ransforms of (a) sin {3( 2π)} L() = 2 ( 2π) (b) M() = ei + i Your soluion HELM (205): Secion 24.2: Properies of he Fourier Transform 2

24 Answer (a) Using he frequency shif propery wih 0 = 2π (b) l() = F {L()} = p 3 ()e i2π Using he ime shif propery wih 0 = m() = e (+) u( + ) m() 4. Furher properies of he Fourier ransform We sae hese properies wihou proof. As usual F () denoes he Fourier ransform of f(). (a) Time differeniaion propery: F{f ()} = if () (Differeniaing a funcion is said o amplify he higher frequency componens because of he addiional muliplying facor.) (b) Frequency differeniaion propery: F{f()} = i df d or F{( i)f()} = df d Noe he symmery beween properies (a) and (b). (c) Dualiy propery: If F{f()} = F () hen F{F ()} = 2πf( ). Informally, he dualiy propery saes ha we can, apar from he 2π facor, inerchange he ime and frequency domains provided we pu raher han in he second erm, his corresponding o a reflecion in he verical axis. If f() is even his laer is irrelevan. For example, we know ha if f() = p () = Then, by he dualiy propery, since p () is even, { < < 0 oherwise, hen F () = 2 sin. F{2 sin } = 2πp ( ) = 2πp (). 22 HELM (205): Workbook 24: Fourier Transforms

25 Graphically: p () P () F P () 2π 2πp () F Figure 6 Task Recalling he Fourier ransform pair { e 2 > 0 f() = e 2 F () = 4 < 0 4 +, 2 obain he Fourier ransforms of (a) g() = (b) h() = cos (a) Use he lineariy and dualiy properies: Your soluion HELM (205): Secion 24.2: Properies of he Fourier Transform 23

26 Answer We have F{f()} F{e 2 } = F{ 2 4 e 2 } = (by lineariy) F{ 4 + } = 2π 2 4 e 2 = π 2 e 2 = G() (by dualiy). f() F F () g() 4 F π 2 G() (b) Use he modulaion propery based on he frequency shif propery: Your soluion Answer We have h() = g() cos 2. F{g() cos 0 } = (G( 2 0) + G( + 0 )), so wih 0 = 2 F{h()} = π { e e 2 +2 } = H() 4 H() HELM (205): Workbook 24: Fourier Transforms

27 Exercises. Using he superposiion and ime delay heorems and he known resul for he ransform of he recangular pulse p(), obain he Fourier ransforms of each of he signals shown. (a) x a () (b) x b () x c () x d () 2 2 (c) (d) Obain he Fourier ransform of he signal f() = e u() + e 2 u() where u() denoes he uni sep funcion. 3. Use he ime-shif propery o obain he Fourier ransform of 3 f() = 0 oherwise Verify your resul using he definiion of he Fourier ransform. 4. Find he inverse Fourier ransforms of (a) F () = 20 sin(5) 5 e 3i (b) F () = 8 sin 3 ei (c) F () = ei i 5. If f() is a signal wih ransform F () obain he Fourier ransform of f() cos( 0 ) cos( 0 ). HELM (205): Secion 24.2: Properies of he Fourier Transform 25

28 Answer. X a () = 4 sin( 2 ) cos(3 2 ) X b () = 4i sin( 2 ) sin(3 2 ) X c () = 2 [sin(2) + sin()] X d () = 2 ( sin( 3 ) 2 ) + sin( e 3i/ F () = 3 + 2i i 3. F () = 2 sin e 2i { 2 2 < < 8 4. (a) f() = 0 oherwise { 4 4 < < 2 (b) f() = 0 oherwise { e + < (c) f() = 0 oherwise (using he superposiion propery) 5. 2 F () + 4 [F ( + 2 0) + F ( 2 0 )] 26 HELM (205): Workbook 24: Fourier Transforms

29 Some Special Fourier Transform Pairs 24.3 Inroducion In his final Secion on Fourier ransforms we shall sudy briefly a number of opics such as Parseval s heorem and he relaionship beween Fourier ransform and Laplace ransforms. In paricular we shall obain, inuiively raher han rigorously, various Fourier ransforms of funcions such as he uni sep funcion which acually violae he basic condiions which guaranee he exisence of Fourier ransforms! Prerequisies Before saring his Secion you should... Learning Oucomes On compleion you should be able o... be aware of he definiions and simple properies of he Fourier ransform and inverse Fourier ransform. use he uni impulse funcion (he Dirac dela funcion) o obain various Fourier ransforms HELM (205): Secion 24.3: Some Special Fourier Transform Pairs 27

30 . Parseval s heorem Recall from 23.2 on Fourier series ha for a periodic signal f T () wih complex Fourier coefficiens c n (n = 0, ±, ±2,...) Parseval s heorem holds: T + T 2 T 2 f 2 T ()d = n= c n 2, where he lef-hand side is he mean square value of he funcion (signal) over one period. For a non-periodic real signal f() wih Fourier ransform F () he corresponding resul is f 2 ()d = 2π F () 2 d. This resul is paricularly significan in filer heory. For reasons ha we do no have space o go ino, he lef-hand side inegral is ofen referred o as he oal energy of he signal. The inegrand on he righ-hand side F () 2 2π is hen referred o as he energy densiy (because i is he frequency domain quaniy ha has o be inegraed o obain he oal energy). Task Verify Parseval s heorem using he one-sided exponenial funcion f() = e u(). Firsly evaluae he inegral on he lef-hand side: Your soluion Answer f 2 ()d = 0 [ e e 2 2 d = 2 ] 0 = 2. Now obain he Fourier ransform F () and evaluae he righ-hand side inegral: Your soluion 28 HELM (205): Workbook 24: Fourier Transforms

31 Answer so Then F () = F{e u()} = + i, F () 2 = ( + i). ( i) = +. 2 F () 2 d = 2π π = π 0 0 F () 2 d + d = [ ] an = 2 π 0 π π 2 = 2. Since boh inegrals give he same value, Parseval s heorem is verified for his case. 2. Exisence of Fourier ransforms Formally, sufficien condiions for he Fourier ransform of a funcion f() o exis are (a) f() 2 d is finie (b) f() has a finie number of maxima and minima in any finie inerval (c) f() has a finie number of disconinuiies. Like he equivalen condiions for he exisence of Fourier series hese condiions are known as Dirichle condiions. If he above condiions hold hen f() has a unique Fourier ransform. However cerain funcions, such as he uni sep funcion, which violae one or more of he Dirichle condiions sill have Fourier ransforms in a more generalized sense as we shall see shorly. 3. Fourier ransform and Laplace ransforms Suppose f() = 0 for < 0. Then he Fourier ransform of f() becomes F{f()} = 0 f()e i d. As you may recall from earlier unis, he Laplace ransform of f() is L{f()} = 0 f()e s d. Comparison of () and (2) suggess ha for such one-sided funcions, he Fourier ransform of f() can be obained by simply replacing s by i in he Laplace ransform. An obvious example where his can be done is he funcion f() = e α u(). () (2) HELM (205): Secion 24.3: Some Special Fourier Transform Pairs 29

32 In his case L{f()} = α + s F{f()} = α + i = F (i). = F (s) and, as we have seen earlier, However, care mus be aken wih such subsiuions. We mus be sure ha he condiions for he exisence of he Fourier ransform are me. Thus, for he uni sep funcion, L{u()} = s, whereas, F{u()}. (We shall see ha F{u()} does acually exis bu is no equal o i i.) We should also poin ou ha some of he properies we have discussed for Fourier ransforms are similar o hose of he Laplace ransforms e.g. he ime-shif properies: Fourier: F{f( 0 )} = e i 0 F () Laplace: L{f( 0 )} = e s 0 F (s). 4. Some special Fourier ransform pairs As menioned in he previous subsecion i is possible o obain Fourier ransforms for some imporan funcions ha violae he Dirichle condiions. To discuss his siuaion we mus inroduce he uni impulse funcion, also known as he Dirac dela funcion. We shall sudy his opic in an iniuiive, raher han rigorous, fashion. Recall ha a symmerical recangular pulse { a < < a p a () = 0 oherwise has a Fourier ransform P a () = 2 sin a. If we consider a pulse whose heigh is raher han (so ha he pulse encloses uni area), hen 2a we have, by he lineariy propery of Fourier ransforms, { } F 2a p sin a a() = a. As he value of a becomes smaller, he recangular pulse becomes narrower and aller bu sill has uni area. 2 a = 4 a = 2 2 a = Figure 7 30 HELM (205): Workbook 24: Fourier Transforms

33 We define he uni impulse funcion δ() as δ() = lim a 0 2a p a() and show i graphically as follows: δ() =0 Then, F{δ()} = F Figure 8 { } { } lim a 0 2a p a() = lim F a 0 2a p a() = lim a 0 sin a a =. Here we have assumed ha inerchanging he order of aking he Fourier ransform wih he limi operaion is valid. Now consider a shifed uni impulse δ( 0 ): δ( 0 ) =0 0 We have, by he ime shif propery F{δ( 0 )} = e i 0 () = e i 0. Figure 9 These resuls are summarized in he following Key Poin: HELM (205): Secion 24.3: Some Special Fourier Transform Pairs 3

34 Key Poin 4 The Fourier ransform of a Uni Impulse F{δ( 0 )} = e i 0. If 0 = 0 hen F{δ()} =. Task Apply he dualiy propery o he resul F{δ()} =. (From he way we have inroduced he uni impluse funcion i mus clearly be reaed as an even funcion.) Your soluion Answer We have F{δ()} =. Therefore by he dualiy propery F{} = 2πδ( ) = 2πδ(). We see ha he signal f() =, < < which is infiniely wide, has Fourier ransform F () = 2πδ() which is infiniesimally narrow. This reciprocal effec is characerisic of Fourier ransforms. f() 2πδ() F () This resul is inuiively plausible since a consan signal would be expeced o have a frequency represenaion which had only a componen a zero frequency ( = 0). 32 HELM (205): Workbook 24: Fourier Transforms

35 Task Use he resul F{} = 2πδ() and he frequency shif propery o obain F{e i 0 }. Your soluion Answer F{e i 0 } = F{e i 0 f()} where f() =, < <. Bu F{f()} = 2πδ(), herefore, by he frequency shif propery F{e i 0 } = 2πδ( 0 ). F{e i 0 } 2πδ( 0 ) 0 Task Obain he Fourier ransform of a pure cosine wave f() = cos 0 < < by wriing f() in erms of complex exponenials and using he resul of he previous Task. Your soluion HELM (205): Secion 24.3: Some Special Fourier Transform Pairs 33

36 Answer { We have f() = cos 0 = 2 e i 0 + e } i 0 so F{cos 0 } = 2 F{ei 0 } + 2 F{e i 0 } = πδ( 0 ) + πδ( + 0 ) F () 0 0 Noe ha because cos 0 d diverges, one of he Dirichle condiions is violaed. Neverheless, as we can see via he use of he uni impulse funcions, he Fourier ransform of cos 0 exiss. By similar reasoning we can readily show F{sin 0 } = π i δ( 0) π i δ( + 0). Noe ha he usual resuls for Fourier ransforms of even and odd funcions sill hold. 5. Fourier ransform of he uni sep funcion We have already poined ou ha alhough L{u()} = s we canno simply replace s by i o obain he Fourier ransform of he uni sep. We proceed via he Fourier ransform of he signum funcion sgn() which is defined as { > 0 sgn = < 0 sgn() We obain F{sgn()} as follows. Figure 0 34 HELM (205): Workbook 24: Fourier Transforms

37 Consider he odd wo-sided exponenial funcion f α () defined as { e α > 0 f α () = e α < 0, where α > 0: f α () Figure By slighly adaping our earlier calculaion for he even wo-sided exponenial funcion we find F{f α ()} = (α i) + (α + i) = (α + i) + (α i) α = 2i α The parameer α conrols how rapidly he exponenial funcion varies: f α () α >α 2 >α 3 α 3 α 2 α Figure 2 As we le α 0 he exponenial funcion resembles more and more closely he signum funcion. This suggess ha F{sgn()} = lim α 0 F{f α ()} ( = lim 2i ) α 0 α = 2i = 2 i. HELM (205): Secion 24.3: Some Special Fourier Transform Pairs 35

38 Task Wrie he uni sep funcion in erms of he signum funcion and hence obain F{u()}. Firs express u() in erms of sgn(): Your soluion Answer From he graphs sgn() u() he sep funcion can be obained by adding o he signum funcion for all and hen dividing he resuling funcion by 2 i.e. u() = ( + sgn()). 2 Now, using he lineariy propery of Fourier ransforms and previously obained Fourier ransforms, find F{u()}: Your soluion 36 HELM (205): Workbook 24: Fourier Transforms

39 Answer We have, using lineariy, F{u()} = 2 F{} + 2 F{sgn()} = 2 2πδ() i = πδ() + i Thus, he Fourier ransform of he uni sep funcion conains he addiional impulse erm πδ() as well as he odd erm i. Exercises. Use Parserval s heorem and he Fourier ransform of a wo-sided exponenial funcion o show ha d (a ) = π 2 2 a 3 2. Using F{sgn()} = 2 i find he Fourier ransforms of (a) f () = (b) f 2 () = Hence obain he ransforms of (c) f 3 () = 2 (d) f 4 () = Show ha Answers F{sin 0 } = iπ[δ( + 0 ) δ( 0 )] Verify your resul using inverse Fourier ransform properies. 2 (a) F{ } = πi sgn() (by he dualiy propery) (b) F{ } = 2 2 (c) F{ 2 } = π sgn() = { π, > 0 π, < 0 (d) F{ } = iπ2 3 2 sgn() (Using ime differeniaion propery in (b), (c) and (d).) HELM (205): Secion 24.3: Some Special Fourier Transform Pairs 37

40 NOTES

41 Index for Workbook 24 Dirac dela funcion 30 Dirichle condiions 29, 34 Dualiy propery 22, 32 Energy 28 Energy densiy 28 Exponenial funcion -2-sided 0 Fourier ransform -definiion 4 Frequency differeniaion propery 22 Frequency domain 4 Frequency shif propery 6 Heaviside uni sep funcion 6 Inverse Fourier ransform 4, 8-22 Impulse funcion Laplace ransform 29 Lineariy properies 5 Magniude specrum 3 Parseval s heorem 28, 37 Phase specurm 3 Polar form 2 Recangular pulse 8 Shif properies 6 Signum funcion Sep funcion 6, 34 Time differenion propery 22 Time domain 4 Time shif propery 6 Uni impulse funcion Uni sep funcion 6, 34 EXERCISES 3, 25, 37

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