24.3. Some Special Fourier Transform Pairs. Introduction. Prerequisites. Learning Outcomes

Transcription

1 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 (28): Secion 24.3: Some Special Fourier Transform Pairs 27

2 . Parseval s heorem Recall from 23.2 on Fourier series ha for a periodic signal f T () wih complex Fourier coefficiens c n (n =, ±, ±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 (ω) he corresponding resul is f 2 ()d = 2π F (ω) 2 dω. 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 (ω) 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 f 2 ()d = [ e e 2 2 d = 2 ] = 2. Now obain he Fourier ransform F (ω) and evaluae he righ-hand side inegral: Your soluion 28 HELM (28): Workbook 24: Fourier Transforms

3 so Then F (ω) = F{e u()} = + iω, F (ω) 2 = ( + iω). ( iω) = + ω. 2 F (ω) 2 dω = 2π π = π F (ω) 2 dω + ω dω = [ ] an ω = 2 π 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() = for <. Then he Fourier ransform of f() becomes F{f()} = f()e iω d. As you may recall from earlier unis, he Laplace ransform of f() is L{f()} = 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 iω in he Laplace ransform. An obvious example where his can be done is he funcion f() = e α u(). () (2) HELM (28): Secion 24.3: Some Special Fourier Transform Pairs 29

4 In his case L{f()} = α + s F{f()} = α + iω = F (iω). = 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 iω iω.) 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( )} = e iω F (ω) Laplace: L{f( )} = e s 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 () = oherwise has a Fourier ransform P a (ω) = 2 sin ωa. ω 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 ωa a() = ωa. 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 3 HELM (28): Workbook 24: Fourier Transforms

5 We define he uni impulse funcion δ() as δ() = lim a 2a p a() and show i graphically as follows: δ() = Then, F{δ()} = F Figure 8 { } { } lim a 2a p a() = lim F a 2a p a() = lim a sin ωa ωa =. 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( ): δ( ) = We have, by he ime shif propery F{δ( )} = e iω () = e iω. Figure 9 These resuls are summarized in he following Key Poin: HELM (28): Secion 24.3: Some Special Fourier Transform Pairs 3

6 Key Poin 4 The Fourier ransform of a Uni Impulse F{δ( )} = e iω. If = 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 We have F{δ()} =. Therefore by he dualiy propery F{} = 2πδ( ω) = 2πδ(ω). We see ha he signal f() =, < < which is infiniely wide, has Fourier ransform F (ω) = 2πδ(ω) which is infiniesimally narrow. This reciprocal effec is characerisic of Fourier ransforms. f() 2πδ(ω) F (ω) ω 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 (ω = ). 32 HELM (28): Workbook 24: Fourier Transforms

7 Task Use he resul F{} = 2πδ(ω) and he frequency shif propery o obain F{e iω }. Your soluion F{e iω } = F{e iω f()} where f() =, < <. Bu F{f()} = 2πδ(ω), herefore, by he frequency shif propery F{e iω } = 2πδ(ω ω ). F{e iω } 2πδ(ω ω ) ω ω Task Obain he Fourier ransform of a pure cosine wave f() = cos ω < < by wriing f() in erms of complex exponenials and using he resul of he previous Task. Your soluion HELM (28): Secion 24.3: Some Special Fourier Transform Pairs 33

8 { We have f() = cos ω = 2 e iω + e } iω so F{cos ω } = 2 F{eiω } + 2 F{e iω } = πδ(ω ω ) + πδ(ω + ω ) F (ω) ω ω ω Noe ha because cos ω 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 ω exiss. By similar reasoning we can readily show F{sin ω } = π i δ(ω ω ) π i δ(ω + ω ). 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 iω o obain he Fourier ransform of he uni sep. We proceed via he Fourier ransform of he signum funcion sgn() which is defined as { > sgn = < sgn() We obain F{sgn()} as follows. Figure 34 HELM (28): Workbook 24: Fourier Transforms

9 Consider he odd wo-sided exponenial funcion f α () defined as { e α > f α () = e α <, where α > : f α () Figure By slighly adaping our earlier calculaion for he even wo-sided exponenial funcion we find F{f α ()} = (α iω) + (α + iω) = (α + iω) + (α iω) α 2 + ω 2 = 2iω α 2 + ω 2. 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 he exponenial funcion resembles more and more closely he signum funcion. This suggess ha F{sgn()} = lim α F{f α ()} ( = lim 2iω ) α α 2 + ω 2 = 2i ω = 2 iω. HELM (28): Secion 24.3: Some Special Fourier Transform Pairs 35

10 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 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 (28): Workbook 24: Fourier Transforms

11 We have, using lineariy, F{u()} = 2 F{} + 2 F{sgn()} = 2 2πδ(ω) iω = πδ(ω) + iω Thus, he Fourier ransform of he uni sep funcion conains he addiional impulse erm πδ(ω) as well as he odd erm iω. Exercises. Use Parserval s heorem and he Fourier ransform of a wo-sided exponenial funcion o show ha dω (a 2 + ω 2 ) 2 = π 2 a 3 2. Using F{sgn()} = 2 iω find he Fourier ransforms of (a) f () = (b) f 2 () = Hence obain he ransforms of (c) f 3 () = 2 (d) f 4 () = Show ha s F{sin ω } = iπ[δ(ω + ω ) δ(ω ω )] Verify your resul using inverse Fourier ransform properies. 2 (a) F{ } = πi sgn(ω) (by he dualiy propery) (b) F{ } = 2 ω 2 (c) F{ 2 } = πω sgn(ω) = { πω, ω > πω, ω < (d) F{ } = iπω2 3 2 sgn(ω) (Using ime differeniaion propery in (b), (c) and (d).) HELM (28): Secion 24.3: Some Special Fourier Transform Pairs 37