Topic From Complex Fourier Series o Fourier Transforms. Inroducion In he previous lecure you saw ha complex Fourier Series and is coeciens were dened by as f ( = n= C ne in! where C n = T T = T = f (e in! d : However, we noed ha his did no exend Fourier analysis beyond periodic funcions and discree frequency specra. Wha would x his?.. Wha happens if we le T ge very large? A hough... Suppose he period T ended <o inniy. The fundamenal frequency! would become so small ha i migh properly be called an inniesimal d!. To reach any nie frequency!, m would have o end o inniy. Then m d!!!, and he C m migh bes be called C(!d!. (This poin is ricky. I is o preserve he inegral as he discree specrum morphs ino a coninuous one! So in his regime, C(!d! = lim T! T T = T = f (e i! d : Bu as T!, T = =d! C(!d! = d! f (e i! d ; C(! = f (e i! d :
/ The expression for f ( has o be changed oo. Firs change C n! C(!d!, hen change from a sum o an inegral: f ( = C(!d!ei! = C(!ei! d! :.. We (NEARLY ge o he Fourier Transform! Our las hough was a good one. By allowing T o end o inniy, we seem o have a mehod of going from he non-periodic ime domain signal f ( o a frequency domain specrum C(!. Everyhing is correc, EXCEPT he acual Fourier ransform is, by modern convenion, C(!. So forge he hinking, le's use he deniion.. The deniion of he Fourier Transform The Fourier Transform of a emporal signal f ( is he frequency specrum F (! = f ( e i! d : Given a frequency specrum, he equivalen emporal signal is given by he Inverse Fourier Transform f ( = F (! ei! d! : Because Fourier Series have a coninuous ime signal bu discree frequency specrum, i is raher naural o hink abou he signal, f (, as he \proper hing", and he coecens A n and B n (or C n as slighly \derived". Wih Fourier Transforms, however, here is a real equaliy beween he represenaions. Indeed he ime signal and frequency specrum are described as a Fourier ransform pair.
/3 A Fourier Transform Pair is denoed by f (, F (! : This means F (! = FT [f (] and f ( = FT [F (!] :.3 Example [Q] Deermine he Fourier ransform F (! of he funcion shown in Figure.(a, f ( = u(e b, where u( is he Heaviside sep funcion, and plo he ampliude specrum jf (!j. [A] The Heaviside sep funcion is zero for < and uniy hereafer, so FT [ u(e b] = e b e i! d = b + e (b+i! = i! b + i! : The frequency specrum F (! is complex, p and so i has ampliude and phase. The ampliude specrum is jf (!j = = b +!, and is ploed in Figure.(b. f( F(ω /b ω (a Figure.: (a f ( = u(e b (b (b Ampliude specrum jf (!j = j=(b + i!j. Noice ha negaive values of! are permied. There is nohing deep happening: recall ha cos(! = cos(! and sin(! = sin(!..4 Example [Q] Deermine he Fourier ransform of he uni \op ha" funcion of widh a: { jj < a= f ( = oherwise
/4 Π( F( ω a ω a / a/ π/ a π/ a Figure.: [A] The Fourier ransform is F (! = a= a= e i! d = i! (e i!a= e i!a= = sin (!a=! sin (!a= sin (!a= = a = a!a=!a= = a sinc N (!a= = a sinc H (!a= The Fourier Transform is real, and he frequency specrum, raher han he ampliude specrum, is shown in Figure.. Why do we wrie he answer in wo ways? Unforunaely he sinc( funcion has wo deniions in common use. The HLT deniion is sinc H (x = sin(x=(x, bu ofen elsewhere you'll see sinc dened as sinc N (x = sin(x=x, where he N sands for No-HLT. Because of he confusion we will ry o avoid using i enirely, bu if i is used we should ry o sick o he HLT deniion. Conversion is easy. If you read sinc( in non-hlt-compliance-mode replace i wih sinc(=.
.5 Example [Q] Deermine he Fourier ransform of he uni \riangle" funcion of widh a: { + =a a < f ( = =a a /5 f( F( ω a a a π/ a π/ a ω (a (b Figure.3: [A] The Fourier ransform is F (! = = ( + =ae i! d + a a a ( =ae i! d ( =a cos! d = grind using inegraion by pars = a sin (!a= (!a=.6 Furher examples There are furher examples of he Fourier ransforms of imporan funcions laer in he noes. Bu a more complee lis is in HLT.
/6.7 More realisic signals In case you had sared o hink ha Fourier Transforms were merely for signals ha could be described in as funcions, Figure.4 shows a compuaional example (using mehods described laer of he ransform of signal which has equal ampliude 5Hz and Hz harmonic componens bu is corruped wih noise. 5 Signal Corruped wih Zero Mean Random Noise 8 Frequency conen of y 7 6 5 4 3 5 3 4 5 ime (milliseconds 3 4 5 frequency (Hz Figure.4:.8 \Rouine" Properies of he FT To sar applying he Fourier ransform o engineering problems will require a few more mahemaical ools o be in place. However, we can immediaely explore some of he basic properies of he Fourier ransform hose ha can be derived from sraighforward mahemaical properies of he inegral. These are all in HLT. Deriving hem is a bi ricky in places, and migh seem a slighly edious, bu i does provide pracice a recognizing he Fourier inegral..8. Propery #: Lineariy Lineariy Propery: If FT [f (] = F (! and FT [g(] = G(!, hen FT [f ( + g(] = F (! + G(! Proof: Inegraion is a linear operaion. Tha is [f ( + g(]e i! d = f (e i! d + g(e i! d.
/7.8. Propery #: Dualiy Dual Propery If FT [f (] = F (!, hen FT [F (] = f (! Yes, weird! Wach ou for he! Proof: We know ha f ( is he FT [F (!] ha is f ( = F (!ei! d! Bu and! are jus symbols, so replace wih (! and! wih. NB! Replacing! wih does NOT ip he limis of inegraion. So f (! = F (ei(! d = F (e i! d f (! = FT [F (] \comme il fau", as Mrs Fourier would say (If you did no like he swapping of symbols in one go, ry! p and!! q, hen p! (! and q! : f ( = F (!ei! d! # # # # f (p = F (qeiqp dq # # # # f (! = hen rearrange... F (ei(! d
/8.8.3 Propery #3: Similariy Propery Parameer Scaling or Similariy Propery: FT [f (a] = (! jaj F a If FT [f (] = F (!, hen Proof: The appearance of jaj hins ha his proof needs o consider he ranges a > and a < separaely. For a > : FT [f (a] = f (ae i! d : Wrie p = a, and noe ha he signs on he limis do NOT change because a is posiive. Then FT [f (a] = a f (pe i(!=ap dp = (! a F a For a < : Subsiue p = jaj, and remember o change signs on he limis when =, p = : FT [f (a] = jaj f (pe i(!=ap dp = jaj f (pe i(!=ap dp = (! jaj F a So, for boh cases, one can wrie FT [f (a] = (! jaj F a
/9.8.4 Propery #3: Example [Q] The Fourier ransform of f ( is F (!. Wha he he FT of g(? f( g( [A] a (a FT [g(] = FT [f (S] = (! jsj F S So he real quesion is wha is he scale S? b VERY emping o say ha g( looks wider so ha S >. Now b > a so ha S = b=a.wrong Somehing ineresing happens o g( when = b. This poin maches o f (a. So Sb = a, and S = a=b. CORRECT.8.5 Propery #4: Parameer Shifing Parameer Shifing Propery: FT [f ( a] = exp( i!af (! (b If FT [f (] = F (!, hen Proof: FT [f ( a] = f ( ae i! d : Subsiue p = a. Then FT [f ( a] = f (pe i!(p+a dp = e i!a f (pe i!p dp = e i!a F (!
/.8.6 Properies #,3,4: Example [Q] Find he Fourier Transform of he funcion shown (a below. f( Π( f( + a / a/ (a (b (c [A] We know he FT of he opha in (b is!a sin(.! The version in (c is ime-shifed \o he lef" and involves + a, so is FT is exp ( i!a (!a sin! a Subrac a copy ime-shifed o he righ... and se a =... = [ exp [ i sin ( i!a (!a ( i!a exp ] (!a sin! ] (!a sin! = i! sin (! Now scale he ampliude by a facor of... Ans = 4i (!! sin We've done he scaling in ime by seing parameer a =. Suppose you applied he scaling propery. Wha would be he correc scaling a or (=a?.8.7 Propery #5: Frequency Shifing Frequency Shifing Propery: If FT [f (] = F (!, hen FT [f ( exp( i! S ] = F (!! S
/ Proof: FT [f ( exp( i! S ] = f (e i!s e i! d = f (e i(!!s d = F (!! S
/.8.8 Propery #6: Ampliude modulaion by a cosine Ampliude modulaion by a cosine If FT [f (] = F (! hen FT [f ( cos! ] = [F (!! + F (! +! ] Proof: Wrie cos! = ( e i! + e i! hen use he Frequency shifing propery..8.9 Propery #7: Diereniaion wr ime Diereniaion Propery in ime: FT [ d n d n f ( ] = (i! n F (! If FT [f (] = F (!, hen Proof: I is so emping o sar by wriing FT [ d n [ d n f ( ] e i! d. Resis. Insead, wrie d n f ( = F (!ei! d! d n f ( ] = Now diereniae w.r... Because he inegral is w.r..!, he dierenaion can move hrough he inegral sign: d n d f ( = d n n d n F (!ei! d! = dn F (! d n ei! d! = F (!(i!n e i! d! This las expression says ha d n d n f ( = FT [F (!(i! n ] FT [ d n d n f ( ] = F (!(i! n :
/3.8. Propery #8: Diereniaion wr frequency Diereniaion Propery in frequency: FT [( i n f (] = dn d! n F (! Proof: For you o colour in. If FT [f (] = F (!, hen.9 The Fourier Transform of Complex Fourier Series This requires knowledge of he -funcion. See Lecure 3.. Summary By allowing he period T o end o inniy, and he fundamenal frequency end o an inniesimal d!, we have given a reasoned argumen for he origin of he Fourier Transform which allows non-periodic funcions o be represened in he coninuous (and complex frequency domain. However, here is no need o reproduce he argumen and you migh jus as well ake he deniion for graned. We have worked ou he Fourier Transform for a couple of signals. We have gone hrough he proofs of several basic properies of he Fourier Transform. We are nearly a he poin of being able o do hings, bu no quie! In he nex lecure we mus learn abou he -funcion and he general echnique of convoluion, and hen applicaions begin o open up for us.