The Fourier Transform

The Fourier Transform As we have seen, an (sufficienl smooh) funcion f() ha is periodic can be buil ou of sin s and cos s. We have also seen ha complex exponenials ma be used in place of sin s and cos s. We shall now use complex exponenials because he lead o less wriing and simpler compuaions, bu e can easil be convered ino sin s and cos s. If f() has period l, is (complex) Fourier series expansion is l f() = c k e ik π l wih c k = l f()e ik π l d () k= No surprisingl, each erm c k e ik π l in his expansion also has period l, because c k e ik π l (+l) = c k e ik π l e ikπ = c k e ik π l. We now develop an expansion for non-periodic funcions, b allowing complex exponenials (or equivalenl sin s and cos s) of all possible periods, no jus l, for some fixed l. So, from now on, do no assume ha f() is periodic. For simplici we ll onl develop he expansions for funcions ha are zero for all sufficienl large. Wih a lile more work, one can show ha our conclusions appl o a much broader class of funcions. Le L > 0 be sufficienl large ha f() = 0 for all L. We can ge a Fourier series expansion for he par of f() wih L < < L b using he periodic exension rick. Define F L () o be he unique funcion deermined b he requiremens ha Then, for L < < L, f() = F L () = k= l i) F L () = f() for L < L ii) F L () is periodic of period L L c k (L)e ik π L where c k (L) = L f()e ik π L d () L If we can somehow ake he limi L, we will ge a represenaion of f ha is is valid for all s, no jus hose in some finie inerval L < < L. This is exacl wha we shall do, b he simple expedien of inerpreing he sum in () as a Riemann sum approximaion o a cerain inegral. For each ineger k, define he k h frequenc o be k = k π L and denoe b = π L he spacing, k+ k, beween an wo successive frequencies. Also define ˆf() = f()e i d. Since f() = 0 for all L L c k (L) = L f()e ik π L d = L f()e i(k π L ) d = L f()e ik d = ˆf( L k ) = ˆf( π k ) L In his noaion, f() = F L () = π k= ˆf( k )e i k (3) for an L < < L. As we le L, he resricion L < < L disappears, and he righ hand side converges exacl o he inegral ˆf()e i π d. To see his, cu he domain of inegraion ino = i π ˆf()e 0 3 c Joel Feldman. 007. All righs reserved. March, 007 The Fourier Transform

small slices of widh and approximae, as in he above figure, he area under he par of he graph of ˆf()e i π wih beween k and k + b he area of a recangle of base and heigh ˆf( π k )e ik. This approximaes he inegral ˆf()e i π d b he sum ˆf( π k= k )e ik. As L he approximaion ges beer and beer so ha he sum approaches he inegral. So aking he limi of (3) as L gives f() = π ˆf()e i d where ˆf() = f()e i d (4) The funcion ˆf is called he Fourier ransform of f. I is o be hough of as he frequenc profile of he signal f(). Example Suppose ha a signal ges urned on a = 0 and hen decas exponeniall, so ha f() = for some a > 0. The Fourier ransform of his signal is ˆf() = f()e i d = 0 e a e i d = e a if 0 0 if < 0 0 e (a+i) d = e (a+i) (a+i) 0 = a+i Since a + i has modulus a + and phase an a, we have ha ˆf() = A()e iφ() wih A() = a+i = and φ() = a + an a. The ampliude A() and phase φ() are ploed below. A() a ϕ() a a a Example Suppose ha a signal consiss of a single recangular pulse of widh and heigh. Le s sa ha i ges urned on a = and urned off a =. The sandard name for his normalized recangular pulse is rec() = if < < 0 oherwise I is also called a normalized boxcar funcion. The Fourier ransform of his signal is rec() = rec() e i d = / / e i d = e i i / / = ei/ e i/ i = sin c Joel Feldman. 007. All righs reserved. March, 007 The Fourier Transform

when 0. When = 0, rec(0) = / / d =. B l Hôpial s rule lim rec() = lim sin 0 0 = lim cos 0 = = rec(0) so rec() is coninuous a = 0. There is a sandard funcion called sinc ha is defined () b sinc = sin. In his noaion rec() = sinc. Here is a graph of rec(). rec() π π Properies of he Fourier Transform Lineari If α and β are an consans and we build a new funcion h() = αf()+βg() as a linear combinaion of wo old funcions f() and g(), hen he Fourier ransform of h is ĥ() = h()e i d = [ ] αf()+βg() e i d = α f()e i d+β = αˆf()+βĝ() g()e i d (L) Time Shifing Suppose ha we build a new funcion h() = f( 0 ) b ime shifing a funcion f() b 0. The eas wa o check he direcion of he shif is o noe ha if he original signal f() has a jump when is argumen = a, hen he new signal h() = f( 0 ) has a jump when 0 = a, which is a = a+ 0, 0 unis o he righ of he original jump. 0 f() h() = f( 0 ) The Fourier ransform of h is a a+ 0 ĥ() = h()e i d = = e i0 f( )e i d = e f( 0 )e i d = i0 ˆf() f( )e i( + 0) d where = + 0, d = d (T) () There are acuall wo differen commonl used definiions. The firs, which we shall use, is sinc = sin. The second is sinc = sinπ. I is someimes called he normalized sinc funcion. π c Joel Feldman. 007. All righs reserved. March, 007 The Fourier Transform 3

Scaling If we build a new funcion h() = f ( α) b scaling ime b a facor of α > 0, hen he Fourier ransform of h is ĥ() = h()e i d = = αˆf(α) f ( α) e i d = α f( )e iα d where = α, d = αd (S) Example 3 Now consider a signal ha consiss of a single recangular pulse of heigh H, widh W and cenre C. H = r HWC () C W C + W The funcion r HWC () = H rec ( C W H and jumps when C W = ±, i.e. = C± ) (wih rec he normalized recangular pulse of Example ) has heigh W and so is he specified signal. B combining properies (L), ) for an H, C and W. We (T) and (S), we can deermine he Fourier ransform of r HWC () = H rec ( C W build i up in hree seps. Firs we consider he special case in which H = and C = 0. Then we have R () = rec ( W). So, according o he scaling proper (S) wih f() = rec(), h() = R () = rec ( ) ( W = f ) ( W = f α) wih α = W we have ˆR () = ĥ() = αˆf(α) = W rec(w) = W W W sin = W sin Nex we allow a nonzero C oo and consider R () = rec ( ) C W = R ( C). B (T), wih f() = R (), h() = R () = R ( C) = f( 0 ), wih 0 = C he Fourier ransform is ˆR () = ĥ() = e i0 ˆf() = e ic ˆR () = We ic rec ( W ) Finall, b (L), wih α = H and β = 0, he Fourier ransform of r HWC = Hrec ( ) C W = HR () is ˆr HWC () = H ˆR () = HWe ic rec(w) = HWe ic W W sin = e ic H sin W Example 4 Now suppose ha we have a signal ha consiss of a number of recangular pulses of various heighs and widhs. Here is an example c Joel Feldman. 007. All righs reserved. March, 007 The Fourier Transform 4

= s() 0 3 4 Wecanwriehissignalasasumofrecangularpulses. Aswesawinhelasexample, r HWC () = H rec ( ) C W is a single recangular signal wih heigh H, cenre C and widh W. So s() = s ()+s ()+s 3 () where s n () = r HWC () wih H =, W =, C =.5 for n = H =, W =, C = for n = H = 0.5, W =, C = 3 for n = 3 So, using (L) and ˆr HWC () = e ic H W sin, which we derived in Example 3, ŝ() = ŝ ()+ŝ ()+ŝ 3 () = 4 ei3 sin + e i sin + e i3 sin Differeniaion If we build a new funcion h() = f () b differeniaing an old funcion f(), hen he Fourierransform of h is ĥ() = h()e i d = f ()e i d Now inegrae b pars wih u = e i and dv = f () d so ha du = ie i d and v = f(). Assuming ha f(± ) = 0, his gives ĥ() = u dv = uv v du = f() ( i)e i d = iˆf() (D) Example 5 The differeniaion proper is going o be useful when we use he Fourier ransform o solve differenial equaions. As an example, le s ake anoher look a he RLC circui + x() R L C + () hinking of he volage x() as an inpu signal and he volage () as an oupu signal. If we feed an inpu signal x() ino an RLC circui, we ge an oupu () which obes he differenial equaion LC ()+RC ()+() = x() (5) c Joel Feldman. 007. All righs reserved. March, 007 The Fourier Transform 5

You should be able o quickl derive his equaion, which is also (6) in he noes Fourier Series, on our own. Take he Fourier ransform of his whole equaion and use ha he Fourier ransform of () is iŷ() and he Fourier ransform of () is i imes he Fourier ransform of () and so is ŷ() So he Fourier ransform of (5) is This is riviall solved for LC ŷ()+ircŷ()+ŷ() = ˆx() ŷ() = ˆx() LC +irc+ (6) which exhibis classic resonan behaviour. The circui acs independenl on each frequenc componen of he signal. The ampliude ŷ() of he frequenc par of he oupu signal is he ampliude ˆx() of he frequenc par of he inpu signal muliplied b A() = LC +irc+ = ( LC ) +R C. A We shall shorl learn how o conver (6) ino an equaion ha expresses he ime domain oupu signal () in erms of he ime domain inpu signal x(). Example 6 The Fourier ransform, rec(), of he recangular pulse funcion of Example decas raher slowl, like for large. We can r suppressing large frequencies b eliminaing he disconinuiies a = ± in rec(). For example 0 if 5 8 4(+ 5 8 ) if 5 8 3 8 g() = if 3 8 3 8 = g() 4( 5 8 ) if 3 8 5 8 0 if 5 8 5 8 3 8 3 8 5 8 I is no ver difficul o evaluae he Fourier ransform ĝ() direcl. Bu i easier o use properies (L) (D), since 0 if 5 8 4 if 5 8 3 3 5 8 g () = 0 if 3 8 3 8 8 8 = s 4 ()+s 5 () 4 if 3 8 5 3 5 8 8 8 0 if 5 = g () 8 H = 4, W = 4 where s n () = r HWC () wih, C = for n = 4 H = 4, W = 4, C = for n = 5 B Example 3, he Fourier ransform of ŝ 4 ()+ŝ 5 () = 8 ei/ sin 8 8 e i/ sin 8 = i8 sin sin 8 c Joel Feldman. 007. All righs reserved. March, 007 The Fourier Transform 6

B Proper (D), he Fourier ransform of g () is i imes ĝ(). So he Fourier ransform of g() is i imes he Fourier ransform of g (): ( ĝ() = i i 8 sin sin ) 8 = 6 sin sin 8 This looks somewha like he he Fourier ransform in Example bu exhibis faser deca for large. ĝ() π π Parseval s Relaion The energ carried b a signal f() is defined () o be f() d = f()f() d We would like o express his energ in erms of he Fourier ransform ˆf(). To do so, subsiue in ha This gives f() = π f() d = π = π ˆf()e i d = π This formula, f() d = π Duali ˆf() e i d = π f()ˆf()e i d d = π ˆf() ˆf() d = π [ ˆf() ˆf() d ˆf() d, is called Parseval s relaion. ˆf()e i d f()e i d ] d The duali proper sas ha if we build a new ime domain funcion g() = ˆf() b exchanging he roles of ime and frequenc, hen he Fourier ransform of g is ĝ() = πf( ) (Du) To verif his, jus wrie down jus he definiion of ĝ() and he Fourier inversion formula (4) for f() and, in boh inegrals, make a change of variables so ha he inegraion variable is s: ĝ() = g()e i d = f() = π ˆf()e i d =s = ˆf()e i d =s = π ˆf(s) e is ds ˆf(s) e is ds So ĝ(), which is given b he las inegral of he firs line is exacl π imes he las inegral of he second line wih replaced b, which is πf( ). () B wa of moivaion for his definiion, imagine ha f() is he volage v() a ime across a resisor of resisance R. Then he power (i.e. energ per uni ime) dissipaed b he resisor is v()i() = R v(). So he oal energ dissipaed b he resisor will be R v() d, which, up o a consan, agrees wih our definiion. c Joel Feldman. 007. All righs reserved. March, 007 The Fourier Transform 7

Example 7 In his example, we shall compue he Fourier ransform of sinc() = sin. Our saring poin is Example, where we saw ha he Fourier ransform of he recangular pulse rec() of heigh one and widh one is rec() = sin = sinc. So, b duali (Du), wih f() = rec() and g() = ˆf() = rec(), he Fourier ransform of g() = sinc is ĝ() = πf( ) = πrec( ) = πrec(), since rec() is even. So we now know ha he Fourier ransform of sinc is πrec(). To find he Fourier ransform of sinc(), we jus need o scale he ou of. So we appl he scaling proper (S) wih f() = sinc and h() = sinc() = f() = f ( ) α where α =. B (S), he Fourier ransform of h() = sinc() is αˆf(α) = α πrec(α) = πrec ( ). Muliplicaion and Convoluion A ver common operaion in signal processing is ha of filering. I is used o eliminae high frequenc noise and also o eliminae six ccle hum, arising from he ordinar household AC curren. I is also used o exrac he signal from an one desired radio or TV saion. In general, he filer is described b a funcion Ĥ() of frequenc. For example ou migh have Ĥ() = for desired frequencies and Ĥ() = 0 for undesirable frequencies. When a signal f() is fed ino he filer, an oupu signal g(), whose Fourier ransform is ˆf()Ĥ() is produced. For example, he RLC circui of Example 5 is a filer wih Ĥ() = LC +irc+. The quesion wha is g()? remains. Of course i is he inverse Fourier ransform g() = π ˆf()Ĥ()ei d of ˆf()Ĥ(), bu we would like o express i more direcl in erms of he original ime domain signal f(). So le s subsiue ˆf() = f(τ)e iτ dτ (which expresses ˆf() in erms f()) and see if we can simplif he resul. g() = π = = [ f(τ) π f(τ)e iτ Ĥ()e i dτd f(τ)h( τ) dτ The las inegral is called a convoluion inegral and is denoed (f H)() = ] Ĥ()e i( τ) d dτ f(τ)h( τ) dτ If we make he change of variables τ =, dτ = d we see ha we can also express (f H)() = f( )H( ) ( d ) = f( )H( ) d We conclude ha ĝ() = ˆf()Ĥ() = g() = (f H)() (C) Example 8 The convoluion inegral gives a sor of moving weighed average, wih he weighing deermined b H. Is value a ime involves he values of f for imes near. I sa sor of because here is no requiremen ha H() d = or even ha H() 0. c Joel Feldman. 007. All righs reserved. March, 007 The Fourier Transform 8

Suppose for example, ha we use he filer wih Ĥ() = sin The graph of his funcion was given in Example. I is a low pass filer in he sense ha i les hrough mos small frequencies and suppresses high frequencies. I is no reall a pracical filer, bu I have chosen i anwa because i is relaivel eas o see he effec of his filer in space. From Example we know ha H() = if < < 0 oherwise So when his filer is applied o a signal f(), he oupu a ime is (f H)() = = + f( τ)h(τ) dτ = f( ) d f( τ) dτ = f( ) d + where = τ, d = dτ which is he area under he par of he graph of f( ) from = o τ = +. To be concree, suppose ha he inpu signal is f( ) = if 0 0 if < 0 Then he oupu signal a ime is he area under he par of his graph from = o τ = +. if + < 0, ha is <, hen f(τ ) = 0 for all < τ < + (see he figure below) so ha (f H)() = 0 = f( ) + if + 0 bu 0, ha is, hen f(τ ) = 0 for < τ < 0 and f( ) = for 0 < < + (see he figure below) so ha (f H)() = + f( ) d = + 0 d = + = f( ) + if > 0, ha is >, hen f(τ ) = for all < τ < + and (f H)() = + f( ) d = + d =. = f( ) All ogeher, he oupu signal is + c Joel Feldman. 007. All righs reserved. March, 007 The Fourier Transform 9

0 if < (f H)() = + if if > = (f H)() Example 9 For a similar, bu more complicaed example, we can hrough he procedure of Example 8 wih H() sill being rec() bu wih f() replaced b he more complicaed signal s() of Example 4. This ime he oupu signal a ime is (s H)() = s( τ)h(τ) dτ = s( τ) dτ = + s( ) d where = τ which is he area under he par of he graph of s( ) from = o τ = +. We can read off his area from he figures below. In each figure, he lef hand verical doed line is = and he righ hand verical doed line is = +. The firs figure has so small ha he enire inerval < τ < + is o he lef of =, where he graph of s( ) sars. Then, in each successive figure, we increase, moving he inerval o he righ. In he final figure is large enough ha he enire inerval < τ < + is o he righ of = 4, where he graph of s( ) ends. = s( ) = s( ) + 0 3 4 0 3 4 Case: + <, ha is < 5 area = 0 Case: +, ha is 5 3 area = ((+ ) ( )) = +5 = s( ) = s( ) 0 3 4 0 3 4 Case: + 0, ha is 3 area = (( ) ( )) = Case: 0 +, ha is area = + = s( ) 0 3 4 0 3 4 Case: +, ha is 3 area = Case: + 3, ha is 3 5 area = [ ( )+ [(+ ) ] = 7 4 c Joel Feldman. 007. All righs reserved. March, 007 The Fourier Transform 0

0 3 4 0 3 4 Case: 3 + 4, ha is 5 7 area = Case: 4 + 5, ha is 7 9 area = [4 ( )] = 9 4 = s( ) 0 3 4 + All ogeher, he graph of he oupu signal is Case: > 4, ha is > 9 area = 0 (s H)() 0 3 4 Impulses The inverse Fourier ransform H() of Ĥ() is called he impulse response funcion of he filer, because i is he oupu generaed when he inpu is an impulse a ime 0. An impulse, usuall denoed δ() (and called a dela funcion ) akes he value 0 for all imes 0 and he value a ime = 0. In fac i is so infinie a ime 0 ha he area under is graph is exacl. Of course here isn an such funcion, in = δ() he usual sense of he word. Bu i is possible o generalize he concep of a funcion (o somehing called a disribuion or a generalized funcion) so as o accommodae dela funcions. One generalizaion involves c Joel Feldman. 007. All righs reserved. March, 007 The Fourier Transform

aking he limi as ε 0 of approximae dela funcions like δ ε () = ε if ε < < ε 0 oherwise ε ε ε = δ ε () A reamen of hese generalizaion procedures is well beond he scope of his course. Forunael, in pracice i suffices o be able o compue he value of he inegral δ()f() d for an coninuous funcion f() and ha is eas. Because δ() = 0 for all 0, δ()f() is he same as δ()f(0). (Boh are zero for 0.) Because f(0) is a consan, he area under δ()f(0) is f(0) imes he area under δ(), which we alread said is. So δ()f() d = f(0) In paricular, choosing f() = e i gives he Fourier ransform of δ(): ˆδ() = δ()e i d = e i =0 = B he ime shifing proper (T), he Fourier ransform of δ( 0 ) (where 0 is a consan) is e i0. We ma come o he same conclusion b firs making he change of variables τ = 0 o give δ( 0 )e i d τ= 0 = δ(τ)e i(τ+0) dτ and hen appling δ(τ)f(τ) dτ = f(0) wih f(τ) = e i(τ+0). Reurning o he impulse response funcion, we can now verif ha he oupu generaed b a filer Ĥ() in response o he impulse inpu signal δ() is indeed (δ H)() = δ(τ)h( τ) dτ = H() where we have applied δ(τ)f(τ) dτ = f(0) wih f(τ) = H( τ). Example 0 We saw in (6) ha for an RLC circui Ĥ() = LC +irc+ To make he numbers work ou cleanl, le s choose R =, L = 6 and C = 5. Then Recall from Example ha Ĥ() = 6 +i5+ = (3i+)(i+) f() = e a if 0 0 if < 0 ˆf() = a+i (7) So we can deermine he impulse response funcion H() for he RLC filer jus b using parial fracions o wrie H() as a linear combinaion of a+i s. Since Ĥ() = (3i+)(i +) = a 3i+ + b i + = a(i+)+b(3i+) = (a+3b)i+(a+b) (3i +)(i+) (3i +)(i+) a+3b = 0, a+b = b = a, a+3( a) = 0 a = 3, b = c Joel Feldman. 007. All righs reserved. March, 007 The Fourier Transform

we have, using (7) wih a = 3 and a =, Ĥ() = which has graph 3 3i+ i + = 3 +i +i H() = e /3 e / if 0 0 if < 0 = H() Example 0 (again) Here is a sneakier wa o do he parial fracion expansion of Example 0. We know ha Ĥ() has a parial fracion expansion of he form Ĥ() = (3i+)(i +) = a 3i+ + b i+ The fas wa o deermine a is o mulipl boh sides of his equaion b (3i +) b(3i+) = a+ i+ i+ and hen evaluae boh sides a 3i+ = 0. Then he b erm becomes zero, because of he facor (3i+) in he numeraor and we ge a = = i + = 3 3 i= 3 Similarl, mulipling b (i +) raher han (3i+), b = 3i+ = i+=0 3i+ i= = = c Joel Feldman. 007. All righs reserved. March, 007 The Fourier Transform 3

Properies of he Fourier Transform Proper Signal Fourier Transform x() = π ˆx()ei d () = π ŷ()ei d ˆx() = x()e i d ŷ() = ()e i d Lineari Ax() + B() Aˆx() + Bŷ() Time shifing x( 0 ) e i0ˆx() Frequenc shifing e i0 x() ˆx( 0 ) Scaling x ( ) α α ˆx(α) Time shif & scaling x ( 0 α ) α e i0ˆx(α) Frequenc shif & scaling α e i0 x(α) ˆx ( ) 0 α Conjugaion x() ˆx( ) Time reversal x( ) ˆx( ) Differeniaion x () iˆx() x (n) () (i) nˆx() Differeniaion x() i dˆx() d ( n x() i d d) nˆx() Convoluion x(τ)( τ)dτ ˆx()ŷ() Muliplicaion x()() π Duali ˆx() πx( ) ˆx(θ)ŷ( θ)dθ Parseval e a 0 if < 0, u() = e a if > 0 e a u( ) = e a if < 0, 0 if > 0 e a x() d = π ˆx() d a+i (a consan, Rea > 0) (a consan, Rea < 0) a+i a a (a consan, Rea > 0) + if < Boxcar in ime rec() = 0 if > H if C < W General boxcar r HWC () = 0 if C > W Boxcar in frequenc π sinc( ) = π sin( ) sinc ( ) = sin( ) HWe ic sinc ( ) W = e ic H W sin rec() = if < / 0 if > / Impulse in ime δ( 0 ) e i0 δ( 0 )x() e i0 x( 0 ) Single frequenc e i0 πδ( 0 ) c Joel Feldman. 007. All righs reserved. March, 007 The Fourier Transform 4