《高等路面设计理论》课程授课教案（讲义）THE FOURIER TRANSFORM（扫描版）

2THE FOURIER TRANSFORMA principal analysis tool in many of today's scientific challenges is theFouriertransform.Possibly themost well-known application of this mathe-matical technique is the analysis of linear time-invariant systems. But asemphasized in Chapter l, the Fourier transform is essentially a universalproblem solving technique. Its importance is based on the fundamentalproperty that one can examine a particular relationship from an entirelydifferent viewpoint.Simultaneous visualization of a function and itsFouriertransformis oftenthekeyto successfulproblem solving2-1THEFOURIERINTEGRALThe Fourier integral is defined bytheexpressionH()n(t)e-j27dt(2-1)If the integral exists for every value of the parameter f then Eq.(2-1) definesH(f),theFourier transform of h(0).Typicallyh(t)istermedafunction of thevariabletimeandH(f)is termedafunctionof thevariablefrequency.Wewillusethis terminology throughout thebook;tis time andfis frequency.Further,a lower casesymbol will represent afunction of time;theFourier transformof this timefunction will be represented by the same upper case symbol as afunction of frequency.In general the Fourier transform is a complex quantity:H()=R()+jI(f)=|H()lejo)(2-2)where R(f) is the real part of the Fourier transform,11

2 THE FOURIER TRANSFORM A principal analysis tool in many of today's scienti challenges is the Fourier transform. Possibly the most well-known application of this mathe￾matical technique is the analysis of linear time-invariant systems. But as emphasized in Chapter 1, the Fourier transform is essentially a universal problem solving technique. Its importance is based on the fundamental property that one can examine a particular relationship from an entirely different viewpoint. Simultaneous visualization of a function and its Fourier transform is often the key to successful problem solving 2-1 THE FOURIER INTEGRAL The Fourier integral is defined by the expression H (f) = 二峭川ft dl (2-1) If the integral exists for every value of the parameter fthen Eq. (2-1) defines H(.f), the Fourier transform of 11(/). Typically l1(t) is termed a function of the variable time and H(.f) is termed a function ofthe variable frequency. We will use this terminology throughout the book; t is time andfisfi"equency. Further, a lower case symbol will represent a function of time; the Fourier transform of this time function will be represen by the same upper case symbol as a function of frequency In general the Fourier transform is a complex quantity H(f) = R(f) + jl(f) = I H (f) I f) where R(f) is the real part of the Fourier transform, (2-2) 11

13THEFOURIERTRANSPORMSee, 2-212THEFOURIERTRANSFORMChap.2HenceI()is the imaginary part of the Fourier transform,BaH(f)lis the amplitude or Fourier spectrum of h(c)and is given byRD=(2NRO)+2元／8(f)is the phaseangle of the Fourier transform and is given by（2元tan-[(O/R()]BIHI=EXAMPLE 2-12+(2元）To illustrate the various defining terms of the Fourier transform consider the6() tan-1[-2n/function of timeoh(n)-βe-al1>0(2-3)Each of these functions is plotted in Fig. 2-1 to illustrate the various forms of01<06(t)Fouriertransformpresentation.From Eg. (2-1)βe"atle-12nftdt=β(a+/20/ dtHO=B2-2THEINVERSEFOURIER TRANSFORM(a1/2元α+j2元斤α+j2元fThe inverse Fourier transform is defined asBa2n/B2+(2元)2(2元）(2-5)h(o)Headftan-l[-2af/a](2-4)2+（2元）Inversion transformation (2-5)allows the determination of a function of timefromits Fouriertransform.If thefunctions h()and H()are relatedbyEqs. (2-1)and (2-5), the two functions are termed a Fourier transform pair,and we indicate this relationship by the notation(2-6)H()H(F)h(o)0EXAMPLE2-2Consider the frequency function determined in the previous example2元/BBa6H()=2元（2元α+/2元o(n)From Eq. (2-5)2元/BBaejaafrdfh0)=（2元（2元Sincee/acos(2fn)+ jsin (2元fim),then「Ba cos (2元/+2βsin 2元/dff()=+（2元）2L+（2元）[Basin (2/)_2/βcos(2n/d(2-7)α+（2元/）202+（2元）3The second integral of Eq. (2-7) is zero since each integrand term is an odd function.This point is clarified by examination of Fig.2-2; the first integrand term in thesecond integral of Eq. (2-7) is illustrated. Note that the function is odd; that isg()-g(-1).Consequently,thearea underthefunction from-f,to fois zero.Therefore, in the limit as foapproaches infinity,the integralof the function remainsFigure 2-1, Real,imaginary,magnitude, and phase presenta-zero; the infinite integral of any odd function is zero.tions of the Fourier transform

12 THE FOURIER TRANSFORM Chap.2 I (f) is the imaginary part of the Fourier transform, I H(f) I is the amplitude or Fourier spectrum of h(t) and is given by 7τF币， 。(f) is the phase allgle of the Fourier transform and is given by tan- I [I(f)fR(f)]. EXAMPLE 2-1 To illustrate the various defining terms of the Fourier transform consider the function of time h(l) ~βeω From Eq. (2-1) h (f刀)~斗Il e- 必如 dl ~ ß f>-川-(.阳川川 叶呐/22.π = "f 川川川.-→巾叩们叫{阳刷川川回 e- 川川忖叽叫叫叫 Jρ山川川叫，.川叫 旷呐f) =盯住?一jd 口布会言f)言川 IH(f)1 Figure 2-1. ReaJ, imaginary, magnitude, and phase presenta￾tions of the Fourier transform. (2-3) (2-4) Sec.2-2 HE FOURIER TRANSfORM 13 Hence (f)= 阿结 移尹 IH(f)1 =万古言f)言 (f) = tan- I [守LJ Each of these functions is plotted in Fig. 2-1 to illustrate the various forms of Fouríer transform presentation. 2-2 THE INVERSE FOURJER TRANSFORM The inverse Fourier transform is defined as =5 (J)ei'.J< df (2-5) Inversion transformation (2-5) allows the determination of a function of time from lts Fourler tran rm. If the funct Eqs. (ρ2-1η) and (ρ2-5匀)， the two functions re termed a Four le trm s._101 patr, and we indicate thi s relationshiip by the notation h(t) g H(f) EXAMPLE 2-2 Consider the frequency function determined in the previous example fh 写了 =dk予一 击忘乎 From Eq. (2-5) =J二[归结芹 jd如]川矿 Since ei2;r:!1 = cos (2πβ) sin (2"fl), then rßαcos (2"fl) , 2"f 豆旦豆豆21 dr h(I)=| [Ff古芹十节 苍可γJ (2-6) í~ 旦旦坐旦丘 旦旦与单黑21 df (2-7) J I I <)G' +否可)2 以十 2π f)' J The h intpur~nd term is an odd function second integral of Eq. (2-7) is zero SlDce each mtegranu term Ui <<11 This point is clarified by examination of Fig. the rst integrand term the nd integral of Eq. (2-7) is illustrated. Note that the function is odd; that 1S, g(l) = -g(二。 Consequently the area under the function from -10 to 10 is zero. Therefore, ln the lImit asJG approaches lnRmty,the Integral of the function remams zero; the in nite integral of any odd functîon is zero

15THE FOURIER TRANSFORMSec,2-3Chap, 214THEFOURIER TRANSFORMBor sin (2mft)2-3 EXISTENCEOFTHEFOURIERINTEGRAL02+(2m)2To this point we have not considered the validity of Eqs. (2-1) and (2-5);the integral equations have been assumed to be well defined for all functions.In gencral, for most functions encountered in practical scientific analysis,the Fouriertransform and its inverse are well defined. We do not intend topresent a highly theoretical discussion of the existence of the Fourier trans-form but rather to point out conditions for its existence and to give examplesfoof these conditions.Our discussion follows that of Papoulis [5].Condition I. If h(o) is integrable in the sense(2-11)0dt00then its Fourier transform H(f) exists and satisfies the inverse Fouriertransform (2-5).Figure 2-2. Integration of an odd function.It is important to note that Condition I is a sufficient but not a necessarycondition for existence of a Fourier transform.There are functions which doEq. (2-7) becomesnot satisfy Condition Ibut have a transform satisfying (2-5).This class offsin (2nif)cOs(2元)2元Baa/2元/2d(2-8)(a2+):functionswill be covered byCondition 2.(2元)2（2元）2From a standard table.of integrals [4]:EXAMPLE2-3To illustrate Condition 1 consider thepulse time waveformfewco5.0xdxma>o+x10dxe-ab2+x(2-12)=0>TeHence Eq- (2-8) can be written aswhich is shown in Fig. 2-3. Equation (2-11) is satisfied for this function; therefore,2元β(2], [ne-(2e0i/2)](2元)[(a/2)e(20(a/2)]the Fourier transform exists and is given by()=Ae-JafidtHU)--e"al=βe-ar1>0(2-9)sin(2元ff)dtcos(2元ft)dtjAAThe time functionh(o) βe-sr1>0The second integral is equal to zero since the integrand is odd;and the frequency function2元) sin (2 /)HUO=BH()=+（2元）2A7,sin(2r7D)(2-13)2元10/are related by both Eqs. (2-1) and (2-5) and hence are a Fourier transform pair;Those terms which obviously can be canceled are retained to emphasize the [sin (af)/β(2-10)Be-ar t>00(af) characteristic of theFouriertransform of a pulse waveform (Fig.2-3)α+（2元）Because this example satisfies Condition 1 then H() as given by (2-13) mustsatisfy Eq. (2-5)

14 THE FOURIER TRANSFORM sin (21Tft) α2 + (2πfJ2 Figure 份2. Integration of an odd function. 11(1) 二{拮熙节矿 iZ 二揣焊轩旷 f~ Fτ言王 Lπm (IX τe fJFτ巾-x' =ne-. Hence Eq. (2-8) can be written as a>O a> 0 h(f)=# -Le{ 小生乌 [π )( 0 The time functÍon 11(1) = ße-.' t and the frequency function H (f) = CO Chap.2 (2-8) (2-9) are related by both Eqs. (2-1) and (2-5) and hence are a Fourier transform paìr; ße-.' t> 0 g F 宇刀刃 (2-10) Sec.2-3 THE FOU ER TRANSFORM 15 2-3 EXISTENCE OF THE FOURIER INTEGRAL To this point we have not considered the validity of Eq乱。-1) and (2-5); the integral equations have heen assumed to be well defined fQr all functions Ingeneral, for most functions encountered in prauleal SCIentIEc artalySIL the Fourier transform and its inverse are well defined. We do not intend to present a hlghIy theoretical discussion of the existence of the FOUrzer trans" form but rather to point out condltmns for Its exlStence and togve examples of these conditions. Our discussion follows that of Papoulis [5] Condition 1. If h(t) is integrable in the sense [二 忡川州附 hκ岭峭(t川)t) then its Fou er transform H(f) Xl and satisfi the inverse Fourler transform (2-5) It is important to note that Condition 1 is a suflicient but not a necessary condition for existence of a Fourier transform. There are functions wh do not satufy CondItion l but have a transform satISfymE(2-5)This ciass of functions will be covered by Condition 2. EXAMPLE 2-3 To il1ustrate Condition 1 consider the pulse time waveform h(t) = A 1 t 1 To (2-12) which is shown in Fig. 2-3. Equation (2-11) is satisfied for this function; therefore, the Fourier transform exists and is given by H(f) = Ae- J2 .r !t dt =A (2πP)dr-JA lrLm(2πβ)dt J -To '" -~， The second integral is equal to zero since the integrand is odd; H (f) = 2~f sin (2时)[。 _._sin(2π To f) k J'i.l. 一五To j' (2-13) Those terms which obviously can be canceled are retained to emphasize the [sin (a[)1 (af)) characteristic of the Four transform of a pulse waveform (F 2-3) Because this example satIdes CondiuonIthen H(f}as given byG-13)must satisfy Eq. (2-5)

17THEFOURIERTRANSFORMSec. 2-3Chap. 216THEFOURIER YRANSFORMBA H(f) = 2AT。sin (2mT。f)Ah(n)2WT.7le2AT-T。To-ToFigure 2-3, Fourier transform of a pulse waveform.十-24A7, sin 2TgDenr dfTaig02元T.J>0, thefunction h(o)/t is absolutelysin (2rax x = 27aintegrable in the sense of Eq. (2-1l) then H(f) exists and satisfies the(2-18)2naxinverse Fourier transform Eq. (2-5).then+4！An important example is the function [sin (af)/(af)) which does not(2-19)(r)=:21T0-12T。+satisfy the integrability requirements of Condition 1.Each term of Eq. (2-19) is illustrated in Fig. 2-4; by inspection these terms add toyieldEXAMPLE 2-4M)-A T,(2-20)2元00

Chap.21819THE FOURIER TRANSFORMSee. 2-3THE FOURIERTRANSFORMBy means of Condition 2, the Fourier transform pairAHIt)Aht)-2Af,sin (2mf,t)2Af,sin(2nf)zfotHU)A(2-27)IfIfoBecause this example satisfies Condition 2, H() [Eq, (2-25], must satisfy theBecause the integrand of the second integral is an odd function, the integral is zero;inverse Fourier transform relationship (2-5)the first integral is meaningless unless it is interpreted in the sense of distributiontheory.From Eq.(A-21),Eq. (2-31) cxists and can be rewritten asAe/zsfrdfh(t)(2-32)e/2nfidf-Kcos(2nfi)df=K8(0)h(t)Ksin(2元ft)/cos (2nft)dfAThese results establish the Fourier transform pair2元(2-33)KS)H)=Ksin (2元fa)(2-26)=2Afo2forwhich is illustrated in Fig. 2-6

21THEFOURIER TRANSFORMSec. 2-Chap,220THEFOURIERTRANSFORM[80+80)ed(0)mAht)-K8(t)AHIf)-K-e-12mfue/2s/er↓(2-37)= A cos (2n fot)KFourier transform pair48f -0) +40(+0)(2-38)Acos (2nfot)1is illustrated in Fig. 2-8.Figure 2-6.Fourier transform of an impulse function.Similarly the Fourier transform pair (Fig.2-9)(2-39)8f0)8fo)Asin (2nfor)Ahtt) -Kcan be established. Note that the Fourier transform is imaginary.+H(f)=KS(f)+AKA H(f) = R(f)h(t) =Acos (2f,t)Ot合5t-f)AStf+fFigure 2-7. Fourier transform ofa constant amplitude waveform.2Similarly the Fourier transform pair (Fig. 2-7)(2-34)K&(f)0)=K(can be established whcre the reasoning process concerning existence is exactly asFigure 2-8. Fourier transform of 4 cos (ar)argued previously.H(t) -11(n)A h(t) = A sin (2rf,t)EXAMPLE2-6To illustrate the Fourier transform of periodic functions consider(2-35)h(0)Acos (2fo)46fe)The Fourier transform is given by-0 6-40Acos(2nfoje-jnfidtH(f)-el2ufu+ e-nfje-nfediFigure 2-9. Fourier transform of 4 sin (at))[e-J2f-fu) +e-/2art+dtEXAMPLE 2-7Without prooft, the Fourier transform of a sequence of equal distant impulse40+8+0)(2-36)functions is another sequence of equal distant impulses;A Papoulis, The Fourier Inregral and Its Applications (NewYork: McGraw-Hill,1962)where arguments identical to those leading to Eq. (2-32) have been employed, Thep, 44.inversion formula yields

23THE FOURIERTRANSFORMSec. 2-4Chap. 222THEFOURIERTRANSFORMSubstitution of H() [Eq. (2-1)] into the inverse Fourier transform (2-5)AH,af)Ah,t)-1+2cos (2mf,u)yields[ H(f)elm df = " e/an af " h(x)e-nx dx(2-41)Since [Eq. (A-21)]-fg I lo-1J, e/a dt = 0()(a)then an interchange of integration in (2-41)givescos(2mkf,t)Ahatti-1KAHgif)[ H()ea dt = (x) dx f eraxrun df(2-42)=(x)(t-x) dx0But by the definition of the impulse function (2-28), Eq. (2-42) simply equals102%g31h(x). This statement is valid only if h(o) is continuous.t However if it is as-(b)sumed thatAhgitl-1cos (2mkf,t)h0) =()+h()(2-43)2AMsinthat is, if h(t) is defined as the mid-value at a discontinuity, then the inversionformula still holds.Note that in the previous examples we carefully definedeach discontinuous function consistent with Eq.(2-43)1000el0Lfo2f,3f,4f。5f,f2-4 ALTERNATEFOURIERTRANSFORMDEFINITIONS(c)It is a well established fact that the Fourier transform is a universallyAHm-1.E5f-nTh.ftaccepted tool of modern analysis.Yet to this day there is not a commondefinition of the Fourier integral and its inversionformula.To be specific theFourier transformpair is defined as(2-44)T2TTo-1h(t)en dt0=2元fH() =a,Id)(2-45)Figure 2-10. Graphical development of the Fourier trans-H()elar da)h(t) = azform ofa sequence of equal distantimpulse functions.where the coefficients a, and a, assume different values depending on theuser. Some set a, 1; a, - 1/2n; others set a, a, 1//2元, or set a,Z-)H)-(-)(2-40)0=1/2元;a,=1.Eqs.(2-44)and (2-45) imposethe requirementthata,a,A graphical development of this Fourier transform pair is illustrated in Fig. 2-10.1/2n.Various users are then concerned with the splitting of the product a,a.The importance ofFouriertransformpair(2-4O)will become obvious in futureTo resolve this question,we must define the relationship desired bctweendiscussionsofdiscreteFouriertransforms.the Fourier transform and the Laplace transform and the definition we wishto assumefor the relationshipbetween the total energy computed in the timeInversion Formula ProoftSee Appendix A.The definition of the impuse response is based on continuity of theBy means of distribution theory concepts it is possible to derive a simpletesting function, h(x).formal proof of the inversion formula (2-5)

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29THE FOURIER TRANSFORMChap. 2THEFOURIER TRANSFORM28f0A1>0(2-46)(Acos(2nfot)h:(0)dt=2naiH)doA1=0d. h(n) =2If the energy computed in is required to be equal to the energy computedlo10transform, universally defined ast=a;tabe.ht)=2(2-47)e-ta+ohdtL[ho)] h(t)e"" dt =loelsewhere1≥0shall reduce to the Fourier transform when the real part of s is set to zero,JAe-ursin(2fot)f.h)=10fkARAsc, J.,Fourier Transforms and theTheory of Distributions.Englewood1=0b.h(0)1:ACliffs,N.J.:Prentice-Hall,1966lo1<0