光学光学光学光学 (7).pdf

-286-Chapter 7 Fourier Optics(Zhaos,Chap 5,1-5;Hechts 7.3,7.4.1;Chap 11:11.2,11.3)Here I only intend to introduce the Fourier optics quite briefly.The detailed derivation and discussion are both out of the scope this course,and should be addressed in a complete course,and out of the range of my expertise.First I shall make a brief discussion on Fourier Transform,simply give out the important results without rigorous math derivation.Then I shall continue to its application in optics,especially focus on the relation with Fraunhoffer diffraction.It is a nice and important example and is sufficient enough to illustrate many important concepts.7-1 Fourier Expansion(Fourier Series)and Fourier integrals(Hecht 7.3,7.4.1,11.2;Zhaos 5,chap 5,Vol II)For an arbitrary function,say,x is some general variable,it is sometimes very useful expanding it into series,i.e.into components of simpler functions.One famous one is Taylor expansion,is expanded into polynomial form,with the coefficient of component corresponds to.()f x()f xmxmmd fdx-287-Such expansion can be useful in physics.As we already mentioned before,that for a potential,close to the equilibrium point,it can be approximated by,a harmonic potential.much like harmonic potential.Another very useful expansion is to express in terms of sinusoidal functions,or,here is some general variable,is the general angular frequency,where,is the generalized frequency.If,then harmonic oscillation.If,the spatial dependent of wave,wave vector If,the general coordinate of a spatial distribution,such as a pattern on the screen,then is the spatial angular frequency of the distribution,where the spatial freq.,d is is the spatial period,Such expansion of generally takes the form of:()V x2Ax()fsincosie2f=ft,ii teex2=,iikxkeex1222ffd=f()f x-288-The expansion is Fourier expansion(see below for detail).7-1-1 Fourier Expansion of a Periodic Function is a periodic function of x,i.e.,is the period,then.For such periodic function,can be expanded into series of,or,the harmonic components with different frequency,the frequency is a multiple of the fundamental frequency.i.e.:(7-1)n is integer.or (7-2)Of course there is one issue whether such series expansion convergent,i.e.the series closely resemble the original function f(x).This question is a complexed mathematical problem and we shall not consider it here.The Strategy is“assume”such series exist and convergent(For the functions 001()2(),01()2/2nnnikxnnnnnCAiBf xC enCAiBCA+=+=()f x()()f xf x+=22kf=()f xsin(),cos()nkxnkxinkxek011()cossin2nnnnAf xAnkxBnkx=+001()2(),01()2/2nnnikxnnnnnCAiBf xC enCAiBCA+=+=-289-you encounter in physics,the above is true.Actually for piece-wise continuous function,Fourier series(or integral)exist).The question now is what are the coefficients of each frequency component is the DC component(direct current).They can be evaluated by applying the orthogonal condition of sine,cosine functions,i.e.:,a,b are integers.(7-3)the Kronecker delta.Then the can be calculated easily:So using orthogonal relation When we find the Fourier series expression for a function,we can use Euler formula to re-express the expansion in terms of imkxe,and here m can take both positive and negative values,the expansion coefficients are,nnA B02A0sincos0akxbkxdx=00sinsin2coscos2ababakxbkxdxakxbkxdx=0()1()abababab=,nnA B00()cos2()sin2mmf xmkxdxAf xmkxdxB=00()2f x dxA=002()cos (74.1)2()sinmmAf xmkxdxBf xmkxdx=-290-in(7-2),but can be more readily expressed using orthogonal relations:(7-4.2)This procedure is really analogous to finding the components of a vector in an orthonormal basis.Here the function f(x)is analogous to arbitrary vector;the sinusoidal functions are“base vectors”,and(7-3)is the orthogonal condition for the base vectors.(7-4)is like finding the components by dot product.If possess certain symmetry,i.e.if is even function with respect to(x),then clearly(only cosine left);if the odd function:(only sine left).The symmetry sometimes simplifies the evaluation.Example 1.For the square wave in Hechts Fig 7.18 2()2i m n kxmnedx=221()inkxnCf x edx=()f x()f x()()f xfx=0mB=()f x0mA=-291-The Fourier expansion of then is(see the Hechts book for detail)Figure below shows the Matlab graphical results,a single harmonic is a poor approximation for the square wave,however,as the number of harmonic increases,the summation looks more like the square wave(the figure is the Matlab result for first 2,3 and 10 sine components)1(0)2()1()2xf xx+=()f x411()(sinsin3sin5.)35f xkxkxkx=+2k=sinkx-292-Example 2.The response of a low pass filter to the square wave input as given in the last example:is the input in forms of square wave,then what is,the RC circuit forms a low pass filter that cuts freq.at.The square wave by Fourier analysis can be treated as superposition of many harmonic components,and the expansion is given in the previous example,basically it contains components.The low pass filter will truncate higher frequency component,say,(i.e.inVoutV0RC=,3,5 m-293-),the then would be a summation of the truncated series and would distort from the original the output,and would be something like given in the figure above.Example 3.Forced oscillator under periodic driving force.Earlier in the course,we discussed the oscillation of a oscillator driven by a harmonic force,i.e.:,where Harmonic driving force.Now,what happens if the is periodic not necessarily harmonic?The solution to the problem is simple,since we can expand into Fourier Series,which has components of.For each,the solution is known,the final solution would be superposition of them.From the discussions above,we can see that the usefulness of Fourier Expansion.The system response to harmonic functions,i.e.sometimes are much easier to evaluate;then for the linear system knowing its harmonic response,(linear here means:if the system response to input signals is;then for the multiple inputs,the response is:)the response to summation of input equals to the summation of response to the individual components.Then using Fourier Expansion,the systems 0m=outVoutV()mxxkxF t+=()cosF tFt=cos()xAt+()F t()F tcos,sin,cos2,sin2,.ttttcosm tsin,cos i xxx or e1S1()R S123123(.)()()().R SSSR SR SR S+=+-294-response to other functions will be known too.For such periodic function,with frequency,the harmonic components in Fourier expansion are discrete,i.e.only,the integer multiple of,another way to say this is that its Fourier Spectrum is discrete.Not only periodic function,but also functions that only defines in a limited space,has discrete Fourier Spectrum.To see this,take a look of the example For such,it can be treated as periodic,with period of L(or mL)(certainly many ways to construct such periodic function,but in the region,is fixed).Then the Fourier Expansion for periodic function applies here too,with fundamental frequency of (or depending on the construction).7-1-2 Non-periodic functions Fourier Transform(Fourier Integral)We shall see in this section that for the non-periodic functions,the Fourier spectrum would become continuous,i.e.for component,its()f xm()f x(|)2()(|)2Lx xf xLundefinedx()f x|2Lx 2,in xe=2=n0(1)nn=+=nCin xe()nCF-296-Mathematically,it can easily be proved by:For non-periodic function,it is a limiting case for.Starting with limited:Fourier Expansion 7.2()f x()in xnnf xC e+=-297-where (7.4.2)Let where where Now as,and we should treat as continuous variable and,the.The above relation becomes:(7.5)(7.6)(These are same as Zhaos 5.5,but with rather than,which)(7-5)and(7-6)are the definition of Fourier Transform between f(x)and F().is called the Fourier Transform of ,knowing,its Fourier Transform can be evaluated by(7-6);Relation(7.5)is the reverse Fourier Transform,i.e.knowing by(7.5).Another useful definition of Fourier Transform is to take symmetric 221()in xnCf x edx=22()()nixnnnCFFf x edx=nn=11()()()2nnixixnnnnnf xFeFe+=12nnn+=0nnndn+=1()()2i xf xFed+=()()i xFf x edx+=ddf2ddf=()F()f x()f x()()Ff x-298-form for Fourier Transform and reverse transform.(7.5)(7.6)(7.5),(7.6)is another definition for the transform,and will be frequently used by me.The two definitions in(7.5),(7.6)and(7.5)are equivalent,except a difference in common factor.Say if is fixed then in(7.5)is in(7.5)Sometimes we shall use short hand:to represent 7.6 for s Fourier Transform.=to represent reverse Fourier Transform.Using(7.5),(7.6),it is straight forward to evaluate the Fourier Transform given or reverse.Some examples of typical function,and its Fourier transform are given in Zhaos book,Table V-3,pg.92,Vol.II and copied below.Noticed that there is a trivial difference in Fourier Transform given here and in Zhaos book,section 5.There he uses frequency,rather than angular freq.as here for the variable in Fourier Transform,the results are same,up to a common factor depending on using(7.5)or(7.5).1()()2ixf xFed+=1()()2ixFf x edx+=()f x()F1()2F()()F kf x=F()f x1()()f xF=F1()f xFF()F()f x()f xf2 f=-299-300-301-302-303-304-The examples(1)-(3)in the Table V-3,we actually already worked them out when dealing with superposition of waves or Fraunhoffer diffraction by the gratings.The,the Fourier Transform in(1)-(3)looks exactly as the Fraunhoffer diffraction pattern.This should be obvious if you recall how we derive the Fraunhoffer diffraction in the previous chapter,and I shall readdress this later.Here,you should realize that such similarity between Fourier Transform and F-diffraction is no coincidence.Example(4),(8)are basically:Gaussian Gaussian(The Fourier Transform of Gaussian is still a Gaussian).Lorentian exponential.What is clear from these examples are:(1)is generally a complex function,i.e.The dotted line in Table V-3 in example 8,9 are.(2)If is the width of is the width of Then constant.The value will depend on definition of,.The proof is:()F()F()F()()|()|iFFe=()x()f x()Fx=x-305-Define in as:in as:Since:-Then Or ()This is the uncertainty relation we referred to a few times earlier.7-1-3.Dirac Delta Function What happens for a square wave,or a cosine wave train as in Table V-3,Zhaos,(1),(2),if the width of increases,clearly,the width of will decrease.What if to an extreme case,that,basically the square wave would become a constant A,the cosine wave train would become a,their the Fourier transform would become with zero width,infinite amplitude,it would become a Dirac x()f x1()(0)xf x dxf+=()F1()(0)FdF+=1()()2i xFf x edx+=1(0)()2Ff x dx+=1()()2i xf xFed+=1(0)()2fFd+=(0)(0)22(0)(0)FfxfF=1x f =2f=()xx()Fx 0cosx()F-306-Delta Function.Dirac Delta Function is a special function describes a singular physical distribution,it is extremely useful in representing some physical idealization,such as mass point,single freq.,point charge etc.(1)Definition of.(7-8)(2)Properties of.(a)even function.(b).(c)convolution.(d)(e),it has property of:(prove with integration by part)These properties are easy to derive from the definition.(3)Other representation of.(7-9-1),using property(b)()x(0)()0(0)xxx=()1x dx+=()x()()xx=()()(0)f xx dxf+=()()()f xxx dxf x+=()()f xx1()()|axxa=()()dxxdx=()()(0)x f x dxf=()x11()()()22i xxFx edx+=F01ie=-307-(7-9-2)(7-9-2)is a useful representation of.The above derivation clearly shows:(or if(7.5),(7.6)are used.This is the results in Zhaos book).It is also easy to show that,if,then.So (or if(7-5),(7-6)are used,this same as Zhaos book,).This means that for the extremely broad distribution in,for example constant,a single frequency in the transform.If the,a single point in,transform to a constant in domain,means infinite broad distribution;Exactly as we stated at the beginning of this session.If,then the (This can be proved either by the definition of Fourier transform or using the phase shift 1()()112212i xi xxFeded+=F()x1()2x()1x1()2f x=1()()()2i xf xedx+=F1()2x12()x1()f1()()2f=()f x()f x=()()()f xx=x0001()cos2ixixf xxee=+00()()()f x +F-308-property we shall discuss)Please refer to Zhaos book Vol.2,Table V-4 on pg.106.(1)(5)in the table are straightforward;the(6),(7)of the table are explained in the book by example 5 on pg.105.(Note,there is an error in Zhaos Table V-4,(7),it should be,not).There are other representation of,by using extreme cases of Gaussian,Lorentian or exponential decay,there are listed in Zhaos book pg.103.7-1-4 Properties of Fourier Transform(Zhaos book,Chap 5,5.3)Many of the properties below are self-evident and I shall discuss a little bit of differential property and convolution in detail.Two functions of variable,their Fourier Transforms are.or as in Zhaos book.(1)Linearity:(7-10)Means Fourier Transform(2)Conservation Theory(Parsevals Formula)(7-11)001()()nnG ffxx=001()()nnG ffxx+=()x(),()g x h xx(),()GH(),()G fH f()()()()ag xbh xaGbH+22|()|gdxGd+=-309-This can be treated as a result from conservation of energy,so that the total energy analyzed in space is same as analyzed in frequency domain.The formula can be easily derived directly from definition of Fourier Transform.A note on this,relation(7-11)is right if we take the definition for Fourier Transforms as given by(7.5),(7.6).If we use that of(7.5),(7.6).recall that(Define by 7.6 and 7.6),then relation(7.11).(3)Expansion of the variable:(7-12)(4)Phase Shift (7-13)Inversely:(7-13)This relation is useful to get quick result of Fourier transform.The proof is straightforward from definition.(5)Relation of complex Conjugate:2*|()()1()()21()()21()()2()()i xi xi xgg x gxgxGedgg dxg xGeddxGg x edx dGGd+=1()()2FF=221|()|()|2g xdxGd+=1()()|g axGaa00()()i xg xxeG00()()ixeg xG-310-(7-14)(6)Fourier Transform of Derivatives If Whats the Fourier Transform for?If as (which is true in many physical problems).Then:(7-15)Similarly:(7-16)Fourier Transform of We can immediately see that Fourier Transform would be useful in solving Differential Equations.This is better to be illustrated with example and one of the homework.Example:Solving wave equation:is a wave.At (is a function of,this is the initial condition).The partial differential equation generally is difficult to solve directly,*()()()()gxGgxG()()g xG1()G()dg xdx11()()21()()21()|()22i xi xi xi xGg x edxdg xGedxdxieg xg x edx+=+()0g x=x 1()()Gi G=()()()nnGiG=()nnd g xdx222221yyxvt=(,)y x t0,(,0)()ty xf x=()f xx-311-but we can apply Fourier Transform to simplify it.The procedures are:Taking Fourier Transform on both side of the equation()is Fourier Transform of with respect to.Solving Differential equation This is a much simpler problem,an ODE instead of partial differential equation.The initial condition for,Do an Inverse Fourier Transform to get from (You can also directly use(7-13),the phase shift to get this)This is exactly the traveling wave form we give at the very beginning of this course.x22222222222211()(,)1(,)i xi xi xyyedxedxxvtiYtyedxvtYtvt+=(,)Yt(,)y x tx2222(,)()(,)YtiYtvt=0,(,0)()ty xf x=1(,0)()()2(,)(,0)()i xiv tiv tYf x edxFYtYeFe+=(,)y x t(,)Yt()1(,)()21()2()iv ti xix vty x tFeedFedf xvt+=-312-This demonstrates a typical application of Fourier transform,which can be summarized by the chart below:(7)Conv