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1、-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 expe
2、rtise.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 sufficien
3、t 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.
4、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
5、 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 g
6、eneral 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 exp
7、ansion 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 fundam
8、ental 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
9、 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 functio
10、n,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 calcul
11、ated 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=00sinsin2coscos2abab
12、akxbkxdxakxbkxdx=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
13、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
14、 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
15、 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.)3
16、5f 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
17、 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 ou
18、tput,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 har
19、monic?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
20、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 indiv
21、idual 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 harm
22、onic 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
23、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 fun
24、ctions 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
25、: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,whic
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