数字图像处理中的数学问题初探

1、 毕 业 设 计（论 文）题 目： 数字图像处理中的数学问题初探 目 录摘 要 02引 言 03一、已有知识03二、所得结果08致谢13参考文献 13 ABSTRACT 14英文文献翻译 15数字图像处理中的数学问题初探吕娟数学与应用数学（师范）专业 06080210南京师范大学数学科学学院摘 要本课题介绍了数字图像处理的概念以及数字图像处理中的一些基本的数学问题,主要包括:平面上的变换（旋转、平移、伸缩变换等）,Matlab在数字图像处理（图像文件的读写与显示操作,图像变换,图像增强等）中的应用.关键词: 数字图像处理 MATLAB 平面上的变换 论文正文引 言随着科学技术的不断发展,数字图

2、像处理也在不断发展,人们对数字图像处理的要求也在不断增加,图像的应用范围也在不断地扩大,因此图像处理的理论也需要不断地补充和发展.本课题旨在研究数字图像中的一些基本的数学问题,能够更好地把所学的数学知识应用于一些实际问题,达到学以致用的目的.1、 已有知识（一）、数字图像处理简介1、 数字图像处理的概念数字图像处理又称计算机图像处理,它是指利用数字计算机或者其他数字硬件,对从图像信息转换而得到的数字电信号进行某些数学运算或处理,以期提高图像的质量或达到人们所预期的结果.2、 数字图像处理的特点数字图像处理再现性好,处理精度高, 适用面宽,灵活性高,数字图像中的各个像素相关性较大.但是数字图像处

3、理处理信息量大,占用频带较宽,因此对计算机的计算速度等要求较高.3、 数字图像处理的主要研究内容（1） 图像变换:采用多种图像变换的方法,如傅里叶变换,离散余弦变换等间接处理技术,将处理空间域转换成处理变换域,一方面可以减少总体的计算量,另一方面,可以更有效地处理图像,从而达到预期.目前研究比较多的小波变换,在图像处理中有着较为广泛的应用.（2） 图像编码与压缩:图像编码与压缩技术可以减少描述数据的数据量,以达到减少占用的贮存空间的目的.编码则是图像压缩中的最为重要的方法,它是数字图像处理技术中发展比较早并且比较成熟的一项技术.（3） 图像增强与复原:图像增强与复原是为了提高图像的质量,抗干扰

4、.图像增强是突出图像中感兴趣的部分,但不考虑图像降质的原因,而图像复原则要求对图像降质原因有一定的了解.（4） 图像分割:图像分割主要是将图像中的有意义的特征提取出来,为进一步研究图形奠定基础.目前虽然已经找出了一些方法,但是还是没有找到一种适用于任何一种图像的方法.（5） 图像描述:图像描述是图像识别的前提.目前二值图像主要有边界描述和区域描述两种方法.随着图像处理的不断发展,已经开始研究三维物体的描述方法了.（6） 图像识别:图像识别主要是提取图像中的信息进行判别.近年来,在图像识别中人们越来越重视刚发展起来的模糊模式和人工神经网络模式.4、 数字图像处理的应用在日常生活中,数字图像处理已

5、经得到了广泛应用.例如:交通管理中车牌的识别,自动售货机钞票的识别.在日常通信中,图像传输、电视电话、电视会议等都运用了数字图像处理技术.随着数字图像处理技术的进一步发展,它已经被广泛应用在在航空航天、生物医学工程、工业检测、机器人视觉、公安司法、军事制导、文化艺术等领域.（二）、平面上的变换给定一个二阶矩阵,就相应的确定了一个变换.反过来,平面中的常见变换也可以用矩阵来表示.下面本文将来介绍几种在数字图像处理中常见的平面图形的几何变换及其矩阵表示.1、 平移变换由一个图形变化成另一个图形,在变化过程中,原来的图形上的所有的点都沿着同一个方向运动,且运动相等的距离,这样的图形变化叫做图形的平移

6、变换,简称为平移.平移变换不改变图形的形状、大小和方向;连结对应点的线段平行且相等.图12、伸压变换沿竖直方向或者水平方向伸长或者压缩的平面图形变换称为垂直伸压变换,简称为伸压变换 单墫.矩阵与变换.江苏教育出版社.当M=时,M对应的变换将平面图形作沿着x轴方向伸长或压缩.当时伸长,当时压缩.M对应的变换不是简单地把平面上的点(向量) 沿x轴方向“向下”压或”向外”伸,它是向x轴方向伸长或压缩.以为例,对于x轴上方的点向下压缩,对于x轴下方的点向上压缩,对于x轴上的点变换前后原地不动当M=时,M对应的变换将平面图形作沿着y轴方向伸长或压缩.当时伸长,当时压缩.经过伸压变换以后,直线仍然变为直线

7、,线段仍然变为线段. 图23、旋转变换图3矩阵 通常叫做旋转变换矩阵 单墫.矩阵与变换.江苏教育出版社.对应的变换就称做旋转变换,其中的角叫做旋转角.点A叫做旋转中心.旋转变换只改变几何图形的位置,不会改变几何图形的形状.旋转中心在旋转过程中不发生改变.图形的旋转是由旋转中心和旋转角度决定的.（三）、MATLAB在数字图像处理中的应用下面就先介绍下MATLAB的大致情况.1、MATLAB的概况MATLAB的名称源自Matrix Laboratory,是由美国MathWorks公司开发的应用于数值计算和图形处理的软件.它是一种科学计算软件,专门以矩阵的形式处理数据. MATLAB非常适合做矩阵运

8、算,这也是它的最大特点之一.目前MATLAB被广泛地应用于科学研究、工程技术应用研究、CAI(Computer Aided Instruct)、数学实验、数学建模等领域,适合多学科,功能非常强大.同时,它也是很多高级课程的基本教学工具.MATLAB拥有许多功能各异的工具箱,用于解决各个领域的专业问题.它的工具箱主要包括图像处理、信号处理、通信、统计、最优化、控制系统、非线性控制设计等.因而,用户可以借助这些工具箱,简单方便的进行分析、计算与设计工作.同时,MATLAB语法比较简单,只用一两个函数就可以代替C语言里需要几十行甚至上百行的语句,并且使用者也不需要重复去编程,只需要简单地调用和使用,

9、因此,相对来说,MATLAB又简单易学,容易掌握.（1）、MATLAB的图像处理工具箱简介MATLAB的图像处理工具箱函数包括以下15类.按其功能可分为:图像显示;图像文件输入与输出;图像的几何操作;图像线性滤波以及二维线性滤波器设计;图像变换;图像邻域及块操作;二进制图像操作;区域处理;颜色映像处理;颜色空间转换等.MATLAB图像处理工具箱支持的四种基本图像处理类型包括:二值图像、RGB图像、索引图像、灰度图像.MATLAB的图像处理工具箱功能十分强大,支持非常丰富的图像处理格式,如*.BMP、*.JEPG、*.GIF、*.TIFF、*.PCX、*.HDF、*.XWD、*.PNG等.（2）

10、、MATLAB的主要功能MATLAB主要功能有:（1）数值计算:可以进行任意精度的数值计算;（2）符号计算:可以进行解析推理;（3）可视计算:丰富的图像图形显示;（4）GUI编程:图形用户界面化;（5）多媒体:灵活的多媒体文件处理功能.（3）、MATLAB的主要特点1） 运算符和库函数极其丰富,语言简洁,编程效率高.2） 图形功能强大 .3） 拥有功能强大的工具箱.4） 比较容易扩充.下面将结合一些例子来介绍利用MATLAB图像处理工具箱进行图像处理的方法.2、MATLAB在数字图像中的应用（1） 数字图像的读取在MATLAB中利用函数imread将图形图像读成一个矩阵的形式.其主要格式如下:

11、A=imread(filename,FMT) ;其中,filename 是一个含有图像文件全名的字符串,指定图像文件的完整路径和文件名.如果在work工作目录下只需要提供文件名.FMT为图像文件的格式对应的扩展名.例如:I=imread(E:1111.jpg);%读入图像（2）数字图像的显示在MATLAB中主要利用函数imshow来实现数字图像的显示,主要有以下三种形式:imshow(I, G)I为要显示的图像矩阵.G表示显示该图像的灰度级数,如果将G省略,则默认灰度级数为256.imshow(I,low high)I为要显示的图像矩阵.low high指定显示灰度图像的灰度范围.高于high

12、的像素被显示为白色;低于low的像素被显示为黑色;介于High和low之间的像素被按照比例拉伸后显示为各种等级的灰色.例如:imshow(10）;title(The Main Pass Part of 11);imshow(I,)可以将变量low设置成数组f的最小值,将变量high设置成数组f的最大值.该函数常用来显示动态范围较小的图像.若需要显示多个图像,就可以在命令窗口加figure函数;需要注意的是只要用逗号或分号正确地分隔开不同的命令,一行中就可以写好几条命令.例如:imshow(J),figure,imshow(K).也可以用subplot命令来实现多幅图像的显示.例如:figure

13、;subplot(i,j,k);imshow(I);其中subplot(i,j,k)含义是:打开一个有i行j列图像位置的窗口,并将焦点位于第k个位置上.（3） 数字图像的保存图像的保存一般使用函数imwrite，其主要形式为:imwrite(f,'filename),其中,Filename必须包含文件的扩展名.例如:imwrite (I, 12.jpg);将数组I存放到文件名为12的图中.二、所得结果（一）、图像的平移不妨设原始图像对应矩阵为,平移后的图像相对应的矩阵为.那么平移前后图像对应的矩阵有以下的关系: =,其中,.例如:function B=translation(A,m,n

14、)m1=length(A(:,1);n1=length(A(1,:);B=zeros(m1+m,n1+n);for i=1:m1 for j=1:n1 B(m+i,n+j)=A(i,j); endendclearA=imread('E:1111.jpg');m=70;n=80;B(:,:,1)=translation( A(:,:,1),m,n);B(:,:,2)= translation( A(:,:,2),m,n);B(:,:,3)= translation( A(:,:,3),m,n);B=uint8(B);figure;imshow(B);imwrite(B,'

15、nir平移后图像.JPG') （二）、图像的旋转一般图像的旋转就是以图像的中心为原点,旋转一定的角度,即将图像上的所有像素都旋转同样的一个角度.不妨设原始图像上的任意一点经过旋转角度 以后到新的位置 ,为方便表示,将采用极坐标形式表示,原始的角度为,如上图所示:原始图像的点的坐标如下:旋转到新的位置的点的坐标如下:因此,图像的旋转前后对应两点的关系可以用矩阵表示如下:=例如:用程序实现图像旋转如下:function B=spin(A,sita) %sita可以取pi/2,pi,3*pi/2,2*pim=length(A(:,1);n=length(A(1,:);M=max(m,n);B

16、1=zeros(2*M+2,2*M+2);for i=1:m for j=1:n i1=i*cos(sita)+j*sin(sita); j1=i*(-sin(sita)+j*cos(sita); B1(1+M+round(i1),1+M+round(j1)=A(i,j); endendB=B1(find(sum(B1')>0),find(sum(B1)>0);clearA=imread('E:1111.jpg');m=3;n=2;sita=-pi/2;B(:,:,1)=spin( A(:,:,1),sita);B(:,:,2)= spin( A(:,:,2

17、),sita);B(:,:,3)= spin( A(:,:,3),sita);B=uint8(B);figure;imshow(B);imwrite(B,'nir旋转后图像.jpg') A 原始图像 B旋转后图像 （三）、图像转置不妨设原始图像对应矩阵为,平移后的图像相对应的矩阵为.那么转置前后图像对应的矩阵有以下的关系:.例如:用程序实现图像转置如下:A=imread('E:1111.JPG');B(:,:,1)= A(:,:,1)'B(:,:,2)= A(:,:,2)'B(:,:,3)= A(:,:,3)'figure;imshow(

18、A);figure;imshow(B);imwrite(B,'nir转置后图像.JPG') a原始图像 b转置后图像（四）、图像的镜像图像的镜像是指将原始图像相对于某一参照面旋转180°后得到的一幅新的图像.不妨设原始图像对应矩阵为 ,发生变换后的图像相对应的矩阵为 .那么变换前后图像对应的矩阵有以下的关系:水平镜像（相对于y轴） 垂直镜像（相对于x轴） 例如:用程序实现镜像如下:垂直镜像:function B=mirror_x(A)B=n=length(A(:,1);for i=1:n B=B;A(n+1-i,:);endA=imread('E:1111.j

19、pg');imshow(A);B(:,:,1)=mirror_x( A(:,:,1);B(:,:,2)=mirror_x( A(:,:,2);B(:,:,3)=mirror_x( A(:,:,3);B=uint8(B);figure;imshow(B);imwrite(B,'nir垂直镜像.JPG')水平镜像:function B=mirror_y(A)B=n=length(A(1,:);for i=1:n B=B A(:,n+1-i);endA=imread('E:1111.jpg');imshow(A);B(:,:,1)=mirror_y( A(:,

20、:,1);B(:,:,2)=mirror_y( A(:,:,2);B(:,:,3)=mirror_y( A(:,:,3);B=uint8(B);figure;imshow(B);imwrite(B,'nir水平镜像图像.JPG')a 原始图像 b 水平镜像 C垂直镜像致 谢本文从课题研究到论文完成是在朱建栋老师的指导下进行的.期间得到了朱老师悉心的指导,宝贵的建议和热情的鼓励,在此谨向朱建栋老师致以诚挚的谢意!同时也感谢大学四年里的老师们,他们的严谨治学使我受益匪浅;感谢同学们,在论文写作过程中给予我的帮助.参 考 文 献【1】姚敏等.数字图像处理.机械工业出版社,2007.【

21、2】杨杰.数字图像处理及matlab实现.北京：电子工业出版社,2010.【3】孟祥光,田萱,王丽丽.数字图像处理;Photoshop.北京:人民邮电出版社,2010.【4】贾永红.数字图像处理,第2版.武汉:武汉大学出版社,2010.【5】阮秋琦.数字图像处理基础.北京:清华大学出版社,2009.【6】张紫潇,史秀璋, 张群力.Photoshop设计与案例教程.北京:清华大学出版社,2011.【7】李了了,邓善熙.MATLAB 在图像处理技术方面的应用.【8】董长虹.MATLAB 图像处理与应用.国防工业出版社,2004.【9】何东健.数字图像处理.西安电子科技大学出版社,2003【10】图

22、像处理中的数学问题（连载）（【11】秦襄培,郑贤中.MATLAB图像处理宝典.北京:电子工业出版社,2011.【12】李国朝.MATLAB基础及应用.北京:北京大学出版社,2011.【13】伍家德.坐标系与坐标变换.湖北教育出版社,1985.【14】单墫.矩阵与变换.江苏教育出版社.【15】Rafael C.Gonzalez.数字图像处理.电子工业出版社.【16】章毓晋.图像处理和分析M.北京：清华大学出版社，1999.【17】飞思科技.MATLAB6.5辅助图像处理.电子工业出版社，2003.The mathematical problem of digital image processi

23、ngLvJuan06080210 Mathematics and Applied Mathematics,School of Mathematical Sciences,Nanjing Normal UniversityABSTRACTThis subject introduces the concept of digital image processing and some of the basic math problems of digital image processing, mainly including:the transform in the plane (such as

24、the the rotation,telescopic, expansion or compression transformation),the application of MATLAB in digital image processing (reading and writing, display, image transformation,image enhancement ,edge detection etc) .Key words: digital image processing MATLAB the transformation in the plane THE FUNDA

25、MENTAL THEOREM OF ALGEBRA VIA LINEAR ALGEBRA KEITH CONRAD Our goal is to use abstract linear algebra to prove the following result,which is called the fundamental theorem of algebra.Theorem 1.Any nonconstant polynomial with complex coefficients has a complex root. We will prove this theorem by refor

26、mulating it in terms of eigenvectors of linear operators.Let have degree ,with .By induction on n,the matrix satisfies .Therefore Theorem 1 is a consequence of Theorem 2. For each ,every square matrix over has an eigenvector.Equivalently, for each ,every linear operator on an -dimensional complex ve

27、ctor space has an eigenvector. Theorem 2 is also consequence of Theorem 1, so the two theorems are equivalent. In fact,the implication Theorem 1 Theorem 2 is usually how one first meets the fundamental theorem of algebra in a linear algebra course: it assures us that any complex square matrix has an

28、 eigenvector because the characteristic polynomial of the matrix has a complex root.But here, we will prove Theorem 2 without assuming Theorem 1, so we can deduce Theorem1 as a consequence of Theorem 2. Our argument is a modification of a proof by H. Derksen1. It uses an interesting induction. Our s

29、tarting point is the following lemma.Lemma 3. Fix an integer and a field . Suppose that, for every -vector space whose dimension is not divisible by , every linear operator on has an eigenvector. Then for every -vector space whose dimension is not divisible by , any pair of commuting linear operator

30、s on has a common eigenvector. The hypothesis of Lemma 3 is quite restrictive, so Lemma 3 does not apply in too many examples. For instance, it does not apply for any when . We will use Lemma3 when is a power of 2. I know no worthwhile applications of Lemma 3 for other values of .Proof. We induct on

31、 the dimension which runs through integers not divisible by . The case is trivial: any nonzero vector in a one-dimensional space is an eigenvector of any linear operator on the space (any two such operators also commute). Assume now that is not divisible by and we have settled all dimensions less th

32、an which are not divisible by . Let and be commuting linear operators on an -vector space of dimension . Since is not divisible by , the hypothesis of the lemma tells us has an eigenvalue in , say . LetEach of these subspaces of is -stable (that is, ,and ) ,and since is an eigenvalue of .Each of and

33、 is also -stable since and commute. (For instance, if, write . Thenis also in .) Note is not divisible by , so one of or has dimension not divisible by . If the subspace or with dimension not divisible by is a proper subspace of , then and have a common eigenvector in that subspace (and thus in ) by

34、 induction. The other case is that or is all of and the other subspace is since the dimensions add up to , in since we already noted that is positive. From the equation we see every vector in is an eigenvector for , and one of them is an eigenvector for since has dimension not divisible by . Corolla

35、ry 4. For every real vector space whose dimension is odd, any pair of commuting linear operators on has a common eigenvector.Proof. In Lemma 3, use and . Any linear operator on an odd-dimensional real vector space has an eigenvector since the characteristic polynomial has odd degree and therefore ha

36、s a real root, which is a real eigenvalue. Any real eigenvalue leads to a real eigenvector. Note Lemma 3 and Corollary 4 are not saying commuting linear operators have a common eigenvalue, but rather a common eigenvector. A common eigenvector does not have to occur with the same eigenvalue for and .

37、Example 5. Let viewed as linear operators on . A direct calculation shows , so there must be a common eigenvector for and in . One common eigenvector iswith eigenvalue 1 for and 3 for . In fact, this is the only common eigenvector for and in , up to scaling.Now we turn to the proof of the Fundamenta

38、l Theorem of Algebra.Proof. (Derksen) Our proof will be by induction on the highest power of 2 dividing . That is, writing , where and is odd, we will prove the theorem by induction on . Let's be concrete about how the induction proceeds, before we carry it out. First we will treat the case , wh

39、ich means is odd. For the base case , let be odd. Pick. To show has an eigenvector in , we will associate to some linear operators on the spacewhere denotes the conjugate-transpose of . (In coordinates, .Note ) Matrices in are called Hermitian. They form a real vector space, with dimension over . In

40、 particular, has odd dimension over . We can decompose any aswhere and are Hermitian. (Symbolically, .) For, take :(0.1) When ,(0.1)becomes (0.2) Looking at (0.2), we are inspired to consider the -linear operators , : defined byThe operators and commute with each other (check!). Since has odd (real)

41、dimension, Corollary 4 implies and have a common eigenvector in . That is, there is some nonzero such thatwhere .ThereforeAny nonzero column vector of is an eigenvector in for . That concludes the proof of the Fundamental Theorem of Algebra for odd , which was the base case. (Notice that the base ca

42、se of a proof by induction is not always trivial!) Now we treat the inductive step. The following is our inductive hypothesis for : for any where the highest power of 2 dividing is less than (that is, is not divisible by ), any linear operator on any -dimensional complex vector space has an eigenvec

43、tor.It follows, by Lemma 3 with , that any pair of commuting linear operators on a vector space of such a dimension has a common eigenvector. The hypothesis of Lemma 3 is precisely our inductive hypothesis. (Derksen's paper 1 has a more involved inductive hypothesis, because he is trying to prov

44、e not only the Fundamental Theorem of Algebra, but also that any finite set of commuting linear operators on a complex vector space -not just two commuting operators as in Lemma 3 - has a common eigenvector.) The case to consider now is linear operators on complex vector spaces with dimension divisi

45、ble by but not by a higher power of 2. By choosing bases to convert operators into matrices, it will suffice to treat the case of matrices acting on concrete spaces of the form instead of linear operators acting on abstract complex vector spaces. However, in the course of our argument for matrix ope

46、rators, we will be using linear operators on vector spaces other than (specifically, we will be using linear operators on the vector space ), so it is important to have the abstract point of view in mind. Let be a positive integer such that the highest power of 2 dividing is . Let be any complex mat

47、rix. We want to show has an eigenvector in . Consider the space of complex symmetric matrices,This is a complex vector space with dimension . Notice that the highest power of 2 dividing is . Since is not divisible by , any pair of commuting linear operators on has a common eigenvector. (This is the

48、application of Lemma 3which we made right after formulating the inductive hypothesis.) We are going to apply this result to a pair of operators on built from the matrix . Define ;byThe maps and are -linear operators and commute (check!). Therefore, by the application of Lemma 3 in the previous parag

49、raph, and have a common eigenvector:some nonzero satisfieswhere. Applying to the first equation on the left, and using the second equation to simplify, we findSO Since any complex number has a complex square root, the quadratic formula shows any quadratic polynomial over splits into linear factors.

50、(This is the first time we have used something about which is not true for . Up until this point, we could have been trying to prove the Fundamental Theorem of Algebra for operators on real vector spaces, and even invoked Lemma 3 with as part of the induction. But the proof would break down now, bec

51、ause real quadratics generally don't have real roots.) Factor the complex polynomial . Then the matrix product is ,soIf ,then any nonzero column vector of is an eigenvector of (with eigenvalue ). If , then any nonzero column vector of is an eigenvector of (with eigenvalue ). That concludes the F

52、undamental Theorem of Algebra. 通过线性代数证明代数的基本定理基思·康拉德 我们的目标是用抽象的线性代数来证明代数的基本定理.定理一.任意有复杂系数的非常数多项式有一个复杂的根. 我们将用线性算子特征值来表述定理,以证明它.假设通过归纳,矩阵满足.因此定理一是下一定理的结论:定理二.在复数域内,每个方阵都有一个特征向量.等价地,在复杂的维向量空间里的线性算子有一个特征向量. 定理二也是定理一的结论,因此这两个结论是等价的.事实上,定理一定理二的推论通常来判断是否满足线性代数课程的代数基本定理:它告诉我们,任何复杂的方阵都有一个特征向量,因为矩阵的特征多项

53、式有一个复杂的根.但在这里,我们将不通过假设定理一来证明定理二,因此,我们可以推断出定理一是定理二的结论.我们的论点是对H.德克森1的证明的调整.它使用了一个有趣的归纳法.我们的出发点是下面的引理.引理三.规定整数和数域.假设每个上不能被整除的线性空间,上的线性算子有一个特征向量.因此,对于每个上不能被整除的线性空间,任意一对交换线性线性算子有一个共同的特征向量. 引理三的假设太严格,因此并不适合于太多例子.例如:它并不适用于当时的任意情况.我们将在是2的乘方时使用引理三.我知道为其他值时引用引理三没有价值.证明.我们在全是不能被整除的整数的维进行归纳.的情况是不重要的:任何非零向量在一维空间上是任何该空间上线性算子的一个特征向量（任意这样的两个线性算子可交换）. 现在假设不能被整除的整数的大于1小于维的情况我们已经解决.规定 和 是数域上维的线性空间的交换线性算子.由于不能被整除,引理的假设告诉我们在数域上有一