Digital Image ProcessingChapter 2 Digital Image Fundame
图像处理Chap 2 Digital Image Foundamentals
找出图象中，具有相同或相似灰度属性的 像素，标记值相同的像素构成了图象中的一个 区域(Region)。
2.2 Basic Relation Between Pixels
2.2.4 Adjacency
2.1 Image Sampling and Quantization
2.1.3 采样、量化参数与数字化图像间的关系
2.1.3.1 数字化方式
分为：均匀采样和量化、非均匀采样和量化。
均匀：指的是采样、量化为等间隔。
图像数字化一般采用均匀采样和均匀量化方式。
非均匀采样根据图像细节的丰富程度改变采样间距。
3.
扫描结束时，所有值为1的点都被打上标记，但这些标记中的一 些也许是等价的，整理所有的等价对成为等价类，然后给每一 个类一个标记。 第二次扫描图象，用所赋予的等价类的标记替换不同的标记。
4.
2.2 Basic Relation Between Pixels
2.2.3 Labeling of Connected Components
1 I 0 1 0 0 1 0 1 0
2.1 Image Sampling and Quantization
灰度图像：
每个像素由一个量化的灰度值来描述，它不 包含彩色信息。 灰度图像中的灰度级数通常取值：256。
0 150 200 I 120 50 180 250 220 100
采样后图像被分割成空间上离散的像素，但 其灰度是连续的，还不能用计算机进行处理。
量化：将像素灰度转换成离散整数值的过程。 一幅数字图像中不同灰度值的个数称为灰度 级数，用L表示。
2.1 Image Sampling and Quantization
数字图像处理_课件_2
研究人员将之植入失明小鼠的眼底，成功使得
小鼠的视觉得到了恢复。
20
数 第 ➢ 相关研究题为《Nanowire arrays restore vision
字二 图章
in blind mice》，2018年3月6日发表在《Nature
像 数 Communications》上。
处字 理图
➢ 这种光感受器使用的是金/氧化钛纳米线阵列，
h表示视网膜图像 中物体的高度
晶状体
视网膜
15 h 100 0.017
中央凹
h=2.55mm
x 2.55mm
100m
17mm
17
数 第 ➢ 视网膜图像主要聚焦在中央凹区域。
字二
图章 像数
➢ 然后，光接收器的相对刺激作用产生感
处 字 知，把辐射能转变为电脉冲，最后由大
理图 像
脑解码。
基
础
18
并触发附近神经元的电活动，从而恢复小鼠对
光的敏感性。
22
数 第 ➢ 这种纳米线光感受器有效地替代了原有视网膜
字二 图章
中的视杆细胞和视锥细胞，因而，在植入后能
像 数 够观察到视觉中枢的神经元也恢复了对光的响
处 字 应，同时，瞳孔对光的反射也有所改善。
理图
像 ➢ 这种新一代可植入人工光感受器为黄斑变性等
基 体的电磁波的波长必须小于等于物体的尺
础 寸。
41
图像成像的途径与方式
数 第 成像以电磁波发射的能量为基础；
字二
图章 像数
物体反射的声波也可用于形成超声波图像；
字二 图章
技术，尝试利用一种可吸收红外光发出可见光
像 数 的上转换纳米材料，导入动物视网膜中以实现
数字图像处理Chapter02
if D(u , v) ≤ D0 if D(u, v) ≥ D0
0 H (u, v) = 1
if D(u, v) ≤ D0 if D(u , v) ≥ D0
Digital Image Processing
Chapter 2 Image Pre-Processing in the Spatial and Frequent Domain 2.2.2 Butterworth filter Butterworth Lowpass Filter:
( s ,t )∈S xy
Digital Image Processing
Chapter 2 Image Pre-Processing in the Spatial and Frequent Domain 2.1.4 Min filter Min filters are the 0th percentile filters that are useful for finding the lowest point in an image. They can reduce salt noise.
The disadvantage of Laplace operator is to respond doubly to some edges in the image.
The Sobel operator approximates the first deritative. The gradient is estimated in eight possible directions, and the convolution result of greatest magnitude indicates the gradient direction. It is often used as a simple detector of horizontality and verticality of edges
digital image processing 引用格式 -回复
digital image processing 引用格式-回复什么是数字图像处理？数字图像处理是指通过计算机技术来对图像进行处理和分析的一门学科。
它涉及到图像的获取、存储、传输、压缩和呈现等多个方面，主要目标是改善图像的质量，提取出图像中的有用信息，并进行各种图像操作，如增强、滤波、分割、识别等。
数字图像处理广泛应用于医学影像分析、图像识别、视频处理、无人驾驶等领域。
数字图像处理主要包括以下几个步骤：1. 图像获取：数字图像通常是通过物理设备（如摄像机）捕获得到的。
在图像获取阶段，需要考虑一些因素，如相机的性能、拍摄环境和拍摄参数，以确保获得高质量的图像。
2. 图像预处理：图像在被处理之前，往往需要经过一系列的预处理步骤。
这些操作包括去噪、亮度调整、对比度增强等。
预处理的目的是提高图像的质量，为后续的处理步骤提供更好的输入。
3. 图像增强：图像增强是改善图像质量的一种技术。
图像增强可以通过调整图像的亮度、对比度、色彩饱和度等参数来改善图像的视觉效果。
常见的图像增强方法包括直方图均衡化、滤波和锐化等。
4. 图像分割：图像分割是将图像划分为不同的区域或对象的过程。
图像分割可以通过阈值分割、边缘检测和区域生长等方法来实现。
图像分割在目标检测、图像识别和医学影像分析中都发挥着重要作用。
5. 特征提取：特征提取是从图像中提取出有用的信息的过程。
图像特征可以是形状、颜色、纹理等方面的特征。
特征提取常用的方法有边缘检测、角点检测和纹理描述符等。
6. 图像识别：图像识别是根据图像的特征将其分类到不同的类别中的过程。
图像识别可以通过机器学习和深度学习等方法来实现。
图像识别在人脸识别、车辆识别和物体识别等领域具有广泛应用。
7. 图像压缩：图像压缩是减少图像文件大小的一种技术。
图像压缩可以通过无损压缩和有损压缩来实现。
无损压缩可以保持图像的质量不变，而有损压缩会引入一定的失真，但可以大幅度减小文件大小，适用于图像传输和存储。
DigitalImageProcessing2ImageProcessingFundamentals
16
of Light and the Electromagnetic Spectrum
58
17
of Light and the Electromagnetic Spectrum
58
Visible light: 0.43~0.79um The electromagnetic spectrum is split up according to the wavelengths of different forms of energy
12
Brightness Adaptation &
of
58
Discrimination (cont…)
Another experiment: a piece of paper on a desk is always white, but can appear totally black when used to shield the eyes while looking directly at a bright sky
4
of
Human Visual System
58
The best vision model we have!
Knowledge of how images form in the eye can help us with processing digital images
We will take just a whirlwind tour of the human visual system
58
Light is just a particular part of the electromagnetic spectrum that can be sensed by the human eye
Chapter2_Digital Image Fundamentals(2014)
--distributed over the retina surface
Harbin Engineering University
2.1 Elements of Visual Perception
2.1.1 Structure of the Human Eye
Distribution of cones and rods
Blind spot ������
the absence of receptors area
Radiant Energy
The principle of Image Formation in the Eye
Harbin Engineering University
2.1 Elements of Visual Perception
2.1.3 Brightness adaptation and discrimination
2.1 Elements of Visual Perception 2.2 Image and Digital Image 2.3 Image Sensing and Acquisition techniques and application areas 2.4 Spatial and Gray-Level Resolution of digital image 2.5 Basic types of Digital image 2.6 Basic Operations of digital image
数字图像处理习题讲解
1 1 1
1 0 1 1 01 1 1
1
0 0 0
0
[ f ] H4[F ]H4
1 1 1 1 2 0 0 21 1 1 1 0 1 1 0
1 1 1 1 10 0 0 0 1 1 1 1 1 0 1 1 0 2 1 1 1 10 0 0 0 1 1 1 1 2 0 1 1 0
1 1 1
采样间隔为Δt ，Δt <D, 即1/Δt >2*f = 1/D,
满足采样定理，所以没有混叠 。
2021/1/2
Digital Image Processing:
8
Problems
第一章第5题
(c) 如果 D=0.3 mm,你能否使用2倍过采样？3倍过采样？. 可以， ∵ Δt < D/2 =0.15mm, Δt < D/3 =0.1mm
2021/1/2
Digital Image Processing:
22
滤波器频响 零响应！
滤波器频响 负响应！
2021/1/2
Digital Image Processing:
23
2021/1/2
Digital Image Processing:
24
第三章习题8 (a)
设: f(g x,( yx ) , -y >) F (u4 , v)f ( x 则, :y ) [ f ( x 1 , y ) f ( x 1 , y ) f ( x , y 1 ) f ( x , y 1 )
2021/1/2
Digital Image Processing:
4
第一章习题2 一台光导摄像管摄像机的靶直径为25mm，感应点直径为35微米。若像素间距与点直 径相同，它数字化一幅正方形图像时的最大行数和列数是多少？若要数字化的图像 为480 ×640像素，靶上的最大像素间距是多少？
Digital Image Processing_中文翻译
数字图像处理萨尔普埃蒂尔克科贾埃利大学简介数字图像处理迅速成为流行在科学和工程应用中有许多用途。
因此，数字图像处理，包括在许多电子和计算机工程计划的研究生课程。
LabVIEW编程和许多并入IMAQ视觉的图像处理功能的易用性使实施简单和高效的数字图像处理算法。
本手册的目的是作为一种辅助课堂演示以及互动研究实验室指南是有用的。
实验2基本的图像处理图像处理图像处理是指操作图像的步骤，。
常用的图像处理的计算机通过在数字域中进行。
数字图像处理涵盖范围广泛的不同的技术来改变的性能或外观的图像。
在最简单的层次上，图像处理，可以通过改变的图像的像素的物理位置。
它可以通过扭转像素的图像的对称性按照一个对称位置。
如图2-1原图对称处理翻转处理图2-1它可以改变通过简单的翻译的图像的像素的位置。
如果所有像素均转向右，左，向上或向下，不改变整个图像将被翻转。
图2-2显示了20个像素的水平和垂直移位的结果。
水平移位可表示为图像2[X][Y]=图像1[X+△X] [Y]和垂直移位可表示成图像2[X] [Y]=图像1[X] [Y+ΔY]其中，Δx和Δy分别以像素为单位的水平和垂直的平移量。
由于翻转原始图像的某些部分将搬出来看，不提供在原始图像中的对应像素作为结果，得到的图像的一部分，而另一些是未知的未知留为空白（对应于像素值的零表示为黑色区域）。
同时可以采用垂直和水平移位。
图2-2可以被应用到图像的另一种变换是旋转。
在这种情况下，图像中的像素的位置是围绕目标确定的旋转角度的原点。
一般被选择的图像的中心为原点，与给出的图像分别旋转。
图2-3表示沿逆时针方向旋转60度的结果。
在翻转时，原始影像的某些部分可能会丢失，而一些空白区域出现在所产生的图象。
需要注意的是由于变换的特征，旋转可能需要插值的像素值。
图2-3算术图像处理虽然基本的图像处理改变图像像素的位置，即像素的图像，并将其移动到另一个位置，操纵图像的另一种方式是进行算术运算图像像素。
digital signal processing第二章答案
SOLUTIONS MANUALto accompanyDigital Signal Processing: A Computer-Based ApproachThird EditionSanjit K. MitraPrepared byChowdary Adsumilli, John Berger, Marco Carli, Hsin-Han Ho, Rajeev Gandhi, Chin Kaye Koh, Luca Lucchese, and Mylene Queiroz de Fariasw ww.k h da w.c omChapter 22.1 (a) ,81.4,1396.9,85.2212111===∞x x x(b) .48.3,1944.7,68.1822212===∞x x x2.2 Hence, Thus, ⎩⎨⎧<≥=µ.0,0,0,1][n n n ⎩⎨⎧≥<=−−µ.0,0,0,1]1[n n n ].1[][][−−µ+µ=n n n x 2.3 (a) Consider the sequence defined by If n < 0, then k = 0 is not includedin the sum and hence, x [n ] = 0 for n < 0. On the other hand, for k = 0 is included in the sum, and as a result, x [n ] =1 for Therefore,.][][∑δ=−∞=nk k n x ,0≥n .0≥n ].[,0,0,0,1][][n n n k n x n k µ=⎩⎨⎧<≥=∑δ=−∞=(b) Since it follows that Hence,⎩⎨⎧<≥=µ,0,0,0,1][n n n ⎩⎨⎧<≥=−µ.1,0,1,1]1[n n n ].[,0,0,0,1]1[][n n n n n δ=⎩⎨⎧≠==−µ−µ2.4 Recall ].[]1[][n n n δ=−µ−µ Hence,]3[4]2[2]1[3][][−δ+−δ−−δ+δ=n n n n n x ])4[]3[(4])3[]2[(2])2[]1[(3])1[][(−µ−−µ+−µ−−µ−−µ−−µ+−µ−µ=n n n n n n n n ].4[4]3[6]2[5]1[2][−µ−−µ+−µ−−µ+µ=n n n n n2.5 (a) },4512302{]2[][−−−=+−=↑n x n c(b) },006310872{]3[][↑−−−=−−=n y n d(c) },003221025{][][↑−−=−=n w n e(d) },27801332154{]2[][][−−−−=−+=↑n y n x n u(e) },040342150{]4[][][−−=+⋅=↑n w n x n v(f) },221000543{]4[][][−−−=+−=↑n w n y n s(g) }.75.248.205.35.1021{][5.3][−−−==↑n y n r2.6 (a) ],3[2]1[3][2]1[]2[5]3[4][−δ+−δ−δ−+δ++δ++δ−=n n n n n n n x],5[2]4[7]3[8]1[][3]1[6][−δ−−δ+−δ+−δ−δ−+δ=n n n n n n n y ],8[5]7[2]5[]4[2]3[2]2[3][−δ+−δ−−δ−−δ+−δ+−δ=n n n n n n n w(b) Recall ].1[][][−µ−µ=δn n n Hence,])[]1[(])1[]2[(5])2[]3[(4][n n n n n n n x µ−+µ++µ−+µ++µ−+µ−= ])4[]3[(2])2[]1[(3])1[][(2−µ−−µ+−µ−−µ−−µ−µ−n n n n n nw ww.k h da w.c om],4[2]3[2]2[3]1[][3]1[4]2[9]3[4−µ−−µ+−µ+−µ−µ−+µ−+µ++µ−=n n n n n n n n2.7 (a)x [n ]y [n ]From the above figure it follows that ].2[]2[]1[]1[][]0[][−+−+=n x h n x h n x h n y(b)x [n ]y [n ]From the above figure we get ])2[]1[][](0[][2111−β+−β+=n x n x n x h n w and].2[]1[][][2212−β+−β+=n w n w n w n y Making use of the first equation in the second we arrive at])2[]1[][](0[][2111−β+−β+=n x n x n x h n y])3[]2[]1[](0[211112−β+−β+−β+n x n x n x h])4[]3[]2[](0[211122−β+−β+−β+n x n x n x h ]2[)(]1[)(][]0[221112211211(−β+ββ+β+−β+β+=n x n x n x h.]4[]3[)()212211222112−ββ+−ββ+ββ+n x n x(c) Figure P2.1(c) is a cascade of a first-order section and a second-order section. The input-output relation remains unchanged if the ordering of the two sections is interchanged as shown below.y [n ]x [n ]w ww.k h domThe second-order section can be redrawn as shown below without changing its input-output relation.x [n ]y [n ]ordering we finally arrive at the structure shown below:x [n ]y [n ]x [x [Analyzing the above structure we arrive at],2[2.0]1[3.0][6.0][−+−+=n x n x n x n s ],2[5.0]1[8.0][][−−−−=n u n u n s n u].[4.0][]1[n y n u n y +=+From Substituting this in the second equation we get after some algebra ].[4.0]1[][n y n y n u −+=].2[8.0]1[18.0][4.0][]1[−+−−−=+n y n y n y n s n y Making use of the first equation in this equation we finally arrive at the desired input-output relation].3[2.0]2[3.0]1[6.0]3[2.0]2[18.0]1[4.0][−+−+−=−−−+−+n x n x n x n y n y n y n y(d) Figure P2.19(d) is a parallel connection of a first-order section and a second-order section. The second-order section can be redrawn as a cascade of two sections as indicated below:x [n ]y [n ]2w ww.k h dInterchanging the order of the two sections we arrive at an equivalent structure shown below:x [n y [n ]2 1]__],2[2.0]1[3.0][−+−=n x n x n q].2[5.0]1[8.0][][222−−−−=ny n y n q n ySubstituting the first equation in the second we have].2[2.0]1[3.0]2[5.0]1[8.0][222−+−=−+−+n x n x n y n y n y (2-1)Analyzing the first-order section of Figure P2.1(d) given belowx [n y [n ]1we get],1[4.0][][−+=n u n x n u].1[6.0][1−=n u n ySolving the above two equations we have].1[6.0]1[4.0][11−=−−n x n y n y (2-2)The output of the structure of Figure P2.19(d) is given by][n y].[][][21n y n y n y += (2-3) From Eq. (2-2) we get ]2[48.0]2[32.0]1[8.011−=−−−n x n y n y and].3[3.0]3[2.0]2[5.011−=−−−n x n y n y Adding the last two equations to Eq. (2-2) we arrive at ]3[2.0]2[18.0]1[4.0][1111−−−+−+n y n y n y n y].3[3.0]2[48.0]1[6.0−+−+−=n x n x n x(2-4) Similarly, from Eq. (2-1) we get].3[08.0]2[12.0]3[2.0]2[32.0]1[4.0222−−−−=−−−−−−n x n x n y n y n y Adding this equation to Eq. (2-1) we arrive at]3[2.0]2[18.0]1[4.0][2222−−−+−+n y n y n y n y ].3[08.0]2[08.0]1[3.0−−−+−=n x n x n x(2-5) Adding Eqs. (2-4) and (2-5), and making use of Eq. (2-3) we finally arrive at the input-output relation of Figure P2.1(d) as:].3[22.0]2[56.0]1[9.0]3[2.0]2[18.0]1[4.0][−+−+−=−−−+−+n x n x n x n y n y n y n yw ww.k h w.c m2.8 (a) },137235241{][*1j j j j j n x −−−−+−−−=↑Therefore }.415223371{][*1j j j j j n x −−−+−−−−=−↑()},....{][][][*,51543545111211j j j j n x n x n x cs −−−+−=−+=↑()}.....{][][][*,521452245252111211j j j j j n x n x n x ca +−+−−++=−−=↑(b) Hence, and thus, Therefore, .][3/2n j e n x π=3/*2][n j e n x π−=].[][23/*2n x e n x n j ==−π()],[][][][/*,n x e n x n x n x n j cs 23222212==−+=π and().][][][*,022212=−−=n x n x n x ca(c) Hence, and thus, Therefore, .][5/3n j e j n x π−=5/*3][n j e j n x π−=].[][35/*3n x e j n x n j −=−=−π−(),][][][*,033213=−+=n x n x n x cs and().][][][][/*,5333213n j ca e j n x n x n x n x π−==−−=2.9 (a) }.2032154{][−−−=↑n x Hence, }.4512302{][−−−=−↑n x Therefore, }2524252{])[][(][2121−−−−−=−+=↑n x n x n x ev}15.21215.21{−−−−−=↑and }6540456{])[][(][2121−−−=−−=↑n x n x n x od}.35.22025.23{−−−=↑(b) }.27801360000{][−−−=↑n y Hence,}.00006310872{][↑−−−=−n yTherefore, }15.3405.235.2045.31{])[][(][21−−−=−+=↑n y n y n y evand }.15.3405.305.3045.31{])[][(][21−−−−=−−=↑n y n y n y od(c) }.52012230000000000{][−−=↑n w Hence,}.00000000003221025{][↑−−=−n w Therefore])[][(][21n w n w n w ev −+=w ww.k h da w.c om}5.2105.0115.10005.1115.0015.2{−−−−=↑and])[][(][21n w n w n w od −−=}.5.2105.0115.10005.1115.0015.2{−−−−−−=↑2.10 (a)Hence, ].2[][1+µ=n n x ].2[][1+−µ=−n n x Therefore, ⎪⎩⎪⎨⎧≤−≤≤−≥=+−µ++µ=,3,2/1,22,1,3,2/1])2[]2[(][21,1n n n n n n x ev and⎪⎩⎪⎨⎧≤−−≤≤−≥=+−µ−+µ=.3,2/1,22,0,3,2/1])2[]2[(][21,1n n n n n n x od(b) Hence, Therefore,].3[][2−µα=n n x n ].3[][2−−µα=−−n n x n ()⎪⎪⎩⎪⎪⎨⎧≤−α≤≤−≥α=−−µα+−µα=−−,3,,22,0,3,]3[]3[][212121,2n n n n n n x n n n n ev and()⎪⎪⎩⎪⎪⎨⎧≤−α≤≤−≥α=−−µα−−µα=−−−.3,,22,0,3,]3[]3[][212121,2n n n n n n x n n n n od(c) Hence, Therefore,].[][3n n n x n µα=].[][3n n n x n −µα−=−−()n n n ev n n n n n n x α=−µα−+µα=−21][)(][][21,3 and().][)(][][2121,3n n n od n n n n n n x α=−µα−−µα=−(d) .][4nn x α= Hence, ].[][44n x n x nn=α=α=−− Therefore,nev n x n x n x n x n x n x α==+=−+=][])[][(])[][(][444214421,4 and.0])[][(])[][(][44214421,4=−=−−=n x n x n x n x n x od2.11 ().][][][21n x n x n x ev −+= Thus, ()].[][][][21n x n x n x n x ev ev =+−=− Hence, is an even sequence. Likewise, ][n x ev ().][][][21n x n x n x od −−=Thus,()].[][][][21n x n x n x n x od od −=−−=− Hence, is an odd sequence.][n x odw ww.k h da w.c om2.12 (a) Thus, ].[][][n x n x n g ev ev =].[][][][][][n g n x n x n x n x n g ev ev ev ev ==−−=− Hence,is an even sequence. ][n g(b) Thus, ].[][][n x n x n u od ev =()].[][][][][][n u n x n x n x n x n u od ev od ev −=−=−−=− Hence, is an odd sequence. ][n u(c) Thus, ].[][][n x n x n v od od =()()][][][][][n x n x n x n x n v od od od od −−=−−=− Hence, is an even sequence. ].[][][n v n x n x od od ==][n v2.13 (a) Since is causal, ][n x .0,0][<=n n x Also, .0,0][>=−n n x Now,().][][][21n x n x n x ev −+= Hence, ()]0[]0[]0[]0[21x x x x ev =+= and.0],[][21>=n n x n x ev Combining the two equations we get⎪⎩⎪⎨⎧<=>=.0,0,0],[,0],[2][n n n x n n x n x ev ev Likewise, ().][][][21n x n x n x od −−= Hence, ()0]0[]0[]0[21=−=x x x od and.0],[][21>=n n x n x od Combining the two equations we get ⎩⎨⎧≤>=.0,0,0],[2][n n n x n x ev(b) Since is causal, ][n y .0,0][<=n n y Also, .0,0][>=−n n y Letwhere and are real causal sequences.],[][][n jy n y n y im re +=][n y re ][n y im Now, ().][][][21n y n y n y ca −−=∗ Hence, ()]0[]0[]0[]0[21im ca jy y y y =−=∗ and.0],[][21>=n n n y y ca Since is not known, cannot be fully recovered from.]0[re y ][n y ][n y ca Likewise,().][][][21n y n y n y cs −+=∗ Hence, ()]0[]0[]0[]0[21re cs y y y y =+=∗ and.0],[][21>=n n n y y cs Since is not known, cannot be fully recovered from.]0[im y ][n y ][n y cs2.14 Since is causal, ][n x .0,0][<=n n x From the solution of Problem 2.13 we have].[][)cos(2,0,0,01,0),cos(2,0,0,0],[,0],[2][n n n n n n n n n n x n n x n x o o ev ev δ−µω=⎪⎩⎪⎨⎧<=>ω=⎪⎩⎪⎨⎧<=>=2.15 (a) where }{]}[{n A n x α=A and α are complex numbers with .1<α Since fornn α<,0 can become arbitrarily large, is not a bounded sequence.]}[{n xw ww.k h da w.c om(b) where ⎩⎨⎧<≥α=µα=,0,0,0,][][n n A n A n y nn A and α are complex numbers with.1<α Here, .0,1≥≤αn nHence A n y ≤][ for all values of Hence, is abounded sequence..n ]}[{n y(c) where and ][]}[{n C n h n µβ=C β are complex numbers with .1>β Since fornn β>,0 can become arbitrarily large, is not a bounded sequence. ]}[{n h(d) Since ).cos(4]}[{n n g o ω=4][≤n g for all values of is a boundedsequence.]}[{,n g n (e) ⎪⎩⎪⎨⎧≤≥⎟⎠⎞⎜⎝⎛−=.0,0,1,1][21n n n v n Since 121<n for and 1>n 121=n for ,1=n 1][<n v for all values of Thus is a bounded sequence..n ]}[{n v2.16 ].1[)1(][1−µ−=+n n n x n Now .1)1(][111∞=∑=∑−=∑∞=∞=+∞−∞=n n n n nn n x Hence is not absolutely summable. ]}[{n x2.17 (a) Now ].1[][1−µα=n n x n∞<α−α=∑α=∑α=∑∞=∞=∞−∞=1][112n n n n n n x , since .1<α Hence, is absolutely summable. ]}[{1n x(b) Now ].1[][2−µα=n n x n∑α=∑α=∑∞=∞=∞−∞=112][n nn nn n n n x ,)1(2∞<α−α=since.12<αHence, is absolutely summable.]}[{2n x(c) Now ].1[][23−µα=n n n x nnn n nn n n n x α∑=∑α=∑∞=∞=∞−∞=12123][K +α+α+α+α=423222432)(5)(3)(543432432K K K +α+α+α++α+α+α++α+α+α+α=w ww.k h da w.c om)(7654K +α+α+α+ =K +α−α+α−α+α−α+α−α1715131432⎟⎟⎠⎞⎜⎜⎝⎛∑α−∑αα−=⎟⎟⎠⎞⎜⎜⎝⎛∑α−α−=∞=∞=∞=111211)12(11n n n n n n n n ⎟⎟⎠⎞⎜⎜⎝⎛α−α−α−αα−=1)1(2112 .)1()1(3∞<α−α+α=Hence, is absolutely summable.]}[{3n x 2.18 (a) ].[21][n n x na µ=Now .2112121][2100∞<=−=∑=∑=∑∞=∞=∞−∞=n n n n n a n x Hence, is absolutely summable. ]}[{n x a(b) ].[)2)(1(1][n n n n x b µ++= Now ∑++=∑∞=∞−∞=0)2)(1(1][n n b n n n x.151414131312121121110∞<=+⎟⎠⎞⎜⎝⎛−+⎟⎠⎞⎜⎝⎛−+⎟⎠⎞⎜⎝⎛−+⎟⎠⎞⎜⎝⎛−=∑⎟⎠⎞⎜⎝⎛+−+=∞=K n n n Hence,is absolutely summable.]}[{n x b2.19 (a) A sequence is absolutely summable if ][n x .][∞<∑∞−∞=n n x By Schwartz inequalitywe have .][][][2∞<⎟⎟⎠⎞⎜⎜⎝⎛∑⎟⎟⎠⎞⎜⎜⎝⎛∑≤∑∞−∞=∞−∞=∞−∞=n n n n x n x n x Hence, an absolutely summablesequence is square summable and has thus finite energy.Now consider the sequence ].1[][1−µ=n n x nThe convergence of an infinite series can be shown via the integral test. Let ),(x f a n = where a continuous, positive anddecreasing function is for all Then the series and the integralboth converge or both diverge. For .1≥x ∑∞=1n n a ∫∞1)(dx x f .)(,11x nn x f a == But ∞=−∞==∫∞∞0)(ln 111x dx x. Hence, ∑=∑∞=∞−∞=11][n nn n x does not converge. As a result, ]1[][1−µ=n n x n is notabsolutely summable.w ww.k h da w.c om(b) To show that is square-summable, we observe that here ]}[{n x 21nn a =, and thus,.)(21x x f = Now, .11111112=+∞−=⎟⎠⎞⎜⎝⎛−=∫∞∞x dx xHence, ∑∞=112n n converges, or in otherwords, ]1[][1−µ=n n x nis square-summable.2.20 See Problem 2.19, Part (a) solution.2.21 ].1[cos ][2−µπω=n n n n x c Now, .1cos ][1221222∑π≤∑⎟⎠⎞⎜⎝⎛πω=∑∞=∞=∞−∞=n n c n nn n n x Since,,62112π=∑∞=n n.61cos 12≤∑⎟⎠⎞⎜⎝⎛πω∞=n c n n Therefore is square-summable. ][2n xUsing the integral test (See Problem 2.19, Part (a) solution) we now show that isnot absolutely summable. Now, ][2n x ∞∞ω⋅ωπω⋅π=∫πω11)int(cos cos cos 1cos x x x xdx x x c c c cwhere is the cosine integral function. Sinceint cos dx xxc ∫πω∞1cos diverges, ∑πω∞=1cos n c n nalso diverges. Hence, is not absolutely summable.][2n x2.22 ()∑+=∑=∞−∞=+∞→−=+∞→K od ev K K K K n K K x n x n x n x 21212121][][lim ][lim P()∑++=−=+∞→K Kn od ev od ev K K n x n x n x n x ][][2][][lim 22121()(][][][][lim 12121n x n x n x n x K Kn K K x x od ev −−∑−+++=−=+∞→P P )od ev n K Kn od ev x x n x n x K K x x P P P P +=++=⎟⎟⎠⎞⎜⎜⎝⎛∑−−∑⋅+∞→∞−∞=−=][][2112122limas Now for the given sequence,∑−∑=−=−=KKn K Kn n x n x ].[][22∑+∞→=+∞→−=+∞→=⎟⎠⎞⎜⎝⎛=∑⎟⎠⎞⎜⎝⎛=∑=K n od K K K n K K K K n od K K x n x 0112163106311212121limlim ][lim P .lim63121121631⎟⎠⎞⎜⎝⎛⎟⎠⎞⎜⎝⎛==++∞→K K K Hence, .1063121⎟⎠⎞⎜⎝⎛−=−=od ev x x x P P Pw ww.k h da w.c om2.23 .10),/2sin(][−≤≤π=N n N kn n x Now )/2(sin ][102102N kn n x N n N n x π∑=∑=−=−=E().)/4cos()/4cos(11212121∑π−=∑π−=−=−=N n N N n N kn N kn Let and∑π=−=10)/4cos(N n N kn C Then .)/4sin(10∑π=−=N n N kn S .011/4410/4=−−=∑=+π−π−−=π−Nk j knj N n N kn j ee e jS C This implies Hence .0=C .2N x =E2.24 (a) Then ].[][n A n x µ=α.1][22222α−=∑α=∑=∞=∞−∞=A A n x n nn a x a E (b) ].1[][21−µ=n n x nb Then ∑=∑=∑∞=∞=∞−∞=111122][n nn nn b x n x b E .904π= 2.25(a) Then average power .)1(][1n n x −=,1)12(121lim][lim 211211=++=∑=∞→−=+∞→K K n x K K K n K K x P and energy .1][211∞=∑=∑=∞−∞=∞−∞=n n x n x E(b) Then average power].[][2n n x µ=,21121lim1121lim ][121lim 0222=++=∑+=∑+=∞→=∞→−=∞→K K K n x K K K n K K K n K x P and energy .1][0222∞=∑=∑=∞=∞−∞=n n x n x E(c) Then average power].[][3n n n x µ=,6)12)(1(lim 121lim ][121lim12233∞=++=∑+=∑+=∞→=∞→−=∞→K K K n K n x K K K n K K K n K x P and energy .][02233∞=∑=∑=∞=∞−∞=n n x n n x E(d) Then average power .][004nj eA n x ω=∑+=−=∞→K Kn K x n x K P 24][121lim 4w ww.k h da w.c om,)12(121lim 121lim 121lim 202020200A K A K A K e A K K K K n K K K n n j K =+⋅+=∑+=∑+=∞→−=∞→−=ω∞→ and energy .][20202303∞=∑=∑=∑=∞−∞=∞−∞=ω∞−∞=n n nj n x A eA n x E(e) .cos ][25⎟⎠⎞⎜⎝⎛φ+=πM nA n x Note is a periodic sequence. Then average power ][5n x .1cos 21cos 1][110242102210255∑⎟⎠⎞⎜⎝⎛+⎟⎠⎞⎜⎝⎛⋅=∑⎟⎠⎞⎜⎝⎛φ+=∑=−=φ+π−=π−=M n Mn M n M n M n x A M A M n x M PLet ∑⎟⎠⎞⎜⎝⎛φ+=−=1042cos M n M πn C and .2cos 104∑⎟⎠⎞⎜⎝⎛φ+=−=M n M πn S Then .011/44210/42124=−−⋅=∑=∑=+ππφ−=πφ−=⎟⎠⎞⎜⎝⎛φ+πMj j j M n Mn j j M n j ee eeee jS C M nHence Therefore .0=C .212121025A A M P M n x =∑⋅=−= Since is a periodic sequence, it has infinite energy.][5n x2.26 In each of the following parts, denotes the fundamental period and N r is a positiveinteger.(a) Here and ).5/2cos(4][~1n n x π=N r must satisfy the relation .252r N π=⋅πAmong all positive solutions for N and r , the smallest values are and 5=N .1=r Hence the average power is given by.8cos 451][~12405210211=∑⎟⎠⎞⎜⎝⎛=∑==π−=n n N n x n x N P (b) Here and ).5/3cos(3][~2n n x π=N r must satisfy the relation .253r N π=⋅πAmong all positive solutions for N and r , the smallest values are and 10=N .3=r Hence the average power is given by.5.4cos 3101][~12905310222=∑⎟⎠⎞⎜⎝⎛=∑==π−=n n N n x n x N P (c) Here and ).7/3cos(2][~3n n x π=N r must satisfy the relation .273r N π=⋅πAmong all positive solutions for and N r , the smallest values are and 14=N .3=r Hence the average power is given by.2cos 2141][~121307310233=∑⎟⎠⎞⎜⎝⎛=∑==π−=n n N n x n x N P w ww.k h da w.c om(d) Here and ).3/5cos(4][~4n n x π=N r must satisfy the relation .235r N π=⋅πAmong all positive solutions for and N r , the smallest values are and 6=N .5=r Hence the average power is given by.8cos 461][~12503510244=∑⎟⎠⎞⎜⎝⎛=∑==π−=n n N n x n x N P (e) ).5/3cos(3)5/2cos(4][~5n n n x π+π= We first determine the fundamental periodof Here and 1N ).5/2cos(n π1N r must satisfy the relation.2152r N π=⋅πAmong all positive solutions for and , the smallest values are 1N r 51=N and We next determine the fundamental period of .1=r 2N ).5/3cos(n π Here and r must satisfy therelation2N .2253r N π=⋅πAmong all positive solutions for and 2N r , the smallest valuesare and The fundamental period of is then given by 102=N .3=r ][~5n x .10)10,5(),(21==LCM N N LCMHence the average power is given by24053521025cos 3cos 4101][~15∑⎟⎠⎞⎜⎝⎛+⎟⎠⎞⎜⎝⎛=∑==ππ−=n n n N n x n x N P .5.1205.48cos cos 24cos 9cos 161011105352531102521102=++≅⎟⎟⎠⎞⎜⎜⎝⎛∑⎟⎠⎞⎜⎝⎛⎟⎠⎞⎜⎝⎛+⎟⎠⎞⎜⎝⎛∑+⎟⎠⎞⎜⎝⎛∑==πππ=π=n n n n n n n(f) ).5/3cos(3)3/5cos(4][~6n n n x π+π= We first determine the fundamental period of Here and r must satisfy the relation1N ).3/5cos(n π1N .2135r N π=⋅πAmong allpositive solutions for and , the smallest values are 1N r 61=N and We next determine the fundamental period of .5=r 2N ).5/3cos(n π Here and r must satisfy therelation2N .2253r N π=⋅πAmong all positive solutions for and r , the smallest valuesare and The fundamental period of is then given by 2N 102=N .3=r ][~6n x .30)10,6(),(21==LCM N N LCM Hence the average power is given by229053351026cos 3cos 4301][~16∑⎟⎠⎞⎜⎝⎛+⎟⎠⎞⎜⎝⎛=∑==ππ−=n n n N n x n x N P .5.1205.48cos cos 24cos 9cos 163012905335533002352902=++≅⎟⎟⎠⎞⎜⎜⎝⎛∑⎟⎠⎞⎜⎝⎛⎟⎠⎞⎜⎝⎛+⎟⎠⎞⎜⎝⎛∑+⎟⎠⎞⎜⎝⎛∑==πππ=π=n n n n n n n2.27 Now , from Eq. (2.38) we have Therefore.][][~∑+=∞−∞=k kN n x n y w ww.k h da w.c om.][][~∑++=+∞−∞=k N kN n x N n y Substituting 1+=k r we get].[~][][~n y rN n x N n y r =∑+=+∞−∞= Hence is a periodic sequence with a period][~n y .N 2.28 (a) Now The portion of in the range .5=N .]5[][~∑+=∞−∞=n p k n x n x ][~n x p 40≤≤n is given by }15400{]5[][]5[−=+++−n x n x n x.40},17432{}00000{}02032{≤≤−−−=+−−+nHence, one period of is given by ][~n x p .40},17432{≤≤−−−n Now The portion of in the range is given by .]5[][~∑+=∞−∞=n p k n y n y ][~n y p 40≤≤n }60000{]5[][]5[=+++−n y n y n y.40},138015{}00002{}78013{≤≤−−=−+−−+nHence, one period of is given by ][~n y p .40},138015{≤≤−−n Now The portion of in the range is given by .]5[][~∑+=∞−∞=n p k n w n w ][~n w p40≤≤n }00000{]5[][]5[=+++−n w n w n w.40},27101{}05201{}22300{≤≤−=−−++nHence, one period of is given by ][~n w p.40},27101{≤≤−n (b) .7=N Now The portion of in the range .]7[][~∑+=∞−∞=n p k n x n x ][~n x p 60≤≤n is given by }1540000{]7[][]7[−=+++−n x n x n x}0000000{}0002032{+−−+.60},1542032{≤≤−−−=n Hence, one period of is given by][~n x p .60},1542032{≤≤−−−nNow The portion of in the range is given by .]7[][~∑+=∞−∞=n p k n y n y ][~n y p 60≤≤n }6000000{]7[][]7[=+++−n x n x n x}0000000{}0278013{+−−−+.60},6278013{≤≤−−−=n Hence, one period of is given by][~n y p .60},6278013{≤≤−−−nNow The portion of in the range is given by .]7[][~∑+=∞−∞=n p k n w n w ][~n w p60≤≤n }0000000{]7[][]7[=+++−n w n w n w}0000052{}0122300{−+−+w ww.k h da w.com.60},0122352{≤≤−−=n Hence, one period of is given by ][~n w p.60},0122352{≤≤−−n 2.29).cos(][~φ+ω=n A n x o(a) }.11111111{][~−−−−=n x Hence .4/,2/,2π=φπ=ω=o A (b) }.30303030{][~−−=n x Hence ,3=A ,2/π=ωo.2/π=φ (c) }.366.1366.01366.1366.01{][~−−−=n x Hence ,2=A ,3/π=ωo.4/π=φ (d) }.02020202{][~−−=n x Hence .0,2/,2=φπ=ω=o A2.30 The fundamental period of a periodic sequence with an angular frequency N o ω satisfies Eq. (2.47a) with the smallest value of and .N r(a) Here Eq. (2.47a) reduces to .5.0π=ωo r N π=π25.0 which is satisfied with.1,4==r N (b) Here Eq. (2.47a) reduces to .8.0π=ωo r N π=π28.0 which is satisfied with.2,5==r N (c) We first determine the fundamental period of In this case, Eq. (2.47a) reduces to 1N ).2.0cos(}Re{5/n e n j π=π1122.0r N π=π which is satisfied with .1,1011==r NWe next determine the fundamental period of In this case, Eq. (2.47a) reduces to 2N ).1.0sin(Im{10/n j e n j π=π2221.0r N π=π which is satisfied with .1,2022==r N Hence the fundamental period of is given byN ][~n x c .20)20,10(),(21==LCM N N LCM(d) We first determine the fundamental period of In this case, Eq. (2.47a) reduces to 1N ).3.1cos(3n π1123.1r N π=π which is satisfied with .13,2011==r N We next determine the fundamental period of 2N ).5.05.0sin(4π+πn In this case, Eq. (2.47a) reduces to which is satisfied with 2225.0r N π=π.1,422==r N Hence thefundamental period of is given by N ][~4n x .20)4,20(),(21==LCM N N LCM (e) We first determine the fundamental period of In this case, Eq. (2.47a) reduces to 1N ).75.05.1cos(5π+πn 1125.1r N π=π which is satisfied with .3,411==r N We next determine the fundamental period of 2N ).6.0cos(4n π In this case, Eq. (2.47a) reduces to which is satisfied with 2226.0r N π=π.3,1022==r N We finallydetermine the fundamental period of 3N ).5.0sin(n π In this case, Eq. (2.47a) reduces to which is satisfied with 3325.0r N π=π.1,433==r N Hence the fundamental periodN of is given by ][~5n x .20)4,10,4(),,(321==LCM N N N LCM 2.31The fundamental period N of a periodic sequence with an angular frequency o ω satisfies Eq. (2.47a) with the smallest value of N and .r w ww.k h da w.c om(a) Here Eq. (2.47a) reduces to .6.0π=ωo r N π=π26.0 which is satisfied with.3,10==r N (b) Here Eq. (2.47a) reduces to .28.0π=ωo r N π=π228.0 which is satisfied with.7,50==r N (c) Here Eq. (2.47a) reduces to .45.0π=ωo r N π=π245.0 which is satisfied with.9,40==r N (d) Here Eq. (2.47a) reduces to .55.0π=ωo r N π=π255.0 which is satisfied with.11,40==r N (e) Here Eq. (2.47a) reduces to .65.0π=ωo r N π=π265.0 which is satisfied with.13,40==r N 2.32Here Eq. (2.47a) reduces to .08.0π=ωo r N π=π208.0 which is satisfied withFor a sequence .1,25==r N )sin(][~22n n x ω= with a fundamental period of 25=N , Eq. (2.47a) reduces to .2252r π=ω For example, for 2=r we haveAnother sequence with the same fundamental period is obtained by setting which leads to .16.025/42π=π=ω3=r .24.025/63π=π=ω The corresponding periodicsequences are therefore )16.0sin(][~2n n x π= and ).24.0sin(][~3n n x π=2.33 The three parameters and ,,o A Ωφ of the continuous-time signal can bedetermined from )(t x a )cos()(][φ+Ω==nT A nT x n x o a by setting 3 distinct values of For example .n,cos ]0[α=φ=A x,sin )sin(cos )cos()cos(]1[β=φΩ+φΩ=φ+Ω−=−T A T A T A x o o o ,.sin )sin(cos )cos()cos(]1[γ=φΩ−φΩ=φ+Ω=T A T A T A x o o oSubstituting the first equation into the last two equations and then adding them we getαγ+β=Ω2)cos(T o which can be solved to determine o Ω. Next, from the second equation we have ).cos(cos )cos(sin T T A A o o Ωα−β=φΩ−β=φ Dividing thisequation by the last equation on the previous page we arrive at TT o o ΩαΩα−β=φsin()cos(tanwhich can be solved to determine .φ Finally, the parameter is determined from the first equation of the last page.w ww.k h da w.c omNow consider the case .22o T TΩ=π=Ω In this case β=φ+π=)cos(][n A n x and ()β=φ+π=φ+π+=+)cos()1(cos ]1[n A n A n x . Since all sample values are equal, the three parameters cannot be determined uniquely.Finally consider the case .22o T TΩ<π=Ω In this case )cos(][φ+Ω=nT A n x oimplying )cos(φ+ω=n A o .π>Ω=ωT o o As explained in Section 2.2.1, a digitalsinusoidal sequence with an angular frequency o ω greater than assumes the identity of a sinusoidal sequence with an angular frequency in the range . Hence, cannot be uniquely determined from π.0π<ω≤o Ω)cos(][φ+Ω=nT A n x o .2.34If is periodic with a period , then).cos(][nT n x o Ω=][n x N ()).cos(][cos ][000nT n x NT nT N n x Ω==Ω+Ω=+ This implies r NT o π=Ω2with r any nonzero positive integer. Hence the sampling rate must satisfy the relationIf i.e., ./2N r T o Ωπ=,20=Ωo ,8/π=T then we must have r N π=π⋅2820. Thesmallest value of and r satisfying this relation are N 4=N and The fundamental period is thus .5=r 4=N .2.35(a) For an input the output is,2,1],[=i n x i .2,1],2[]1[]2[]1[][][21210=−+−+−+−+=i n y a n y a n x b n x b n x b n y i i i i i i Then, for an input the output is ],[][][21n Bx n Ax n x +=])[][(][210n Bx n Ax b n y +=])1[]1[(])2[]2[(])1[]1[(211212211−+−+−+−+−+−+n By n Ay a n Bx n Ax b n Bx n Ax b ])2[]2[(211−+−+n By n Ay a ]1[]2[]1[][(11121110−+−+−+=n y a n x b n x b n x b A ])2[]1[]2[]1[][(])1[222123222122−+−+−+−++−+n x a n x a n x b n x b n x b B n y a ].[][21n By n Ay += Hence, the system of Eq. (2.18) is linear.(b) For an input the output is ,2,1],[=i n x i ⎩⎨⎧±±==otherwise.,0,2,,0],/[][LL L n L n x n y ii For an input ],[][][21n Bx n Ax n x += the output for K ,2,,0L L n ±±= is][][]/[]/[]/[][2121n By n Ay L n Bx L n Ax L n x n y +=+==. For all other values of Hence the system of Eq. (2.20) is linear. .000][,=⋅+⋅=B A n y n(c) For an input the output is ,2,1],[=i n x i .2,1],/[][==i M n x n y i i Then, for an inputthe output is ],[][][21n Bx n Ax n x +=].[][]/[]/[][2121n By n Ay M n Bx M n Ax n y +=+= Hence the system of Eq. (2.21) is linear.w ww.k h da w.c om(d) For an input the output is ,2,1],[=i n x i ∑=−=−=10.2,1],[1][M k i i i k n x M n y Then, foran input the output is ],[][][21n Bx n Ax n x +=()∑−+−=−=121][][1][M k k n Bx k n Ax M n y].[][][1][12110211n By n Ay k n x M B k n x MA M k M k +=⎟⎟⎠⎞⎜⎜⎝⎛∑−+⎟⎟⎠⎞⎜⎜⎝⎛∑−=−=−= Hence the system of Eq. (2.61) is linear.(e) The first term on the RHS of Eq. (2.65) is the output of a factor-of-2 up-sampler. The second term on the RHS of Eq. (2.65) is simply the output of an unit delayfollowed by a factor-of-2 up-sampler, whereas, the third term is the output of an unit advance operator followed by a factor-of-2 up-sampler. We have shown in Part (b) that the up-sampler is a linear system. Moreover, the unit delay and the unit advance operator are linear systems. A cascade of two linear systems is linear and the linear combination of linear systems is also linear. Hence, the factor-of-2 interpolator of Eq. (2.65) is a linear system.(f) Following the arguments given in Part (e), we can similarly show that the factor-of-3 interpolator of Eq. (2.66) is a linear system.2.36(a) For an input ].[][3n x n n y =,2,1],[=i n x i the output isThen, for an input .2,1],[][3==i n x n n y i i ],[][][21n Bx n Ax n x += the output is Hence the system is linear. ()][][][213n Bx n Ax n n y +=].[][21n By n Ay +=For an input the output is the impulse response As for and the system is causal.],[][n n x δ=].[][3n n n h δ=0][=n h ,0<n Let1 for all values of Then ][=n x .n 3][n n y = and as Since a bounded input results in an unbounded output, the system is not BIBO stable.∞→][n y .∞→n Finally, let and be the outputs for inputs and respectively. If ][n y ][1n y ][n x ],[1n xthen However, ][][1o n n x n x −=].[][][3131o n n x n n x n n y −===−][o n n ySince ].[)(3o o n n x n n −−],[][1o n n y n y −≠ the system is not time-invariant.(b) For an input .])[(][5n x n y =,2,1],[=i n x i the output is.2,1,])[(][5==i n x n y i i Then, for an input ],[][][21n Bx n Ax n x += the output isHence the system is nonlinear.()521][][][n Bx n Ax n y +=.])[(])[(5251n x B n x A +≠For an input the output is the impulse response As for and the system is causal.],[][n n x δ=.])[(][5n n h δ=0][=n h ,0<n w ww.k h da w.c om。
digital image processing 引用格式 -回复
digital image processing 引用格式-回复关于数字图像处理的引用格式数字图像处理是一门研究图像获取、处理、压缩和分析的学科。
在现代科技的发展中，数字图像处理已经成为了计算机视觉、图像识别和模式识别等领域中必不可少的一部分。
本文将介绍数字图像处理的引用格式，以帮助读者更好地引用相关研究工作。
1. 引言首先，引言部分是文章的开头，可以在其中介绍数字图像处理的背景和重要性。
可以引用一些相关的研究工作，如“根据[1]的研究，数字图像处理在医学影像分析中起着重要的作用。
”2. 文献综述文献综述部分可以对已有研究进行回顾和分析，以支持自己的研究工作。
在这一部分，应该引用已有研究的相关内容，例如：“Smith等人在[2]中提出了一种基于深度学习的数字图像处理方法，该方法在图像语义分割任务中取得了显著的性能提升。
”3. 方法部分在方法部分，应该详细介绍自己的研究方法，并引用已有工作以支持自己的方法选择。
例如：“为了对图像进行去噪处理，我们采用了基于小波变换的方法，该方法已被广泛研究和应用[3]。
”4. 实验部分在实验部分，应该详细介绍自己的实验设置和结果，并引用相关的实验结果以对比自己的工作。
例如：“与之前的工作相比，我们的方法在图像去噪任务上取得了更好的效果，与[4]的实验结果相吻合。
”5. 结论在结论部分，可以总结自己的研究工作，并提出一些建议和展望。
例如：“本研究结果表明，数字图像处理在图像质量增强和分析中具有巨大的潜力。
未来的研究可以考虑将深度学习方法应用于数字图像处理领域。
”6. 参考文献在参考文献部分，应该列出所有在文章中引用的文献，并按照特定的引用格式进行排版。
在数字图像处理领域，常用的引用格式包括IEEE格式和APA格式等。
例如：IEEE格式：[1] Author name(s), "Title of the article," Title of the Journal or Journal Abbreviation, vol.(issue), pp. page number, Month, Year. [2] Author name(s), "Title of the conference paper," Title of the Conference, pp. page number, Month, Year.APA格式：[1] Author name(s). (Year). Title of the article. Title of the Journal or Journal Abbreviation, vol.(issue), page number.[2] Author name(s). (Year). Title of the conference paper. In Title of the Conference (pp. page number).以上是关于数字图像处理的引用格式的一般步骤和示例。
数字图像处理英语课件chapter2
– If (x0,y0) =(xn, yn) , the path is a closed path.
• Define 4-, 8-, or m-paths (see example above)
© 2009 Lianxi Yuan
Digital Image Processing, 2nd ed.
© 2009 Lianxi Yuan
Digital Image Processing, 2nd ed.
Region
Let R be a subset of pixels in an image. • Region: if R is a connected set. • Two regions, Ri and Rj are said to be adjacent if their union forms a connected set. • Regions that are not adjacent are said to be disjoint • 4- and 8-adjacency are considered when referring to regions
数字图像处理DigitalImageProcessing
C
A
B
F
C
A
FB
D D
• 几何变换的应用举例
• 图像在生成过程中，由于系统本身具有非线性或拍摄角
度不同，会使生成的图像产生几何失真。几何失真一般 分为系统失真和非系统失真。系统失真是有规律的、能 预测的；非系统失真则是随机的。
• 但对图像作定量分析时，就要对失真的图像进行几何校
2.4 离散Gabor变换
2.4.1 加窗傅立叶变换 2.4.2 Gabor变换的基本概念 2.4.3 离散Gabor变换
2.5 小波变换
2.5.1 连续小波变换 2.5.2 二进小波变换 2.5.3 离散小波变换 2.5.4 二维离散小波变换 2.5.5 小波变换的应用
2.6 PCA变换
2.6.1 PCA的基本概念及问题描述 2.6.2 PCA变换的应用
a (x ,y ) cx b (x ,y ) dy
a(x, y) c 0 0x b(x, y) 0 d 0y 1 0 0 11
• 旋转
a(x,y)xcos()ysin() b(x,y)xsin()ycos()
a(x,y) cos() sin() 0x
• 1. 加法运算
C (x ,y ) A (x ,y ) B (x ,y )
• 2. 减法运算（差分）
C (x ,y ) A (x ,y ) B (x ,y )
=
+
=
—
（a）原图
（b）梯度运算
• 2.2.2 几何运算
• 几何运算可以改变图像中物体之间的空间关系。
这种运算可以看成是图像内的各物体在图像内移动 的过程。例如，物体的转动、扭曲、倾斜、拉伸等 等，都是几何运算的结果。
最新数字图像处理的应用-Digital-Image-Processing教学讲义ppt课件
12 of
数字水印技术的技术特性
36
自恢复性
由于经过一些操作或变换后，可能会使原图产生较大 的破坏，如果只从留下的片段数据，仍能恢复隐藏信 号，而且恢复过程不需要宿主信号，我们就说这样的 算法具有自恢复性
高通滤波
直方图均衡化
锐化处理
13 of 36
自恢复性
提取水印的算法
在水印提取时，选取相同的DCT系数 并根据系数之间的关系抽取特定信息。
算法分析
数据改变的幅度比较小，透明性好。 抵抗几何变换攻击的能力会比较弱。
18 of
数字水印的典型算法
36
压缩域算法
基于JPEG、MPEG标准。 水印信号的嵌入、提取、检测直接在压缩域数据中进 行。 节省了解码和重新编码的过程
输入图像
生成 查询要求
相似性 匹配
返回 初步结果
返回结果图像
特征调整
逐步筛选 、求精
25 of
基于内容的图像检索系统
36
基于内容的图像检索系统的4种检索方式
利用图片样本检索（Query By Example） • 可以由用户准备图片样本 • 可以在图片库中浏览
数字图像处理的应用-DigitalImage-Processing
2 of
数字图像处理的应用
36
本节课我们学习：
图像处理的主要应用领域 图像处理在数字水印中的应用 基于内容的图像检索
3 of
图像处理的主要应用领域
36
航天和航空技术方面的应用 生物医学工程方面的应用 通信工程方面的应用 工业和工程方面的应用 军事公安方面的应用 文化艺术方面的应用
但对于数字水印系统来说，隐藏信息的丢失，意味着 版权信息的丢失，从而失去了版权保护的功能，因此， 这样的系统是失败的
Digital Image Processing_2
Digital Image Processing Computer Project Report II Image EnhancementStudent ID: 20091613310032Name: Xiaopeng JiDate: March 19, 2012A. Objectivesi. Familiar with the general method of digital image enhancement;ii. Familiar with the command of the Matlab image enhancement;B. MethodsPractice the command of Matlab image enhancement andfamiliar with the following module functions.Image enhancement.histeq - histogram equalization.imadust - Adjust image intensity values or colormap.Image noising.imnoise - Add noise to an image.Image filteringmedfilt2 - Perform 2-D median filtering.ordfilt2 - Perform 2-D order-statistic filtering.wiener2 - Perform 2-D adaptive noise-removal filtering.C. Resultsa. Load cameraman.tif image from your hard disk (using function imread).At the command prompt enter:>>A=imread('cameraman.tif');Then we get the image data by matrix A.b. Show the image in a figure window.At the command prompt enter:>>figure;imshow(A);The command shows cameraman.tif in a windows and has always been used with function imread in the reportI.Figure 1:Load cameraman.tif imagec. Show the histogram of the image (using function imhist). At the command prompt enter:>>figure;imhist(A);Figure 2:The histogram of cameraman.tifFunctionimhist display histogram of image data. As the cameraman.tif is a grayscale image, function imhistuses 256 bins as a default value.d. Enhance the contrast of the image using histogram equalization.At the command prompt enter:>>I=histeq(A);figure;imshow(I);Figure 3:After histogram equalizationFrom the result compared with figure 1, the image isbrighter and not as good as before.It appears to lose some information.e. Show the histogram of the image after processing.At the command prompt enter:>>figure;imhist(I);Figure 4:Histogram of the image after equalizationFrom the result compared with figure 2, we can find the difference easily and what the function histeq is used to do.Histogram of the original image becomesevenly distributed in all gray-scale range from a concentratedgray-scale range. So we can find that the histogram of the image is sparser than before.f. Compare the qualities of two images and makes a discussion about them. Comparing figure 1 with figure 3, the image becomes brighter. By this way, we may enhance the image contrast to get more details that hidden in the original image. Comparing figure 2 with figure 4, we can find that the gray-scale rangebecome evenly distributed from a concentratedgray-scale range. Absolutely the image lost some information after using histogram equalization.g. Add noises, such as gaussian, salt&pepper, speckle noise into the image respectively. Compare with the influence of the different Means and Variance. At the command prompt enter:>>figure;subplot(2,2,1);imshow(A);title('Original Image');L=imnoise(A,'salt & pepper');subplot(2,2,2);imshow(L);title('Add salt & pepper noise');M=imnoise(A,'gaussian');subplot(2,2,3);imshow(M);title('Add gaussian noise');N=imnoise(A,'speckle');subplot(2,2,4);imshow(N);title('Add speckle noise');Figure 5:Add different noisesSalt & pepper noiseisa random"on and off" pixels.Salt & pepper noiseis added to the image Iby using imnoise(I,'salt & pepper',D), where D is the noise density and default value for D is 0.05.Gaussian noise is a noise with constantmean and variance. Gaussian noise of mean M andvariance V is added to the image I by using imnoise (I,'gaussian',M,V) . When unspecified, default value of M and V are 0 and0.01 respectively.Speckle noise is a Multiplicative noise. Speckle noise isaddedto the image I by using imnoise (I,'speckle',V). The default value for V is 0.04.The image is added with different noise of default value in the experiment. From the results, we can find that different noise have different effect. Such as Salt & pepper noise isa random"on and off" pixels.h. Remove the added noise from the image by function medfilt2, ordfilt2 and wiener2 respectively. Compare the qualities of the original images with the processed images and discuss the effect of the methods.In order to compare the effect of different filter, remove the different noise in order.First of all, remove the Salt & pepper noise from the image by different functions.At the command prompt enter:>>figure;subplot(2,2,1);imshow(L);title('Add salt & pepper noise');subplot(2,2,2);L1=medfilt2(L);imshow(L1);title('Remove the noise by medfilt2');subplot(2,2,3);L2=ordfilt2(L,5,ones(3,3));imshow(L2)title('Remove the noise by ordfilt2');subplot(2,2,4);L3=wiener2(L,[5,5]);imshow(L3)title('Remove the noise by wiener2');Figure 6:Remove the added salt & pepper noise by different functionsFrom the results, we can find that the function medfilt2 and ordfilt2 performbetter. The effect of function wiener2 is not well.Then,remove the Gaussian noise from the image by different functions.At the command prompt enter:>>figure;subplot(2,2,1);imshow(M);title('Add gaussian noise');subplot(2,2,2);M1=medfilt2(M);imshow(M1);title('Remove the noise by medfilt2');subplot(2,2,3);M2=ordfilt2(M,5,ones(3,3));imshow(M2)title('Remove the noise by ordfilt2');subplot(2,2,4);M3=wiener2(M,[5,5]);imshow(M3)title('Remove the noise by wiener2');Figure 7:Remove the added Gaussiannoise by different functionsFrom the results, we can find that the function medfilt2 and ordfilt2perform notwell. Even though the image usingthe function wiener2 is fuzzy compared with original image, wiener2almost remove all the added noise from the image. Finally, remove the speckle noise from the image by different functions.At the command prompt enter:>>figure;subplot(2,2,1);imshow(N);title('Add speckle noise');subplot(2,2,2);N1=medfilt2(N);imshow(N1);title('Remove the noise by medfilt2');subplot(2,2,3);N2=ordfilt2(N,5,ones(3,3));imshow(N2)title('Remove the noise by ordfilt2');subplot(2,2,4);N3=wiener2(N,[5,5]);imshow(N3)title('Remove the noise by wiener2');From the results, we can find that all the functionsperform not wellcompared with original image.The function wiener2 perform better than others.Figure 8:Remove the speckle noise by different functionsFrom the above, we can find that different filters have different effects on special noise. So we can try different methods to remove the added noise in the experiment.D.ConclusionsImage enhancement is the improvement of digital image quality. It can enhance the useful information in the image and may be a distortion. Such as enhance the contrast of the image using histogram equalization will change the gray-scale and get some useful details. At the same time, the image will lose some details.Image processing is often accompanied by some noise. Such as Gaussian noise. Different noises often have different effect on the images. So we should different filters to remove the added noise. Different noises often have different feature. Although the noise is random, we can still build a noise model by statistical methods and add noises into the image. Then remove the added noise bydifferent methods.There are other methods to realize image enhancement except MATLAB. For example, the Photoshop is also a better tool. Comparing with Photoshop, MATLAB is more professional by writing a program rather than usinga program.。
- 1、下载文档前请自行甄别文档内容的完整性，平台不提供额外的编辑、内容补充、找答案等附加服务。
- 2、"仅部分预览"的文档,不可在线预览部分如存在完整性等问题,可反馈申请退款(可完整预览的文档不适用该条件!)。
- 3、如文档侵犯您的权益，请联系客服反馈,我们会尽快为您处理(人工客服工作时间：9:00-18:30)。
Simultaneous Contrast
(Images from Rafael C. Gonzalez and Richard E. Wood, Digital Image Processing, 2nd Edition.
Optical illusion
(Images from Rafael C. Gonzalez and Richard E. Wood, Digital Image Processing, 2nd Edition.
(Images from Rafael C. Gonzalez and Richard E. Wood, Digital Image Processing, 2nd Edition.
Mach Band Effect (Cont)
B
Intensity
A
Position In area A, brightness perceived is darker while in area B is brighter. This phenomenon is called Mach Band Effect.
Image Sensors : Line Sensor
Fingerprint sweep sensor Computerized Axial Tomography
(Images from Rafael C. Gonzalez and Richard E. Wood, Digital Image Processing, 2nd Edition.
Brightness Adaptation of Human Eye : Mach Band Effect
Intensity
Position
Mach Band Effect
Intensities of surrounding points effect perceived brightness at each point. In this image, edges between bars appear brighter on the right side and darker on the left side.
f(x,y)
w An image: a multidimensional function of spatial coordinates. w Spatial coordinate: (x,y) for 2D case such as photograph, (x,y,z) for 3D case such as CT scan images (x,y,t) for movies w The function f may represent intensity (for monochrome images) or color (for color images) or other associated values.
4. Blind spot is the region of emergence of the optic nerve from the eye.
Range of Relative Brightness Sensation
Simutaneous range is smaller than Total adaptation range
Brightness Adaptation of Human Eye : Simultaneous Contrast
Simultaneous contrast. All small squares have exactly the same intensity but they appear progressively darker as background becomes lighter.
Digital Image
Digital image = a multidimensional array of numbers (such as intensity image) or vectors (such as color image)
Each component in the image called pixel associates with the pixel value (a single number in the case of intensity images or a vector in the case of color images).
CCD KAF-3200E from Kodak. (2184 x 1472 pixels, Pixel size 6.8 microns2)
Gate Vertical Transport Register
Horizontal Transportation Register
Image Sensor: Inside Charge-Coupled Device
(Images from Rafael C. Gonzalez and Richard E. Wood, Digital Image Processing, 2nd Edition.
Image Sensors : Single Sensor
(Images from Rafael C. Gonzalez and Richard E. Wood, Digital Image Processing, 2nd Edition.
Visual Perception: Human Eye (cont.)
1. 2. 3.
The lens contains 60-70% water, 6% of fat. The iris diaphragm controls amount of light that enters the eye. Light receptors in the retina - About 6-7 millions cones for bright light vision called photopic - Density of cones is about 150,000 elements/mm2. - Cones involve in color vision. - Cones are concentrated in fovea about 1.5x1.5 mm2. - About 75-150 millions rods for dim light vision called scotopic - Rods are sensitive to low level of light and are not involved color vision.
Visible Spectrum
(Images from Rafael C. Gonzalez and Richard E. Wood, Digital Image Processing, 2nd Edition.
Image Sensors
Single sensor
Line sensor
Array sensor
DБайду номын сангаасgital Image Processing
Chapter 2: Digital Image Fundamental
6 June 2007
What is Digital Image Processing ?
Processing of a multidimensional pictures by a digital computer
การประมวลผลสัญญาณรูปภาพโดยใช้ดิจิตอลคอมพิวเตอร์
Why we need Digital Image Processing ?
1. เพื่อบันทึกและจัดเก็บภาพ 2. เพื่อปรับปรุงภาพให้ดีขึ้นโดยใช้ กระบวนการทางคณิตศาสตร์ 3. เพื่อช่วยในการวิเคราะห์รปภาพ ู 4. เพื่อสังเคราะห์ภาพ 5. เพื่อสร้างระบบการมองเห็นให้กับ คอมพิวเตอร์
(Images from Rafael C. Gonzalez and Richard E. Wood, Digital Image Processing, 2nd Edition.
Distribution of Rods and Cones in the Retina
(Images from Rafael C. Gonzalez and Richard E. Wood, Digital Image Processing, 2nd Edition.
Digital Images
Digital image: an image that has been discretized both in Spatial coordinates and associated value. w Consist of 2 sets:(1) a point set and (2) a value set w Can be represented in the form I = {(x,a(x)): x X, a(x) F} where X and F are a point set and value set, respectively. w An element of the image, (x,a(x)) is called a pixel where - x is called the pixel location and - a(x) is the pixel value at the location x
10 10 16 28 9 65 70 56 43 6 26 37 32 99 70 56 78 54 96 67 25 13 22 15 60 90 96 67 21 54 47 42 32 158587853943 92 54 65 65 39 32 65 87 99