1、1. Give a single intensity transformation function for spreading the intensities of an image so the lowest intensity is 0 and the highest is L-1. 为了扩展一幅图像的灰度，使其最低灰度为 0、最高灰度为 L-1，请给出一个单调的变换函数。Answer: Let f denote the original image. First subtract the minimum value of f denoted fmin from f to yield a
2、 function whose minimum value is 0: 1mingfNext divide g1 by its maximum value to yield a function in the range 0,1 and multiply the result by L 一 1 to yield a function with values in the range 0, L 一 11 minin()max()ax()LgffKeep in mind that fmin is a scalar and f is an image.让 f 表示原始图像。首先从图像函数 f 中减掉
3、 f 的最小值 fmin, 然后生成一个新的函数 g1，它的最小值为 0： 1mingf接下来让 g1 的最大值除 g1 得到另一新的函数，它的值域在0,1区间，然后再乘上 L 一 1，得到值域为0, L 一 1的新函数。请注意 fmin 是一个标量，而 f 是一个图像。2.Explain why the discrete histogram equalization technique does not,in general,yield a flat histogram. 请解释为什么离散直方图均衡化技术一般不能得到平坦的直方图。Answer: All that histogram equa
4、lization does is remap histogram components on the intensity scale. To obtain a uniform (flat) histogram would require in general that pixel intensities actually be redistributed so that there are L groups of n/L pixels with the same intensity, where L is the number of allowed discrete intensity lev
5、els and n=MN is the total number of pixels in the input image. The histogram equalization method has no provisions for this type of (artificial) intensity redistribution process.直方图均衡化就是把直方图重新映射到灰度尺度上。为了得到平坦的直方图，通常需要图像中像素的灰度值重新分配，这样会产生 L 个组，每个组中包含相同灰度值的像素的数量是n/L，其中 L 是允许的离散灰度值的个数， n=MN 是输入图像中总的像素的个数
6、。而离散直方图的均衡化没有提供这样一个重新分配的过程。3.In a given application an averaging mask is applied to input image to reduce noise, and then a Laplacian mask is applied to enhance small details. Whould the result be the same if the order of these operations were reserved? 在给定的应用中，一个均值模板被用于输入图像以减少噪声，然后再用一个拉普拉斯模板来增强细节。如
7、果交换一下两个操作步骤的顺序，能否得到相同的结果？Answer: The student should realize that both the Laplacian and the averaging process are linear operations, so it makes no difference which one is applied first.学生应该认识到拉普拉斯和平均化的处理过程都是线性运算过程，所以无论先应用哪一个模板对结果是没有影响的。4. You saw in Fig. 2.18 that the Laplacian with a -8 in the cen
8、ter yields sharper results than the one with a -4 in the center. Explain the reason in detail. 在图 2.18 中所看到的中心系数为-8 的拉普拉斯模板所得到的结果，要比中心系数为-4 的模板所得到的结果清晰一些。请详细说明原因。Answer: The Laplacian mask with a -4 in the center performs an operation proportional to differentiation in the horizontal and vertical di
9、rections. Consider for a moment a 3 x 3 “Laplacian“ mask with a -2 in the center and 1 above and below the center. All other elements are 0. This mask will perform differentiation in only one direction, and will ignore intensity transitions in the orthogonal direction. An image processed with such a
10、 mask will exhibit sharpening in only one direction. A Laplacian mask with a -4 in the center and 1 in the vertical and horizontal directions will obviously produce an image with sharpening in both directions and in general will appear sharper than with the previous mask. Similarly, and mask with a
11、-8 in the center and is in the horizontal, vertical, and diagonal directions will detect the same intensity changes as the mask with the -4 in the center but, in addition, it will also be able to detect changes along the diagonals, thus generally producing sharper-looking results.中心系数为-4 的拉普拉斯模板执行的是
12、与水平和垂直方向的差分成比例的运算。考虑一个中心系数为-2 的拉普拉斯模板，2 的上下都是 1。模板中其它所有的系数都是 0。这个模板只会在一个方向上（垂直方向）执行差分计算，并会忽略掉水平方向上灰度值的变化。用这样的模板处理图像，只会在一个方向上展示出图像的锐化。若采用中心系数为-4，上下左右的系数都为 1 的拉普拉斯模板对图像进行处理，则会在水平和垂直的两个方向上展示图像的锐化，并且锐化的效果比前一个模板的要好。类似，若采用中心系数为-8，水平，垂直和对角线方向的系数都为 1 的模板对图像进行处理，与前一个模板一样，能检测到水平和垂直方向上的灰度值的变化，此外，它还能检测到对角线方向的灰度
13、值的变化，所以看起来结果要更清晰一些。5.Write an expression for 2-D discrete convolution. 请写出一个 2 维离散卷积的表达式。10(,)(,)(,),).1;0.1MNeeeemnfxygfgxmynxN分别是 的周期化函数。(,)(,)eef和 (,)(,)fy和6.The two fourier spectra shown are of the same image. The spectrum on the left corresponds to the original image, and the spectrum on the ri
14、ght was obtained after the image was padded with zeros. Explain the significant increase in signal strength along the vertical and horizontal axes of the spectrum shown on the right. 同一幅图像的两个傅立叶频谱如图所示。 左边的频谱对应于原图像，右边的频谱图像使用 0 值填充后所得。请解释右图所示的谱沿垂直轴和水平轴方向的信号强度显著增强的原因。Answer: Unless all borders on of an
15、 image are black, padding the image with 0 introduces significant discontinuities (edges) at one or more borders of the image. These can be strong horizontal and vertical edges. These sharp transitions in the spatial domain introduce high-frequency components along the vertical and horizontal axes o
16、f the spectrum.除非图像中所有的边界都是黑色的，否则图像中填充 0 值会在一个或多个边界产生灰度值上的明显的不连续性，这样能够增强图像中水平和垂直的边缘。这样空间域上图像锐化的改变能在频谱中水平和垂直轴方向上引入高频分量。7. Can you think of a way to use the Fourier transform to compute (or partially compute) the magnitude of the gradient for use in image differentiation? If your answer is yes, give a
17、 method to do it. If your answer is no, explain why. 你能想出一种使用傅立叶变换计算（或部分计算）用于图像差分的梯度幅度的方法吗？如果您的回答是可以，那么请给出一种方法去实现它。如果你的回答是不可以，那么请解释原因。Answer: The answer is no. The Fourier transform is a linear process, while the square and square roots involved in computing the gradient are nonlinear operations. Th
18、e Fourier transform could be used to compute the derivatives as differences, but the squares, square root, or absolute values must be computed directly in the spatial domain.答案是否定的。傅里叶变换是一个线性的运算过程，而涉及到计算梯度的平方和平方根是非线性的运算过程。傅立叶变换能够用来计算微分和差分，但是若涉及到平方，平方根和绝对值的计算，则必须在空间域内直接进行计算。rk P(rk) Sk Sk舍入 Sk P(sk)0 0.29 0.29 2/7 s0 0.291/7 0.24 0.53 4/7 s1 0.242/7 0.17 0.7 5/7 s2 0.173/7 0.12 0.82 6/74/7 0.09 0.91 6/7 s3 0.215/7 0.06 0.97 16/7 0.02 0.99 11 0.01 1 1s4 0.09下题两种方式都正确，第一种是宋同学的作业
