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MATLABtuwensuanfa
说明： 关于matlab数字图像处理算法的实验原理及其报告(Matlab digital image processing algorithms on the experimental principle and the report)
- 2010-04-26 20:10:29下载
- 积分：1
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10
Converting an image into blocks of 8 x 8 and apply DCT on any one of the blocks
- 2014-01-22 13:28:41下载
- 积分：1
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extensionm
该程序是MATLAB源程序，可以用来对图像进行延拓，方便进行EMD分解。(MATLAB source code of the procedure can be used to image extension, facilitate the EMD decomposition.)
- 2009-05-09 21:05:35下载
- 积分：1
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Fuzzy_part1
Fuzzy logic functions by matlab
- 2009-12-22 00:57:27下载
- 积分：1
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Prueba
Is a good way to use
geodesic active contours give you more comfortable
you will be able to apply this code around the world
- 2013-10-04 16:44:13下载
- 积分：1
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Untitled.m
high patch filtering
- 2013-11-02 17:44:01下载
- 积分：1
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JPEG_BaseLine_Encoder
The JPEG compression scheme is divided into the following stages:
1. Transform the image into an optimal color space.
2. Adjust Aspect Ratio 16:9
3. Digitization Scheme 4:2:0.
4. Apply a Discrete Cosine Transform (DCT) to blocks of pixels, thus removing redundant image data.
5. Quantize each block of DCT coefficients using weighting functions optimized for the human eye.
6. Encode the resulting coefficients (image data) using a Huffman variable word-length algorithm to remove redundancies in the coefficients.
7. Byte Stuffing.
8. Header JFIF
9. JPG Data Store
- 2013-05-04 23:30:46下载
- 积分：1
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Code-on-MATLAB
相关ca码在matlab软件平台下的仿真和编程相关理论知识(
Related ca code simulation in matlab software platform and the relevant theoretical knowledge)
- 2013-05-13 13:37:04下载
- 积分：1
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_FuzzyPID
MATLAB算法 自己设计的 希望能帮助你 模糊PID(MATLAB algorithm design their own hope to help you fuzzy PID)
- 2015-04-20 21:48:44下载
- 积分：1
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szfxsy1
实验一：三次样条插值（P56，例6）
一、实验目的：
1) 掌握三次样条插值的运用
2) 了解拉格朗日插值在高次上的误差
二、实验环境：Matlab6.5
三、实验内容：
1) 给定函数f(x)=1/(1+x2)，-5<=x<=5，节点xk=-5+k，(k=0,1,2…10)，用三次样条插值求S10(x)，S10’(-5)=f’(-5)，S10’(5)=f’(5)。
2）作原函数f(x)，拉格朗日插值函数Ln(x)，三次样条差值函数Sn(x)。画出三个函数的图像，比较它们的区别。
(Experiment I: cubic spline interpolation (P56, cases of 6) First, the experiment was: 1) cubic spline interpolation to master the use of 2) understanding of Lagrange interpolation in high-error second, experimental environment: Matlab6 .5 c, experimental elements: 1) a given function f (x) = 1/(1+ x2),-5 <= x <= 5, node xk =- 5+ k, (k = 0,1,2 ... 10), using cubic spline interpolation for S10 (x), S10 (-5) = f (-5), S10 (5) = f (5). 2) for the original function f (x), the Lagrange interpolation function Ln (x), cubic spline difference function Sn (x). Draw the three functions of images, compare the difference between them.)
- 2009-01-18 23:06:15下载
- 积分：1
