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用3330+2020两芯片设计DVB的电原理图,可以直接用
用3330+2020两芯片设计DVB的电原理图,可以直接用-With 3330+ 2020 two DVB chip design electrical schematics, can be directly used
- 2022-07-24 20:11:53下载
- 积分:1
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数字图像处理中,图像匹配是目前研究的一个主要的大方向。这里提供一个图像配准的代码,用于实现图像拼接。...
数字图像处理中,图像匹配是目前研究的一个主要的大方向。这里提供一个图像配准的代码,用于实现图像拼接。-Digital image processing, image matching is a major principle. Here to provide an image registration code, used to achieve image mosaic.
- 2022-02-04 20:50:57下载
- 积分:1
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非常全面的生物信号分析工具包,能够详细分析信号时域频域、相位等特性,针对于生物信号有良好效果。...
非常全面的生物信号分析工具包,能够详细分析信号时域频域、相位等特性,针对于生物信号有良好效果。-an excellent signal processing toolbox, it could give a detailed analysis on time series, frequency and phase. it is really useful on biomedical signal processing.
- 2023-04-13 12:00:04下载
- 积分:1
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several image filtering algorithm, spent many great efforts to find, in the hope...
matlab的十几种图像滤波算法,费好大力气才找到的,希望能给于借鉴-several image filtering algorithm, spent many great efforts to find, in the hope that they will learn from
- 2022-03-19 20:36:27下载
- 积分:1
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该函数采用kmeans法进行阈值分割 二值化kmeans.c
该函数采用kmeans法进行阈值分割 二值化kmeans.c-The function kmeans method using threshold segmentation kmeans.c Binarization
- 2022-03-11 12:00:18下载
- 积分:1
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基于fisher准则的阈值分割方法,陈果文章中的方法
基于fisher准则的阈值分割方法,陈果文章中的方法-thresholding method based on fisher algorithm,
- 2023-08-07 06:50:04下载
- 积分:1
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opencv+ vs2008 watershed algorithm to achieve
opencv+vs2008实现分水岭算法-opencv+ vs2008 watershed algorithm to achieve
- 2022-08-05 12:32:31下载
- 积分:1
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一款不错的烟花程序,带有声音和美术字,已调试通过
一款不错的烟花程序,带有声音和美术字,已调试通过-A good fireworks procedures, with voice and word art, have been debugging through
- 2023-08-28 05:20:04下载
- 积分:1
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图像处理。。
histogram specification
图像处理。。
histogram specification-image processing. . Histogram specification
- 2022-01-25 23:57:20下载
- 积分:1
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geometry calculation of correlation algorithm can determine the function of the...
计算几何学的相关算法 可以根据这个函数确定两条线段在交点处的转向,比如确定p0 p1和p1 p2在p1处是左转还是右转,只要求(p2-p0)*(p1-p0),若0则右转,=0则共线-geometry calculation of correlation algorithm can determine the function of the two segments to the intersection, such as identifying p0 p1 p1 and p2 p1 is the left or the right, only requirements (p2- p0)* (p1- p0), and if 0 right, a total = 0 line
- 2022-07-13 05:45:28下载
- 积分:1