TDI CMOS图像传感器中随机电平跳变噪声特性

研究论文
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"这篇研究论文探讨了在时间延迟积分（TDI）CMOS图像传感器中随机电传噪声（RTS噪声）的特性。通过建立基于大量样本模拟的统计模型，研究了RTS噪声与TDI阶段数量之间的关系。Matlab仿真结果显示，随着TDI阶段数M的增加，RTS噪声的平均值增加的比例大约为M的0.5次方，且当TDI阶段更大时，这一比例接近于M的0.5次方。在噪声直方图中，当TDI阶段超过特定值时，RTS噪声呈现出高斯分布的特性。这些发现对于TDI阶段的设计和噪声分析提供了指导。" 本文是关于CMOS图像传感器技术中的一种特定噪声现象——随机电传信号噪声（Random Telegraph Signal Noise，简称RTS噪声）的研究。随着CMOS技术的不断微缩，RTS噪声变得越来越显著，对图像传感器的性能产生了重要影响。作者建立了基于大样本模拟的统计模型，旨在理解RTS噪声与TDI（Time Delay Integration）阶段之间的相互作用。 TDI CMOS图像传感器是一种广泛应用于航天、遥感和工业检测等领域的成像技术，它通过连续的像素行积累光信号来提高信噪比。RTS噪声主要由晶体管中的缺陷引起，表现为信号在两个稳定状态之间随机切换，从而导致图像质量下降。论文指出，随着TDI阶段数的增加，RTS噪声的平均强度会以大于M的0.5次方的比例增长，这意味着噪声的累积效应随着TDI的深度而加剧。 进一步的，论文的仿真结果还揭示了一个关键点：当TDI阶段数超过某个特定阈值时，RTS噪声的分布呈现出高斯分布的特征。这种分布模式对于理解和预测噪声在图像中的影响非常重要，因为高斯分布通常与均方误差最小化和信号处理中的其他优化问题相关联。 此外，这些研究结果对设计者有直接的指导意义。它们可以帮助工程师优化TDI CMOS图像传感器的设计，比如通过调整TDI阶段的数量来平衡噪声和信噪比，或者通过噪声抑制算法来减少RTS噪声的影响。同时，对于图像数据的后处理和分析，了解RTS噪声的统计特性也能帮助开发更有效的噪声过滤策略。 这篇论文深入研究了RTS噪声在TDI CMOS图像传感器中的行为，提供了关于噪声控制和图像质量提升的重要理论依据，对未来的传感器设计和技术发展具有重要的参考价值。

Telegraph-Image-Hosting:利用Telegraph搭建免费图床-源码

2021-05-09 上传
Telegraph-Image-Hosting利用Telegraph搭建免费图床写在前面利用Telegraph搭建的图床，有什么优势和缺陷呢？优势：免费、无限流量(区别于你的VPS)、无限容量、无审查(懂得都懂哈:)、理论上可以上传任意小于5MB的文件...

写一段Java实现Input The first line contains the number n of fighters in the division (2 ≤ n ≤ 50000). The second line contains ten integers in the range from 1 to 10000 separated with a space written in the nonascending order. These are the times of sending a message from one telegraph to another if the length of their common prefix is zero, one, two, …, nine. The next n lines contain the numbers of telegraphs given to the fighters of the division. The number of Anka's telegraph is described first, and the number of Chapaev's telegraph is described last. All the numbers of telegraphs are different. Output Output the only line “-1” if it is impossible to deliver the message to Chapaev. Otherwise, in the first line output the minimal time required to deliver the message. In the second line output the number of fighters in the delivery path, and in the third line output their numbers separated with a space in the order from Anka to Chapaev. The fighters of the 25th Division are numbered from 1 to n in the order in which their mobile telegraphs are described in the input. If there are several ways to deliver the message in minimal time, output any of them.

2023-05-30 上传
Scanner sc = new Scanner(System.in); int n = sc.nextInt(); int[] t = new int[10]; for (int i = 0; i ; i++) { t[i] = sc.nextInt(); } int[] telegraphs = new int[n]; for (int i = 0; i ; i++) { ...

用c++解决Time limit: 3.0 second Memory limit: 64 MB. Input The first line contains the number n of fighters in the division (2 ≤ n ≤ 50000). The second line contains ten integers in the range from 1 to 10000 separated with a space written in the nonascending order. These are the times of sending a message from one telegraph to another if the length of their common prefix is zero, one, two, …, nine. The next n lines contain the numbers of telegraphs given to the fighters of the division. The number of Anka's telegraph is described first, and the number of Chapaev's telegraph is described last. All the numbers of telegraphs are different. Output Output the only line “-1” if it is impossible to deliver the message to Chapaev. Otherwise, in the first line output the minimal time required to deliver the message. In the second line output the number of fighters in the delivery path, and in the third line output their numbers separated with a space in the order from Anka to Chapaev. The fighters of the 25th Division are numbered from 1 to n in the order in which their mobile telegraphs are described in the input. If there are several ways to deliver the message in minimal time, output any of them.

2023-05-27 上传
下面是完整的 C++ 代码，实现了上述算法： ```c++ #include #include #include #include #include using namespace std; ...const int INF = 0x3f3f3f3f;...2.然后根据每个电报的编号构建图，对于每个电报，根据其...

用c++解决Input The first line contains the number n of fighters in the division (2 ≤ n ≤ 50000). The second line contains ten integers in the range from 1 to 10000 separated with a space written in the nonascending order. These are the times of sending a message from one telegraph to another if the length of their common prefix is zero, one, two, …, nine. The next n lines contain the numbers of telegraphs given to the fighters of the division. The number of Anka's telegraph is described first, and the number of Chapaev's telegraph is described last. All the numbers of telegraphs are different. Output Output the only line “-1” if it is impossible to deliver the message to Chapaev. Otherwise, in the first line output the minimal time required to deliver the message. In the second line output the number of fighters in the delivery path, and in the third line output their numbers separated with a space in the order from Anka to Chapaev. The fighters of the 25th Division are numbered from 1 to n in the order in which their mobile telegraphs are described in the input. If there are several ways to deliver the message in minimal time, output any of them.

2023-05-27 上传
这是一道经典的图论最短路径问题，可以使用 Dijkstra 算法或者 Floyd 算法来解决。 以下是使用 Dijkstra 算法的示例代码： ```c++ #include #include #include #include #include using namespace std;...

图像随机电报噪声是什么

2023-05-20 上传
图像随机电报噪声（Image random telegraph noise, RTN）是一种影响图像质量的噪声，它是由图像传感器中的单个像素元件（pixel）在一定温度下的随机热运动引起的。这种噪声与传感器的工艺技术、材料和温度等因素有关...

What is a plugboard? For what purpose was it used?

2023-05-13 上传
A plugboard, also known as a patch panel, was an early form of manual switchboard used in early telephone and telegraph communication systems. It consisted of a board with a set of sockets or jacks ...

解释以下这段代码并说出他的计算复杂度import java.util.*; public class 1806 { static int n; static int[] t = new int[10]; static int[] telegraph = new int[50005]; static int[] dis = new int[50005]; static int[] pre = new int[50005]; static boolean[] vis = new boolean[50005]; static ArrayList<Integer> path = new ArrayList<Integer>(); public static void main(String[] args) { Scanner sc = new Scanner(System.in); n = sc.nextInt(); for (int i = 0; i < 10; i++) { t[i] = sc.nextInt(); } for (int i = 1; i <= n; i++) { telegraph[i] = sc.nextInt(); } dijkstra(1); if (dis[n] == Integer.MAX_VALUE) { System.out.println("-1"); } else { System.out.println(dis[n]); getPath(n); System.out.println(path.size()); for (int i = 0; i < path.size(); i++) { System.out.print(path.get(i) + " "); } } } private static void dijkstra(int s) { Arrays.fill(dis, Integer.MAX_VALUE); dis[s] = 0; for (int i = 1; i <= n; i++) { pre[i] = i; } for (int k = 0; k < n; k++) { int u = -1; int minDis = Integer.MAX_VALUE; for (int i = 1; i <= n; i++) { if (!vis[i] && dis[i] < minDis) { u = i; minDis = dis[i]; } } if (u == -1) { break; } vis[u] = true; for (int i = 1; i <= n; i++) { if (!vis[i]) { int w = getWeight(telegraph[u], telegraph[i]); if (dis[u] + w < dis[i]) { dis[i] = dis[u] + w; pre[i] = u; } } } } } private static int getWeight(int a, int b) { int weight = 0; String s1 = String.valueOf(a); String s2 = String.valueOf(b); int len = Math.min(s1.length(), s2.length()); for (int i = 0; i < len; i++) { if (s1.charAt(i) != s2.charAt(i)) { weight = t[i]; break; } } return weight; } private static void getPath(int u) { if (u != 1) { getPath(pre[u]); } path.add(u); } }

2023-06-03 上传
其中，telegraph数组存储了n个节点的电报号码，t数组存储了电报号码每一位的开销，dis数组存储了每个节点到源节点的最短距离，pre数组存储了每个节点的前驱节点，vis数组表示每个节点是否已经被访问过。getPath方法...

cv2.INPAINT_TELEA

2023-08-20 上传
它是基于 Telegraph Equation 的图像修复算法，用于填补图像中的缺失部分。这种方法通常用于去除图像中的噪点、水印或其他不想要的区域。它通过从周围的像素中进行推断来填补缺失的区域，以使修复后的图像看起来更加...

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2023-09-23 上传
CRC16（Cyclic Redundancy Check）是一种错误检测码，CCITT则是国际电信联盟（International Telegraph and Telephone Consultative Committee，简称CCITT）制定的一个CRC算法的标准。 在Java编程语言中，可以通过...

python selenium 怎么通过div获取 selenium源码

2023-09-02 上传
【clsxpath = driver.find_element_by_xpath('//div[@class="b-c-e6e7ea telegraph-list"][' str(count-i) ']')】 【leachText = clsxpath.get_attribute('innerHTML')】 这样，你就可以通过div获取Selenium源码了...
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