搜索资源列表
ST7787
- ST7787 芯片的SPEC，比亚迪2.4inchLCM的SPEC。The ST7787 is a single-chip controller/driver for 262K-color, graphic type TFT-LCD. It consists of 720 source line and 320 gate line driving circuits. This chip is capable of connecting directly to an external micr
lloyd_max
- dct- computes de discrete cosine transform for a matriz 8*8 from a picture.It receaves the point of the block where begin calculate, the tam of the block and the original picture... lLoyd-max- computes de lLoyd-max optimum quantizer for N bits.
BM
- The Berlekamp-Massey algorithm is an efficient algorithm for determining the linear complexity of a finite binary sequence sn of length n. The algorithm takes n iterations, with the Nth iteration computing the linear complexity of the subsequence sN
SDMX512appg
- 同步DMX512 简介和控制器的设计 在灯具调光和控制中，人们大量采用了DMX512 的控制协议，它是由美国剧 场技术协会（United State Institute for Theatre Technology，Inc）于1986 年8 月提出的一个能在一对线上传送512 路可控硅调光亮度信息的标准. DMX512 通信方式是采用了异步通信格式,每个调光点由11 位组成,其中一个 是起始位,8 位调光数据,两个停止位.每一次传输能512 个调光点.-DMX512
sumador2
- basic exemple. Do an ADD of two 4 bits numbers
doc1116
- The AT24C512 provides 524,288 bits of serial electrically erasable and programmable read only memory (EEPROM) organized as 65,536 words of 8 bits each. The device鈥檚 cascadable feature allows up to four devices to share a common two-wire bus. The
QPSK_TX
- qpsk_tx, transmision qpsk de 10 bits en matlab , el qpsk es igual al 4-psk
ac
- 最长上升子序列 非常经典的问题。 给定的整数序列，请求出其最长上升子序列的长度。 有多个测试用例。每个测试用例由一个行整数组成，其中第一个是一个整数 n (1 ≤ n ≤ 10000)，表示给定序列的长度。然后是 n 个 32-bits 带符号整数表示序列中的元素。 输入直至没有数据或者遇到长度 n 是非正数为止。 -Increase in the longest sequence is a classic problem. A given integer sequen
SAE-J1939-11-2006.Physical-layer-250K-bits-s-twis
- SAE J1939-11-2006.Physical layer,250K bits-s,twisted shielded pair
Division
- this the code to make a division of 8 bits using the avr mega16. The program is made in avrstudio 4-this is the code to make a division of 8 bits using the avr mega16. The program is made in avrstudio 4
Mscomm
- 2.MSComm 控件的常用属性 MSComm控件有很多重要的属性，但首先必须熟悉几个属性。 CommPort 设置并返回通讯端口号。 Settings 以字符串的形式设置并返回波特率、奇偶校验、数据位、停止位。 PortOpen 设置并返回通讯端口的状态。也可以打开和关闭端口。 Input 从接收缓冲区返回和删除字符。 Output 向传输缓冲区写一个字符串。-Common attributes 2.MSComm control MSComm cont
200991323298249
- VB开发的串口通信源码。对端口速度、奇偶校验、停止位等都可以设置。用于串口通信的开发-VB source code developed by serial communication. On the port speed, parity, stop bits and so can be set. Serial Communication for Development
pai_dui
- 介绍了飞机从到达起飞位，到选择起飞位，排队起飞状态，-Describes the plane took off from the arrival place, to select off bits, queuing off state,
eeprom
- The AT24C128/256 provides 131,072/262,144 bits of serial electrically erasable and programmable read only memory (EEPROM) organized as 16,384/32,768 words of 8 bits each
54379
- We investigate the case of independent Rayleigh faded paths between antenna elements and find that with high probability extraordinary capacity is available. Compared to the baseline n = 1 case, which by Shannon’s classical formula scales as one
ARM
- 基于ARM 体系结构的32 位微处理器已得到了广泛应用。该文通过具体开发实例，从硬件和软件2 个方面讨论了以32 位ARM9 微处理器S3C2410 为核心的商标识别系统的设计，结合实际应用说明了嵌入式操作系统Mizi Linux 的使用，描述了底层硬件驱动的编写、上层应用程序商标识别算法的设计，以及硬件平台的搭建和系统的移植等内容。-】ARM architecture-based 32-bit microprocessor is applied widely. The design of
doc3443
- The AT45DB041B is an SPI compatible serial interface Flash memory ideally suited for a wide variety of digital voice-, image-, program code- and data-storage applications. Its 4,325,376 bits of memory are organized as 2048 pages of 264 bytes each
SpaceCV
- Descr iption of a color space converter YCrCb to RGB with 16 bits high precision coeff.
test-scheme-of-nand-for-bits-flip
- nand flash由于本身的特性会存在位反转的现象，为了研究位反转现象产生的原因，对位反转出现的问题进行解析，所以设计了本方案。-Phenomenon of bits flip，because of Nand flash the nature of itself.In order to study the causes of this phenomenon ，and the scheme is designed
dlab-handout
- 这个任务的目的是更熟悉位级表示和操作。你会通过解决一系列的编程难题来解决这个问题。“很多这样的谜题都是人为的，但你会发现的。你自己想的更多的是通过你的方式通过他们，你将学到很多关于如何计算机表示整型数据。(The purpose of this assignment is to become more familiar with bit-level representations and manipulations. You'll do this by solving a series of p