c语言版的多项式曲线拟合。 用最小二乘法进行曲线拟合. 用p－1 次多项式进行拟合，p<= 10 x,y 的第0个域x[0],y[0],没有用，有效数据从x[1],y[1] 开始 nNodeNum,有效数据节点的个数。 b,为输出的多项式系数，b[i] 为b[i-1]次项。b[0],没有用。 b,有10个元素ok。
上传时间： 2014-01-12
上传用户：变形金刚
Wavelets have widely been used in many signal and image processing applications. In this paper, a new serial-parallel architecture for wavelet-based image compression is introduced. It is based on a 4-tap wavelet transform, which is realised using some FIFO memory modules implementing a pixel-level pipeline architecture to compress and decompress images. The real filter calculation over 4 · 4 window blocks is done using a tree of carry save adders to ensure the high speed processing required for many applications. The details of implementing both compressor and decompressor sub-systems are given. The primarily analysis reveals that the proposed architecture, implemented using current VLSI technologies, can process a video stream in real time.
标签： applications processing Wavelets widely
上传时间： 2014-01-22
上传用户：hongmo
crc任意位生成多项式 任意位运算 自适应算法 循环冗余校验码(CRC，Cyclic Redundancy Code)是采用多项式的 编码方式，这种方法把要发送的数据看成是一个多项式的系数 ，数据为bn-1bn-2…b1b0 (其中为0或1)，则其对应的多项式为： bn-1Xn-1+bn-2Xn-2+…+b１X+b０ 例如：数据“10010101”可以写为多项式 X７+X４+X２+1。 循环冗余校验CRC 循环冗余校验方法的原理如下： (1) 设要发送的数据对应的多项式为P(x)。 (2) 发送方和接收方约定一个生成多项式G(x)，设该生成多项式 的最高次幂为r。 (3) 在数据块的末尾添加r个0，则其相对应的多项式为M(x)=XrP(x) 。(左移r位) (4) 用M(x)除以G(x)，获得商Q(x)和余式R(x)，则 M(x)=Q(x) ×G(x)+R(x)。 (5) 令T(x)=M(x)+R(x)，采用模2运算，T(x)所对应的数据是在原数 据块的末尾加上余式所对应的数据得到的。 (6) 发送T(x)所对应的数据。 (7) 设接收端接收到的数据对应的多项式为T’(x)，将T’(x)除以G(x) ，若余式为0，则认为没有错误，否则认为有错。
上传时间： 2014-11-28
上传用户：宋桃子
硬件设计指南(PDF格式)，主要包括：Low Voltage Interfaces；Grounding in Mixed Signal Systems；Digital Isolation Techniques； Power Supply Noise Reduction and Filtering； Dealing with High Speed Logic
上传时间： 2015-08-31
上传用户：阿四AIR
Absolute Database 5.12 src. Absolute Database lets you forget the Borland Database Engine (BDE). This BDE replacement is the compact, high-speed, robust and easy-to-use database engine. With Absolute Database you will not need special installation and configuration, it compiles right into your EXE. Make your application faster and smaller with Absolute Database BDE alternative!
标签： Database Absolute Borland forget
上传时间： 2013-12-19
上传用户：海陆空653
crc任意位生成多项式 任意位运算 自适应算法 循环冗余校验码(CRC，Cyclic Redundancy Code)是采用多项式的 编码方式，这种方法把要发送的数据看成是一个多项式的系数 ，数据为bn-1bn-2…b1b0 (其中为0或1)，则其对应的多项式为： bn-1Xn-1+bn-2Xn-2+…+b１X+b０ 例如：数据“10010101”可以写为多项式 X７+X４+X２+1。 循环冗余校验CRC 循环冗余校验方法的原理如下： (1) 设要发送的数据对应的多项式为P(x)。 (2) 发送方和接收方约定一个生成多项式G(x)，设该生成多项式 的最高次幂为r。 (3) 在数据块的末尾添加r个0，则其相对应的多项式为M(x)=XrP(x) 。(左移r位) (4) 用M(x)除以G(x)，获得商Q(x)和余式R(x)，则 M(x)=Q(x) ×G(x)+R(x)。 (5) 令T(x)=M(x)+R(x)，采用模2运算，T(x)所对应的数据是在原数 据块的末尾加上余式所对应的数据得到的。 (6) 发送T(x)所对应的数据。 (7) 设接收端接收到的数据对应的多项式为T’(x)，将T’(x)除以G(x) ，若余式为0，则认为没有错误，否则认为有错
上传时间： 2014-01-16
上传用户：hphh
Jitter is extremely important in systems using PLL-based clock drivers. The effects of jitter range from not having any effect on system operation to rendering the system completely non-functional. This application note provides the reader with a clear understanding of jitter in high-speed systems. It introduces the reader to various kinds of jitter in high-speed systems, their causes and their effects, and methods of reducing jitter. This application note will concentrate on jitter in PLL-based frequency synthesizers.
标签： extremely PLL-based important drivers
上传时间： 2014-11-25
上传用户：asddsd
bulk endpoint endless source/sink firmware. EP2OUT will always accept a bulk OUT EP4OUT will always accept a bulk OUT EP6IN will always return a 512 byte packet. The packet contains an incrementing byte starting at 0x02. Since EP6 always returns a 512 byte packet, this endpoint should never be accessed except with a high-speed host controller. EP8IN will continuously return the packet most recently written to EP4OUT
标签： bulk will endpoint firmware
上传时间： 2016-01-14
上传用户：sk5201314
We have a group of N items (represented by integers from 1 to N), and we know that there is some total order defined for these items. You may assume that no two elements will be equal (for all a, b: a<b or b<a). However, it is expensive to compare two items. Your task is to make a number of comparisons, and then output the sorted order. The cost of determining if a < b is given by the bth integer of element a of costs (space delimited), which is the same as the ath integer of element b. Naturally, you will be judged on the total cost of the comparisons you make before outputting the sorted order. If your order is incorrect, you will receive a 0. Otherwise, your score will be opt/cost, where opt is the best cost anyone has achieved and cost is the total cost of the comparisons you make (so your score for a test case will be between 0 and 1). Your score for the problem will simply be the sum of your scores for the individual test cases.
标签： represented integers group items
上传时间： 2016-01-17
上传用户：jeffery
The PCI Local bus concept was developed to break the PC data I/O bottleneck and clearly opens the door to increasing system speed and expansion capabilities. The PCI Local bus moves high speed peripherals from the I/O bus and places them closer to the system’s processor bus, providing faster data transfers between the processor and peripherals. The PCI Local bus also addresses the industry’s need for a bus standard which is not directly dependent on the speed, size and type of system processor. It represents the first microprocessor independent bus offering performance more than adequate for the most demanding applications such as full-motion video.
标签： bottleneck developed the concept
上传时间： 2014-12-03
上传用户：ikemada
