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[Mathimatics-Numerical algorithmscodeing000

Description: 从空格(ASCII码32)到~(ASCII码126)。表内的第一行与表头相同,下面的每一行都与上一行的内容相同,只是字符相左移动了一个位置。这样,下一行的最后一个字符与上一行的第一个字符相同。 为了进行文本编码,可以任意选择一个字符串,称之为编码字符串,也就是常说的密钥。为解释编码方法,我们假设密钥是Walrus,待编码的文本(即常说的明文)是: Meet me in St. Louis 我们在待编码的文本之上重复书写上述密钥,使得其长度与待编码文本相同: WalrusWalrusWalrusWa Meet me in St. Louis 从上述两行文本中按列对应方式依次提取一个字符,可得到多个字符对:WM、ae、le等,这些字符对可用作上表的索引。这样,依次以这些字符对作为索引可从上表查到一系列字符,这些字符就构成了文本编码,即常说的密文。例如,第W行第M列队应得字符是%,因此编码的第一个字符就是%;第a行第e列对应的字符是G;第l行第e列对应的是R。依次进行上述查找操作,可以得到完整的密文 %Grgua=aVauGLol?eiAU 进行相反的操作就可对该文本解码。 编写编码/解码程序,可以对文本文件或键盘输入的字符串进行编码/解码,在选择编码解码后,需要提示用户输入密钥。 -from space (ASCII 32) to ~ (ASCII 126). The schedule of the first line with the same head table, each of the rows below the previous line with the same content. Character movement is a curator position. So, the next line of characters with a final party on the first character of the same. To text encoding, can choose an arbitrary string, the string as coding, which is often said that the key. To explain the encoding method, we assume that key is Walrus when coding the text (often said that an express) : Meet me in St.. Louis coding in question above the text written above key repeat, making its length and text encoding the same question : WalrusWalrusWalrusWa Meet me in St.. Louis from the above two OK text counterpart way out by the extraction followed a character, a number of characters
Platform: | Size: 1879 | Author: 梁清华 | Hits:

[Other resource2222

Description: (1)利用多项式拟合的两个模块程序求解下题: 给出 x、y的观测值列表如下: x 0 1 2 3 4 5 y 2.08 7.68 13.8 27.1 40.8 61.2 试利用二次多项式y=a0+a1x+a2x2进行曲线拟合。 (1)多项式拟合方法:假设我们收集到两个相关变量x、y的n对观测值列表: x x0 x1 x2 x3 x4 x5 y y0 y1 y2 y3 y4 y5 我们希望用m+1个基函数w0(x),w1(x),…,wm(x)的一个线形组合 y=a0w0(x)+a1w1(x)+…+amwm(x) 来近似的表达x、y间的函数关系,我们把几对测量值分别代入上式中,就可以得到一个线形方程组: a0w0(x0)+a1w1(x0)+…+amwm(x0)=y0 a0w0(x1)+a1w1(x1)+…+amwm(x1)=y1 …   … a0w0(xn)+a1w1(xn)+…+amwm(xn)=yn 只需要求出该线形方程组的最小二乘解,就能得到所构造的的多项式的系数,从而解决问题。
Platform: | Size: 1398 | Author: lz | Hits:

[Mathimatics-Numerical algorithmscodeing000

Description: 从空格(ASCII码32)到~(ASCII码126)。表内的第一行与表头相同,下面的每一行都与上一行的内容相同,只是字符相左移动了一个位置。这样,下一行的最后一个字符与上一行的第一个字符相同。 为了进行文本编码,可以任意选择一个字符串,称之为编码字符串,也就是常说的密钥。为解释编码方法,我们假设密钥是Walrus,待编码的文本(即常说的明文)是: Meet me in St. Louis 我们在待编码的文本之上重复书写上述密钥,使得其长度与待编码文本相同: WalrusWalrusWalrusWa Meet me in St. Louis 从上述两行文本中按列对应方式依次提取一个字符,可得到多个字符对:WM、ae、le等,这些字符对可用作上表的索引。这样,依次以这些字符对作为索引可从上表查到一系列字符,这些字符就构成了文本编码,即常说的密文。例如,第W行第M列队应得字符是%,因此编码的第一个字符就是%;第a行第e列对应的字符是G;第l行第e列对应的是R。依次进行上述查找操作,可以得到完整的密文 %Grgua=aVauGLol?eiAU 进行相反的操作就可对该文本解码。 编写编码/解码程序,可以对文本文件或键盘输入的字符串进行编码/解码,在选择编码解码后,需要提示用户输入密钥。 -from space (ASCII 32) to ~ (ASCII 126). The schedule of the first line with the same head table, each of the rows below the previous line with the same content. Character movement is a curator position. So, the next line of characters with a final party on the first character of the same. To text encoding, can choose an arbitrary string, the string as coding, which is often said that the key. To explain the encoding method, we assume that key is Walrus when coding the text (often said that an express) : Meet me in St.. Louis coding in question above the text written above key repeat, making its length and text encoding the same question : WalrusWalrusWalrusWa Meet me in St.. Louis from the above two OK text counterpart way out by the extraction followed a character, a number of characters
Platform: | Size: 2048 | Author: 梁清华 | Hits:

[Windows DevelopSysKey

Description: 目标 设置组合键来触发某一菜单命令。例如,用A l t + D组合键删除视图中所选项目。 策略 首先,使用Developer Studio提供的加速键编辑器(Accelerator Editor)创建一项资源,当视 图被创建、使用、拥有输入焦点时该资源被装入。在视图没有输入焦点时或当我们想定制键 的处理时,我们也可以手工处理W M _ K E Y D O W N窗口命令。-Goal Setting key combination to trigger a menu command. For example, A lt+ D key combination to delete the selected item view. Strategy First, use Developer Studio to provide the acceleration of key editor (Accelerator Editor) to create a resource, when the view is created, use, possession input focus when the resources are loaded. In view of the focus has not been entered or when we want to deal with custom keys, we can also deal with WM _ KEYDOWN manual command window.
Platform: | Size: 1024 | Author: 旺老财 | Hits:

[Algorithm2222

Description: (1)利用多项式拟合的两个模块程序求解下题: 给出 x、y的观测值列表如下: x 0 1 2 3 4 5 y 2.08 7.68 13.8 27.1 40.8 61.2 试利用二次多项式y=a0+a1x+a2x2进行曲线拟合。 (1)多项式拟合方法:假设我们收集到两个相关变量x、y的n对观测值列表: x x0 x1 x2 x3 x4 x5 y y0 y1 y2 y3 y4 y5 我们希望用m+1个基函数w0(x),w1(x),…,wm(x)的一个线形组合 y=a0w0(x)+a1w1(x)+…+amwm(x) 来近似的表达x、y间的函数关系,我们把几对测量值分别代入上式中,就可以得到一个线形方程组: a0w0(x0)+a1w1(x0)+…+amwm(x0)=y0 a0w0(x1)+a1w1(x1)+…+amwm(x1)=y1 …   … a0w0(xn)+a1w1(xn)+…+amwm(xn)=yn 只需要求出该线形方程组的最小二乘解,就能得到所构造的的多项式的系数,从而解决问题。-err
Platform: | Size: 1024 | Author: lz | Hits:

[Technology Managementshellresapps.dll.0804.mui

Description: 手机WM系统 6.5 shellresapps.dll.0804.mui 汉化软件-Mobile WM system software 6.5 shellresapps.dll.0804.mui汉化
Platform: | Size: 14336 | Author: 吴迦锋 | Hits:

[GUI Developwmpsample

Description: wm media player 的MFC实现,可播放多种音视频格式的界面-Using MFC to make windows media player come true.It can play several kinds of music or video.
Platform: | Size: 1120256 | Author: hanqiushi | Hits:

[MultiLanguage23071mui

Description: WM6.5 VGA 中文mui wm6.5 23071 chinese MUI-wm6.5 23071 ENG MUI
Platform: | Size: 600064 | Author: leo | Hits:

[Software EngineeringDesignofcontrollerofPMSMutilizingdspanditsrealizat

Description: :介绍了采用 T MS 3 2 0 L F 2 4 0 7 A芯片实现对永磁 同步 电机 ( P MS M)的设 计与控制 ,讨论 了空间矢量脉宽调制 ( S V P WM)控制方法 ,并给出了控制系统的硬件设计和 软件实现。试验结果表 明该系统具有较好的跟踪性能 ,稳态精 度较高 ;T MS 3 2 0 L F 2 4 0 7 A作为 D S P控制器 2 4 x系列 的新成员 ,是电机数字化控制的升级产品。-The d e s i g n a n d c on t r o l o f PMS M a t e r e a l i z e d u t i l i z i ng TMS 3 2 02 4 07A,a n d s pa c e v e c t o r p ul s e w d t h mo d u l a t on s d】 c u s s e d.Th e e x De r i me n t r e s uI t i n di c a t es t h a t t h e s y s t e m h a s be t t e r f ol l o wi n g p e r f o r ma n c e a n d h i g h e r s t e a d y p r e c i s o n, a s a n e wme Ⅱ l be o t t h e 24 x s e r i e s o f DSP c o n t r o l l e r s,TMS 3 2 02 4 07A i s a up g r a d e pr o d u c t i o n f o r mo t o r d i g i t a l c o nt r o l ·
Platform: | Size: 133120 | Author: 张国辉 | Hits:

[OtherdisSVPWMAlg

Description: : 在分析了 S V P WM原理的基础上 , 详细介绍 了广义的不连续空间矢量 D P WM算法及其在 S i m u l i n k环 境下实现的方法, 并且对 S V P WM、 四种 D P WM算法的相电压调制波和逆变器输出线电压的谐波特征进行了比较分 析, 最后将 D P WM应用于异步电动机调速系统, 仿真结果验证了方法的可行性。-: An analysis based on the principles of SVP WM, detailing the broad discrete space vector DP WM algorithm in S imulink method realized in the environment, and on the SVP WM, the four DP WM algorithm for phase voltage modulation wave and inverter output line voltage harmonic characteristics of a comparative analysis, and finally DP WM induction motor drive system used in the simulation results demonstrate the feasibility of the method.
Platform: | Size: 376832 | Author: Wangjg | Hits:

[matlabnon-linear-information

Description: 这是一个关于非线性信息处理技术的文档;里面包括logistic映射、lorenz、rossler等混沌经典方程的程序和matlab仿真结果;还包括求分形w-m函数的维数的半方差法忽然R/S方法-This is a non-linear information processing on the document which includes logistic map, lorenz, rossler chaotic classical equations such as the procedures and matlab simulation results also includes wm seeking function of the fractal dimension of a sudden half of variance R/S method
Platform: | Size: 1628160 | Author: 李阳 | Hits:

[Program doc2fsk

Description: XG_FSK.m将fsk信号封装成函数,目的是为了计算相干解调的误码率 函数中,i为噪声功率(程序中,经调整处理,i/2才是噪声功率) wm设为0。例如,可在commander窗口中调用 XG_FSK(0.02,0), 0.02代表着功率为0.01的噪声, 就能产生fsk信号的调制、相干解调过程-XG_FSK.m FSK signal encapsulated into function, the purpose is to calculate the coherent demodulation BER function, i is the noise power (the program, after adjustment processing, i/2 is the noise power) WM is set to 0. For example, the commander window the call XG_FSK (0.02,0) 0.02 represents 0.01 power of the noise, can produce the fsk signal modulation, the coherent demodulation process
Platform: | Size: 7168 | Author: 王嘉博 | Hits:

[Otherfilter

Description: M=63 J=(M-1)/2 Wp=0.5*pi Ws=0.6*pi Np=100 Ns=100 weightp=1 weights=100 通带和阻带的加权系数 WpN=linspace(0,Wp,Np) WsN=linspace(Ws,pi-0.05,Ns) Wm=[WpN WsN] p=[ones(1,50),ones(1,50)/2] d=[p,zeros(1,Ns)] W=diag([weightp*ones(1,Np),weights*ones(1,Ns)]) C=cos(Wm*[0:J]) Q=diag(cos(Wm*0.5)) g=(W*Q*C)\(W*d) g=g h=(g(J:-1:2)+g(J+1:-1:3))/4 h=[g(J+1)/4 h g(1)*0.5+g(2)/4] h=[h fliplr(h)] W=linspace(0,pi-0.1,512) H=freqz(h,1,W) plot(W/pi,20*log10(abs(H))) -filter design
Platform: | Size: 1024 | Author: 李天 | Hits:

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