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[Software Engineering3.5GHz--channel-measurements

Description: 在3.49-3.51GHZ频段内的典型中国城市环境下,地面信号衰落仍服从经典的指数对数分布。在中度误差(σx =10.291)下,路损指数在3.1-3.7之间。半径2公里的小区内,RMS时延在3-4微秒之间;典型的办公室环境中,存在直达径时RMS时延小于0.1微秒。在城市环境中,多径现象很明显,主要由高层建筑物引起,并以簇的方式存在。在城市中低速移动条件下(接收机移动速度小于100 Km/h)某一强径的生存时间通常超过10分钟,但相对到达时间会发生滑动。单一小区内的传播环境具有一定相关性,如果存在直达径,相关性会上升到0.4-0.5。此外,室外室内的测量结果都可以证明Jake模型关于入射角分布假设的合理性。当天线间距在1个载波波长增大到10个载波波长时,天线相关性在0.05-0.4之间。 SIMO信道探测系统能够揭示B3G信道的很多特征,但不能够全面、充分的分析和评估B3G信道。在后续工作中,我们将进一步改进这个测试系统,对B3G系统作更深入的研究。 参考-Aided by a 1-Tx and 2-Rx system working in 3.5GHz frequency and 20MHz bandwidth, JSI-MTLab studied the classic B3G urban wireless propagation environment of China and the classic indoor channel propagation environment through a recent indoor and outdoor wireless propagation measurement. A new path loss exponent is proposed in such band while the general outdoor model of path loss is verified. Also, the statistic results of RMS excess delay, max excess delay and mean excess delay in both indoor and outdoor scenarios are given. The lag-time domain and time-frequency domain channel responses are analyzed. At the end, the results of correlation of different propagation environment in the singular cell and antenna correlation are given as well as the phenomenon being analyzed and explained.
Platform: | Size: 159744 | Author: xuchen | Hits:

[Algorithmaitkenroot

Description: 埃特金求解C语言代码 //函数名:aitkenroot //功能描述:求出[a,b]区间内的非线性方程f(x)的一个实根 //输入参数:*x0(指向迭代初值和终值的指针) // f(迭代函数) // eps(精度要求) // max(最大迭代次数) //返回值:0(迭代失败),1(迭代成功)-Aitken Solution C language code// function name: aitkenroot// Function Description: Find the [a, b] interval of the nonlinear equation f (x) a real root// input parameters:* x0 (pointing Diego on behalf of the initial and final values of the pointer)// f (IFS)// eps (precision)// max (the maximum number of iterations)// return value: 0 (iteration failure), 1 (iteration success )
Platform: | Size: 1024 | Author: 穆欢 | Hits:

[AI-NN-PRgenetic-algorithm

Description: 采用遗传算法,求函数max f(x1,x2) = xsin(10π*x)+1的最大值。用C语言实现-Using genetic algorithms, find a function max f (x1, x2) = xsin (10π* x)+1 maximum. Using C language
Platform: | Size: 7168 | Author: 苏倩 | Hits:

[DirextXxskinexp_max2009

Description: xskinexp for max 2009 .x 导出插件-xskinexp for max 2009. x export plugin
Platform: | Size: 35840 | Author: 李金峰 | Hits:

[GIS programUtilidades

Description: Utilidades en ArcMap 9.2. Generar reporte de vecindad de un Layer de poligonos Reporte de coordenadas minimas y maximas (X,Y Min X, Y Max) de los feature de un FClass de poligonos Reporte de coordenadas de los centroides de los feature de poligonos.-Utilidades en ArcMap 9.2. Generar reporte de vecindad de un Layer de poligonos Reporte de coordenadas minimas y maximas (X,Y Min X, Y Max) de los feature de un FClass de poligonos Reporte de coordenadas de los centroides de los feature de poligonos.
Platform: | Size: 91136 | Author: jimypz | Hits:

[AI-NN-PRf(x)=xP10sin(5x)P7cos(4x)--max

Description: 遗传求解详例,其中包括程序,以及一些文字说明。-genertic algorithm
Platform: | Size: 4096 | Author: 周涛 | Hits:

[Chess Poker gamesrjxq-mygame

Description: 人机下棋游戏,用C++语言实现,其中包含一个.h和一个.cpp文件.int n=0,max=0,temp=0 char ch //输出棋盘原始状态 Print() cout<<"You First or Second(F/S)?" cin>>ch if((ch== S )||(ch== s )) { status[4]= X counter=counter+1 Print() cout<<"Computer plays 5!"<<endl }-it is include two files,one is .h file,another one is .cpp file. int n=0,max=0,temp=0 char ch //输出棋盘原始状态 Print() cout<<"You First or Second(F/S)?" cin>>ch if((ch== S )||(ch== s )) { status[4]= X counter=counter+1 Print() cout<<"Computer plays 5!"<<endl }
Platform: | Size: 121856 | Author: 园子 | Hits:

[Data structsSeqList

Description: 针对带附加头结点的单链表,试编写下列函数。 (1)定位函数Locate:在单链表中寻找第i个结点。若找到,则函数返回第i个结点的地址;若找不到,则函数返回NULL。 (2)求最大值函数max:通过一趟遍历在单链表中确定值最大的结点。 (3)统计函数number:统计单链表中具有给定值x的所有元素。 (4)建立函数create:根据一维数组a[n]建立一个单链表,使单链表中各元素的次序与a[n]中各元素的次序相同,要求该程序的时间复杂性为O(n)。 -Singly linked list with additional head node, try to write the following function. (1) orientation function Locate is: in a single linked list to find the i-th node. If found, the function returns the i-node address if no, then the function returns NULL. (2) the maximum function max: traversing through a trip to determine the value of the largest node in a singly linked list. (3) The number of statistical functions: statistical singly linked list with all the elements of a given value of x. (4) the establishment of the create function: one-dimensional array a [n] to establish a single linked list, singly linked list in the order of the elements a [n] is the order of the elements of the same requirements of the program s time complexity is O (n).
Platform: | Size: 1024 | Author: zby | Hits:

[AlgorithmFunc_Ga

Description: Finds a maximum of a function of several variables. fmaxga solves problems of the form: max F(X) subject to: LB <= X <= UB-Finds a maximum of a function of several variables. fmaxga solves problems of the form: max F(X) subject to: LB <= X <= UB
Platform: | Size: 3072 | Author: nanoc | Hits:

[matlabGeneticAlgorithm

Description: Test function : f= x+10*sin(x.*5)+7*cos(x.*4) Aim to get its max value- Test function: f = ' x+10* sin (x.* 5)+7* cos (x.* 4)' Aim to get its max value
Platform: | Size: 104448 | Author: 王银峰 | Hits:

[3D Graphicconv3ds

Description: 装换工具 将3ds文件装换为x文件 这样c++可以直接调用-3ds max
Platform: | Size: 47104 | Author: 李明 | Hits:

[matlab[M]

Description: 原始单纯形法(大M法,无需给出初始基变量) Programmed by Liyang(faruto s Studio~!) BNU MATH Email:liyangbnu@mail.bnu.edu.cn QQ:516667408 last modified 2008.4.27 求解标准型线性规划:max C*X s.t. A*X=b (b>=0) X>=0 输入:C是n维行向量,A是m*n的系数矩阵,b是m维列向量- Of the original simplex method (big M method, without giving the initial basic variable) Programmed by Liyang (faruto' s Studio ~!) BNU MATH Email: liyangbnu@mail.bnu.edu.cn QQ: 516667408 last modified 2008.4.27 solving standard linear programming: max C* X st A* X = b (b> = 0) X> = 0 input: C is the n-dimensional row vector, A is the coefficient matrix of m* n, b is the m-dimensional column vector
Platform: | Size: 1024 | Author: hongye | Hits:

[matlabget_ft

Description: input: fun x(t) t min & t max a df T = t_max-t_min fs = a*1/T 输入1:方程,xt;输入2,3:信号起始时间;输入4:选取的采样周期是信号长度倒数的倍数, 默认为50;输入5:频域分辨率,默认为0.3 输出频域信号和频域范围 为了提高fft的速度,频域信号长度是2的次方- Input: fun x (t) t min & t max a df T = t_max-t_min fs = a* 1/T Input 1: Equation XT input 2,3: signal start time input 4: select the sampling period is the reciprocal multiples of the signal length, default is 50 input 5: frequency domain resolution, default is 0.3 of the output frequency domain signal and the frequency domain in order to improve the speed of the FFT, the frequency domain signal length is a power of 2
Platform: | Size: 1024 | Author: 庞岩 | Hits:

[Software Engineeringgdfg

Description: duilib 源 码 分 析 文 档 - 1 stdAfx.h * lengthof(x) 计算数组长度宏 * MAX 两值最大 * MIN 两值最小 * CLAMP(x,a,b) x在a,b之间则取x否则取最小 2 UIBase * UI_WNDSTYLE_ 窗口风格属性 * UI_CLASSSTYLE_ 窗口类属性 * TRACE(LPCTSTR pstrFormat, ...) 功能 - 调试信息 参考 - 参考printf-duilib source analysis document
Platform: | Size: 9216 | Author: 1010841065 | Hits:

[CSharporiginalsimple

Description: 原始单纯形法(需直接给出初始的基变量) 求解标准型线性规划:max C*X s.t. A*X=b (b>=0) X>=0 输入:C是n维行向量,A是m*n的系数矩阵,b是m维列向量,XB承装初始基变量的下标。输出:x最优解(如果有的话),fval最优值,flag解的状态说明,interation求解时的循环次数。-Original simplex method (to be given directly to the initial base variable) Solving the standard linear programming: max C* X st A* x = b (b> = 0) X> = 0 Input: C is an n-dimensional row vector, A is a coefficient matrix of m* n, b is the m-dimensional column vector, XB Holds initial base variable subscript Output: x optimal solution (if any) fval optimal value of the state flag solution illustrates the interation solution cycles.
Platform: | Size: 1024 | Author: kangkankan | Hits:

[VC/MFCXfile-outport

Description: 讲述在3Dmax中怎么导出.x文件,为DirectX使用,希望对大家有所用。-how to export Xfile in the 3D max,and use in the DriectX
Platform: | Size: 736256 | Author: 陈海全 | Hits:

[Other Embeded programWS0010.ZIP

Description: FEATURES • CMOS technology • Low power consumption • Microprocessor Interface -- High-speed 8-bit parallel bi-directional interface with 6800-series or 8080-series -- Serial interface available • 4-bit or 8-Bit MPU interface • High speed MPU interface: 2MHz (VDD=5V) • 128 x 8-bit display RAM (128 characters max.) • Auto reset function • 5 x 8 and 5 x 10 dot matrix • Built-in oscillator with external resistors • Programmable duty cycle: - 1/8 duty: (1 display line, 5 x 8 dots with cursor) - 1/11 duty: (1 display line, 5 x 10 dots with cursor) - 1/16 duty: (2 display lines, 5 x 8 dots with cursor) -FEATURES • CMOS technology • Low power consumption • Microprocessor Interface -- High-speed 8-bit parallel bi-directional interface with 6800-series or 8080-series -- Serial interface available • 4-bit or 8-Bit MPU interface • High speed MPU interface: 2MHz (VDD=5V) • 128 x 8-bit display RAM (128 characters max.) • Auto reset function • 5 x 8 and 5 x 10 dot matrix • Built-in oscillator with external resistors • Programmable duty cycle: - 1/8 duty: (1 display line, 5 x 8 dots with cursor) - 1/11 duty: (1 display line, 5 x 10 dots with cursor) - 1/16 duty: (2 display lines, 5 x 8 dots with cursor)
Platform: | Size: 10356736 | Author: Denis | Hits:

[Game Engine0767

Description: 3D MAX中导出X模型文件,将Panda插件安装到3D Max中,并掌握X文件的导出方法; -how to use 3D MAX to make X file
Platform: | Size: 593920 | Author: lvlingxuan | Hits:

[Otherdaishu2

Description: 设代数系统V1=<A, .>、V2=<A, .>、 V3=<A, ⊙>、 V4=<A, △>,A={ x|x∈Z,1≤x≤20}, a.b= 2a+b,a.b=max(a, b),a⊙b=(a+b) mod8 + 2,a△b=min(a,b),试设计算法并编程实验, 要求: (1)试判断V1、V2、V3、V4 是否为代数系统。 (2)试判断这些代数系统是否有左单位元与右单位元、零元,如存在,请求出。 -Let Algebra System V1 = <A, .> , V2 = <A, .> , V3 = <A, ⊙> , V4 = <A, △> , A = {x | x ∈ Z, 1 ≤ x ≤ 20}, ab = 2a+b, ab = max (a, b), a ⊙ b = (a+b) mod8+ 2, a △ b = min (a, b ), try to design algorithms and programming experiment requirements: (1) trial judge V1, V2, V3, V4 is an algebraic system. (2) whether the trial judge these algebraic systems have left and right units per unit, nil, if present, the request out.
Platform: | Size: 158720 | Author: 梁涛 | Hits:

[matlabpsoToolbox

Description: Help psoToolbox V1.0 psoToolbox provides an interective GUI based Toolbox to solve optimization problems using particle swarm optimization. Creat a fitness function in M-file. Inputs: Function : Function handle of fitness function. Nvars : Number of variable to be optimized. LB : Lower Bound of Nvars (1 X Nvars) UB : Upper Bound of Nvars (1 X Nvars) Parameters: C1 : Cognative Attraction C2 : Social Attraction W : Inertial Population Size : Number of Swarms Max Iterations : Maximum number of epochs. Click on " RUN PSO " button to start PSO search. You will get the out put at Edit box below the axes.-Help psoToolbox V1.0 psoToolbox provides an interective GUI based Toolbox to solve optimization problems using particle swarm optimization. Creat a fitness function in M-file. Inputs: Function : Function handle of fitness function. Nvars : Number of variable to be optimized. LB : Lower Bound of Nvars (1 X Nvars) UB : Upper Bound of Nvars (1 X Nvars) Parameters: C1 : Cognative Attraction C2 : Social Attraction W : Inertial Population Size : Number of Swarms Max Iterations : Maximum number of epochs. Click on " RUN PSO " button to start PSO search. You will get the out put at Edit box below the axes.
Platform: | Size: 99328 | Author: kiumars | Hits:
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