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Description: 该程序是非特定人语音识别程序，原来在SPCE500A下运行，能识别并应答16个命令，存储空间为128K字节。 //我们将其进行了修改，使其能在SPCE061A下运行，由于存储空间为32K字，只能识别并应答9个命令。 //在这个例程当中，我们示范了如何使用语音识别引擎对8个已经经过训练的命令进行识别。 //第一个命令 \"Snoopy\"，可以作为使能命令，也就是说使用者只能在该命令之后发出其他命令。 //例如，用户只有在发出 \"Snoopy\"命令并且听到响应\"Snoopy here\"之后才能继续下达其他命令 //如\"How are you\", \"I love you\",等等。 //当\"Snoopy\"命令响应之后，程序将等待8秒钟，如果没有其他命令发出， //系统将跳回到第0组，并且再次等待触发命令\"Snoopy\"。当用户的命令无法被识别时， //程序将响应为\"Execuse me?\", \"I can t hear you!\"或 \"Say again!\"这时用户可以再一次下达语音命令。-the procedures were non-specific voice recognition procedure, the original run in SPCE500A can identify and Response 16 orders, storage space for 128K bytes. / / We had to amending it so that it can run in SPCE061A, storage space for 32K words, recognize only nine responses and orders. / / This routine, we have demonstrated how to use voice recognition engine has eight pairs of trained orders for identification. / / The first order "Jack Champion", as it can make an order, users can only say in the order issued after other orders. / / For example, users only the issue of "Jack Champion" command and response hear "Jack Champion here" can be issued to other orders / / as "How are you" and "I love you" and so on. / / When "Jack Champion&q
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Size: 121445 |
Author: 胡珈 |
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Description: 十个小孩围成一圈分糖果，老师分给第一个小孩10块，第二
个小孩2块，第三个小孩8块，第四个小孩22块，第五个小孩16块
，第六个小孩4块，第七个小孩10块，第八个小孩6块，第九个小
孩14块，第十个小孩20块。然后所有的小孩同时将手中的糖分一
半给右边的小孩；糖块数为奇数的人可向老师要一块。问经过这
样几次后大家手中的糖的块数一样多？每人各有多少块糖？-10 points .... candy children, teachers added to a child 10 and the second two children, the third child eight, the fourth child 22, the fifth child 16, a sixth four children, the seventh child 10, the eighth child six, nine children 14, 10, 20 child. Then all of the children to the hands of sugar to the right half of the children; Retrievers few odd people to be a teacher. Q After this we After several hands of the few pieces of sugar as much? The number of blocks each have sugar?
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Size: 1291 |
Author: 姚紫欣 |
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Description: 这是一个字母或数制之间的转化程序，5为主模块，调用模块1和模块7
模块1又调用模块2、3、4和6四个模块，其中：
模块2实现小写字母向大写字母的转换
模块3实现大写字母向小写字母的转换
模块4实现二进制数向十六进制数的转换
模块6实现十六进制数向二进制数的转换
模块7实现十六进制数向十进制数的转换
按“q”键退出。
使用时，需将7个文件分别汇编，连接的方法为：
5＋1＋2＋3＋4＋6＋7
生成可执行文件“5” 即可运行。
-This a letter or number into the system procedures, five main modules, a module called seven modules and module also a module called 2,3,4 and 6 4 module, which : module 2 lowercase letters to the realization of capital letters conversion modules to achieve three capital letters to lowercase letters conversion module is 4 now the binary number to hexadecimal number of conversion modules to achieve 6 hexadecimal number to the number of binary conversion modules into seven achieve 16 decimal system for a few to several conversion by "q" key to exit. When used, the need to document seven separate compilation, linking methods : 5 +1 +2 +3 +4 +6 +7 generate executable file "5" can run.
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Size: 848 |
Author: 严岩 |
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Description: 这是一个字母或数制之间的转化程序，5为主模块，调用模块1和模块7
模块1又调用模块2、3、4和6四个模块，其中：
模块2实现小写字母向大写字母的转换
模块3实现大写字母向小写字母的转换
模块4实现二进制数向十六进制数的转换
模块6实现十六进制数向二进制数的转换
模块7实现十六进制数向十进制数的转换
按“q”键退出。
使用时，需将7个文件分别汇编，连接的方法为：
5＋1＋2＋3＋4＋6＋7
生成可执行文件“5” 即可运行。-This a letter or number into the system procedures, five main modules, Call module and a module is a 7 module called Blocks 2, 3, 4 and 6 4 module, which : Module 2 lowercase letters to the realization of capital letters conversion module to achieve three capital letters to lowercase letters conversion module is 4 is the binary number to hexadecimal number of conversion modules to achieve 6 hexadecimal number to the number of binary conversion modules into seven achieve 16 to the metric system for a few number of conversion by the "q" key to exit. Use, the need to document seven separate compilation, linking method : 5 +1 +2 +3 +4 +6 +7 generating executable files "5" will be running.
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Size: 4893 |
Author: 李浪 |
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Description: 这是一个字母或数制之间的转化程序，5为主模块，调用模块1和模块7
模块1又调用模块2、3、4和6四个模块，其中：
模块2实现小写字母向大写字母的转换
模块3实现大写字母向小写字母的转换
模块4实现二进制数向十六进制数的转换
模块6实现十六进制数向二进制数的转换
模块7实现十六进制数向十进制数的转换
按“q”键退出。
使用时，需将7个文件分别汇编，连接的方法为：
5＋1＋2＋3＋4＋6＋7
生成可执行文件“5” 即可运行。-This a letter or number into the system procedures, five main modules, Call module and a module is a 7 module called Blocks 2, 3, 4 and 6 4 module, which : Module 2 lowercase letters to the realization of capital letters conversion module to achieve three capital letters to lowercase letters conversion module is 4 is the binary number to hexadecimal number of conversion modules to achieve 6 hexadecimal number to the number of binary conversion modules into seven achieve 16 to the metric system for a few number of conversion by the "q" key to exit. Use, the need to document seven separate compilation, linking method : 5 +1 +2 +3 +4 +6 +7 generating executable files "5" will be running.
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Size: 4442 |
Author: richard |
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Description: 用FFT分别计算Xa(n) (p=8, q=2)与Xb(n) (a =0.1,f =0.0625)的16点循环卷积和线性卷积。
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Size: 10018 |
Author: 万宁宁 |
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Description: 游戏说明
用法：
用户名和密码都是1234
1，输入后，会验证，通过后会出现全屏幕的状态，无论当前是什么模式，都会统一设置成800*600，16位色深，60HZ的刷新率模式，如果当前系统不允许修改显示模式(linux下测试会出现此情况)，则会使用当前的分辩率，这样可以会有一些显示方面的问题
2，押分键为主键盘的1至9，按回车键起动，跑完后，会根据你押的分数算出你的得分，如果你中了，则会重复播放音乐，如果没有中，则不播音乐
3，音乐会一直播下去，直到你按回车键把分移回去为止，此时你可以再按一次回车键续押上一把的情况，也可以重新押分。另外也可以根据你的得分，进行猜大小，按B键出现两个小人物，再按B键消除，你猜是左边赢还是右边赢，按Q猜左边赢，按P猜右边赢，如果是平的话，则按QP都赢。输完了后，你还可以用向左方向键把你已有的分移到左边来，继续玩，也可以把左边的分用右方向键移到右边去，如果按回车，左边有分的话，则所有的分会自动到右边去。
4 ，一开始进去，必须得押分后，按回车键才有效，当当前分数小于你上次押的分时，按回车键将不能续押你上次的分数，此时必须重新押分。
5，退出游戏请按ESC
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Size: 24121 |
Author: 希望 |
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Description: 实现最优二叉树的构造；在此基础上完成哈夫曼编码器与译码器。 假设报文中只会出现如下表所示的字符：
字符 A B C D E F G H I J K L M N
频度 186 64 13 22 32 103 21 15 47 57 1 5 32 20 57
字符 O P Q R S T U V W X Y Z ， ．
频度 63 15 1 48 51 80 23 8 18 1 16 1 6 2
要求完成的系统应具备如下的功能：
1．初始化。从终端（文件）读入字符集的数据信息，。建立哈夫曼树。
2．编码：利用已建好的哈夫曼树对明文文件进行编码，并存入目标文件（哈夫曼码文件）。
3．译码：利用已建好的哈夫曼树对目标文件（哈夫曼码文件）进行编码，并存入指定的明文文件。
4．输出哈夫曼编码文件：输出每一个字符的哈夫曼编码。
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Size: 132535 |
Author: 张娟 |
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Description: 仿真1：首先把网络温度参数T固定在100，按工作规则共进行1000次状态更新,把这1000次状态转移中网络中的各个状态出现的次数Si（i＝1，2，…，16）记录下来 按下式计算各个状态出现的实际频率： Pi=Si/∑i=1,NSi=Si/M 同时按照Bo1tzmann分布计算网络各个状态出现概率的理论值： Q（Ei）=（1/Z）exp（-Ei/T） 仿真2：实施降温方案，重新计算 采用快速降温方案：T（t）= T0/（1+t） T从1000降到0.01，按工作规则更新网络状态 当T=0.01时结束降温，再让T保持在0.01进行1000次状态转移，比较两种概率-a simulation : First of all network parameters temperature T fixed at 100 and, according to the rules for a total of 1000 to update the state, this state of the 1000 network transfer of all states for the number of Si (i = 1, 2, ..., 16) all recorded determined by the formula state-of the actual frequency : Pi = Si / i = 1, NSi = Si / M in accordance with Bo1tzmann distributed computing network of states all probability the theoretical value : Q (Ei) = (1 / Z) exp (- Ei / T) Simulation 2 : implementation of cooling, re-using rapid cooling programs : T (t) = T0 / (1 t) T dropped to 0.01 from 1000 and, according to the rules updated network state when T = 0.01 at the end of cooling, let T at 0.01 for the 1000 state transfer, the probability of two more
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Author: 韵子 |
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Description: 第一部分 基础篇 第1章 绪论 3 1.1 统一建模语言UML 3 1.1.1 UML的背景 3 1.1.2 UML的发展 3 1.1.3 UML的内容 5 1.1.4 UML的主要特点 5 1.1.5 UML的功能 6 1.1.6 UML的组成 7 1.2 Rational统一过程（Rational Unified Process） 9 1.2.1 RUP的发展 9 1.2.2 什么是RUP 10 1.2.3 过程概览 11 1.2.4 时间轴 12 1.2.5 迭代 14 1.2.6 工作流（Workflows） 15 1.2.7 微过程的划分 16 1.3 工具 20 1.4 小结 20 第2章 面向对象分析与设计方法 21 2.1 OOA／OOD方法 21 2.1.1 面向对象分析（OOA） 23 2.1.2 面向对象设计（OOD） 24 2.2 OMT方法 25 2.2.1 分析 26 2.2.2 系统设计 28 2.2.3 对象设计（Object Design） 29 2.2.4 实现（Implementation） 30 2.2.5 测试（Testing） 30 2.2.6 模型 30 2.3 Booch方法 31 2.3.1 宏过程 32 2.3.2 微过程 32 2.4 OOSE方法 34 2.4.1 分析阶段 35 2.4.2 构造阶段 35 2.4.3 测试阶段 36 2.5 Fusion 方法 36 2.5.1 分析阶段 37 2.5.2 设计阶段 38 2.5.3 实现阶段 39 2.6 小结 39 第3章 UML的关系 40 3.1 依赖关系（Dependency Relationship） 40 3.2 类属关系（Generalization Relationship） 43 3.3 关联关系（Association Relationship） 45 3.3.1 角色（Role）与阶元（Multiplicity） 45 3.3.2 导航（Navigation） 46 3.3.3 可见性（Visibility） 47 3.3.4 限定符（Q……
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Author: superhyi@gmail.com |
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Description: asm源码
汇编引导程序 源程序
汇编初见面 名称 解释 格式
a (Assemble) 逐行汇编 a [address]
c (Compare) 比较两内存块 c range address
d (Dump) 内存16进制显示 d [address]或 d [range]
e (Enter) 修改内存字节 e address [list]
f (fin) 预置一段内存 f range list
g (Go) 执行程序 g [=address][address...]
h (Hexavithmetic) 制算术运算 h value value
i (Input) 从指定端口地址输入 i pataddress
l (Load) 读盘 l [address [driver seetor>
m (Move) 内存块传送 m range address
n (Name) 置文件名 n filespec [filespec...]
o (Output) 从指定端口地址输出 o portadress byte
q (Quit) 结束 q
r (Register) 显示和修改寄存器 r [register name]
s (Search) 查找字节串 s range list
t (Trace) 跟踪执行 t [=address] [value]
u (Unassemble) 反汇编 u [address ]或range
w (Write) 存盘 w [address[driver sector secnum>
? 联机帮助 ?
debug小汇编a命令
debug小汇编a命令是一个很有用的功能，许多的小程序都要他来做。
编一些小程序比汇编要来得方便，快洁。
在Debug中，中断是非常有用的，首先，让我们先了解一下中断。
所谓中断，其实，就是，当你做某事时，有人过来找你有其他事，你先放下手中的事（计算机中，称为保护现场）
，再去与叫你的那个人办事去，等完了，你又回，接着做刚才的事。这是个很通俗的讲法。
计算机在运行时，也会出现这种情况，我们叫之中断。
下面是他的一些常用中断向量的入口值详解：(记住哦，很用的...呵呵)
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Size: 17612 |
Author: 525836987@qq.com |
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Description: 仿真1：首先把网络温度参数T固定在100，按工作规则共进行1000次状态更新,把这1000次状态转移中网络中的各个状态出现的次数Si（i＝1，2，…，16）记录下来 按下式计算各个状态出现的实际频率： Pi=Si/∑i=1,NSi=Si/M 同时按照Bo1tzmann分布计算网络各个状态出现概率的理论值： Q（Ei）=（1/Z）exp（-Ei/T） 仿真2：实施降温方案，重新计算 采用快速降温方案：T（t）= T0/（1+t） T从1000降到0.01，按工作规则更新网络状态 当T=0.01时结束降温，再让T保持在0.01进行1000次状态转移，比较两种概率-a simulation : First of all network parameters temperature T fixed at 100 and, according to the rules for a total of 1000 to update the state, this state of the 1000 network transfer of all states for the number of Si (i = 1, 2, ..., 16) all recorded determined by the formula state-of the actual frequency : Pi = Si/i = 1, NSi = Si/M in accordance with Bo1tzmann distributed computing network of states all probability the theoretical value : Q (Ei) = (1/Z) exp (- Ei/T) Simulation 2 : implementation of cooling, re-using rapid cooling programs : T (t) = T0/(1 t) T dropped to 0.01 from 1000 and, according to the rules updated network state when T = 0.01 at the end of cooling, let T at 0.01 for the 1000 state transfer, the probability of two more
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Size: 5120 |
Author: 韵子 |
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Description: 用matlab做的qam的基本原理仿真，清晰明了，有相应结果，可作为qam调制的入门教程。-QAM matlab to do with the basic principles of simulation, clarity, have corresponding results can be used as QAM Modulation Tutorial.
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Size: 1024 |
Author: 阿飞 |
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Description: 采用硬件：
MP416（16位AD，5V/10V/±5V/±10V量程）
设计思路：
1. 使用MP416_AD()函数，采用通道1测试，基准板输出为±200mV,±2.5V,±10V,GND
等，考虑测量精度，使用MP416的±5V量程测量基准板输出±200mV,±2.5V电压；使用MP416的±10V量程测量基准板输出±10V电压。即通过键盘输入gain值2或3来选择合适量程。
2. 不采用MP416_AD()函数测量平均值，故取测量1次。
3. 采样十次（便于分析测量误差）将值存入数组以便处理。
4. 测量值包括AD采样最大值，最小值；电压最大值、最小值，平均值；最大跳字，及对应最大电压变化值。
5. 按回车键，可以继续测量；G键位重新选择量程；ESC或Q键退出。
- use Hardware:
MP416 (16 Wei AD, 5V/10V/± 5V/± 10V range)
design ideas:
1. Use MP416_AD () function, using channel 1 test, the base plate output is ± 200mV, ± 2.5V, ± 10V, GND
And so on, consider the measurement accuracy, use the MP416 measuring range of ± 5V output of the base plate ± 200mV, ± 2.5V voltage the use of MP416 measuring range of ± 10V output ± 10V voltage of the base plate. Input gain value through the keyboard 2 or 3 to select the appropriate range.
2. Do not use MP416_AD () function is measured, on average, so they chose a time measurement.
3. Sampled 10 times (to facilitate analysis of measurement error) will be deposited into an array of values for processing.
4. Measured values, including AD sample maximum, minimum voltage maximum, minimum, average the biggest jump characters, and the corresponding changes in the value of maximum voltage.
5. Press Enter key, you can continue to measure G keys re-select
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Size: 389120 |
Author: whb |
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Description: this is simmulation 16-qam modulation in rayleigh channel
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Size: 8192 |
Author: baimengcian |
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Description: An 8-QAM communications channel simulation in Simulink, constructed from a 16-QAM model and using an I,Q correlation receiver.
QAM (quadrature amplitude modulation) is a method of combining two amplitude-modulated (AM) signals into a single channel, thereby doubling the effective bandwidth. QAM is used with pulse amplitude modulation (PAM) in digital systems, especially in wireless applications. In a QAM signal, there are two carriers, each having the same frequency but differing in phase by 90 degrees (one quarter of a cycle, from which the term quadrature arises). One signal is called the ‘I’ signal, and the other is called the ‘Q’ signal. Mathematically, one of the signals can be represented by a sine wave, and the other by a cosine wave. The two modulated carriers are combined at the source for transmission. At the destination, the carriers are separated, the data is extracted from each, and then the data is combined into the original modulating information. -An 8-QAM communications channel simulation in Simulink, constructed from a 16-QAM model and using an I,Q correlation receiver.
QAM (quadrature amplitude modulation) is a method of combining two amplitude-modulated (AM) signals into a single channel, thereby doubling the effective bandwidth. QAM is used with pulse amplitude modulation (PAM) in digital systems, especially in wireless applications. In a QAM signal, there are two carriers, each having the same frequency but differing in phase by 90 degrees (one quarter of a cycle, from which the term quadrature arises). One signal is called the ‘I’ signal, and the other is called the ‘Q’ signal. Mathematically, one of the signals can be represented by a sine wave, and the other by a cosine wave. The two modulated carriers are combined at the source for transmission. At the destination, the carriers are separated, the data is extracted from each, and then the data is combined into the original modulating information.
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Size: 11264 |
Author: Griffin Wright |
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Description: the OFDM PHY is adaptive therefore it supports
multiple schemes BPSK, QPSK, 16-QAM and 64-QAM for
data carriers’ modulation. The constellation diagrams are
gray mapped and shows the magnitudes I and Q (In-phase
and Quadrature) components of each incoming bit(s)
combination along with their normalization factor C to
calculate magnitude of each model
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Size: 1497088 |
Author: san |
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Description: FPGA经典100问之＜下载验证16问＞。介绍了FPGA在下载验证过程中的常见问题，对FPGA常见配置电路进行了讲解。-FPGA asked the classic 100 < Download verified 16 Q> . FAQ introduced FPGA verification process the download of FPGA configuration circuit common were explained.
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Author: |
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Description: 16 QAM (Q amplitude Modulation)
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Size: 68608 |
Author: zay |
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Description: 典型测试系统3
(1) 如下图所示的IEEE 9节点电力系统：
2．元件参数
（1） 线路参数
线路号 长度（km） ro(Ω/km) xo(Ω/km) bo(1/Ωkm)
L1 25 0.131 0.4 2.98×10-6
L2 50 0.131 0.4 2.98×10-6
L3 45 0.105 0.393 3.04×10-6
L4 30 0.131 0.4 2.98×10-6
L5 35 0.105 0.393 3.04×10-6
L6 20 0.131 0.4 2.98×10-6
（2） 变压器参数
变压器 容量（MVA） 变比 Pk(kW) P0(kW) Uk% i0%
T1 50 121/10.5 250 55 10.5 1.5
T2 80 121/10.5 410 98 10.5 1.5
T3 50 121/10.5 250 55 10.5 1.5
(3) 节点参数
节点名 类 型 P (MW) Q (MVar) 电压 相角
1 PQ 35 18
2 PQ 30 16
3 PQ 40 20
4 PQ 25 12
5 PQ 20 10
6 PQ 15 6
7 PV 40 1.05
8 PV 45 1.05
9 平衡点 1.05 0.0(Typical test system 3
(1) The IEEE 9-node power system shown in the following figure:
2. Component parameters
(1) Line parameters
Line number (km) ro (_/km) XO (_/km) Bo (1/_km)
L1 25 0.131 0.4 2.98 x 10-6
L2 50 0.131 0.4 2.98 x 10-6
L3 45 0.105 0.393 3 3.04 x 10-6
L4 30 0.131 0.4 2.98 x 10-6
L5 35 0.105 0.393 3 3.04 x 10-6
L6 20.131 0.4 2.98 x 10-6
(2) Transformer parameters
Transformer Capacity (MVA) Variable Ratio Pk (kW) P 0 (kW) Uk% i0%
T1 50 121/10.5 250 55 10.5 1.5
T2 80 121/10.5 410 98 10.5 1.5
T3 50 121/10.5 250 55 10.5 1.5
(3) Node parameters
Node Name Type P (MW) Q (MVar) Voltage Phase Angle
1 PQ 3518
2 PQ 3016
3 PQ 4020
4 PQ 2512
5 PQ 2010
6 PQ 156
7 PV 40 1.05
8 PV 45 1.05
9 Balance Point 1.05 0.0)
Platform: |
Size: 19456 |
Author: GKen |
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