Lecture Notes for PAMC (07) - Ch4_L2
Lecture Notes for A Mathematical Introduction to Robotic Manipulation
Lecture Notes for A Mathematical Introduction to Robotic Manipulation
By Z.X. Li∗ and Y.Q. Wu
Dept. of ECE, Hong Kong University of Science & Technology School of ME, Shanghai Jiaotong University
Chapter 4 Robot Dynamics and Control
4.2 Intertial Properties of Rigid Body ◻ Kinetic energy of a rigid body:
Chapter Robot Dynamics and Control Lagrangian Equations Inertial Properties of Rigid Body Dynamics of an Open-chain Manipulator Newton-Euler Equations Coordinateinvariant algorithms for robot dynamics Lagrange’s Equations with Constraints
July
Chapter 4 Robot Dynamics and Control
Summer School-Math. Methods in Robotics@TU-BS.DE
-
July
Chapter Robot Dynamics and Control Lagrangian Equations Inertial Properties of Rigid Body Dynamics of an Open-chain Manipulator Newton-Euler Equations Coordinateinvariant algorithms for robot dynamics Lagrange’s Equations with Constraints
Lecture Notes for PAMC (03) - Ch2_L2-201415(1)
…
22H 21H 20H
字节地址
…
10H 11H 12H 13H 14H 15H 16H 17H 08H 09H 0AH 0BH 0CH 0DH 0EH 0FH 00H 01H 02H 03H 04H 05H 06H 07H
LSB 位地址 MSB
B (C 20H ) 08H
位地址B 对应的字节地址C 及偏移d 分别为
通用寄存器区 (32个字节地址): 00H~1FH 通用位变量区 (16个字节地址): 20H~2FH; 128个位地址: 00H~7FH
高128字节RAM (地址范围: 80H~FFH)
一般数据存储器, 堆栈等; 只能采用间接地址方式访问 特殊功能单元的编程控制; 只能采用直接地址方式访问 只能采用MOVX指令(即外部数据存储器和I/O地址)访问
总结
MCS-51单片机存储器系统的组成 程序和数据存储器空间的使用 片内数据存储器功能分区及其作用 不同功能分区的访问方法 特殊功能寄存器(和特殊功能单元的关系) MCS-51单片机地址空间的编址 总结MCS-51单片机的基本资源
问题和讨论
4个通用寄存器组在使用时有什么特殊作用? 为什么特殊功能位的起始地址是从80H, 而不 是从00H? MCS-51单片机的地址空间编址应该遵循什么 原则? MCS-51单片机具有布尔(位)处理器，所以可 以单独控制单个I/O位的状态，在哪些场合这 种操作有优点？
端口 5个中断源 1个串行端口 3个16位定时/计数器 等.
DPTR 请问: 为什么8052比8051 多出6个SFRs?
其它衍生的类型可能集成了更多的特殊功能单元(如ADC和DAC, 串行端 口等), 所以将具有更多的SFRs
Lecture4
4.1.2 Laws of Large Numbers and Central Limit Theorem Weak Laws of Large Numbers (WLLN): Let xn = ∑ i =1 xi / n , where
n
{ xi }
is a
sequence of i.i.d. random variables with E ( xi ) = µ and V ( xi ) = σ 2 . Then
d
p lim g ( xn ) = g ( p lim xn ) .
Based on the Slutsky theorem, we can derive the following rules for probability limits if xn and yn are random variables with plim xn = c and plim yn = d : (1) p lim( xn + yn ) = c + d
The Slutsky theorem highlights a comparison between the expectation of a random variable and its probability limit. The Slutsky Theorems is very helpful in deriving the asymptotics of functions of sample moments.
(5) If xn and yn are random matrices with p lim xn = A and p lim yn = B , then
p lim xn yn = AB .
MIT基础数学讲义(计算机系)lecture4
2 De nitions
A nuisance in rst learning graph theory is that there are so many de nitions. They all correspond to intuitive ideas, but can take a long time to absorb. Worse, the same thing often has several names and the same name sometimes means slightly di erent things to di erent people! It's a big mess, but muddle through.
2.2 Not-So-Simple Graphs
There are actually many variants on the de nition of a graph. The de nition in the preceding section really only describes simple graphs. There are many ways to complicate matters.
2.1 Simple Graphs
A graph is a pair of sets (V E ). The elements of V are called vertices. The elements of E are called edges. Each edge is a pair of distinct vertices. Graphs are also sometimes called networks. Vertices are also sometimes called nodes. Edges are sometimes called arcs. Graphs can be nicely represented with a diagram of dots and lines as shown in Figure 2 As noted in the de nition, each edge (u v ) 2 E is a pair of distinct vertices u v 2 V . Edge (u v ) is said to be incident to vertices u and v . Vertices u and v are said to be adjacent or neighbors. Phrases like, \an edge joins u and v " and \the edge between u and v " are comon. A computer network is can be modeled nicely as a graph. In this instance, the set of vertices V represents the set of computers in the network. There is an edge (u v) if there is a direct communication link between the computers corresponding to u and v .
lecture_7_Geant4_2
除了轻子、玻色子还包括： G4MesonConstructor G4BaryonConstructor G4IonConstructor G4ShortlivedConstructor 如果对过程中可能需要的粒 子不确定，可以用这种办法 把所有粒子都构造出来。
void ExN01PhysicsList::ConstructParticle() { ConstructLeptons();//构造轻子 ConstructBosons(); //构造玻色子 ... }
2012-4-12 13
G4ParticleGun
//参数n_particle表示一次发射的粒子数目 G4ParticleGun* particleGun = new G4ParticleGun(n_particle); G4ParticleGun有很多设置函数，如： SetParticleDefinition(G4ParticleDefinition*); //粒子类型 SetParticleMomentumDirection(G4ThreeVector); //动量方向 SetParticleEnergy(G4double); //能量 SetParticlePosition(G4ThreeVector); //发射位置 ... 粒子枪的属性设置好之后，才调用generatePrimaryVertex()函数，产生 事例的主顶点。 粒子枪本身不提供随机性，发射的粒子都是指定的。如果需要按照某分布随 即发射粒子，需要在调用generatePrimaryVertex()之前，利用Geant4提 供的随机数产生子自己写出需要的分布。 参见ExN01PrimaryGeneratorAction::generatePrimaries(G4Event*) 函数
Twisted vertex representations via spin groups and the McKay correspondence
TWISTED VERTEX REPRESENTATIONS VIA SPIN GROUPS AND THE MCKAY CORRESPONDENCE
IGOR B. FRENKEL, NAIHUAN JING, AND WEIQIANG WANG Abstract. We establish a twisted analog of our recent work on vertex representations and the McKay correspondence. For each finite group Γ and a virtual character of Γ we construct twisted vertex operators on the Fock space spanned by the super spin characters of the spin wreath products Γ ≀ Sn of Γ and a double cover of the symmetric group Sn for all n. When Γ is a subgroup of SL2 (C) with the McKay virtual character, our construction gives a group theoretic realization of the basic representations of the twisted affine and twisted toroidal algebras. When Γ is an arbitrary finite group and the virtual character is trivial, our vertex operator construction yields the spin character tables for Γ ≀ Sn .
Lecture 7 网易课教程字幕 有机化学 可汗学院
Lecture 7I’ve pre-drawn some molecules here. Let’s see if we can use what we know to name them. So what do we have? This first molecule right here, I have a bunch of rings. This is a one, two, three, four, five, six carbon ring. These are each four carbon rings: one, two, three, four, so the largest ring is essentially going to be our backbone. It’s going to be this six-carbon ring right here: one, two, three, four, five, six carbons. So that is a cyclohexane. All double bonds. That’s where we get the –ane. Six carbons hex-, it’s in a cycle, cyclo-, and then we have two of these four-carbon rings. So four carbons, we’re dealing with the prefix but-, but- for four. We’ve got two of them, so both of these are butyl groups. But they’re in a cycle, so they’re actually cyclobutyl groups. We have two things attached to this ring right there. If we only had one thing attached to it, you wouldn’t have to number it. But when you have two things, you start numbering at one of them. Let’s say we start numbering here, and you go in the direction so that the next group has the lowest number. So in this case, you want to go in the counterclockwise direction. If we went this way, it would be one, two, three, four, five. This guy would be a five. If we go in the counterclockwise, it would be one, two, three. This guy will only be a three. So this right here, we have two cyclobutyls, so it’s dicyclobutyl. We have two of them and they are at the one and three positions. So at the one and three position, I have two cyclobutyls on my cyclohexane main ring.Let’s try this one right here. I have a five-carbon ring, one, two, three, four, five right there, and then I have a one, two, three, four, five, six-carbon ring right there, so this is going to be the main ring. That is a cyclohexane. It has six carbons on them in a cycle, all single bonds. Attached to that, I have a cyclopentyl group, -yl for the group. This is a cyclopentyl group on it. We don’t have to number it because it’s only one group attached to the main ring. If there was another group, we would have to number it like we did up here. This is cyclopentyl. That’s this part right here. Cyclopentyl attached to cyclohexane: cyclopentylcyclohexane.Let’s try this one over here. The first thing we want to do, there’s no cycles here, but we have to identify the longest chain. To do that, let’s just count it out. It could be one, two, three, four, five, six, seven. That’s not the longest chain. Maybe it’s one, two, three, four, five, six, seven, eight. That looks like the longest chain, so let’s make that the longest chain. Let’s make that the longest chain right over there. We want to start numbering in the direction so we encounter the first attached groups first. We do want to start numbering down here, because we have groups attached right on the two carbon. If we started over here, we’d have to to pretty far until something’s attached. So we go one, two, three, four, five, six, seven, eight. We know we’re dealing with an octane, all single bonds. It’s not a cyclooctane. It’s not in a cycle, so we know we’re dealing with an octane. Now we just have to add the groups to it. So what do we have here? This right here is just one carbon attached to the main chain. This is another carbon attached to the main chain. Both of these right here are methyl groups. Meth- is for one carbon. Those are methyl groups. If you look at all of them, these are the only methyl groups. These two up here aren’t methyl groups, so we have two methyl groups on our entire chain, so it’s going to be dimethyl. Both of the methyl groups our at our two position, so this is going to be 2,2-dimethyl. This part right here is 2,2. That right there is 2,2-dimethyl. We’re going to decide what order to write it after we figure out what these are called because it has to be in alphabetical order. This is 2,2-dimethyl. The whole chain is an octane. What are these over here? How many carbons do we have here? One, two, three. They actually look the same. We have one, two, three here. We have one, two, three there, so these areboth propyl groups. If we deal with common names, this is kind of that Y shape. You could call it sec-propyl because this carbon right here that’s attached to the main chain is attached to two other carbons. But the more common one, because it forms this Y shape, is isopropyl. We have two isopropyl groups. These are both isopropyl. We could have disopropyl. They ‘re occurring at the four and five positions, two isopropyls at four and five. This is 4,5-disopropyl. That’s that group and that group right there are accounted for with this. Now we have to just figure out the order that we write it in. You ignore the di- or the tri- out front and you just look at them in alphabetical order. We have an I for isopropyl. We have an M for methyl. Let’s write the isopropyl first. I’ve actually seen some people want to go for the p, but the main thing I ignore is just the di- or the tri- in front of the isopropyl. You shouldn’t involve that. But everything after that, you do involve. So I’ll write the isopropyl first. I comes before M, so this is going to be—if we were going to write the whole thing, this is going to be 4,5-disopropyl, 2,2-dimethyl. Actually, this should be a comma here, 4,5-disopropyl-2,2-dimethyloctane. And we’re done. But this was just the common name. you might remember that when we deal with iso- or sec- or tert- butyl or propyl, that’s the common name. if we want the systematic name, we can start at where we are attached to the main chain and view that as one and then make the longest chain with that so one. And so you could say that we have a chain there and this would be both of these cases. So this is one, two carbons. two carbons, we’re dealing with an ethyl. And on the first carbon, you have a methyl attached to it. So you could also call each of these groups a 1-methylethyl instead of an isopropyl. So you can either say isopropyl for each of these groups or you could call each of them a methylethyl if you do systematic naming. We have two of these 1-methylethyl groups, just like we had two isopropyl groups. If you’re using common naming, you can say disopropyl to say you have two of these groups. When you’re using systematic naming, you don’t say di-1-methylethyl, although that would probably get the point across. You use bis-. Since we have two of them, instead of writing di-, you write bis-. That means you have two of these things right there, and it’s still in the four and the five position. If you look at it in alphabetical order now, methylethyl comes after methyl, right? So the order will now change. So now if you want to write it with systematic naming, it would be written as 2,2-dimethyl. That’s these two guys, 2,2-dimethylethyl. And then you would write this guy, so the order changed for the two groups just based on how they’re named. Bis-, and then over here you have two 1-methyl ethyl groups. I know it’s confusing, but when you just break it down, it actually makes a reasonable amount of sense. You have two of these methyl ethyl groups. Oh, sorry, I forgot where they’re located. So we have them at the four and five position, so in the four and five position, we have two, so bis-(1-methylethyl) groups. I know it’s a little daunting now, but it all makes sense when you break it down. Methylethyl groups. And then we can just add the octane at the end. Let me scroll over to the right a little bit. Octane. Now this might seem more confusing, but when you break it down, it makes sense. We have octane as be backbone. We have two methyls. They’re both sitting on the two position. So you have two methyls sitting on the two position, and then you have two 1-methylethyl groups sitting at the four and five positions. So in the four and five position, 1-methylethyl. You have an ethyl, and in the one position you have a methyl, so that’s all it’s saying. Or another way to think about it, just in case this doesn’t confuse you enough, you could call it that, or you could say 4,5-disopropyl. These two things are the same thing: common naming, systematic. Hopefully, you found that useful.。
Lecture Notes for PAMC (12) - Ch6_L2-201415(1)
数据移动
#define pout ((unsigned char *)0x018000) #define pin ((unsigned char *)0x014000) void main(void) { unsigned int _len=8192 unsigned int i; for(i=0;i<_len;i++) { *(pin+i)=*(pout+i); *(pout+i)＝0x0 } while(1); }
P11
P10
1
EA/VP
PSEN
RST
ALE
P27 80C51
P26
P25
P24
P23
P22
P21
P20
P07
P06
P05
P04
P03
P02
P01
P00
R
9
3
2
2
2
2
2
2
2
2
2
3
3
3
3
3
3
3
3
3
1
9
8
7
6
5
4
3
2
1
0
2
3
4
5
6
7
8
9
1
AB12
AB11
AB10
AB9
AB8
ALE
DB7
DB6
DB5
DB4
要求: 掌握并行接口SRAM存储器的扩展方法和地址计算方法 难点: 地址分配和计算, 二次译码技术
本讲的主要内容
常见的数据存储器及其分类, 特点
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“微机原理及应用”讲义
第7讲 IAR EW8051的流程控制语句和数据类型 (教材4.3-4.5节)
教学目的: 掌握IAR EW8051环境的C语言流程控制和数据类型，如何访问
格式
do { 循环题的语句组; } while (条件表达式);
流程
循环体语句
表达式
真
示例
见教材p.102例4.5
假 下一条语句
请问: 执行while和do…while两种 循环时, 有什么不同?
2010-11-9 9
C流程控制语句 循环语句: for
格式
for (表达式1:表达式2:表达式3) { 循环体的语句组; }
示例
if (a>b) c=a; else c=b;
2010-11-9
条件
语句 1
语句 2
执行下一条语句
4
C流程控制语句
格式
判断语句:条件运算符‘?’
变量或表达式=(条件表达式)?(条件为真时的表达式):(条件为假时的表达式);
示例
maxVal = (a>b)?(a):(b); 等价于 if (a>b) c=a; else c=b;
存储器位置 MCS‐51 单片机内部 RAM MCS‐51 单片机扩展 RAM 程序存储器，ROM 使用 MOVX @R0 寻址
2010-11-9
13
IAR EW8051的变量
变量是按前面定义的类型的实体数据, 需占用一定存储空间 变量命名规范和作用范围(生命周期)和一般的C语言一样 引入变量可以通过定义和声明语句实现, 定义变量时可以对 其初始化, 而且每个工程中, 一个特定变量只能定义一次, 而 声明则可以多次, 且不能对其初始化 为了保证代码的效率, IAR EW8051建议变量格式为
Memory type MSB of offset LSB of offset
存储器类型的定义见下页 ☺ 由于这些特殊性, 在移植某些程序代码时需要注意!
2010-11-9 12
存储器类型和变量存储空间的定义
FFFFH
Extended
FFFFH
On-chip 0100H 00FFH
流程
表达式 1
假
表达式 2 真 循环体语句 表达式 3
示例
见教材p.103例4.6
下一条语句
2010-11-9 10
IAR EW8051支持的基本数据类型
字节(比特) bit 1 sfr 1 char 1 signed char 1 unsigned char 1 short 2 int 2 unsigned short 2 unsigned int 2 long 4 unsigned long 4 pointer 1/2/3 float 4 double 4 long double 4
注释 位类型 等价于无符号字符型 字符型 有符号字符型 无符号字符型 有符号短整型 有符号整型 无符号短整型 无符号整型 有符号长整型 无符号长整型 指针类型 浮点型 双精度型 长双精度型
注: 教材中(p.106)此表这里有误
2010-11-9 11
IAR EW8051数据类型的特殊性
ANSI C的位类型是Boolean, 而且赋值为True或False. 与之 不同的是, IAR EW8051支持真正意义的位变量类型, 并存储 在8051内部的可位寻址区 支持sfr类型, 等价于unsigned char 类型, 这些变量只能存储 在8051的SFR区, 并只能采用直接寻址方式访问 使用IAR EW8051可以将char类型配置为unsigned char 支持特殊指针类型, IAR EW8051采用3个字节存储指针类型 数据(一般C编译器使用4个字节来存储指针). 存储格式
enum 模板名 {枚举值列表};
示例
enum week{mon=1, tue, wed, thu, fri, sat, sun}; // 第1个模板 enum capital{A=0x41, B,C,D,E,F}; // 第2个模板 enum week today; // 引用模板定义一个week型枚举变量 enum capital ABC; // 引用模板定义一个capital型枚举变量 today = mon; ABC = B; // 访问枚举类型变量
2010-11-9 1
本讲的主要内容
C语言流程控制语句
判断语句: if, if…else, 条件运算符‘?’, switch 循环控制语句: while, do…while, for
C语言数据类型
基本数据类型, 变量及其存储空间 复合数据类型
数组类型 指针类型 结构体类型 枚举类型
2010-11-9 2
MCS-51单片机的SFR、扩展的数据存储器和I/O端口, 中断程序格式, 以及C程序模块化设计方法
要求:掌握C语言程序设计的基本语法及其特点，掌握C程序模块化设计方法
难点: 数据变量类型的操作, 存储空间的访问方法，扩展数据存储器和I/O端口的访
问方法
FFH
sfr
idata
FFH
↑
80H
↑
80H 7FH
pdata
data/idata bit
↑
30H 20H~2FH 1FH
0000H
0000H
register xdata data / idata
↑
00H
code
存储类型说明符 Memory type data、idata 0 xdata code pdata 1 2 3
指针类型变量的运算
指针类型变量的“加”,“减”运算等同于指针移动 示例: int abc; int *pt=&abc; pt++; pt+=3; pt-=3; pt--;
指针和数组
数组名本身可以看作一个指针, 指针可以指向数组中的任意元素 示例: int buf[5], a,b,c; int *pt=buf; a=*pt; b=*(pt+2); pt++; c=*pt;
C流程控制语句 判断语句: if
格式
if (条件表达式) C语句;
流程
真 条件 假
示例
a=10; c=0; if (a>b) { c=a; }
2010-11-9
语句
执行下一条语句
3
C流程控制语句 判断语句: if..else
格式
if (条件表达式) 语句1; else 语句2;
流程
真 假
30H 31H 32H 33H 34H 35H 36H 37H
xx
Buf[0]
Buf[1]
Buf[2]
Buf[3]
Buf[4]
Buf[5]
xx
多维数组
数据类型说明符 [<存储器类型说明符>] 数组名[常量表达式1]…[常量表达式n];
请问: “unsigned int idata buf[5];”占用多少个字节型存储单元? 画图表示之.
使用时注意
这种?运算符简称三目操 作符 替代简单的if…else语句
2010-11-9
5
C流程控制语句 判断语句: if嵌套
格式
if (条件表达式1) { 语句1; } else { if (条件表达式2) { if (条件表达式3) 语句2; else 语句3; } else 语句4; }
2010-11-9
相同的数据类型, 允许多个成员函数 (C++中的类概念是结构体的延伸 )
枚举类型: 用它可以使无符号整型变量取界定的值
2010-11-9
15
复合数据类型 数组类型
一维数组
定义: 数据类型说明符 [<存储器类型说明符>] 数组名[常量表达式]; 示例: unsigned char idata buf[6] = {‘0’, ‘1’, ‘2’,’3’,’4’,’5’}; 访问数组元素: 数组名[index]用来访问数组中的第index个元素 存储形式:
示例
见教材p.98例4.2 (将范围为0x00~0x0F的十六进制数转换为ASCII字符)
2010-11-9
7
C流程控制语句 循环语句: while
格式
while (条件表达式) { 循环体的语句组; }
流程
示例
见教材p.99例4.3
假
表达式 真 语句
下一条语句
2010-11-9
8
C流程控制语句 循环语句: do…while
2010-11-9 19
示例
见教材p.96的示例 “根据按键执行程序”
6
C流程控制语句 判断语句: switch
格式
switch (表达式) { case 常量表达式1: 语句组1; case 常量表达式2: 语句组2; … case 常量表达式2: 语句组n; default: 语句组n+1; break; } break; break; break;
2010-11-9 14
IAR EW8051的复合数据类型
除了前面定义的基本数据类型外, IAR EW8051还支 持一般C语言相同的复合数据类型
数组类型: 变量集合, 其中的所有变量具有相同的基本类型, 按块保
存在特定的存储空间
指针类型: 用来引用变量, 或称为变量存储单元的别名(alias) 结构体类型: 变量和函数的集合, 其中的所有变量(成员)不一定具有
2010-11-9 16
复合数据类型 指针类型
指针的定义和声明 (注意: 指针可以指向任意类型的目标)