麦特客 http://www.matke.com Matlab课堂 Fri, 20 Aug 2010 16:33:23 +0000 en hourly 1 http://wordpress.org/?v=3.0 利用MATLAB制作霓虹灯效果动画的教程 http://www.matke.com/neonball/ http://www.matke.com/neonball/#comments Fri, 20 Aug 2010 16:23:00 +0000 admin http://www.matke.com/neonball/ 如今繁华大都市的夜色中，霓虹闪烁，煞是好看，调用MATLAB中的spinmap函数可以做出这种效果的动画，它是通过旋转颜色映像的方式来呈现这种霓虹闪烁的动画效果。spinmap函数的调用格式如下：
spinmap         % 旋转颜色映像约5秒钟
spinmap(t)      % 旋转颜色映像约t秒钟，具体时间取决于硬件
spinmap(t,inc)  % 旋转颜色映像约t秒钟，并设置增量参数inc，该参数用来调整闪烁频率
spinmap(‘inf’)  % 不限时旋转颜色映像，若需终止，请按Ctrl+C键
【例】霓虹闪烁的球体。
源代码：

  1. sphere;         % 绘制单位球面
  2. axis equal;     % 设置坐标显示比例相同
  3. axis off;       % 隐藏坐标轴
  4. spinmap(20,1);  % 设置增量参数为1，旋转颜色映像约20秒

逐条运行以上命令即可看到霓虹闪烁效果的球体，效果如下图：
利用MATLAB制作霓虹灯效果动画

用camorbit函数加了旋转效果如下：利用MATLAB制作霓虹灯效果动画

来源：http://www.matlabsky.com/thread-9255-1-1.htmlMatlabsky

]]> http://www.matke.com/neonball/feed/ 0 Matlab计算阶乘的程序（if，for，while） http://www.matke.com/matlab-factor-function/ http://www.matke.com/matlab-factor-function/#comments Fri, 20 Aug 2010 15:30:00 +0000 admin http://www.matke.com/?p=17 1、while语句版，首先建立函数并保存为jiecheng.m

function sum=jiecheng(n)
n=input(‘please enter a positive number’);
if n<0
    display(‘输入的是负数’)
else
   sum=1;
   m=1;
   while(m<=n)
    sum=sum*m;
    m=m+1;
   end
end

2、for语句。建立函数并保存为jiecheng2.m

function sum=jiecheng2(n)
n=input(‘please enter a positive number’);
sum=1;
m=1;
if (n<0)
    display(‘输入的是负数’)
else     if  (m<=n)
         for i=1:n
         sum=sum*m;
         m=m+1;
         end
         end
end

3、IF语句建立的函数factor.m

function f=factor(n)
if n<=1
   f=1;
else
   f=factor(n-1)*n;  
end

]]> http://www.matke.com/matlab-factor-function/feed/ 0 Matlab实现两个变量交换数值（不使用第三个变量） http://www.matke.com/matlab-exchange-two-variable-value/ http://www.matke.com/matlab-exchange-two-variable-value/#comments Fri, 20 Aug 2010 15:18:00 +0000 admin http://www.matke.com/?p=16 以前一个教授在听Ｃ语言公开课时对一个老师提出来的，现在把Ｃ换成Matlab软件，看大家如何解决。

程序：

syms x y
x=input(‘please enter a number x:’);
y=input(‘please enter a number y:’);
x=x+y;
y=x-y;
x=x-y;

]]> http://www.matke.com/matlab-exchange-two-variable-value/feed/ 0 Regression using SPSS – A suburban hotel derives its gross income from its hotel and restaurant operations. http://www.matke.com/regression-using-spss-a-suburban-hotel-derives-its-gross-income-from-its-hotel-and-restaurant-operations/ http://www.matke.com/regression-using-spss-a-suburban-hotel-derives-its-gross-income-from-its-hotel-and-restaurant-operations/#comments Thu, 29 Jul 2010 15:21:00 +0000 admin http://www.matke.com/?p=14 A suburban hotel derives its gross income from its hotel and restaurant operations. The owners are interested in the relationship between the number of rooms occupied on a nightly basis and the revenue per day in the restaurant. Below is a sample of 25 days (Monday through Thursday) from last year showing the restaurant income and number of rooms occupied.

Day

Income

Occupied

1

$1,452

23

2

1,361

47

3

1,426

21

4

1,470

39

5

1,456

37

6

1,430

29

7

1,354

23

8

1,442

44

9

1,394

45

10

1,459

16

11

1,399

30

12

1,458

42

13

1,537

54

14

$1,425

27

15

1,445

34

16

1,439

15

17

1,348

19

18

1,450

38

19

1,431

44

20

1,446

47

21

1,485

43

22

1,405

38

23

1,461

51

24

1,490

61

25

1,426

39

Use a statistical software package to answer the following questions.

a. Does the breakfast revenue seem to increase as the number of occupied rooms increases? Draw a scatter diagram to support your conclusion.

b. Determine the coefficient of correlation between the two variables. Interpret the value.

c. Is it reasonable to conclude that there is a positive relationship between revenue and occupied rooms? Use the .10 significance level.

d. What percent of the variation in revenue in the restaurant is accounted for by the number of rooms occupied?

Answer

(A)

OUTPUT0

(B)
r = 0.44, there is a WEAK positive correlation.
(C)
df = n-2 = 25-2 = 23
α = 0.05
one-tailed critical value T = 1.319
Test statistic T = r*Sqrt[(n-2)/(1-r²)] = 0.44*Sqrt[(25-2)/(1-0.44²)] = 2.35
Conclusion: Since 2.35 > 1.319 we conclude that there is a positive relationship between the variables.
(D)
r² = 0.44² = 0.1936, so approximately 19% of the variation in revenue is explained by the variation in the number of rooms occupied.

]]> http://www.matke.com/regression-using-spss-a-suburban-hotel-derives-its-gross-income-from-its-hotel-and-restaurant-operations/feed/ 0 Prove that square root of three is an irrational number. http://www.matke.com/prove-that-square-root-of-three-is-an-irrational-number/ http://www.matke.com/prove-that-square-root-of-three-is-an-irrational-number/#comments Mon, 10 May 2010 15:09:08 +0000 admin http://www.matke.com/?p=11 via

The Irrationality of square root of three

Problem:
Prove that square root of three is an irrational number.

Solution:
The number, square root of three, is irrational, ie., it cannot be expressed as a ratio of integers a and b. To prove that this statement is true, let us assume that square root of three is rational so that we may write

square root of three = a/b                          1.

for a and b = any two integers. We must then show that no two such integers can be found. We begin by squaring both sides of eq. 1:

3 = a2/b2                                                   2.

or

3b2 = a2                                                     2a.

If b is odd, then b2 is odd; in this case, a2 and a are also odd. Similarly, if b is even, then b2, a2, and a are even. Since any choice of even values of a and b leads to a ratio a/b that can be reduced by canceling a common factor of 2, we must assume that a and b are odd, and that the ratio a/b is already reduced to smallest possible terms. With a and b both odd, we may write

a = 2m + 1                                   3.

and

b = 2n +1                                      4.

where we require m and n to be integers (to ensure integer values of a and b). When these expressions are substituted into eq. 2a, we obtain

3(4n2 + 4n + 1) = 4m2 + 4m + 1     5.

Upon performing some algebra, we acquire the further expression

6n2 + 6n + 1 = 2(m2 + m)               6.

The Left Hand Side of eq. 3a is an odd integer. The Right Hand Side, on the other hand, is an even integer. There are no solutions for eq. 3a. Therefore, integer values of a and b which satisfy the relationship square root of three = a/b cannot be found. We are forced to conclude that square root of three is irrational.

The other way: from Yahoo anser

We prove that by contradiction.
This means, we suppose √3 is rational and then get to something that contradicts this assumption.
If √3 is rational, we can express it as a/b, a fraction in its lowest terms (i.e., gcd(a,b) = 1). Also, b is not 1, because that would mean √3 is an integer, which it obviously is not.
Now square both sides.
√3 = a/b
3 = a²/b²
since gcd(a,b) = 1, then gcd(a²,b²) = 1 (has to be).
But from 3 = a²/b² we can express a² = 3b².
So a²/b² = 3b²/b² -> now this fraction is obviously NOT in its lowest terms (remember, b is not 1) -> hence the contradiction.
=> √3 is not rational. It is irrational.

The third way:like sqrt two

sqrt (3) = m/n   (a fraction, reduced to its lowest terms)
     3 = m²/n²   m² = 3n²   (So, m is a multiple of 3, call it 3q)
     9q² = 3n²   3q² = n²    (So, n is a multiple of 3)
]]> http://www.matke.com/prove-that-square-root-of-three-is-an-irrational-number/feed/ 0 Coins, dice, probability http://www.matke.com/coins-dice-probability/ http://www.matke.com/coins-dice-probability/#comments Sun, 09 May 2010 13:40:00 +0000 admin http://www.matke.com/?p=9 1)Three dice are thrown simultaneously. The probability of obtaining a sum of 6 is?

1)1/24         2)5/108        3)1/54          4)7/216

2)A coin is tossed 3 times. The probability of getting head and tail alternately is?

1)1/6           2)1/8            3)1/4         4)1/3

3)Three dice are thrown simultaneously. The probability of obtaining a multiple of 5?

1)1/36         2)5/108        3)1/54          4)43/216

Answer：

1）

6=1+1+4，3 possible outcomes
6=1+2+3，6possible outcomes
6=2+2+2，1possible outcomes
(3+6+1)/6*6*6=10/216 = 5/108

2）Here are the possible outcomes:
HHH, HHT, HTH, HTT, TTT, TTH, THT, THH
There are two outcomes that satisfy the alternate condition HTH and THT. Head and tail come alternatively in these two outcomes,
Total possibilities=8
Number of requirea outcome=2
Probability=2/8=1/4
So number 3) is correct answer

3/）

Maximum of three dice and 18
So that may occur in multiples of 5 to 5,10,15
As and when the 5: 5 = 1 +1 +3 will make arrangement with three kinds of 113
5 = 1 + 2 + 2 will be ordered to do 122 with three kinds of
And the time when 10 of 10 = 1 +3 +6 ordered to do 136 with 6
10 = 1 +4 +5 ordered to do 145 with 6
10 = 2 +2 +6 there are three kinds of ordered to do 226
10 = 2 +3 +5 ordered to do 235 with 6
10 = 2 +4 +4 there are three kinds of ordered to do 244
10 = 3 +3 +4 there are three kinds of ordered to do 334
When and when 15 = 3 +6 +6 15 366 do have three kinds of arrangement
15 = 4 +5 +6 ordered to do 456 with 6
15 = 5 +5 +5 555 ordered to do only one
In summary there and in multiples of 5 3 +3 +3 +6 +6 +6 +6 +3 +3 +3 +1 = 43
The vote occurred a total of three dice combination of a 6 * 6 * 6 = 216
The probability is 43/216 = 43/216

right or wrong?

]]> http://www.matke.com/coins-dice-probability/feed/ 0 Prove that square root of two is an irrational number. http://www.matke.com/sqrt2-is-an-irrational/ http://www.matke.com/sqrt2-is-an-irrational/#comments Sun, 09 May 2010 11:31:00 +0000 admin http://www.matke.com/?p=7 Here’s the proof. We start by assuming the square root of two, shown as sqrt(2) here, is equal to some fraction m/n. We intend to contradict this assumption: 

sqrt (2) = m/n   (a fraction, reduced to its lowest terms)
     2 = m²/n²   m² = 2n²   (So, m is a multiple of 2, call it 2q)
     4q² = 2n²   2q² = n²    (So, n is a multiple of 2)

So, both m and n are multiples of two, which is impossible, because m/n was reduced to its lowest terms. So, we have proved that the square root of two cannot be expressed as a fraction, i.e. it is irrational. via.Note:There are other methods of proof.

It follows that 2 = a2/b2,  or  a2 = 2 * b2. If we substitute a = 2k into the original equation 2 = a2/b2, this is what we get:

2= (2k)2/b2

2= 4k2/b2

2*b2= 4k2

b2= 2k2.

This means b2 is even, so we obtain the result.via

]]> http://www.matke.com/sqrt2-is-an-irrational/feed/ 0 MATLAB Toolboxes http://www.matke.com/matlab-toolboxes/ http://www.matke.com/matlab-toolboxes/#comments Sat, 08 May 2010 12:51:00 +0000 admin http://www.matke.com/?p=5 New

Misc.
Astronomy

Audio

Bio Medical Informatics (Top)

Chemometrics (providing PCA) (Top)

Chaos (Top)

Chemistry (Top)
Classification


Coding (Top) 

Control (Top)

Communications (Top)

Engineering (Top)

Data Mining (Top) 

Excel (Top) 

Finite Element Modeling (FEM) (Top) 

Finance (Top)

Fuzzy

Genetic Algorithms (Top)

Graph

Graphics

Images  (Top)

ICA / KICA and KPCA  (Top)

Kernel  (Top)

Markov  (Top)

Medical (Top)

MIDI  (Top)

Misc. (Top)

MPI (Matlab Parallel Interface) (Top)

Neural Networks  (Top)

Oceanography  (Top)

Optimization  (Top)

Plot (Top)

Signal Processing  (Top)

Support Vector Machine (Top)

Statistics  (Top)

Wavelets (Top)
Web (Top)

  • XML Toolbox for Matlab (v3.2.0)

来源：http://www.tech.plym.ac.uk/spmc/links/matlab/matlab_toolbox.html

]]> http://www.matke.com/matlab-toolboxes/feed/ 0 MATLAB工具箱下载地址总汇 http://www.matke.com/matlab-toolboxes-download/ http://www.matke.com/matlab-toolboxes-download/#comments Thu, 06 May 2010 12:54:00 +0000 admin http://www.matke.com/?p=6 MATLAB工具箱下载地址总汇
1.平面操作工具箱
http://vib.hit.edu.cn/vibbbs/dispbbs.asp?BoardID=35&ID=750&replyID=&skin=1
2.SimMechanics 工具箱 (这个好像不是免费的)
http://www.mathworks.com/access/helpdesk/help/toolbox/physmod/mech/mech.shtml
3.遗传算法工具箱
http://vib.hit.edu.cn/vibbbs/dispbbs.asp?BoardID=25&ID=655&replyID=&skin=1
http://vib.hit.edu.cn/vibbbs/dispbbs.asp?BoardID=35&replyID=1430&id=1430&skin=0#f2
http://vib.hit.edu.cn/vibbbs/dispbbs.asp?BoardID=54&ID=3040&replyID=&skin=1
4.Optimization Toolbox 2.2
http://www.ttdown.com/SoftDown.asp?ID=25782
5.Vibration Toolbox
http://www.cs.wright.edu/people/faculty/jslater/vtoolbox/vtoolbox.html
6.MATLAB机器视觉工具箱
http://vib.hit.edu.cn/vibbbs/dispbbs.asp?BoardID=35&ID=751&replyID=&skin=1
7.Polynomial Toolbox
http://www.polyx.com/frm-main-download.htm
8.TIME-FREQUENCY TOOLBOX
http://crttsn.univ-nantes.fr/~auger/tftb.html
9.Uvi_Wave Wavelet Toolbox
http://www.gts.tsc.uvigo.es/~wavelets/
10.THE NNSYSID TOOLBOX – Version 2
http://kalman.iau.dtu.dk/research/control/nnsysid.html
11.Camera calibration toolbox
http://www.ee.oulu.fi/~jth/calibr/
12.Virtual Control Lab
http://www.esr.ruhr-uni-bochum.de/VCLab/
13.有限元工具箱
http://vib.hit.edu.cn/vibbbs/dispbbs.asp?boardid=35&id=3466#f2
14.最速梯度法
http://vib.hit.edu.cn/vibbbs/dispbbs.asp?boardid=35&id=3466#f3

来源：http://mech.cndev.org/articles/3576.aspx

]]> http://www.matke.com/matlab-toolboxes-download/feed/ 0 Matlab R2010a发布 http://www.matke.com/matlab-r2010a-release/ http://www.matke.com/matlab-r2010a-release/#comments Sat, 13 Mar 2010 16:53:00 +0000 admin http://www.matke.com/?p=4 Release 2010a 的新功能

Release 2010a 包括 MATLAB 和 Simulink 的若干新功能、一款新产品以及对其它 85 款产品的更新和缺陷修复。已经购买 MathWorks 软件维护服务的用户可以下载产品更新。从 R2010a 起，PolySpace 产品需要激活。访问许可中心下载产品、激活软件并管理许可证和用户信息。
MATLAB 产品系列的新功能包括：

  • 增加更多多线程数学函数，增强文件共享、路径管理功能以及改进 MATLAB 桌面
  • 针对 50 多个函数提供多核支持并增强性能，并对图像处理工具箱中的大型图像提供更多支持
  • 在 SimBiology 中提供随机近似最大期望 (SAEM) 算法和药动学给药方案支持

Simulink 产品系列的新功能包括：

  • 在 Simulink 中提供可调参数结构、触发模型块以及用于大型建模的函数调用分支
  • 在 IEC 认证工具包中提供对 Real-Time Workshop Embedded Coder 和 PolySpace 产品的 ISO 26262 认证

电驴下载：ed2k://|file|[%E7%9F%A9%E9%99%A3%E5%AF%A6%E9%A9%97%E5%AE%A4].Mathworks.Matlab.R2010a.ISO-TBE.iso|5614561280|B0825B401987DC5D42A688D41B1512A0|h=22Q2JTJPBASQWDEUAE5ABZ7IZIJ5BY52|/

]]> http://www.matke.com/matlab-r2010a-release/feed/ 0