一被控对象 ，给定为阶跃给定，幅值为500，设计一个两维模糊PI型控制器，输入语言变量和输出语言变量均取7个值{NB，NM，NS，ZE，PS，PM，PB}，模糊论域为{-6，-5，-4，-3，-2，-1，0，1，2，3，4，5，6}，用matlab编程仿真研究。
标签： 对象
上传时间： 2013-12-16
上传用户：大融融rr
private double PointToAngle(Point AOrigin, Point APoint) { if (APoint.X == AOrigin.X) if (APoint.Y > AOrigin.Y) return Math.PI * 0.5f else return Math.PI * 1.5f else if (APoint.Y == AOrigin.Y) if (APoint.X > AOrigin.X) return 0 else return Math.PI else {
标签： AOrigin APoint Point PointToAngle
上传时间： 2016-10-31
上传用户：zhyiroy
private double PointToAngle(Point AOrigin, Point APoint) { if (APoint.X == AOrigin.X) if (APoint.Y > AOrigin.Y) return Math.PI * 0.5f else return Math.PI * 1.5f else if (APoint.Y == AOrigin.Y) if (APoint.X > AOrigin.X) return 0 else return Math.PI else {
标签： AOrigin APoint Point PointToAngle
上传时间： 2016-10-31
上传用户：sunjet
private double PointToAngle(Point AOrigin, Point APoint) { if (APoint.X == AOrigin.X) if (APoint.Y > AOrigin.Y) return Math.PI * 0.5f else return Math.PI * 1.5f else if (APoint.Y == AOrigin.Y) if (APoint.X > AOrigin.X) return 0 else return Math.PI else {
标签： AOrigin APoint Point PointToAngle
上传时间： 2013-12-18
上传用户：rocketrevenge
For solving the following problem: "There is No Free Lunch" Time Limit: 1 Second Memory Limit: 32768 KB One day, CYJJ found an interesting piece of commercial from newspaper: the Cyber-restaurant was offering a kind of "Lunch Special" which was said that one could "buy one get two for free". That is, if you buy one of the dishes on their menu, denoted by di with price pi , you may get the two neighboring dishes di-1 and di+1 for free! If you pick up d1, then you may get d2 and the last one dn for free, and if you choose the last one dn, you may get dn-1 and d1 for free. However, after investigation CYJJ realized that there was no free lunch at all. The price pi of the i-th dish was actually calculated by adding up twice the cost ci of the dish and half of the costs of the two "free" dishes. Now given all the prices on the menu, you are asked to help CYJJ find the cost of each of the dishes.
标签： Limit following solving problem
上传时间： 2014-01-12
上传用户：362279997
it is a simulation about ML synchronization algorithm in OFDM systems,you can slao see a function picture in its output,that s useful for a beginner
标签： synchronization simulation algorithm function
上传时间： 2013-12-17
上传用户：yulg
js 异步加载 tree 采用ajax技术，在点击tree的node以后从数据库中读取其下面的分支并进行显示，适合于tree数据量较大的情况下异步从数据库中查询并显示于tr
上传时间： 2013-12-29
上传用户：gtf1207
XHTML标准参考手册,XHTML是The Extensible HyperText Markup Language可扩展标识语言的缩写。目前推荐遵循的是W3C于2000年1月26日推荐XML1.0（参考http://www.w3.org/TR/xhtml1）。XML虽然数据转换能力强大，完全可以替代HTML，但面对成千上万已有的站点，直接采用XML还为时过早。因此，我们在HTML4.0的基础上，用XML的规则对其进行扩展，得到了XHTML。简单的说，建立XHTML的目的就是实现HTML向XML的过渡。
标签： XHTML Extensible HyperText Language
上传时间： 2013-12-02
上传用户：huangld
Euler函数: m = p1^r1 * p2^r2 * …… * pn^rn ai >= 1 , 1 <= i <= n Euler函数： 定义：phi(m) 表示小于等于m并且与m互质的正整数的个数。 phi(m) = p1^(r1-1)*(p1-1) * p2^(r2-1)*(p2-1) * …… * pn^(rn-1)*(pn-1) = m*(1 - 1/p1)*(1 - 1/p2)*……*(1 - 1/pn) = p1^(r1-1)*p2^(r2-1)* …… * pn^(rn-1)*phi(p1*p2*……*pn) 定理：若(a , m) = 1 则有 a^phi(m) = 1 (mod m) 即a^phi(m) - 1 整出m 在实际代码中可以用类似素数筛法求出 for (i = 1 i < MAXN i++) phi[i] = i for (i = 2 i < MAXN i++) if (phi[i] == i) { for (j = i j < MAXN j += i) { phi[j] /= i phi[j] *= i - 1 } } 容斥原理：定义phi(p) 为比p小的与p互素的数的个数 设n的素因子有p1, p2, p3, … pk 包含p1, p2…的个数为n/p1, n/p2… 包含p1*p2, p2*p3…的个数为n/(p1*p2)… phi(n) = n - sigm_[i = 1](n/pi) + sigm_[i!=j](n/(pi*pj)) - …… +- n/(p1*p2……pk) = n*(1 - 1/p1)*(1 - 1/p2)*……*(1 - 1/pk)
上传时间： 2014-01-10
上传用户：wkchong
计算全息close all clc clear A=zeros(64) A(15:20,20:40)=1 A(15:50,20:25)=1 A(45:50,20:40)=1 A(30:34,20:35)=1 ％ ppp=exp(rand(64)*pi*2*i) A=A.*ppp ％ Author s email: zjliu2001@163.com figure imshow(abs(A),[]) Fa=fft2(fftshift(A)) Fs=fftshift(Fa) Am=abs(Fs) ％ amplitude Ph=angle(Fs) ％ phase s=11 ％ 这表示边长吗？ cgh=zeros(64*s) th=max(max(abs(Fs)))
上传时间： 2014-10-13
上传用户：wweqas
