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PI-TR

一被控對(duì)象

一被控對(duì)象，給定為階躍給定，幅值為500，設(shè)計(jì)一個(gè)兩維模糊PI型控制器，輸入語(yǔ)言變量和輸出語(yǔ)言變量均取7個(gè)值{NB，NM，NS，ZE，PS，PM，PB}，模糊論域?yàn)閧-6，-5，-4，-3，-2，-1，0，1，2，3，4，5，6}，用matlab編程仿真研究。

標(biāo)簽： 對(duì)象

上傳時(shí)間： 2013-12-16

上傳用戶：大融融rr
private double PointToAngle(Point AOrigin, Point APoint) { if (APoint.X == AOrigin.X) if (APoin

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 {

標(biāo)簽： AOrigin APoint Point PointToAngle

上傳時(shí)間： 2016-10-31

上傳用戶：zhyiroy
private double PointToAngle(Point AOrigin, Point APoint) { if (APoint.X == AOrigin.X) if (APoin

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 {

標(biāo)簽： AOrigin APoint Point PointToAngle

上傳時(shí)間： 2016-10-31

上傳用戶：sunjet
private double PointToAngle(Point AOrigin, Point APoint) { if (APoint.X == AOrigin.X) if (APoin

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 {

標(biāo)簽： AOrigin APoint Point PointToAngle

上傳時(shí)間： 2013-12-18

上傳用戶：rocketrevenge
For solving the following problem: "There is No Free Lunch" Time Limit: 1 Second Memory Limit: 3

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.

標(biāo)簽： Limit following solving problem

上傳時(shí)間： 2014-01-12

上傳用戶：362279997
it is a simulation about ML synchronization algorithm in OFDM systems,you can slao see a function pi

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

標(biāo)簽： synchronization simulation algorithm function

上傳時(shí)間： 2013-12-17

上傳用戶：yulg
js 異步加載 tree 采用ajax技術(shù)

js 異步加載 tree 采用ajax技術(shù)，在點(diǎn)擊tree的node以后從數(shù)據(jù)庫(kù)中讀取其下面的分支并進(jìn)行顯示，適合于tree數(shù)據(jù)量較大的情況下異步從數(shù)據(jù)庫(kù)中查詢并顯示于tr

標(biāo)簽： ajax tree js

上傳時(shí)間： 2013-12-29

上傳用戶：gtf1207
XHTML標(biāo)準(zhǔn)參考手冊(cè),XHTML是The Extensible HyperText Markup Language可擴(kuò)展標(biāo)識(shí)語(yǔ)言的縮寫。目前推薦遵循的是W3C于2000年1月26日推薦XML1.0（

XHTML標(biāo)準(zhǔn)參考手冊(cè),XHTML是The Extensible HyperText Markup Language可擴(kuò)展標(biāo)識(shí)語(yǔ)言的縮寫。目前推薦遵循的是W3C于2000年1月26日推薦XML1.0（參考http://www.w3.org/TR/xhtml1）。XML雖然數(shù)據(jù)轉(zhuǎn)換能力強(qiáng)大，完全可以替代HTML，但面對(duì)成千上萬(wàn)已有的站點(diǎn)，直接采用XML還為時(shí)過(guò)早。因此，我們?cè)贖TML4.0的基礎(chǔ)上，用XML的規(guī)則對(duì)其進(jìn)行擴(kuò)展，得到了XHTML。簡(jiǎn)單的說(shuō)，建立XHTML的目的就是實(shí)現(xiàn)HTML向XML的過(guò)渡。

標(biāo)簽： XHTML Extensible HyperText Language

上傳時(shí)間： 2013-12-02

上傳用戶：huangld
Euler函數(shù): m = p1^r1 * p2^r2 * …… * pn^rn ai >= 1 , 1 <= i <= n Euler函數(shù)：定義：phi(m) 表示小于等

Euler函數(shù): m = p1^r1 * p2^r2 * …… * pn^rn ai >= 1 , 1 <= i <= n Euler函數(shù)：定義：phi(m) 表示小于等于m并且與m互質(zhì)的正整數(shù)的個(gè)數(shù)。 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 則有 a^phi(m) = 1 (mod m) 即a^phi(m) - 1 整出m 在實(shí)際代碼中可以用類似素?cái)?shù)篩法求出 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：定義phi(p) 為比p小的與p互素的數(shù)的個(gè)數(shù) 設(shè)n的素因子有p1, p2, p3, … pk 包含p1, p2…的個(gè)數(shù)為n/p1, n/p2… 包含p1*p2, p2*p3…的個(gè)數(shù)為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)

標(biāo)簽： Euler lt phi 函數(shù)

上傳時(shí)間： 2014-01-10

上傳用戶：wkchong
計(jì)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,

計(jì)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 ％這表示邊長(zhǎng)嗎？ cgh=zeros(64*s) th=max(max(abs(Fs)))

標(biāo)簽： 20 close clear zeros

上傳時(shí)間： 2014-10-13

上傳用戶：wweqas