91精品免费,国产精品一区电影,涩涩视频网站在线观看

pi-sigam

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，輸入語言變量和輸出語言變量均取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
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
// 入口參數(shù)： // l: l = 0, 傅立葉變換 l = 1, 逆傅立葉變換 // il: il = 0,不計(jì)算傅立葉變換或逆變換模和幅角；il = 1,計(jì)算模和幅角 // n: 輸入的

// 入口參數(shù)： // l: l = 0, 傅立葉變換 l = 1, 逆傅立葉變換 // il: il = 0,不計(jì)算傅立葉變換或逆變換模和幅角；il = 1,計(jì)算模和幅角 // n: 輸入的點(diǎn)數(shù)，為偶數(shù)，一般為32，64，128，...,1024等 // k: 滿足n=2^k(k>0),實(shí)質(zhì)上k是n個(gè)采樣數(shù)據(jù)可以分解為偶次冪和奇次冪的次數(shù) // pr[]: l=0時(shí)，存放N點(diǎn)采樣數(shù)據(jù)的實(shí)部 // l=1時(shí), 存放傅立葉變換的N個(gè)實(shí)部 // pi[]: l=0時(shí)，存放N點(diǎn)采樣數(shù)據(jù)的虛部 // l=1時(shí), 存放傅立葉變換的N個(gè)虛部 // // 出口參數(shù)： // fr[]: l=0, 返回傅立葉變換的實(shí)部 // l=1, 返回逆傅立葉變換的實(shí)部 // fi[]: l=0, 返回傅立葉變換的虛部 // l=1, 返回逆傅立葉變換的虛部 // pr[]: il = 1,i = 0 時(shí)，返回傅立葉變換的模 // il = 1,i = 1 時(shí)，返回逆傅立葉變換的模 // pi[]: il = 1,i = 0 時(shí)，返回傅立葉變換的輻角 // il = 1,i = 1 時(shí)，返回逆傅立葉變換的輻角

標(biāo)簽： il 傅立葉變換計(jì)算模

上傳時(shí)間： 2017-01-03

上傳用戶：ynsnjs
OFDM的發(fā)射端

OFDM的發(fā)射端，包括pi/4DQPSK的調(diào)制，ifft，加幀頭以及組幀

標(biāo)簽： OFDM 發(fā)射端

上傳時(shí)間： 2013-12-26

上傳用戶：stvnash