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人工<b>蜂群算法</b>

  • 印刷電路板設(shè)計(jì)原則

    減小電磁干擾的印刷電路板設(shè)計(jì)原則 內(nèi) 容 摘要……1 1 背景…1 1.1 射頻源.1 1.2 表面貼裝芯片和通孔元器件.1 1.3 靜態(tài)引腳活動(dòng)引腳和輸入.1 1.4 基本回路……..2 1.4.1 回路和偶極子的對(duì)稱(chēng)性3 1.5 差模和共?!?.3 2 電路板布局…4 2.1 電源和地…….4 2.1.1 感抗……4 2.1.2 兩層板和四層板4 2.1.3 單層板和二層板設(shè)計(jì)中的微處理器地.4 2.1.4 信號(hào)返回地……5 2.1.5 模擬數(shù)字和高壓…….5 2.1.6 模擬電源引腳和模擬參考電壓.5 2.1.7 四層板中電源平面因該怎么做和不應(yīng)該怎么做…….5 2.2 兩層板中的電源分配.6 2.2.1 單點(diǎn)和多點(diǎn)分配.6 2.2.2 星型分配6 2.2.3 格柵化地.7 2.2.4 旁路和鐵氧體磁珠……9 2.2.5 使噪聲靠近磁珠……..10 2.3 電路板分區(qū)…11 2.4 信號(hào)線……...12 2.4.1 容性和感性串?dāng)_……...12 2.4.2 天線因素和長(zhǎng)度規(guī)則...12 2.4.3 串聯(lián)終端傳輸線…..13 2.4.4 輸入阻抗匹配...13 2.5 電纜和接插件……...13 2.5.1 差模和共模噪聲……...14 2.5.2 串?dāng)_模型……..14 2.5.3 返回線路數(shù)目..14 2.5.4 對(duì)板外信號(hào)I/O的建議14 2.5.5 隔離噪聲和靜電放電ESD .14 2.6 其他布局問(wèn)題……...14 2.6.1 汽車(chē)和用戶(hù)應(yīng)用帶鍵盤(pán)和顯示器的前端面板印刷電路板...15 2.6.2 易感性布局…...15 3 屏蔽..16 3.1 工作原理…...16 3.2 屏蔽接地…...16 3.3 電纜和屏蔽旁路………………..16 4 總結(jié)…………………………………………17 5 參考文獻(xiàn)………………………17  

    標(biāo)簽: 印刷電路板 設(shè)計(jì)原則

    上傳時(shí)間: 2013-10-22

    上傳用戶(hù):a6697238

  • 數(shù)字運(yùn)算

    數(shù)字運(yùn)算,判斷一個(gè)數(shù)是否接近素?cái)?shù) A Niven number is a number such that the sum of its digits divides itself. For example, 111 is a Niven number because the sum of its digits is 3, which divides 111. We can also specify a number in another base b, and a number in base b is a Niven number if the sum of its digits divides its value. Given b (2 <= b <= 10) and a number in base b, determine whether it is a Niven number or not. Input Each line of input contains the base b, followed by a string of digits representing a positive integer in that base. There are no leading zeroes. The input is terminated by a line consisting of 0 alone. Output For each case, print "yes" on a line if the given number is a Niven number, and "no" otherwise. Sample Input 10 111 2 110 10 123 6 1000 8 2314 0 Sample Output yes yes no yes no

    標(biāo)簽: 數(shù)字 運(yùn)算

    上傳時(shí)間: 2015-05-21

    上傳用戶(hù):daguda

  • c語(yǔ)言版的多項(xiàng)式曲線擬合。 用最小二乘法進(jìn)行曲線擬合. 用p-1 次多項(xiàng)式進(jìn)行擬合

    c語(yǔ)言版的多項(xiàng)式曲線擬合。 用最小二乘法進(jìn)行曲線擬合. 用p-1 次多項(xiàng)式進(jìn)行擬合,p<= 10 x,y 的第0個(gè)域x[0],y[0],沒(méi)有用,有效數(shù)據(jù)從x[1],y[1] 開(kāi)始 nNodeNum,有效數(shù)據(jù)節(jié)點(diǎn)的個(gè)數(shù)。 b,為輸出的多項(xiàng)式系數(shù),b[i] 為b[i-1]次項(xiàng)。b[0],沒(méi)有用。 b,有10個(gè)元素ok。

    標(biāo)簽: 多項(xiàng)式 曲線擬合 c語(yǔ)言 最小二乘法

    上傳時(shí)間: 2014-01-12

    上傳用戶(hù):變形金剛

  • We have a group of N items (represented by integers from 1 to N), and we know that there is some tot

    We have a group of N items (represented by integers from 1 to N), and we know that there is some total order defined for these items. You may assume that no two elements will be equal (for all a, b: a<b or b<a). However, it is expensive to compare two items. Your task is to make a number of comparisons, and then output the sorted order. The cost of determining if a < b is given by the bth integer of element a of costs (space delimited), which is the same as the ath integer of element b. Naturally, you will be judged on the total cost of the comparisons you make before outputting the sorted order. If your order is incorrect, you will receive a 0. Otherwise, your score will be opt/cost, where opt is the best cost anyone has achieved and cost is the total cost of the comparisons you make (so your score for a test case will be between 0 and 1). Your score for the problem will simply be the sum of your scores for the individual test cases.

    標(biāo)簽: represented integers group items

    上傳時(shí)間: 2016-01-17

    上傳用戶(hù):jeffery

  • 用游標(biāo)的方法實(shí)現(xiàn)對(duì)稱(chēng)差的計(jì)算

    用游標(biāo)的方法實(shí)現(xiàn)對(duì)稱(chēng)差的計(jì)算,即 (A-B)+(B-A)

    標(biāo)簽: 對(duì)稱(chēng) 計(jì)算

    上傳時(shí)間: 2016-05-23

    上傳用戶(hù):遠(yuǎn)遠(yuǎn)ssad

  • 詞法分析器 對(duì)輸入一個(gè)函數(shù)

    詞法分析器 對(duì)輸入一個(gè)函數(shù),并對(duì)其分析main() { int a,b a = 10 b = a + 20 }

    標(biāo)簽: 分析器 函數(shù) 輸入

    上傳時(shí)間: 2013-12-20

    上傳用戶(hù):hfmm633

  • 函數(shù)再現(xiàn)機(jī)構(gòu)設(shè)計(jì) 試設(shè)計(jì)一曲柄搖桿機(jī)構(gòu)

    函數(shù)再現(xiàn)機(jī)構(gòu)設(shè)計(jì) 試設(shè)計(jì)一曲柄搖桿機(jī)構(gòu),再現(xiàn)函數(shù) 要求: 輸入構(gòu)件的轉(zhuǎn)角范圍180°,輸出構(gòu)件擺角范圍30°,即: 當(dāng)輸入構(gòu)件從a轉(zhuǎn)至a+90時(shí),輸出構(gòu)件從b轉(zhuǎn)至b+30 當(dāng)輸入構(gòu)件從a+90轉(zhuǎn)至a+180時(shí),輸出構(gòu)件從b+30轉(zhuǎn)至b

    標(biāo)簽: 機(jī)構(gòu) 函數(shù) 曲柄

    上傳時(shí)間: 2013-12-17

    上傳用戶(hù):英雄

  • 這個(gè)連接池是直接從JIVE中取出來(lái)的

    這個(gè)連接池是直接從JIVE中取出來(lái)的,進(jìn)行了一下修改,使得連接參數(shù)直接在程序中設(shè)定而不是從屬性文件中讀取。 [b]用法:[/b] 先設(shè)定自己的連接參數(shù),在DbConnectionDefaultPool.java文件的loadProperties方法中。注意你也需要設(shè)定連接池的log文件的存放位置。

    標(biāo)簽: JIVE 連接

    上傳時(shí)間: 2016-11-21

    上傳用戶(hù):TF2015

  • 漢諾塔?。?! Simulate the movement of the Towers of Hanoi puzzle Bonus is possible for using animation

    漢諾塔!!! Simulate the movement of the Towers of Hanoi puzzle Bonus is possible for using animation eg. if n = 2 A→B A→C B→C if n = 3 A→C A→B C→B A→C B→A B→C A→C

    標(biāo)簽: the animation Simulate movement

    上傳時(shí)間: 2017-02-11

    上傳用戶(hù):waizhang

  • PIC16C63單片機(jī)UART通信——A機(jī)讀取時(shí)鐘芯片DS1302獲得當(dāng)前時(shí)間

    PIC16C63單片機(jī)UART通信——A機(jī)讀取時(shí)鐘芯片DS1302獲得當(dāng)前時(shí)間,通過(guò)UART通信傳給B機(jī),B機(jī)使用LCD1602顯示當(dāng)前時(shí)間

    標(biāo)簽: 1302 UART PIC 16C

    上傳時(shí)間: 2013-11-30

    上傳用戶(hù):shanml

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