Verilog and VHDL狀態(tài)機(jī)設(shè)計(jì)，英文pdf格式 State machine design techniques for Verilog and VHDL Abstract : Designing a synchronous finite state Another way of organizing a state machine (FSM) is a common tASK for a digital logic only one logic block as shown in engineer. This paper will discuss a variety of issues regarding FSM design using Synopsys Design Compiler . Verilog and VHDL coding styles will be 2.0 Basic HDL coding presented. Different methodologies will be compared using real-world examples.
標(biāo)簽： Verilog VHDL and 狀態(tài)
上傳時(shí)間： 2013-12-19
上傳用戶：change0329
Our approach to understanding mobile learning begins by describing a dialectical approach to the development and presentation of a tASK model using the sociocognitive engineering design method. This analysis synthesises relevant theoretical approaches. We then examine two field studies which feed into the development of the tASK model.
標(biāo)簽： approach understanding dialectical describing
上傳時(shí)間： 2014-11-28
上傳用戶：comua
Real-Time Kernel ，簡(jiǎn)易型REAL-TEME SYSTEM 源碼,可用于嵌入Muti tASK學(xué)習(xí)
上傳時(shí)間： 2014-01-14
上傳用戶：kbnswdifs
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：jeffery
As all of you know, MATLAB is a powerful engineering language. Because of some limitation, some tASKs take very long time to proceed. Also MATLAB is an interpreter not a compiler. For this reason, executing a MATLAB program (m file) is time consuming. For solving this problem, Mathworks provides us C Math Library or in common language, MATLAB API. A developer can employ these APIs to solve engineering problems very fast and easy. This article is about how can use these APIs.
標(biāo)簽： some engineering limitation language
上傳時(shí)間： 2013-12-06
上傳用戶：huql11633
定時(shí)中斷程序，源碼的注釋十分詳細(xì)，具體功能如下： 1.Frame 實(shí)現(xiàn)能有效降低VxWorks 內(nèi)存管理內(nèi)部/外部碎片的機(jī)制。 2. Frame 實(shí)現(xiàn)為系統(tǒng)提供軟定時(shí)器功能的機(jī)制，定時(shí)器timeout 信息以message 或其他快捷有效方式通知定時(shí)器申請(qǐng)者（tASK）。 3. 參考實(shí)驗(yàn)一要求，系統(tǒng)中每個(gè)tASK 擁有自己的Message Queue，以此方式作為系統(tǒng)的消息驅(qū)動(dòng)基礎(chǔ)。 4. 系統(tǒng)中各tASK 應(yīng)使用同一類型框架，即統(tǒng)一的tASK 框架。 5. 系統(tǒng)內(nèi)實(shí)體（tASK/ISR）間傳遞的消息應(yīng)有統(tǒng)一格式（消息頭+消息體），可分短消息和長(zhǎng)消息，但消息頭須至少包含消息ID。系統(tǒng)內(nèi)所有消息均有其唯一ID 標(biāo)識(shí)。
上傳時(shí)間： 2016-04-02
上傳用戶：BOBOniu
The IA-32 Software Developer’s Manual, Volume 3: System Programming Guide (Order Number 245472), is part of a three-volume set that describes the architecture and programming environment of all IA-32 Intel® Architecture processors. The IA-32 Software Developer’s Manual, Volume 3, describes the operating-system support environment of an IA-32 processor, including memory management, protection, tASK management, interrupt and exception handling, and system management mode. It also provides IA-32 processor compatibility information. This volume is aimed at operating- system and BIOS designers and programmers.
標(biāo)簽： Programming Developer Software 245472
上傳時(shí)間： 2013-12-23
上傳用戶：小碼農(nóng)lz
在了解實(shí)時(shí)嵌入式操作系統(tǒng)內(nèi)存管理機(jī)制的特點(diǎn)以及實(shí)時(shí)處理對(duì)內(nèi)存管理需求的基礎(chǔ)上，練習(xí)并掌握有效處理內(nèi)存碎片的內(nèi)存管理機(jī)制，同時(shí)理解防止內(nèi)存泄漏問(wèn)題的良好設(shè)計(jì)方法。使用預(yù)先規(guī)劃的思想，構(gòu)建自己的私有內(nèi)存管理機(jī)制，在系統(tǒng)內(nèi)存池中申請(qǐng)內(nèi)存，并將其納入私有內(nèi)存管理機(jī)制中，形成靜態(tài)預(yù)分配內(nèi)存池； 靜態(tài)預(yù)分配內(nèi)存池支持一種以上固定長(zhǎng)度內(nèi)存池，如16 字節(jié)內(nèi)存池和256 字節(jié)內(nèi)存池。固定長(zhǎng)度內(nèi)存池的單塊長(zhǎng)度應(yīng)考慮體系結(jié)構(gòu)開(kāi)銷，并盡量減少內(nèi)部碎片；固定長(zhǎng)度內(nèi)存池?cái)?shù)量應(yīng)可配置； 靜態(tài)預(yù)分配內(nèi)存池與系統(tǒng)內(nèi)存池的統(tǒng)一管理機(jī)制。向用戶分配內(nèi)存時(shí)應(yīng)保證長(zhǎng)度最佳匹配原則。當(dāng)申請(qǐng)內(nèi)存的長(zhǎng)度超過(guò)靜態(tài)預(yù)分配長(zhǎng)度或資源不足時(shí)，自動(dòng)向系統(tǒng)內(nèi)存池申請(qǐng)； 管理機(jī)制包括： a) 初 始化函數(shù)； b) 內(nèi) 存申請(qǐng)/釋放函數(shù)。并特別要保證釋放安全； c) 告 警機(jī)制； d) 管 理監(jiān)視機(jī)制。 5. 利用可能的互斥機(jī)制或代碼可重入設(shè)計(jì)，保證以上管理機(jī)制的操作安全性； 6. 創(chuàng)建多tASK 環(huán)境測(cè)試及演示以上內(nèi)容
標(biāo)簽： 內(nèi)存管理 實(shí)時(shí)嵌入式 實(shí)時(shí)處理 操作系統(tǒng)
上傳時(shí)間： 2016-04-12
上傳用戶：lizhen9880
JRemoteControl is a simple Java™ driven bluetooth remote control.It allows you to initiate virtually any tASK on your PC from a J2ME enabled device.
標(biāo)簽： JRemoteControl bluetooth initiate control
上傳時(shí)間： 2016-04-22
上傳用戶：1583060504
北京大學(xué)ACM比賽題目 In 1742, Christian Goldbach, a German amateur mathematician, sent a letter to Leonhard Euler in which he made the following conjecture: Every even number greater than 4 can be written as the sum of two odd prime numbers. For example: 8 = 3 + 5. Both 3 and 5 are odd prime numbers. 20 = 3 + 17 = 7 + 13. 42 = 5 + 37 = 11 + 31 = 13 + 29 = 19 + 23. Today it is still unproven whether the conjecture is right. (Oh wait, I have the proof of course, but it is too long to write it on the margin of this page.) Anyway, your tASK is now to verify Goldbach s conjecture for all even numbers less than a million.
標(biāo)簽： mathematician Christian Goldbach Leonhard
上傳時(shí)間： 2016-04-22
上傳用戶：wangchong
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