点击切换搜索课件文库搜索结果(50)
文档格式:DOC 文档大小:19KB 文档页数:1
一、名词解释 市场营销 目标市场 营销渠道 市场调查 二、简答题 1、产品的整体概念是什么?有哪些特点? 2、简述市场营销的“4PS”组合。 3、试分析产品生命周期四个阶段的特点及企业应采取的基本策略
文档格式:PPT 文档大小:1.78MB 文档页数:85
9.18255A可编程外围接口芯片 9.2PS-2304数字量I/0接口板简介 9.3BCD码并行数字信号的采集 9.4车速脉冲信号的采集计数
文档格式:PDF 文档大小:80.46KB 文档页数:3
REMINDER: Quiz #2 will be held from 7: 30-9: 30 p. m. Thursday, November 13 in Walker Memorial. The quiz will cover materials in Chapters 4-7(through Section 7. 4)of O&w, Lectures and Recitations through October 29, Problem Sets #4-6, and that part of Problem Set 7 involving problems from Chapter 7 Reading Assignments
文档格式:PDF 文档大小:188.57KB 文档页数:15
Problem set SolutioN Exercise for home study o&W8.35 (a) From the system diagram, we see that a(t)=a(t)cos(wet) Using the multiplication property ZGu)= o(X(u)* FT(cos wct)) FT of cos wt is two impulses with area T at +wc. Therefore, Z(w)is the spectrum
文档格式:PDF 文档大小:208.4KB 文档页数:17
The lowpass filter H(u) has a cutoff frequency wc=205T rad/ sec. Thus, c(t)is r(t) where all terms with frequency above we are removed by the lowpass filter. The terms which are kept have kwol 205T rad /sec k|< 10.25, so the output, ac(t),is r(t)= To obtain n, we sample c(t) every T=5 10-3 seconds with an impulse train The sampling frequency is 400T=2 x maximum frequency in c(t). Therefore
文档格式:PDF 文档大小:77.63KB 文档页数:6
REMINDER: Quiz 1 will be held from 7: 30 to 9: 30 p. m. Tuesday, October 14 The quiz will cover material in Chapters 1-3 of o&w Lectures and Recitations through september 26, Problem Sets 1-3 and that part of Problem Set 4 involving problems from Chapter 3 Reading assignments
文档格式:PDF 文档大小:62.2KB 文档页数:4
REMINDER: Computer Lab #1 is due on October 8 REMINDER: Quiz 1 will be held from 7: 30-9: 30 p. m. Tuesday, October 14 in Walker Memorial. The quiz will cover material in Chapters 1-3 of o W, Lectures and Recitations
文档格式:DOC 文档大小:27KB 文档页数:2
Ps:细写字母为修改后噶答案,具体位置见红色处 毛邓三学习指导书——选择题答案 第258页:(社会主义初级阶段理论) (一)1~5: DDaBA(第3题十四大:邓小平同志建设有中国特色社会主义理论
文档格式:DOC 文档大小:234.5KB 文档页数:11
(1)已知空气的干燥温度为60℃,湿球温度为30℃,试计算空气的湿含量H, 相对湿度叩,焓I和露点温度。 解:查表得t=30C时p=4247KkPa H,.=0622P2A(P-P2)=00272 [(t-t/rl( 30°C时r=2427a/Kn=1.09 ∴H=0.0137 t=60Cp,=19923kPa 由H=00137求得此时p=218kPa =p/ps =(1.01+188×0.0158)×60+2490×0.0158 =9644/kg千空气
文档格式:PPT 文档大小:4.08MB 文档页数:127
6.1 C55x C/C++语言概述 6.1.1 C/C++语言概况 6.1.2 C55x C/C++语言概况 6.2 C55x C/C++语言编程基础 6.2.2 关键字 6.2.3 寄存器变量和参数 6.2.4 asm指令 6.2.5 Pragma指令 6.2.6标准ANSIC语言模式的改变(-pk,-pr和-ps选项) 6.2.7 存储器模式 6.2.8 存储器分配 6.2.9 中断处理 6.2.10 运行时间支持算法及转换程序 6.2.11 系统初始化 6.3 C55x C/C++编译器的使用 6.3.1 编译器外壳程序cl55简介 6.3.2 cl55程序的选项 6.3.3 编译器和CCS 6.4 TMS320C55x的C代码优化 6.5 C55x C和汇编语言混合编程 6.5.1 C和汇编语言混合编程概述 6.5.2 寄存器规则 6.5.3 函数结构和调用规则 6.5.4 C和汇编语言的接口
上页12345
热门关键字
搜索一下,找到相关课件或文库资源 50 个  
©2008-现在 cucdc.com 高等教育资讯网 版权所有