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stepper_motor
control stepper motor by fpga card with vhdl program
- 2012-01-08 02:34:17下载
- 积分:1
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fft
FPGA实现FFT算法的源代码及工程文件,此工程为ISE工程项目。有详细的说明,可以运行。(FPGA Implementation of FFT algorithm source code and project files, this works for the ISE project. There are detailed instructions, you can run.)
- 2013-10-12 17:21:32下载
- 积分:1
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vhdl classical source code
vhdl经典源代码――ps2接口设计,入门者必须掌握-vhdl classical source code-- ps2 interface design, beginners must master
- 2022-04-07 18:12:38下载
- 积分:1
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DUC
说明: 在FPGA内利用verilog实现数字上变频(apply the verilog to implement the digital up frequency)
- 2021-04-09 09:58:59下载
- 积分:1
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verilog实现自动售货机
能实现输入0.5 1 5块钱的累加,然后对应购买的商品价格进行比较,显示找的钱数或错误灯(MY English is very good)
- 2019-01-09 13:35:02下载
- 积分:1
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clk_generator
时钟分频代码,PWM产生 RTL 源代码。(clock divider,PWM generator RTL Source Code)
- 2013-08-18 09:29:42下载
- 积分:1
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Harris-algorithm-based-on-FPGA
在利用FPGA的并行处理能力应对高速数据和去做复杂的数据处理时,对一些较为复杂或者重复性工作模块多的情况下,算法资源就需要进行预评估。有效的资源预评估不仅可以在芯片选型上有益,还可以对程序有较详细的估计,在硬件不变的前提下能够选择更好的算法优化。本文着重在Harris算法在FPGA的实现以及在移植之前对其占用的FPGA资源进行预评估。(Response to high-speed data and do complex data processing in the FPGA parallel processing capabilities, to cope with some of the more complex or repetitive tasks module,it is necessary to pre-assessment algorithm resources. Resources pre-assessment can not only be useful in the chip selection, but also be a more detailed estimate of the program to be able to choose a better algorithm optimization in the same premise hardware. This article focuses on the pre-assessment in the Harris algorithm in the FPGA implementation and its FPGA resources occupied prior to transplantation.)
- 2013-02-28 15:41:39下载
- 积分:1
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Altera USB声卡
altera usb 下载线DIY完全资料-altera usb blaster
- 2022-04-29 21:31:03下载
- 积分:1
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viterbi213
说明: 编码方式为213的Viterbi卷积码编码器和译码器的FPGA的实现,包含整个QuartusII的工程文件,解码方式为寄存器交换法(Encoding for the 213 convolutional code encoder and Viterbi decoder FPGA realization of the project file that contains the entire QuartusII, decoding method for the register exchange)
- 2020-12-27 21:19:02下载
- 积分:1
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基数 4 蝴蝶算法的实现
给定的序列
x(n) 被摧毁成 4 序列的长度 N/4 每个。而不是划分
用密度泛函理论计算成两半在 RAD2,种四分。N 点
输入的序列被分成四个的子序列,x(4n),x(4n+1),,x(4n+2),和
x (4n + 3),其中 n = 0,1,...N/4-1.Radix-4 使用日志4N
阶段,每个阶段有 N/4 蝴蝶。N/4 蝴蝶涉及每年
阶段和阶段数目是日志4N N 点序列。因此,
复杂的乘法次数是 3N/4 日志4N 和数目
复杂的加法是 12N/4 日志4n。在基数 2 FFT 的比较
复杂的乘法次数是减少 25%
- 2023-08-02 05:25:03下载
- 积分:1