`收到一块板子要改，看了一下上面芯片型号，UST2030网上找了半天没找到资料，带AD的芯片，哪位大神有资料`
2021-06-02 13:07
UST_Cs_nr // UST_Xcs_r UST_Xwe_r UST_Addr_r UST_Data结束否则开始
2019-01-21 13:50
最近在做一个数字图像的项目。求购AD71124。有货的麻烦联系我：eexdzhou@ust.hk
2018-08-14 06:57
谢谢。* ECE HKUST如有可能，请通过电子邮件联系我们：mzhaoab@ust.hk 以上来自于谷歌翻译 以下为原文*Dear Keysight Engineer:* *Our university
2018-11-09 10:20
to 128 to get a tickspeed of // and set Timer 1 to free running mode 0000 1101 10usT1CTL = (T1CTL
2018-06-06 02:16
哪位有CS+ for CC的license？打算近期要用瑞萨开发几个东西，CS+提示“W0511179:The evaluation version is valid for the remaining 60 days”
2018-05-09 18:38
中断和T3中断T3CTL |= T3_PSC_32;//32分频,32/16000000=2us//T3CTL &= ~0x03;T3CTL |= 0x01;//倒计数模式T3CC0= 0x64;//100*2usT3CTL |= 0x10;//启动EA = 1;//开总中断}
2019-06-17 04:35
嗨，我下载了最新的ISE 12.1安装版（M.53，用于Windows），并收到消息抱怨丢失文件：“... \ idata \ drop_0324_edk.zip.xz。请检查文件权限”。确实不存在该文件。附件是截图。谢谢，OutputLogic以上来自于谷歌翻译以下为原文Hi, I downloaded the latest ISE 12.1 install (M.53 , for Windows) and got the message complaining about a missing file: "...\idata\drop_0324_edk.zip.xz. Please check file permissions". The file is indeed not there. Attached is the screenshot. Thanks,OutputLogic
2018-11-14 10:35
请问大佬XR806鸿蒙开发板怎么样？
2021-12-29 07:06
嗨，我写了下面的uart接收器，但我无法从输入“rx”中删除所有噪声;我使用了去抖动器和移动平均器，但是，下面的uart接收器有时候无法接收到正确的字节。下面的示例将其参数CLOCKCYCLESPERBIT设置为868，如果使用100 MHz时钟，则为每位的时钟周期数。请帮助解决从输入信号中消除噪声的最佳方法。模块uartreceiver（输入clk，输入rx，输出reg收到，输出reg [8 -1：0]数据）;参数CLOCKCYCLESPERBIT = 868;参数LOG2CLOCKCYCLESPERBIT = 10;//使用的值//寄存器rxstate。//接收器正在等待//在传输上参数RXIDLE = 0;//接收器正在接收比特。参数RXRECEIVE = 1;// Receiver正在检查//传输结束参数RXSTOP = 2;//注册哪个持有//接收者的状态reg [2 -1：0] rxstate;//注册将用于保留//计数时钟周期数。reg [LOG2CLOCKCYCLESPERBIT -1：0]计数器;//注册使用的//跟踪号码//留下来接收的比特。reg [3 -1：0] bitcount;总是@（posedge clk）开始//启用寄存器时钟//只为每一点if（counter == CLOCKCYCLESPERBIT）开始if（rxstate == RXRECEIVE）开始//在最后一点接收，//“bitcount”将为0，并且//将自动重置//减少时为7。位计数以上来自于谷歌翻译以下为原文Hi, I have written the following uart receiver, but I am not able to remove all noise from the input "rx";I have used a debouncer and a moving average, but still, the following uart receiver sometime fail to receive the correct byte.The example below has its parameter CLOCKCYCLESPERBIT set to 868, which is the number of clock cycles per bit if a 100 MHz clock is used. Please help on what is the best way to remove noise from an input signal. module uartreceiver(input clk,input rx,output reg received,output reg[8 -1 : 0] data);parameter CLOCKCYCLESPERBIT = 868;parameter LOG2CLOCKCYCLESPERBIT = 10;// Values used with// the register rxstate.// Receiver is waiting// on a transmission.parameter RXIDLE = 0;// Receiver is receiving bits.parameter RXRECEIVE = 1;// Receiver is checking for// the end of a transmission.parameter RXSTOP = 2;// Register which hold// the state of the receiver.reg[2 -1 : 0] rxstate;// Register that will be used to keep// a count of the number of clock cycles.reg[LOG2CLOCKCYCLESPERBIT -1 : 0] counter;// Register which is used// to keep track of the number// of bits left to receive.reg[3 -1 : 0] bitcount;always @(posedge clk) begin// Enable clocking of registers// only for every bit.if (counter == CLOCKCYCLESPERBIT) beginif (rxstate == RXRECEIVE) begin// At the last bit to receive,// "bitcount" will be 0, and// will be automatically reset// to 7 when decremented.bitcount
2019-07-08 06:59
