题解 | #使用握手信号实现跨时钟域数据传输# VCS

使用握手信号实现跨时钟域数据传输

https://www.nowcoder.com/practice/2bf1b28a4e634d1ba447d3495134baac

使用vcs verdi环境,包含testbench和rtl.

处理跨时钟域信号打两拍同步,第三拍用于取上升沿产生控制信号,

当时钟频率相差过大,ack信号可能采不到,所以不能只产生一个clk的pulse,这里我做了一个可扩展长度的ack,req只有当检测到ack才会拉低所以不需要做这种处理.

`timescale 1ns/10ps

module testbench();

reg clk_a,clk_b,rst_n;

wire data_req,data_ack;

wire [3:0] data;

initial begin

clk_a = 1;

clk_b = 1;

rst_n = 1;

$fsdbDumpfile("out.fsdb");

$fsdbDumpvars(0,testbench);

#30

rst_n = 0;

#100

rst_n = 1;

#500_00 $finish();

end

always #15 clk_a = ~clk_a;

always #10 clk_b = ~clk_b;

data_driver dut_1

( .clk_a(clk_a),

.rst_n(rst_n),

.data(data),

.data_ack(data_ack),

.data_req(data_req)

);

data_receiver #(1) dut_2

( .clk_b(clk_b),

.rst_n(rst_n),

.data(data),

.data_ack(data_ack),

.data_req(data_req)

);

endmodule

module data_driver

(

input clk_a,

input rst_n,

input data_ack,

output reg [3:0] data,

output reg data_req

);

reg data_ack_dff1;

reg data_ack_dff2;

reg data_ack_dff3;

always @(posedge clk_a or negedge rst_n)begin

if(!rst_n)begin

data_ack_dff1 <= 1'b0;

data_ack_dff2 <= 1'b0;

data_ack_dff3 <= 1'b0;

end

else begin

data_ack_dff1 <= data_ack;

data_ack_dff2 <= data_ack_dff1;

data_ack_dff3 <= data_ack_dff2;

end

end

wire data_ack_rp = data_ack_dff2 & ~data_ack_dff3;

reg [2:0] data_cnt;

wire data_cnt5 = data_cnt==3'd4;

always @(posedge clk_a or negedge rst_n)begin

if(!rst_n)

data_cnt <= 3'b0;

else if(data_ack_rp)

data_cnt <= 3'b0;

else if(data_req)

data_cnt <= data_cnt;

else

data_cnt <= data_cnt + 1'b1;

end

always @(posedge clk_a or negedge rst_n)begin

if(!rst_n)

data <= 4'b0;

else if(data_ack_rp)

data <= {1'b0,(data[2:0] + 1'b1)};

else

data <= data;

end

always @(posedge clk_a or negedge rst_n)begin

if(!rst_n)

data_req <= 1'b0;

else if(data_ack_rp)

data_req <= 1'b0;

else if(data_cnt5)

data_req <= 1'b1;

end

endmodule

module data_receiver

#(

parameter DATA_ACK_EXT_LEN = 1

)

(

input clk_b,

input rst_n,

input [3:0] data,

input data_req,

output reg data_ack

);

reg data_req_dff1;

reg data_req_dff2;

reg data_req_dff3;

always @(posedge clk_b or negedge rst_n)begin

if(!rst_n)begin

data_req_dff1 <= 1'b0;

data_req_dff2 <= 1'b0;

data_req_dff3 <= 1'b0;

end

else begin

data_req_dff1 <= data_req;

data_req_dff2 <= data_req_dff1;

data_req_dff3 <= data_req_dff2;

end

end

wire data_req_rp = data_req_dff2 & ~data_req_dff3;

reg [3:0] data_rec;

always @(posedge clk_b or negedge rst_n)begin

if(!rst_n)

data_rec <= 4'b0;

else if(data_req_rp)

data_rec <= data;

else

data_rec <= data_rec;

end

reg [15:0] data_ack_cnt;

always @(posedge clk_b or negedge rst_n)begin

if(!rst_n)

data_ack_cnt <= 16'b0;

else if(data_req_rp)

data_ack_cnt <= 16'b0;

else if(data_req_dff3)

data_ack_cnt <= data_ack_cnt + 1'b1;

else

data_ack_cnt <= data_ack_cnt;

end

wire data_ack_ext = data_req_dff3 & (data_ack_cnt < DATA_ACK_EXT_LEN);

always @(posedge clk_b or negedge rst_n)begin

if(!rst_n)

data_ack <= 1'b0;

else if(data_req_rp | data_ack_ext)

data_ack <= 1'b1;

else

data_ack <= 1'b0;

end

endmodule

#练习时长两年半#
全部评论

相关推荐

1.自我介绍2.介绍一下mcp,&nbsp;skills3.了解react哪些状态管理库4.对话是sse还是什么?是用fetch还是EventSource?5.ts中的any&nbsp;和&nbsp;unknown讲一讲6.是直接用组件库的组件还是自己封装了一些别的7.代码输出题1function&nbsp;main()&nbsp;{{var&nbsp;a&nbsp;=&nbsp;1let&nbsp;b&nbsp;=&nbsp;2}console.log(a);console.log(b);}main()console.log(a);8.什么是块级作用域&nbsp;全局作用域&nbsp;函数作用域9.代码输出题2for&nbsp;(var&nbsp;i&nbsp;=&nbsp;0;i&nbsp;&amp;lt;&nbsp;5;i++)&nbsp;{setTimeout(()&nbsp;=&amp;gt;&nbsp;{console.log(i);},&nbsp;100);}10.代码输出题3for&nbsp;(var&nbsp;i&nbsp;=&nbsp;0;&nbsp;i&nbsp;&amp;lt;&nbsp;5;&nbsp;i++){function&nbsp;printText(temp)&nbsp;{setTimeout(()&nbsp;=&amp;gt;&nbsp;{console.log(temp);},&nbsp;100);}printText(i)}11.代码输出题4for(var&nbsp;i&nbsp;=&nbsp;0;i&nbsp;&amp;lt;&nbsp;5;i++){function&nbsp;printText(temp)&nbsp;{var&nbsp;temp&nbsp;=&nbsp;isetTimeout(()&nbsp;=&amp;gt;&nbsp;{console.log(temp);},&nbsp;100);}printText(i)}12.代码输出题5for(var&nbsp;i&nbsp;=&nbsp;0;i&nbsp;&amp;lt;&nbsp;5;i++){function&nbsp;printText(temp)&nbsp;{setTimeout(()&nbsp;=&amp;gt;&nbsp;{var&nbsp;temp&nbsp;=&nbsp;iconsole.log(temp);},&nbsp;100);}printText(i)}13.点击控制台输出题export&nbsp;default&nbsp;function&nbsp;App()&nbsp;{const&nbsp;[count,&nbsp;setCount]&nbsp;=&nbsp;useState(0)console.log('render',count)return&nbsp;(&lt;div&gt;&lt;h1&gt;{count}&lt;/h1&gt;{setCount(count&nbsp;+&nbsp;1)setTimeout(()&nbsp;=&amp;gt;&nbsp;console.log('setTimeout',&nbsp;count),&nbsp;1000)}}&amp;gt;+1&lt;/div&gt;)}//这个组件点击按钮后,控制台的输出顺序和值如下://&nbsp;1.&nbsp;render&nbsp;1&nbsp;(组件重新渲染,&nbsp;count&nbsp;更新为&nbsp;1)//&nbsp;2.&nbsp;setTimeout&nbsp;0&nbsp;(1秒后输出,注意这里是&nbsp;0&nbsp;而不是&nbsp;1)14.算法:给有序数组arr&nbsp;=&nbsp;[-4,&nbsp;-1,&nbsp;0,&nbsp;3,&nbsp;5],返回平方后的排序//&nbsp;有序数组平方后排序const&nbsp;arr&nbsp;=&nbsp;[-4,&nbsp;-1,&nbsp;0,&nbsp;3,&nbsp;5]function&nbsp;solution(arr)&nbsp;{const&nbsp;len&nbsp;=&nbsp;arr.lengthconst&nbsp;result&nbsp;=&nbsp;new&nbsp;Array(len)let&nbsp;left&nbsp;=&nbsp;0let&nbsp;right&nbsp;=&nbsp;len&nbsp;-&nbsp;1let&nbsp;index&nbsp;=&nbsp;len&nbsp;-&nbsp;1while&nbsp;(left&nbsp;&amp;lt;=&nbsp;right)&nbsp;{if&nbsp;(arr[left]&nbsp;*&nbsp;arr[left]&nbsp;&amp;gt;&nbsp;arr[right]&nbsp;*&nbsp;arr[right])&nbsp;{result[index]&nbsp;=&nbsp;arr[left]&nbsp;*&nbsp;arr[left]left++}&nbsp;else&nbsp;{result[index]&nbsp;=&nbsp;arr[right]&nbsp;*&nbsp;arr[right]right--}index--}return&nbsp;result}console.log(solution(arr));15.反问
查看14道真题和解析
点赞 评论 收藏
分享
评论
点赞
收藏
分享

创作者周榜

更多
牛客网
牛客网在线编程
牛客网题解
牛客企业服务