round robin 5路verilog实现
时间: 2023-08-24 11:06:11 浏览: 47
以下是一个简单的 round robin 5路的 Verilog 实现示例:
```
module round_robin_5(input clk, input reset, input [4:0] in_data, output reg [4:0] out_data);
reg [2:0] select;
always @ (posedge clk or posedge reset) begin
if (reset) begin
select <= 3'd0;
out_data <= 5'd0;
end else begin
select <= select + 1'b1;
if (select >= 3'd5) begin
select <= 3'd0;
end
out_data <= in_data[select];
end
end
endmodule
```
这个模块使用了一个 3 位计数器来实现五个输入端口的 round robin 调度。每当时钟上升沿到来时,计数器加 1,然后根据计数器的值选择一个输入端口。当计数器达到 5 时,它会重新回到 0。在模块被重置时,计数器和输出数据都会被清零。
相关问题
round robin 5路verilog实现 与tb文件
以下是Round Robin 5路 Verilog 实现的代码和测试台文件:
```
// Round Robin 5路 Verilog 实现
module round_robin_5(input clk,
input reset,
input [4:0] request,
output reg [4:0] grant);
reg [4:0] priority;
reg [2:0] round;
always @(posedge clk) begin
if (reset) begin
priority <= 5'b11111;
round <= 3'b0;
end else begin
priority <= {priority[3:0], priority[4]};
if (request != 0) priority[0] <= 1'b0;
if (priority == 5'b00000) priority <= 5'b11111;
if (round == 3'b0) begin
if (priority[0]) grant <= 1'b1;
round <= 3'b1;
end else if (round == 3'b1) begin
if (priority[1]) grant <= 1'b1;
round <= 3'b2;
end else if (round == 3'b2) begin
if (priority[2]) grant <= 1'b1;
round <= 3'b3;
end else if (round == 3'b3) begin
if (priority[3]) grant <= 1'b1;
round <= 3'b4;
end else begin
if (priority[4]) grant <= 1'b1;
round <= 3'b0;
end
end
end
endmodule
```
```
// Round Robin 5路 Verilog 测试台文件
`timescale 1ns / 1ns
module round_robin_5_tb();
reg clk, reset;
reg [4:0] request;
wire [4:0] grant;
round_robin_5 dut(
.clk(clk),
.reset(reset),
.request(request),
.grant(grant)
);
initial begin
clk = 1'b0;
reset = 1'b1;
request = 5'b11111;
#10 reset = 1'b0;
#10 request = 5'b00001;
#10 request = 5'b00010;
#10 request = 5'b00100;
#10 request = 5'b01000;
#10 request = 5'b10000;
#10 request = 5'b00000;
#10 request = 5'b10101;
#10 request = 5'b01010;
#10 request = 5'b00101;
#10 request = 5'b10010;
#10 request = 5'b01001;
#10 request = 5'b00000;
#10 request = 5'b10101;
#10 request = 5'b01010;
#10 request = 5'b00101;
#10 request = 5'b10010;
#10 request = 5'b01001;
#10 request = 5'b00000;
#10 $finish;
end
always #5 clk = ~clk;
endmodule
```
用case实现round robin 5路verilog实现 与tb文件
以下是Round Robin 5路Verilog的代码实现:
```
module round_robin_5(input clk, input rst, input [4:0] req, output reg [4:0] grant);
reg [2:0] cnt;
always @ (posedge clk or posedge rst) begin
if (rst) begin
cnt <= 3'b0;
grant <= 5'b0;
end
else begin
cnt <= cnt + 1;
if (cnt == 3'b100) begin
cnt <= 3'b0;
end
end
end
always @ (posedge clk or posedge rst) begin
if (rst) begin
grant <= 5'b0;
end
else begin
case (cnt)
3'b000: grant <= req[0] ? 5'b00001 : grant;
3'b001: grant <= req[1] ? 5'b00010 : grant;
3'b010: grant <= req[2] ? 5'b00100 : grant;
3'b011: grant <= req[3] ? 5'b01000 : grant;
3'b100: grant <= req[4] ? 5'b10000 : grant;
endcase
end
end
endmodule
```
以下是对应的testbench文件:
```
module round_robin_5_tb;
reg clk;
reg rst;
reg [4:0] req;
wire [4:0] grant;
round_robin_5 dut (.clk(clk), .rst(rst), .req(req), .grant(grant));
initial begin
clk = 0;
rst = 1;
req = 5'b0;
#10 rst = 0;
#10 req = 5'b00001;
#10 req = 5'b00010;
#10 req = 5'b00100;
#10 req = 5'b01000;
#10 req = 5'b10000;
#10 $finish;
end
always #5 clk = ~clk;
endmodule
```
在testbench文件中,我们使用了一个时钟以及一个复位信号。我们在第10个时钟周期后取消了复位信号,并在随后的每个时钟周期设置了一个请求信号。您可以根据您的实际需要更改这些操作。
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