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状态机专题练习
要求
1、根据以下描述功能用verilog编写一段代码,并用状态机来实现该功能。
(1)状态机:实现一个测试过程,该过程包括启动准备状态、启动测试、停止测试、查询测试结果、显示测试结果、测试结束返回初始化6个状态;用时间来控制该过程,90秒内完成该过程;
(2)描述状态跳转时间;
(3)编码实现。
- 画出可以检测10010串的状态图, 并用verilog编程实现之。
(选做,FPGA实训同学必做)完成作业并提交MD文档和博客地址。
一、
1. 新建一个项目
2. 新建一个Verilog HDL文件
每个状态保持15s。每个状态对应的led状态各不相同。led的状态值为当前状态值。
module training_1(
input wire clk,
input wire rst_n,
output reg [3:0]led
);
localparam [2:0] S1 = 1;
localparam [2:0] S2 = 2;
localparam [2:0] S3 = 3;
localparam [2:0] S4 = 4;
localparam [2:0] S5 = 5;
localparam [2:0] S6 = 6;
parameter [29:0] MAX_COUNT = 750_000_000;//15s
reg [29:0] count;
reg [2:0] current_state;
reg [2:0] next_state;
//15s 计时器 以及 状态转移
always@(posedge clk or negedge rst_n) begin
if(!rst_n) begin
current_state <= S1;
count <= 30'd1;
end
else if (count == MAX_COUNT) begin
count <= 30'd1;
end
else begin
count <= count + 30'd1;
current_state <= next_state;
end
end
// 状态判断
always @(posedge clk or negedge rst_n) begin
if(!rst_n) begin
next_state <= S1;
end
else begin
case (current_state)
S1: begin
if(count == MAX_COUNT) begin
next_state <= S2;
end
else begin
next_state <= next_state;
end
end
S2: begin
if(count == MAX_COUNT) begin
next_state <= S3;
end
else begin
next_state <= next_state;
end
end
S3: begin
if(count == MAX_COUNT) begin
next_state <= S4;
end
else begin
next_state <= next_state;
end
end
S4: begin
if(count == MAX_COUNT) begin
next_state <= S5;
end
else begin
next_state <= next_state;
end
end
S5: begin
if(count == MAX_COUNT) begin
next_state <= S6;
end
else begin
next_state <= next_state;
end
end
S6: begin
if(count == MAX_COUNT) begin
next_state <= S1;
end
else begin
next_state <= next_state;
end
end
default: next_state <= S1;
endcase
end
end
//状态输出 根据当前状态点亮不同的灯
always @(posedge clk or negedge rst_n) begin
if (!rst_n) begin
led <= 4'd0;
end
else begin
case