用EDA:verilog的状态机、当步行街上的行人要穿过主干道是,通过设置的按钮发出请求;(LED1车行道黄灯、LED2车行道红灯、LED1车行道绿灯;LED6人行道黄灯,LED7人行道红灯,LED8人行道绿灯)2、按钮被按下时,车行道变为黄灯闪烁,设置计数器计时时间3秒;3、计时结束,车行道变为红灯、人行道绿灯,计数器继续计时(计时时间为30秒),在30秒内若有人再次按按钮(S5),计数器不重新计时(数码管7-数码管8显示倒计时30秒);4、人行道黄灯时间为5秒(LED6闪烁),5秒后车行道道变为绿灯、人行道红灯,车辆通行。在30秒内行人按钮无效。通行时间超过30秒后,若有行人按下按钮,则回到(2
时间: 2023-05-30 17:06:13 浏览: 107
)状态。如果没有行人按下按钮,则状态机回到初始状态,车行道变为绿灯、人行道红灯,等待下一次行人请求。
以下是用Verilog实现的状态机代码:
```
module pedestrian_crossing(
input clk,
input btn,
output reg [7:0] led,
output reg [3:0] seg
);
parameter INIT = 2'b00;
parameter YELLOW_BLINK = 2'b01;
parameter RED_GREEN = 2'b10;
parameter YELLOW = 2'b11;
reg [1:0] state, next_state;
reg [3:0] count;
reg [3:0] timer;
reg btn_pressed;
always @ (posedge clk) begin
if (btn && !btn_pressed) begin
btn_pressed <= 1;
end else begin
btn_pressed <= 0;
end
case (state)
INIT: begin
led <= 8'b00000010; // car green, pedestrian red
seg <= 4'b1111;
if (btn_pressed) begin
next_state <= YELLOW_BLINK;
count <= 3'b000;
timer <= 4'b1111; // 30 sec
end else begin
next_state <= INIT;
end
end
YELLOW_BLINK: begin
led <= {3'b000, count[2], 3'b000, count[2]};
seg <= timer;
if (count == 3'b111) begin
next_state <= RED_GREEN;
count <= 4'b0000;
timer <= 4'b1111; // 30 sec
end else begin
next_state <= YELLOW_BLINK;
count <= count + 1;
end
end
RED_GREEN: begin
led <= 8'b00000101; // car yellow, pedestrian green
seg <= timer;
if (btn_pressed) begin
next_state <= RED_GREEN;
seg <= {7'b0000000, timer};
end else if (timer == 4'b0000) begin
next_state <= YELLOW;
timer <= 4'b0101; // 5 sec
end else begin
next_state <= RED_GREEN;
timer <= timer - 1;
end
end
YELLOW: begin
led <= 8'b00001001; // car red, pedestrian yellow
seg <= timer;
if (timer == 4'b0000) begin
next_state <= INIT;
end else begin
next_state <= YELLOW;
timer <= timer - 1;
end
end
endcase
state <= next_state;
end
endmodule
```
在这个状态机中,使用了4个状态,分别表示初始状态、黄灯闪烁状态、红绿灯状态和黄灯状态。计数器用于计时,数码管显示倒计时时间。按钮按下后,状态机进入黄灯闪烁状态,计时3秒后进入红绿灯状态,计时30秒,如果在这30秒内有人再次按下按钮,则不重新计时,数码管显示倒计时时间。在红绿灯状态下,如果计时结束,进入黄灯状态,计时5秒后进入初始状态,等待下一次行人请求。如果通行时间超过30秒,则回到初始状态。
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