uart接收三个input,把三个数据依次在电脑上显示,Verilog
时间: 2023-10-08 18:10:12 浏览: 76
以下是一个简单的Verilog代码示例,用于从UART接收三个输入,并将它们依次显示在计算机上:
```
module uart_receiver (
input clk,
input reset,
input rx,
output reg [7:0] data_out
);
reg [7:0] buffer [0:2];
reg [1:0] index = 2;
// UART receive state machine
reg [3:0] state = 0;
always @(posedge clk) begin
case (state)
0: if (!reset) begin
state <= 1;
end
1: if (rx == 0) begin
state <= 2;
buffer[index] <= 0;
end
2: if (rx == 1) begin
state <= 3;
end else begin
buffer[index] <= buffer[index] << 1;
buffer[index][0] <= rx;
end
3: if (index == 0) begin
state <= 4;
end else begin
index <= index - 1;
state <= 1;
end
4: begin
data_out <= buffer[0];
$display("Received data: %d", buffer[0]);
data_out <= buffer[1];
$display("Received data: %d", buffer[1]);
data_out <= buffer[2];
$display("Received data: %d", buffer[2]);
state <= 0;
end
default: state <= 0;
endcase
end
endmodule
```
该模块具有一个时钟输入(clk),一个复位输入(reset),一个UART接收器输入(rx)和一个8位数据输出(data_out)。状态机接收串行数据,并将其存储在一个3个元素的缓冲区中。当所有三个数据都接收完毕时,状态机会将它们依次输出到计算机上。在每个数据输出时,$display语句将其打印到控制台上。注意,这只是一个示例,并且可能需要根据您的具体要求进行修改。
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