PS/2 packet parser and datapath
1.Decade counter2.Four-bit binary counter3.Decade counter again4.Slow decade counter5.Counter 1-126.Counter 10007.4-digit decimal counter8.12-hour clock9.Hdlbits博文分布10.4-bit shift register11.Left/right rotator12.Left/right arithmetic shift by 1 or 813.5-bit LFSR14.3-bit LFSR15.32-bit LFSR16.Shift register17.Shift register(2)18.3-input LUT19.Rule 9020.Rule 11021.Conway's Game of Life 16x1622.Simple FSM1(asynchronous reset)23.Simple FSM1(synchronous reset)24.Simple FSM2(asynchronous reset)25.Simple FSM2(synchronous reset)26.Simple state transition 327.Simple one-hot state transition 328.Simple FSM 3(asynchronous reset)29.Simple FSM 3(synchronous reset)30.Design a Moore FSM31.Lemmings 132.Lemmings 233.Lemmings 334.Lemmings 435.One-hot FSM36.PS/2 packet parser
37.PS/2 packet parser and datapath
38.Serial receiver39.Serial receiver and datapath40.Serial receiver with parity checking41.Sequence recognition42.Q8:Design a Mealy FSM43.Q5a:Serial two's complementer(Moore FSM)44.Q5b:Serial two's complementer(Moore FSM)45.Q3a:FSM46.Q3b:FSM47.Q3c:FSM logic48.Q6b:FSM next-state logic49.Q6c:FSM next-state logic50.Q6:FSM51.Q2a:FSM52.Q2:One-hot FSM equations53.Q2a: FSM54.Q2b:Another FSM55.Counter with period 100056.4-bit shift register and down counter57.FSM:Sequence 1101 recognizer58.FSM:Enable shift register59.FSM:The complete FSM60.The complete timer61.FSM:One-hot logic equations62.UARTSee also: PS/2 packet parser.
Now that you have a state machine that will identify three-byte messages in a PS/2 byte stream, add a datapath that will also output the 24-bit (3 byte) message whenever a packet is received (out_bytes[23:16] is the first byte, out_bytes[15:8] is the second byte, etc.).
out_bytes needs to be valid whenever the done signal is asserted. You may output anything at other times (i.e., don't-care).
另请参阅:PS/2 数据包解析器。
现在,您已经拥有了一台状态机,该状态机将识别 PS/2 字节流中的三字节消息,请添加一个数据路径,该数据路径也将在收到数据包时输出 24 位(3 字节)消息(out_bytes[23:16] 是第一个字节,out_bytes[15:8] 是第二个字节,依此类推)。
每当断言完成信号时,out_bytes都需要有效。您可以在其他时间输出任何内容(即,不在乎)。
题目网站
module top_module(
input clk,
input [7:0] in,
input reset, // Synchronous reset
output [23:0] out_bytes,
output done); //
parameter idle=4'b0001,
first=4'b0010,
second=4'b0100,
third=4'b1000;
reg [3:0]state,nstate;
always@(*)begin
case(state)
idle:begin
if(in[3])begin
nstate=first;
end
else begin
nstate=idle;
end
end
first:nstate=second;
second:nstate=third;
third:begin
if(!in[3])begin
nstate=idle;
end
else begin
nstate=first;
end
end
endcase
end
always@(posedge clk)begin
if(reset)begin
state<=idle;
end
else begin
state<=nstate;
end
end
reg [23:0]data;
assign done=(state==third);
always @(posedge clk) begin
if (reset) begin
data <= 24'd0;
end
else begin
data[23:16] <= data[15:8];
data[15:8] <= data[7:0];
data[7:0] <= in;
end
end
assign out_bytes = (done) ? data : 24'd0;
endmodule
在上一题的基础上,增加一个数据流录入的模块
合集:
Verilog学习
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