HDLBits答案(17)_Verilog有限状态机(4)_verilog期末复习题有限机

Verilog有限状态机(4)

HDLBits链接

前言

今天继续更新状态机小节的习题。

题库

题目描述1：

One-hot FSM

独热编码，根据状态转移图输出下一状态与结果。

Solution1：

module top_module(
    input in,
    input [9:0] state,
    output [9:0] next_state,
    output out1,
    output out2);
    
    parameter S0 = 4'd0;
    parameter S1 = 4'd1;
    parameter S2 = 4'd2;
    parameter S3 = 4'd3;
    parameter S4 = 4'd4;
    parameter S5 = 4'd5;
    parameter S6 = 4'd6;
    parameter S7 = 4'd7;
    parameter S8 = 4'd8;
	parameter S9 = 4'd9;
    
    assign next_state[0] = ~in & (state[S0] | state[S1] | state[S2] | state[S3] | state[S4] | state[S7] | state[S8] | state[S9]);
    assign next_state[1] = in & (state[S0] | state[S8] | state[S9]);
    assign next_state[2] = in & state[S1];
    assign next_state[3] = in & state[S2];
    assign next_state[4] = in & state[S3];
    assign next_state[5] = in & state[S4];
    assign next_state[6] = in & state[S5];
    assign next_state[7] = in & (state[S6] | state[S7]);
    assign next_state[8] = ~in & state[S5];
    assign next_state[9] = ~in & state[S6];
    
    assign out1 = (state[S8] | state[S9]);
    assign out2 = (state[S7] | state[S9]);
    
endmodule
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题目描述2：

PS/2 packet parser

首先，这道题目中作者表明in的第3位为1时，状态机启动，其他两位可能为1或0，注意，这里不允许重叠检测，根据作者给出的时序图，大家应该就可以写出状态机了。

Solution2：

module top_module(
    input clk,
    input [7:0] in,
    input reset,    // Synchronous reset
    output done); //

    parameter BYTE_FIRST = 2'd0;
    parameter BYTE_SECOND = 2'd1;
    parameter BYTE_THIRD = 2'd2;
    parameter WAIT = 2'd3;
    
    reg [1:0] state,next_state;
    // State transition logic (combinational)
    always @(*)begin
        case(state)
            BYTE_FIRST:begin
                next_state = BYTE_SECOND;
            end
            BYTE_SECOND:begin
                next_state = BYTE_THIRD;
            end
            BYTE_THIRD:begin
                if(in[3] == 1'b1)begin
                    next_state = BYTE_FIRST;
                end
                else begin
                    next_state = WAIT;
                end
            end
            WAIT:begin
                if(in[3] == 1'b1)begin
                    next_state = BYTE_FIRST;
                end
                else begin
                    next_state = WAIT;
                end
            end
        endcase
    end
    
    // State flip-flops (sequential)
    always @(posedge clk)begin
        if(reset)begin
            state <= WAIT;
        end
        else begin
            state <= next_state;
        end
    end
    
    // Output logic
    assign done = (state == BYTE_THIRD);

endmodule
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题目描述3：

PS/2 packet parser and datapath

这道题目相比上一道多了数据位输出，当done信号为1时，输出24bit的数据，这24bit的数据高8位，中8位，低8位分别从in[3]为1开始计起，依次输出。done信号为0的时候不关心数据信号。高8位输出仅当下一状态为BYTE_SECOND才开始，此处可以简化判断逻辑，大家可以注意一下。

Solution3：

module top_module(
    input clk,
    input [7:0] in,
    input reset,    // Synchronous reset
    output [23:0] out_bytes,
    output done); //

    parameter BYTE_FIRST = 2'd0;
    parameter BYTE_SECOND = 2'd1;
    parameter BYTE_THIRD = 2'd2;
    parameter DONE = 2'd3;
    
    reg [1:0] state,next_state;
    reg [23:0] out_bytes_reg;
    // FSM from fsm_ps2
    always @(posedge clk)begin
        if(reset)begin
            state <= BYTE_FIRST;
        end
        else begin
            state <= next_state;
        end
    end
    
    always @(*)begin
        case(state)
            BYTE_FIRST:begin
                if(in[3])begin
                    next_state = BYTE_SECOND;
                end
                else begin
                    next_state = BYTE_FIRST;
                end
            end
            BYTE_SECOND:begin
                next_state = BYTE_THIRD;
            end
            BYTE_THIRD:begin
                next_state = DONE;
            end
            DONE:begin
                if(in[3])begin
                    next_state = BYTE_SECOND;
                end
                else begin
                    next_state = BYTE_FIRST;
                end
            end
        endcase
    end

    always @(posedge clk)begin
        if(next_state == BYTE_SECOND)begin
            out_bytes_reg[23:16] <= in;
        end
        else if(next_state == BYTE_THIRD)begin
            out_bytes_reg[15:8] <= in;
        end
        else if(next_state == DONE)begin
            out_bytes_reg[7:0] <= in;
        end
        else begin
            out_bytes_reg <= 24'd0;
        end
    end
    // New: Datapath to store incoming bytes.
    assign done = (state == DONE);
    assign out_bytes = out_bytes_reg;

endmodule
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结语

今天就先更新这三题吧，今天上了数分课发现拉下好多课程，希望自己期末能过吧，哈哈哈，之后要好好复习了。若代码有错误希望大家及时指出，我会尽快改正。