JK 触发器状态图实现的错误输出

Question

我正在尝试在 verilog 中实现 JK 触发器的简单 FSM。但是我看到输出 'q' 和 'q_not' 在多个时间点都是错误的。我在下面展示代码和输出。有人可以让我知道代码有什么问题吗？我特别想知道这个实现有什么问题，尽管还有其他方法可以实现 JK 触发器。

JK触发器和测试台模块

`timescale 1ns/100ps

module jk_ff(j, k, clk, reset, q, q_not);

input j, k, clk, reset;

output reg q, q_not;

reg present_state, next_state;

parameter state_a = 1'b0;
parameter state_b = 1'b1;

always @ (present_state or j or k)
begin:comb_logic

 next_state = state_a;
 //next_state = 0;

 case(present_state)

  state_a: begin
   if (j == 1'b0 && k == 1'b0) begin
     next_state = state_a;
     end
   else if (j == 1'b0 && k == 1'b1) begin
     next_state = state_a;
     end
   else if (j == 1'b1 && k == 1'b0) begin
     next_state = state_b;
     end
   else if (j == 1'b1 && k == 1'b1) begin
     next_state = state_b;
     end
   end

  state_b: begin
   if (j == 1'b0 && k == 1'b0) begin
     next_state =  state_b;
   end
   else if (j == 1'b0 && k == 1'b1) begin
     next_state = state_a;
   end
   else if (j == 1'b1 && k == 1'b0) begin
     next_state = state_b;
   end
   else if (j == 1'b1 && k == 1'b1) begin
     next_state = state_a;
   end
  end
  default: next_state = state_a;
 endcase 
end

always @ (posedge clk or reset)
begin: seq_logic

if (reset) begin
 q <= 1'b0;
 q_not <= 1'b1;
 present_state <= state_a;
 end
 else begin
 present_state <= next_state;

 case(present_state)

   state_a: begin
    q <= 1'b0;
    q_not <= 1'b1;
   end
   state_b: begin
    q <= 1'b1;
    q_not <= 1'b0;
   end
   default: present_state <= state_a;

  endcase
  end
 end

endmodule

//testbench

module jk_ff_tb;

reg j, k, clk, reset;

wire q, q_not;

jk_ff DUT(.j(j), .k(k), .clk(clk), .reset(reset), .q(q), .q_not(q_not));

initial begin
clk =0;
forever #5 clk = !clk;
end

initial begin
 $monitor("j = %b, k = %b, q = %b, q_not = %b", j, k, q, q_not);
 $dumpfile("jk_ff_wave.vcd");
 $dumpvars;
 reset = 1;
 j=1'b0;
 k=1'b1;
#10 reset = 0;

 #15 j=1'b1;
 #15 k=1'b0;
 #15 j=1'b0;
 #15 k=1'b1;
 #15 j=1'b1;
 #15 k=1'b1;


 #10 $finish;  

end

endmodule

显示输入值和主要输出值的测试台仿真输出

j = 0, k = 1, reset = 1, q = 0, q_not = 1
j = 0, k = 1, reset = 0, q = 0, q_not = 1
j = 1, k = 1, reset = 0, q = 0, q_not = 1
j = 1, k = 1, reset = 0, q = 1, q_not = 0
j = 1, k = 0, reset = 0, q = 1, q_not = 0
j = 1, k = 0, reset = 0, q = 0, q_not = 1
j = 0, k = 0, reset = 0, q = 1, q_not = 0
j = 0, k = 1, reset = 0, q = 1, q_not = 0
j = 1, k = 1, reset = 0, q = 0, q_not = 1
j = 1, k = 1, reset = 0, q = 1, q_not = 0
j = 1, k = 1, reset = 0, q = 0, q_not = 1

enter code here

谢谢！

Answer 1

你遇到了各种各样的问题：

在 seq_logic 中，您使用阻塞赋值 present_state，下一条语句是 case(present_state)。这将测试 present_state 的 old 值，这不是您想要的
您的 'comb_logic' 进程对 present_state 敏感，但您的 seq_logic 进程在时钟上升沿上发生变化 present_state。乍一看，这似乎是正确的做法，但事实并非如此——把它拉出来。按照你的写法，comb_logic 应该只对 J 和 K

这两个足以获得正确的结果，但这对于 JK 来说太复杂了 - 重新开始，将所有内容放在一个时钟进程中，转储下一个逻辑进程，只需使用 JK 的行为 - 加载、设置或切换。您还应该将当前时间添加到 $monitor.

JK 触发器状态图实现的错误输出

Faulty outputs for JK flip flop state diagram implementation

verilog

iverilog