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下面是预置的jk触发器，编译后清除没有错。但是在模拟之后，我发现我的输出（QA、QB、QC、QD）只是保持为 0 并且没有变化。

我的案例陈述有什么问题吗？或者我以错误的方式使用negedge？

module one(inputA, inputB, R01, R02, R91, R92, QA, QB, QC, QD);
input inputA, inputB, R01, R02, R91, R92;
output QA, QB, QC, QD;

reg qa = 1'b0, qb = 1'b0, qc = 1'b0, qd = 1'b0;
reg r0, r9, pre1, clr1, clr2, clr3, pre4, clr4, j2, j4, k4;


//your code~~
initial begin
r0 = ~(R01 & R02);
r9 = ~(R91 & R92);
end

always @(negedge inputA) begin
pre1 = r0;
clr1 = r9;
case({pre1, clr1})
2'b01 : qa = 1'b1;
2'b10 : qa = 1'b0;
2'b11 : qa = ~qa;//toggle is the original result of jk
endcase//jk1
end

always @(negedge inputB) begin
j2 = ~qd;
clr2 = ~(r0 | r9);
if (j2 == 0 && clr2 == 0) qb = 0;//clr=0 means must clear it
if (j2 == 1 && clr2 == 0) qb = 0;//clr=0 means must clear it
if (j2 == 1 && clr2 == 1) qb = ~qb;//toggle
//jk2
end

always @(negedge qb) begin
qc = ~qc;
clr3 = ~(r0 | r9);
if (clr3 == 0) qc = 0;//clr=0 means must clear it
else qc = ~qc;
//jk3
end

always @(negedge inputB) begin
j4 = qb & qc;
case({j4, k4, pre4, clr4})
4'b0011 : qd = qd; //HOLD
4'b0001 : qd = 1'b1; //PRESET
4'b0010 : qd = 1'b0; //CLEAR

4'b1011 : qd = 1'b1; //SET
4'b1001 : qd = 1'b1; //PRESET
4'b1010 : qd = 1'b0; //CLEAR

4'b0111 : qd = 1'b0; //RESET
4'b0101 : qd = 1'b1; //PRESET
4'b0110 : qd = 1'b0; //CLEAR

4'b1111 : qd = ~qd; //TOGGLE
4'b1101 : qd = 1'b1; //PRESET
4'b1110 : qd = 1'b0; //CLEAR
endcase
//jk4
end

begin
assign QA = qa;
assign QB = qb;
assign QC = qc;
assign QD = qd;
end

endmodule

下面是测试台

module one_tb;
reg clk, r0, r9;
wire [3:0] Q;
one test (.inputA(clk), .inputB(Q[0]), .R01(r0), .R02(r0), .R91(r9), .R92(r9), .QA(Q[0]), .QB(Q[1]), .QC(Q[2]), .QD(Q[3]));

always
#5 clk=~clk;

initial
begin
    clk=1'b0;
    //0,1,2,3,4,5,6,7,8,9,0,1,2,3,4,5,0,1,2,3,9,0,1,2,3,4,5,6,7,8
    r0 = 1'b1; r9 = 1'b0; #3 ;r0 = 1'b0; r9 = 1'b0; #157; 
    r0 = 1'b1; r9 = 1'b0; #3 ;r0 = 1'b0; r9 = 1'b0; #37;
    r0 = 1'b0; r9 = 1'b1; #3; r0 = 1'b0; r9 = 1'b0; #97;
    $stop;
end

endmodule

这是 jk fkip-flop 的照片

4

3 回答 3

0

r0并且r9在模拟 ( ) 中总是未知的，X因为您只在时间 0 将它们分配给值一次。

您可能打算在每次“R”信号更改时更改它们。

改变：

initial begin
r0 = ~(R01 & R02);
r9 = ~(R91 & R92);
end

到：

always @* begin
r0 = ~(R01 & R02);
r9 = ~(R91 & R92);
end

这样做可以解决 X 问题并允许QA输出切换。

于 2020-06-08T10:58:05.687 回答
0

现在我知道为什么我的 QB、QC、QD 仍然为 0。

预置和清除的功能取决于我输入的 r0、r9 而不是 inputA、inputB。

在我将代码更改为以下内容后，结果看起来不错：

module one(inputA, inputB, R01, R02, R91, R92, QA, QB, QC, QD);
input inputA, inputB, R01, R02, R91, R92;
output QA, QB, QC, QD;

reg qa = 1'b0, qb = 1'b0, qc = 1'b0, qd = 1'b0;
reg r0, r9, pre1, clr1, clr2, clr3, j2, j4, k4;

always@(*) begin
r0 = ~(R01 & R02);
r9 = ~(R91 & R92);
clr2 = r0 | r9;
clr3 = r0 | r9;
end

always @(negedge inputA) begin
qa = ~qa;
//jk1
end

always @(negedge inputB) begin
j2 = ~qd;
if (j2 == 0 ) qb = 0;
if (j2 == 1 ) qb = ~qb;
//jk2
end

always @(negedge qb) begin
qc = ~qc;
//jk3
end

always @(negedge inputB) begin
j4 = qb & qc;
k4 = qd;
case({j4, k4})
2'b01 : qd = 1'b0;//reset
2'b10 : qd = 1'b1;//set
2'b00 : qd = qd;//hold
2'b11 : qd = ~qd;//toggle
endcase
//jk4
end

always @(negedge r9)begin
qa = 1'b1;
//preset of jk1
end

always @(negedge r0)begin
qa = 1'b0;
//clear of jk1
end

always @(negedge clr2)begin
qa = 1'b0;
//clear of jk2
end

always @(negedge clr3)begin
qa = 1'b0;
//clear of jk3
end

always @(negedge r9)begin
qa = 1'b1;
//preset of jk4
end

always @(negedge r0)begin
qa = 1'b0;
//clear of jk4
end

begin
assign QA = qa;
assign QB = qb;
assign QC = qc;
assign QD = qd;
end

endmodule
于 2020-06-08T12:09:48.233 回答
0

我将 JK-FF 更改为模块实例化，现在它按预期工作：

module jk_ff_example (/*AUTOARG*/
   // Outputs
   QA, QB, QC, QD,
   // Inputs
   inputA, inputB, R01, R02, R91, R92
   );

  //..ports
  input inputA, inputB, R01, R02, R91, R92;
  output QA, QB, QC, QD;

  //..regs and wires declaration
  wire qa, qna, qb, qnb, qc, qnc, qd, qnd;
  wire r0, r9;

  //..r0 and r9 inputs comb logic
  assign r0 = ~(R01 & R02);
  assign r9 = ~(R91 & R92);

  //..qa jk-ff logic
  jk_ff qa_ff (
      .preset_i (r9),
      .clear_i  (r0),
      .clk_i    (inputA),
      .j_i      (1'b1),
      .k_i      (1'b1),
      .q_o      (qa),
      .qn_o     (qna)
    );

  //..qb jk-ff logic
  jk_ff qb_ff (
      .preset_i (1'b1),
      .clear_i  (r0 | r9),
      .clk_i    (inputB),
      .j_i      (qnd),
      .k_i      (1'b1),
      .q_o      (qb),
      .qn_o     (qnb)
    );

  //..qc jk-ff logic
  jk_ff qc_ff (
      .preset_i (1'b1),
      .clear_i  (r0 | r9),
      .clk_i    (qb),
      .j_i      (1'b1),
      .k_i      (1'b1),
      .q_o      (qc),
      .qn_o     (qnc)
    );

  //..qd jk-ff logic
  jk_ff qd_ff (
      .preset_i (r9),
      .clear_i  (r0),
      .clk_i    (inputB),
      .j_i      (qb & qc),
      .k_i      (qd),
      .q_o      (qd),
      .qn_o     (qnd)
    );

  assign QA = qa;
  assign QB = qb;
  assign QC = qc;
  assign QD = qd;

endmodule // jk_ff_example

这是具有预设和清除低电平有效输入的 JK-FF 的代码：

/*
 * JK-Flip Flop with Preset and Clear inputs Truth Table
 *  Preset, Clear and CLK with active-low logic
 *
 * -------------------------------------------------------------
 * | Preset | Clear | CLK | J | K | Output         | Qo  | ~Qo |
 * -------------------------------------------------------------
 * | 0      | 0     | x   | x | x | Invalid        | 1*  | 0*  |
 * | 0      | 1     | x   | x | x | Preset         | 1   | 0   |
 * | 1      | 0     | x   | x | x | Clear          | 0   | 1   |
 * | 1      | 1     | x   | x | x | No change      | Qo  | ~Qo |
 * | 1      | 1     | NEG | 0 | 0 | No change      | Qo  | ~Qo |
 * | 1      | 1     | NEG | 0 | 1 | Reset          | 0   | 1   |
 * | 1      | 1     | NEG | 1 | 0 | Set            | 1   | 0   |
 * | 1      | 1     | NEG | 1 | 1 | Toggle         | ~Qo | Qo  |
 * -------------------------------------------------------------
 */

module jk_ff (/*AUTOARG*/
   // Outputs
   q_o, qn_o,
   // Inputs
   preset_i, clear_i, clk_i, j_i, k_i
   );

  //..ports
  input preset_i, clear_i, clk_i, j_i, k_i;
  output q_o, qn_o;

  //..regs and wires
  reg q = 0;

  //..jk ff w/ preset and clear logic
  always @ (negedge preset_i, negedge clear_i, negedge clk_i) begin
    case({preset_i, clear_i})
      2'b00: q <= 0; //..invalid
      2'b01: q <= 1; //..preset
      2'b10: q <= 0; //..clear
      2'b11: begin
        if(clk_i) //..no change
          q <= q;
        else begin
          case({j_i, k_i})
            2'b00: q <= q;  //..no change
            2'b01: q <= 0;  //..reset
            2'b10: q <= 1;  //..set
            2'b11: q <= ~q; //..toggle
          endcase
        end
      end
    endcase
  end

  //..output assignment
  assign q_o = q;
  assign qn_o = ~q;

endmodule // jk_ff

模拟的输出是这样的：

 [Cycle:   0] [Q value:  0]
 [Cycle:  10] [Q value:  1]
 [Cycle:  20] [Q value:  2]
 [Cycle:  30] [Q value:  3]
 [Cycle:  40] [Q value:  4]
 [Cycle:  50] [Q value:  5]
 [Cycle:  60] [Q value:  6]
 [Cycle:  70] [Q value:  7]
 [Cycle:  80] [Q value:  8]
 [Cycle:  90] [Q value:  9]
 [Cycle: 100] [Q value:  0]
 [Cycle: 110] [Q value:  1]
 [Cycle: 120] [Q value:  2]
 [Cycle: 130] [Q value:  3]
 [Cycle: 140] [Q value:  4]
 [Cycle: 150] [Q value:  5]
 [Cycle: 160] [Q value:  6]
 [Cycle: 170] [Q value:  7]
 [Cycle: 180] [Q value:  8]
 [Cycle: 190] [Q value:  9]
 [Cycle: 200] [Q value:  9]
 [Cycle: 210] [Q value:  0]
 [Cycle: 220] [Q value:  1]
 [Cycle: 230] [Q value:  2]
 [Cycle: 240] [Q value:  3]
 [Cycle: 250] [Q value:  4]
 [Cycle: 260] [Q value:  5]
 [Cycle: 270] [Q value:  6]
 [Cycle: 280] [Q value:  7]
 [Cycle: 290] [Q value:  8]
End simulation
 [Cycle: 300] [Q value:  9]

我已将此作为示例上传到此存储库中，我已包含一个 Makefile 以简化仿真和 FPGA 综合，请随意使用。

于 2020-06-09T14:44:06.273 回答