Question

我正在进行计算，其中输出是下一个输入之一。但是，mux输出只提供X并导致所有其余计算出错。如何克服这个问题？这是由于时钟吗？

这是我的代码：

module all2(input sel,clk,
           input signed [15:0]x,d,
           output signed [15:0]w,e,y);
localparam u = 16'd2;
wire [15:0]w1;
reg [15:0]y1,e1,u1,wk;
assign w = wk;
assign e = e1;
assign y = y1;
assign w1 = (sel)?wk:16'd0;
    always @(posedge clk or negedge clk)
    y1 <= x * w1;
    always @(posedge clk or negedge clk)
    e1 <= d - y1;
    always @(posedge clk or negedge clk)
    u1 <= e1 * x * u;
    always @(posedge clk or negedge clk)
    wk <= u1 + w1;    
endmodule

这是测试平台：

module all2_tb();
reg sel, clk;
reg signed [15:0] x, d;
wire signed [15:0] w, e, y;

all2 U1(.sel(sel),.clk(clk),.x(x),.d(d),.w(w),.e(e),.y(y));

initial begin
    clk = 1'b0;
    forever
    #1 clk = ~clk;
end

initial begin
sel <= 1'b0;
x <= 16'd0;
d <= 16'd0;
#2;
sel <= 1'b1;
x <= 16'd1;
d <= 16'd2;
#1;
x <= 16'd3;
d <= 16'd4;
#1;
x <= 16'd5;
d <= 16'd6;
#1;
$stop;
end
endmodule

Answer 1

输出X应该不会导致计算的其余部分出错，我会说它是另一种方式。在模块内部使用X使输出变为X.

看看你的方程式：

w = wk;
e = e1;
y = y1;
w1 = (sel)?wk:16'd0;

always @(posedge clk or negedge clk) begin
  y1 <= x * w1;
  e1 <= d - y1;
  u1 <= e1 * x * u;
  wk <= u1 + w1;  
end

请注意，y1，e1，u1和wk在零时将为x，因为您尚未指定重置或初始条件。我不确定你为什么要在时钟的两个边沿触发触发器，但通常在negedge上触发一个有效的低电平复位。

always @(posedge clk or negedge rst_n) begin //<- negedge rst_n
  if (~rst_n) begin
    y1 <= 'd0;
    e1 <= 'd0;
    u1 <= 'd0;
    wk <= 'd0;
  end
  else begin
    y1 <= x * w1;
    e1 <= d - y1;
    u1 <= e1 * x * u;
    wk <= u1 + w1;  
  end 
end

一旦这样做，我就不再看到你模块的x输出了。

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