一、佔空比50%的奇數分頻

1、實現思路

　　實現奇數（N）分頻，分別用上升沿計數到（N-1)/2，再計數到N-1；用下降沿計數到（N-1)/2，再計數到N-1，得到兩個波形，然後把它們相或即可得到N分頻。

2、程式碼

 1 module fp_odd(clk_in,rst_n,cnt_p,cnt_n,clk_p,clk_n,clk_out);
 2 parameter N=5;
 3 input clk_in;
 4 input rst_n;
 5 output[2:0] cnt_p;
 6 output[2:0] cnt_n;
 7 output clk_p;
 8 output clk_n;
 
 9 output clk_out;
10 
11 reg[2:0] cnt_p;
12 reg[2:0] cnt_n;
13 reg clk_p;
14 reg clk_n;
15 
16 always @(posedge clk_in or negedge rst_n)
17 begin
18     if(!rst_n) cnt_p <= 0;
19     else 
20     begin
21         if(cnt_p == N-1) cnt_p <= 0;
22         else cnt_p <= cnt_p + 1'b1;
23     end
24 end
 

25 
26 always @(posedge clk_in or negedge rst_n)
27 begin
28     if(!rst_n) clk_p <= 0;
29     else
30     begin
31         if(cnt_p == (N-1)/2) clk_p <= !clk_p;
32         else if(cnt_p == N-1) clk_p <= !clk_p;
33         else clk_p <= clk_p;
34     end
35 end
36 
37 always @(negedge clk_in or
 
 negedge rst_n)
38 begin
39     if(!rst_n) cnt_n <= 0;
40     else 
41     begin
42         if(cnt_n == N-1) cnt_n <= 0;
43         else cnt_n <= cnt_n + 1'b1;
44     end
45 end
46 
47 always @(negedge clk_in or negedge rst_n)
48 begin
49     if(!rst_n) clk_n <= 0;
50     else
51     begin
52         if(cnt_n == (N-1)/2) clk_n <= !clk_n;
53         else if(cnt_n == N-1) clk_n <= !clk_n;
54         else clk_n <= clk_n;
55     end
56 end
57 
58 assign clk_out = clk_p | clk_n;
59 
60 endmodule

3、模擬波形