题解 | #时钟分频(偶数)#
时钟分频(偶数)
https://www.nowcoder.com/practice/49a7277c203a4ddd956fa385e687a72e
分解可知:
clk_out8:11110000
clk_out4:11001100
clk_out2:10101010
可以看出{clk_out8,clk_out4,clk_out2}是111开始的递减序列
`timescale 1ns/1ns
module even_div
(
input wire rst ,
input wire clk_in,
output wire clk_out2,
output wire clk_out4,
output wire clk_out8
);
//*************code***********//
reg [2:0]cnt;
always@((487950916)posedge clk_in or negedge rst)begin
if(~rst)begin
cnt<=0;
end
else begin
cnt<=cnt-1;
end
end
assign clk_out2=cnt[0];
assign clk_out4=cnt[1];
assign clk_out8=cnt[2];
//*************code***********//
endmodule