面试的美科艺,开始去了,自我介绍,然后就开始笔试,当时的题目6页代码,我不记得了,人家也不让我带出来,姑且现在随便找一道题目来请教大家把。
下面这段程序,已知输入的波形,给出 输出的波形图;
类似这样的题目 做题的思路是什么? 怎么入手去做?
module divfreq(clk, clk1x, rst, clk1xpose, clk1xnege, coutpose, coutnege);
input clk;
input rst;
output clk1x;
output clk1xpose;
output clk1xnege;
output[2:0] coutpose;
output[2:0] coutnege;
reg clk1xpose;
reg clk1xnege;
reg[2:0] coutpose;
reg[2:0] coutnege;
parameter div1 = 2 , div2 = 4; // div1 ? 5 / 2, div2 = 5 - 1
assign clk1x = clk1xpose | clk1xnege;
always@(posedge clk or negedge rst)
begin
if(!rst)
clk1xpose = 0;
else if(coutpose == div1)
clk1xpose = ~clk1xpose;
else if(coutpose == div2)
clk1xpose = ~clk1xpose;
else
clk1xpose = clk1xpose;
end
always@(negedge clk or negedge rst)
begin
if(!rst)
clk1xnege = 0;
else if(coutnege == div1)
clk1xnege = ~clk1xnege;
else if(coutnege == div2)
clk1xnege = ~clk1xnege;
else
clk1xnege = clk1xnege;
end
always@(posedge clk or negedge rst)
begin
if(!rst)
coutpose = 0;
else if(coutpose == div2)
coutpose = 0;
else
coutpose = coutpose + 1;
end
always@(negedge clk or negedge rst)
begin
if(!rst)
coutnege = 0;
else if(coutnege == div2)
coutnege = 0;
else
coutnege = coutnege + 1;
end
endmodule