因為是5分頻　　所以pcnt=2 　為1　 pcnt=4 為0

如果是7分頻　　則　pcnt=3 為1　 pcnt=6 為0

還想著用怎麼pos_flag　作上升沿　　　neg_flag　　作下降沿　　　沒想到最後運用了或運算 。。。。。

always@(posedge    clk)
begin
    if (rst) begin
        pos_flag<=0;
    end
    else if((pcnt==3)&(pcnt==4)) begin
        pos_flag<=1;
    end
    else begin
        pos_flag
 
<=0;
    end
end

module    fenpin_5(
input    wire    clk,
input    wire    rst,
output    wire        clk5);

reg    [2:0]    pcnt;
reg    [2:0]    ncnt;
reg    pos_flag;
reg    neg_flag;

always@(posedge    clk)
begin
    if(rst)
        pcnt<=0;
    else if(pcnt==4)    
            pcnt<=0;                
    
 
else 
        pcnt<=pcnt+1'b1;
    
end

always @(negedge clk)
begin
    if(rst)
        ncnt<=0;
    else if(ncnt==4)
            ncnt<=0;            
    else 
        ncnt<=ncnt+1'b1;    
end

always@(posedge    clk)
begin
    if (rst) begin
        pos_flag<=0;
    end
    else if(pcnt==2
 
) begin
        pos_flag<=1;
    end
    else if(pcnt==4) begin
        pos_flag<=0;
    end
end

always@(negedge    clk)
begin
    if (rst) begin
        pos_flag<=0;
    end
    else if(ncnt==2) begin
        neg_flag<=1;
    end
    else if (ncnt==4)
        begin
        neg_flag<=0;
        end
end

assign clk5 =pos_flag|neg_flag ;

endmodule

`timescale 1ns/1ns
module    tb_fenpin_5();
reg    clk,rst;
wire clk5;

initial    begin
    clk=0;
    rst=1;
    #100    rst=0;

end

always #10    clk=~clk;

fenpin_5 tb_fenpin_5_inst(
    .clk(clk),
    .rst(rst),
    .clk5(clk5));
endmodule