数码管同步蜂鸣器的问题跪求大神指教，麻烦大神们了！

通过Verilog HDL用数码管设计了一个40s倒计时，可以麻烦各位大神帮我加上一个10s之后的蜂鸣器吗，我初学FPGA，试了编辑一个之后，调试后蜂鸣器不响，下面是倒计时的代码，求大神们指教，急急急！
module led_rxd(clk,rst_n,SOS_En_Sig,led,led_seg);
input clk,rst_n;
output [7:0]led;
output [5:0]led_seg;
output SOS_En_Sig;
parameter seg_num0=8'hc0,
          seg_num1=8'hf9,
             seg_num2=8'ha4,
             seg_num3=8'hb0,
             seg_num4=8'h99,
             seg_num5=8'h92,
             seg_num6=8'h82,
             seg_num7=8'hf8,
             seg_num8=8'h80,
             seg_num9=8'h90;
parameter seg_en0=6'b111110,
          seg_en1=6'b111101,
             seg_en2=6'b111011,
             seg_en3=6'b110111,
             seg_en4=6'b101111,
             seg_en5=6'b011111;
reg [26:0]count;
reg [3:0] count1;
reg [3:0] count2;
reg [7:0] led_reg;
reg [5:0] led_seg_reg;
always@(posedge clk or negedge rst_n)
if(！rst_n) count<=27'd0;
else if(count==27'd49_999_999) count<=27'd0;
else count<=count+1'b1;
wire clk_div=(count==27'd49_999_999);
always@(posedge clk_div or negedge rst_n)
if(！rst_n)
begin
count1<=4'd0;
count2<=4'd4;
end
else if((count1==4'd0)&&(count2==4'd0))
begin
count1<=4'd0;
count2<=4'd4;
end
else if(count1==4'd0)
begin
count2<= count2-1'b1;
count1<=4'd9;
end
else count1<=count1-1'b1;
reg [26:0]count_1ms;//
always@(posedge clk or negedge rst_n)
if(！rst_n) count_1ms<=27'd0;  
else if(count_1ms==27'd49_999) count_1ms<=27'd0;
else count_1ms<=count_1ms+1'b1;
wire clk_dis=(count_1ms==27'd49_999);//
//
reg [1:0]state;
always@(posedge clk_dis or negedge rst_n)
if(！rst_n)
begin
led_reg<=8'hff;
led_seg_reg<=6'b111111;
state<=2'b00;
end
else if(state==2'b00)
begin
state<=2'b01;
led_seg_reg<=6'b111101;
case(count2)
4'd0: led_reg<=seg_num0;  
4'd1: led_reg<=seg_num1;  
4'd2:led_reg<=seg_num2;
4'd3: led_reg<=seg_num3;  
4'd4: led_reg<=seg_num4;  
4'd5: led_reg<=seg_num5;  
4'd6: led_reg<=seg_num6;  
4'd7: led_reg<=seg_num7;  
4'd8: led_reg<=seg_num8;   
4'd9: led_reg<=seg_num9;  
default: led_reg<=seg_num0;  
endcase
end
else if(state==2'b01)
begin
state<=2'b00;
led_seg_reg<=6'b111110;
case(count1)
4'd0:led_reg<=seg_num0;
4'd1:led_reg<=seg_num1;
4'd2:led_reg<=seg_num2;
4'd3:led_reg<=seg_num3;
4'd4:led_reg<=seg_num4;
4'd5:led_reg<=seg_num5;
4'd6:led_reg<=seg_num6;
4'd7:led_reg<=seg_num7;
4'd8:led_reg<=seg_num8;
4'd9:led_reg<=seg_num9;
default:led_reg<=seg_num0;
endcase
end
reg isEn;
always@(posedge clk or negedge rst_n)
if(！rst_n)
begin
isEn<=1'b0;
end
else
begin
isEn<=1'b1;
end
assign led=led_reg;
assign led_seg=led_seg_reg;
assign SOS_En_Sig=isEn;
endmodule

你那assign SOS_En_Sig=isEn;应该是让蜂鸣器使能端吧，所以我加了一个频率输出端
至于那个10s才响，你就多判断几次吧，if(10s)~~~~balabala
可以：if(10s) 使能端赋1，
if(使能端) 输出频率

11111111111111111111111

  wire carry;
reg[13:0]divider;
output reg sp;
assign carry=(divider==16383);  
  always @(posedge count[2])
    begin  
         if(carry)//当divider为16383时，将origin的值赋给divider
            divider=11272;
         else     
             divider=divider+1;
    end
always@(posedge carry)       sp =~sp;

音乐频率

对不起，我还以为没人看我的帖子，所以也回你，很抱歉。

其实我的主要疑问是，我那个10s要如何定义。就是if（10s）            这个10s我要如何表示出来，想了很久也没有想出来

我初学FPGA，太菜了。能麻烦你帮我写一下吗，谢谢您了，

if((count2==4'h1)&&(count1==4'h0));
最简单的方法。

嘻嘻，对不起啊，麻烦你了