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標題: Verilog狀態機實現序列檢測 [打印本頁]

作者: 51黑黑黑    時間: 2016-2-23 01:18
標題: Verilog狀態機實現序列檢測
經常在筆試的時候出現這樣的題:檢測序列為11001101,輸出為1,否則輸出為0。抽空寫了下代碼,僅供參考。

FSM1:一段式
module machine_test1(clk,rst,din,dout);
input clk,rst;
input din;
output dout;


//11001101
parameter idle=0,
    st0=1,
    st1=2,
    st2=3,
    st3=4,
    st4=5,
    st5=6,
    st6=7;


reg [3:0] next_state;
reg dout;



always @ (posedge clk or negedge rst)
begin
if(!rst)
begin
  next_state=idle;
  dout=0;
end
else
begin
  case(next_state)
  idle:
   if(din)
   begin
    next_state=st0;
    dout=0;
   end
   else
   begin
    next_state=idle;
    dout=0;
   end
  st0:
   if(din)
   begin
    next_state=st1;
    dout=0;
   end
   else
   begin
    next_state=idle;
    dout=0;
   end
  st1:
   if(!din)
   begin
    next_state=st2;
    dout=0;
   end
   else
   begin
    next_state=st1;
    dout=0;
   end
  st2:
   if(!din)
   begin
    next_state=st3;
    dout=0;
   end
   else
   begin
    next_state=st0;
    dout=0;
   end
  st3:
   if(din)
   begin
    next_state=st4;
    dout=0;
   end
   else
   begin
    next_state=idle;
    dout=0;
   end
  st4:
   if(din)
   begin
    next_state=st5;
    dout=0;
   end
   else
   begin
    next_state=idle;
    dout=0;
   end
  st5:
   if(!din)
   begin
    next_state=st6;
    dout=0;
   end
   else
   begin
    next_state=st1;
    dout=0;
   end
  st6:
   if(din)
   begin
    next_state=idle;
    dout=1;
   end
   else
   begin
    next_state=idle;
    dout=0;
   end
  default:
   begin
    next_state=idle;
    dout=0;
   end
  endcase
end
end  
  

endmodule


FSM2:兩段式
module machine_test(clk,rst,din,dout);
input clk,rst;
input din;
output dout;

//11001101
parameter idle=0,
    st0=1,
    st1=2,
    st2=3,
    st3=4,
    st4=5,
    st5=6,
    st6=7;

reg [3:0] current_state,next_state;
reg dout;


always @ (posedge clk or negedge rst)
begin
if(!rst)
  current_state<=idle;
else
  current_state<=next_state;
end

always @ (next_state or current_state or din or dout)
begin
case(current_state)
idle:
  if(din)
  begin
   next_state=st0;
   dout=0;
  end
  else
  begin
   next_state=idle;
   dout=0;
  end
st0:
  if(din)
  begin
   next_state=st1;
   dout=0;
  end
  else
  begin
   next_state=idle;
   dout=0;
  end
st1:
  if(!din)
  begin
   next_state=st2;
   dout=0;
  end
  else
  begin
   next_state=st1;
   dout=0;
  end
st2:
  if(!din)
  begin
   next_state=st3;
   dout=0;
  end
  else
  begin
   next_state=st0;
   dout=0;
  end
st3:
  if(din)
  begin
   next_state=st4;
   dout=0;
  end
  else
  begin
   next_state=idle;
   dout=0;
  end
st4:
  if(din)
  begin
   next_state=st5;
   dout=0;
  end
  else
  begin
   next_state=idle;
   dout=0;
  end
st5:
  if(!din)
  begin
   next_state=st6;
   dout=0;
  end
  else
  begin
   next_state=st1;
   dout=0;
  end
st6:
  if(din)
  begin
   next_state=idle;
   dout=1;
  end
  else
  begin
   next_state=idle;
   dout=0;
  end
default:
  begin
   next_state=idle;
   dout=0;
  end
endcase
end  

endmodule




FSM3:三段式
module machine_test2(clk,rst,din,dout);
input clk,rst;
input din;
output dout;

//11001101
parameter idle=0,
    st0=1,
    st1=2,
    st2=3,
    st3=4,
    st4=5,
    st5=6,
    st6=7,
    st7=8;

reg [3:0] current_state,next_state;
reg dout;


always @ (posedge clk or negedge rst)
begin
if(!rst)
  current_state<=idle;
else
  current_state<=next_state;
end

always @ (next_state or current_state or din or dout)
begin
case(current_state)
idle:
  if(din)
   next_state=st0;
  else
   next_state=idle;
st0:
  if(din)
   next_state=st1;
  else
   next_state=idle;
st1:
  if(!din)
   next_state=st2;
  else
   next_state=st1;
st2:
  if(!din)
   next_state=st3;
  else
   next_state=st0;
st3:
  if(din)
   next_state=st4;
  else
   next_state=idle;
st4:
  if(din)
   next_state=st5;
  else
   next_state=idle;
st5:
  if(!din)
   next_state=st6;
  else
   next_state=st1;
st6:
  if(din)
   next_state=st7;
  else
   next_state=idle;
st7:
  if(din)
   next_state=st1;
  else
   next_state=idle;
default:
  next_state=idle;
endcase
end  

always @ (posedge clk)
begin
if(next_state==st7)
  dout<=1'd1;
else
  dout<=1'd0;
end
   
endmodule

仿真結果:








作者: LIJIAPUA    時間: 2020-6-28 18:49
二段式具體的波形信號輸入的是多少呀




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