Exams/ece241 2013 q4

 蓄水池问题

S3 S2 S1 

例如：000  代表 无水 ，需要使FR3, FR2, FR1 都打开（111）

S3 S2 S1     FR3 FR2 FR1

000                  111

001                011

011               001

111               000

fr代表水变深为0 ，水变少为1

module top_module (input clk,input reset,input [3:1] s,output fr3,output fr2,output fr1,output dfr
); parameter B_S1 = 3'b000,  B_S1_S2 = 3'b001,B_S2_S3 = 3'b011,A_S3 = 3'b111;reg [2:0] state;reg [2:0] next_state;always@ (posedge clk) beginstate <= next_state;   //next_state 变化较快end always@ (*)if(reset)next_state <= B_S1;else case({s[3], s[2], s[1]})B_S1: if(s[1]+s[2]+s[3] == 0)   next_state <= B_S1;else  next_state <= B_S1_S2;B_S1_S2: if(s[1]&& s[2] && ~s[3] )  next_state <= B_S2_S3;else if(s[1]+s[2]+s[3] == 0)next_state <= B_S1;elsenext_state <= B_S1_S2;B_S2_S3:if(s[1] & s[2] && s[3]) next_state <= A_S3;else if(s[1] && ~s[2] && ~s[3])   next_state <= B_S1_S2;elsenext_state <= B_S2_S3;A_S3:     if(&s) next_state <= A_S3;else  next_state <= B_S2_S3;default: next_state <= B_S1;endcasealways@ (posedge clk)if(reset) begin{fr3, fr2, fr1} <= 3'b111;endelse case ({s[3], s[2], s[1]})B_S1: {fr3, fr2, fr1} <= 3'b111;B_S1_S2: {fr3, fr2, fr1} <= 3'b011;B_S2_S3: {fr3, fr2, fr1} <= 3'b001;A_S3: {fr3, fr2, fr1} <= 3'b000;default:{fr3, fr2, fr1} <= 3'b111;endcasealways@ (posedge clk)if(reset)  dfr <= 1;else if(state < next_state)  //水变深dfr <= 0;else if(state > next_state)   dfr <= 1;else dfr <= dfr;endmodule