题解 | #根据状态转移表实现时序电路#
根据状态转移表实现时序电路
https://www.nowcoder.com/practice/455c911bee0741bf8544a75d958425f7
`timescale 1ns/1ns
module seq_circuit(
input A ,
input clk ,
input rst_n,
output wire Y
);
reg [1:0]cs;
reg [1:0]ns;
always@(posedge clk or negedge rst_n)begin
if(!rst_n)
cs <= 0;
else
cs <= ns;
end
always@(*)begin
if(A == 0)
begin
case(cs)
2'b00: ns = 2'b01;
2'b01: ns = 2'b10;
2'b10: ns = 2'b11;
2'b11: ns = 2'b00;
endcase
end
else
begin
case(cs)
2'b00: ns = 2'b11;
2'b01: ns = 2'b00;
2'b10: ns = 2'b01;
2'b11: ns = 2'b10;
endcase
end
end
assign Y = (cs == 2'b11)?1'b1:1'b0;
endmodule
module seq_circuit(
input A ,
input clk ,
input rst_n,
output wire Y
);
reg [1:0]cs;
reg [1:0]ns;
always@(posedge clk or negedge rst_n)begin
if(!rst_n)
cs <= 0;
else
cs <= ns;
end
always@(*)begin
if(A == 0)
begin
case(cs)
2'b00: ns = 2'b01;
2'b01: ns = 2'b10;
2'b10: ns = 2'b11;
2'b11: ns = 2'b00;
endcase
end
else
begin
case(cs)
2'b00: ns = 2'b11;
2'b01: ns = 2'b00;
2'b10: ns = 2'b01;
2'b11: ns = 2'b10;
endcase
end
end
assign Y = (cs == 2'b11)?1'b1:1'b0;
endmodule
