题解 | #Johnson Counter#
Johnson Counter
https://www.nowcoder.com/practice/7ee6e9ed687c40c3981d7586a65bc22d
//方法1
`timescale 1ns/1ns
module JC_counter(
input clk ,
input rst_n,
output reg [3:0] Q
);
always@(posedge clk or negedge rst_n)
begin
if(!rst_n)
begin
Q <= 4'd0;
end
else
begin
Q <= {!Q[0],Q[3:1]};//低位取反右移
end
end
endmodule
//方法2
`timescale 1ns/1ns
module JC_counter(
input clk ,
input rst_n,
output reg [3:0] Q
);
reg [2:0] cnt;
always@(posedge clk or negedge rst_n)
begin
if(!rst_n)
cnt <= 3'd0;
else
cnt <= cnt + 1'b1;
end
always@(*)
begin
if(!rst_n)
Q <= 3'd0;
else
case(cnt)
3'd0:Q <= 4'b0000;
3'd1:Q <= 4'b1000;
3'd2:Q <= 4'b1100;
3'd3:Q<= 4'b1110;
3'd4:Q <= 4'b1111;
3'd5:Q <= 4'b0111;
3'd6:Q <= 4'b0011;
3'd7:Q <= 4'b0001;
default:Q <= 4'b0000;
endcase
end
endmodule
