题解 | #使用3-8译码器①实现逻辑函数#(真值表的运用)
使用3-8译码器①实现逻辑函数
https://www.nowcoder.com/practice/be81e76ebade445baca7257aa4eca8f2
`timescale 1ns/1ns
module decoder_38(
input E1_n ,
input E2_n ,
input E3 ,
input A0 ,
input A1 ,
input A2 ,
output wire Y0_n ,
output wire Y1_n ,
output wire Y2_n ,
output wire Y3_n ,
output wire Y4_n ,
output wire Y5_n ,
output wire Y6_n ,
output wire Y7_n
);
wire E ;
assign E = E3 & ~E2_n & ~E1_n;
assign Y0_n = ~(E & ~A2 & ~A1 & ~A0);
assign Y1_n = ~(E & ~A2 & ~A1 & A0);
assign Y2_n = ~(E & ~A2 & A1 & ~A0);
assign Y3_n = ~(E & ~A2 & A1 & A0);
assign Y4_n = ~(E & A2 & ~A1 & ~A0);
assign Y5_n = ~(E & A2 & ~A1 & A0);
assign Y6_n = ~(E & A2 & A1 & ~A0);
assign Y7_n = ~(E & A2 & A1 & A0);
endmodule
module decoder0(
input A ,
input B ,
input C ,
output wire L
);
wire Y0_n, Y1_n, Y2_n, Y3_n, Y4_n, Y5_n, Y6_n, Y7_n;
assign L = ~Y1_n | ~Y3_n | ~Y6_n | ~Y7_n;
decoder_38 decoder_38(
. E1_n (1'b0) ,
. E2_n (1'b0) ,
. E3 (1'b1) ,
. A0 (C) ,
. A1 (B) ,
. A2 (A) ,
. Y0_n (Y0_n) ,
. Y1_n (Y1_n) ,
. Y2_n (Y2_n) ,
. Y3_n (Y3_n) ,
. Y4_n (Y4_n) ,
. Y5_n (Y5_n) ,
. Y6_n (Y6_n) ,
. Y7_n (Y7_n)
);
endmodule
与上一题一致解题思路一致：
画出真值表
A B C L
0 0 0 0
0 0 1 1
0 1 0 0
0 1 1 1
1 0 0 0
1 0 1 0
1 1 0 1
1 1 1 1
与 三八译码器的真值表进行对比
最后的得出 L 与 三八译码器的关系为
L = ~y1_n | ~y3_n |~y6_n | ~y7_n
