CODEWITHSUNDEEP
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HarryYNG
HarryYNG

Reputation: 1

Why does the s2 signal have the opposite value as the s3 signal after the nanosecond 20?

I have this code in VHDL:

library IEEE;
use IEEE.std_logic_1164.all;

entity crono is
end entity crono;

architecture crono of crono is
    signal x1, x2, x3, s1, s2, s3, s4: std_logic;
begin
    x1 <= '1',
          '0' after 15 ns,
          '1' after 20 ns,
          '0' after 35 ns,
          '1' after 50 ns;
    x2 <= '0',
          '1' after 25 ns,
          '0' after 30 ns;
    x3 <= '0',
          '1' after 10 ns,
          '0' after 20 ns;
    s1 <= x1 after 5 ns;
    Proc1: process (x1, x2, x3)
        variable v1: std_logic;
    begin
        s2 <= x1 xor x2;
        v1 := x1 xor x2;
        s3 <= s2;
        s4 <= v1;
    end process;
end architecture crono;

And it generates this waves:

enter image description here

Dont know why s2 and s3 signals are opposite when they should be the same after the 20 nanosecond mark, I really have no clue.

Edit: I just noticed my misunderstanding, the signal doesnt update until the end of the process block. This means that the signal "s3" doesn't take the new "s2" value.

Upvotes: -1

Views: 54

Answers (1)

AlaBek
AlaBek

Reputation: 28

Adding s2 to sensitivity list of Proc1 makes s2=s3=s4 enter image description here

"Dont know why s2 and s3 signals are opposite when they should be the same after the 20 nanosecond mark" -> s2 and s3 are the same at time 0, since you have in your code x1 initialized to 1 and x1 is Proc1 sensitivity list.

Upvotes: 0

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