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我在使用 lfsr 方法创建 prng 时遇到了一些麻烦。这是我的代码：

library IEEE;
use IEEE.STD_LOGIC_1164.ALL;

entity pseudorng is
Port ( clock : in STD_LOGIC;
       reset : in STD_LOGIC;
       Q : out STD_LOGIC_VECTOR (7 downto 0);
       check: out STD_LOGIC);

       constant seed: STD_LOGIC_VECTOR(7 downto 0) := "00000001";
end pseudorng;

architecture Behavioral of pseudorng is

signal temp: STD_LOGIC;
signal Qt: STD_LOGIC_VECTOR(7 downto 0);

begin

PROCESS(clock)
BEGIN

IF rising_edge(clock) THEN
IF (reset='1') THEN Qt <= "00000000";
ELSE Qt <= seed; 
END IF;
temp <= Qt(4) XOR Qt(3) XOR Qt(2) XOR Qt(0);
--Qt <= temp & Qt(7 downto 1);

END IF;
END PROCESS;

check <= temp;
Q <= Qt;

end Behavioral;

这是我运行的模拟： prng sim

首先，检查输出就在那里，所以我可以监控温度信号的输出。其次，被注释掉的行是导致问题的原因。

从仿真中可以看出，在时钟的第一个上升沿，Qt 信号读取种子。但是，这是我的问题，由于某种原因，临时信号仅在时钟的第二个上升沿对 Qt 信号的位进行异或。它在第一个时钟脉冲上保持未定义。这是为什么？如果它在 Qt 信号读取种子后的第一个上升沿上运行，那么我可以取消注释移动位的行，它会解决我的问题。任何帮助将非常感激！

如果有人关心，这里是测试台：

library IEEE;
use IEEE.STD_LOGIC_1164.ALL;

entity tb_pseudorng is
end tb_pseudorng;

architecture bench of tb_pseudorng is

COMPONENT pseudorng
      Port ( clock : in STD_LOGIC;
      reset : in STD_LOGIC;
      Q : out STD_LOGIC_VECTOR (7 downto 0);
      check: out STD_LOGIC);
END COMPONENT;

signal clock1: STD_LOGIC;
signal reset1: STD_LOGIC;
signal Q1: STD_LOGIC_VECTOR(7 downto 0);
signal check1: STD_LOGIC;

begin

mapping: pseudorng PORT MAP(
clock => clock1,
reset => reset1,
Q => Q1,
check => check1);

clock: PROCESS
BEGIN
clock1<='0'; wait for 50ns;
clock1<='1'; wait for 50ns;
END PROCESS;

reset: PROCESS
BEGIN
reset1<='0'; wait for 900ns;
END PROCESS; 

end bench;
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1 回答 1

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我对你所拥有的做了一些轻微的修改（虽然你几乎在那里）；我不认为 LFSR 会正确地采取其他措施。我向 LFSR 添加了一个启用信号，以便您可以有效地控制何时需要它步进。生成的 sim 就在这里。

顺便说一句，如果您想使用不同的值（而不是使其成为 const）为 LFSR 播种，您还可以包含 aload和输入。seed

library IEEE;
use IEEE.STD_LOGIC_1164.ALL;

entity pseudorng is
Port ( clock : in STD_LOGIC;
       reset : in STD_LOGIC;
       en : in STD_LOGIC;
       Q : out STD_LOGIC_VECTOR (7 downto 0);
       check: out STD_LOGIC);

--       constant seed: STD_LOGIC_VECTOR(7 downto 0) := "00000001";
end pseudorng;

architecture Behavioral of pseudorng is

--signal temp: STD_LOGIC;
signal Qt: STD_LOGIC_VECTOR(7 downto 0) := x"01";

begin

PROCESS(clock)
variable tmp : STD_LOGIC := '0';
BEGIN

IF rising_edge(clock) THEN
   IF (reset='1') THEN
   -- credit to QuantumRipple for pointing out that this should not
   -- be reset to all 0's, as you will enter an invalid state
      Qt <= x"01"; 
   --ELSE Qt <= seed;
   ELSIF en = '1' THEN
      tmp := Qt(4) XOR Qt(3) XOR Qt(2) XOR Qt(0);
      Qt <= tmp & Qt(7 downto 1);
   END IF;

END IF;
END PROCESS;
-- check <= temp;
check <= Qt(7);
Q <= Qt;

end Behavioral;

和结核病：

library IEEE;
use IEEE.STD_LOGIC_1164.ALL;

entity tb_pseudorng is
end tb_pseudorng;

architecture bench of tb_pseudorng is

COMPONENT pseudorng
      Port ( clock : in STD_LOGIC;
      reset : in STD_LOGIC;
      en : in STD_LOGIC;
      Q : out STD_LOGIC_VECTOR (7 downto 0);
      check: out STD_LOGIC);
END COMPONENT;

signal clock1: STD_LOGIC;
signal reset1: STD_LOGIC;
signal Q1: STD_LOGIC_VECTOR(7 downto 0);
signal check1: STD_LOGIC;
signal en : STD_LOGIC;

begin

mapping: pseudorng PORT MAP(
clock => clock1,
reset => reset1,
en => en,
Q => Q1,
check => check1);

clock: PROCESS
BEGIN
   clock1 <= '0'; wait for 50 ns;
   clock1 <= '1'; wait for 50 ns;
END PROCESS;

reset: PROCESS
BEGIN
   reset1 <= '0';
   en <= '1';
   wait for 900 ns;
END PROCESS;

end bench;
于 2017-03-28T23:41:31.557 回答