目前我正在开发一个名为Valyado的小型项目,一个Mealy FSM。程序必须检测6位序列001011,并输出" 1"当检测到序列时。
关于序列检测的代码做得很好,但除此之外,它还必须使用三个触发器:JK,D和T.
有关如何添加它们的任何建议或建议?
感谢您的时间。
这是FSM代码:
library ieee;
use ieee.std_logic_1164.all;
use ieee.std_logic_arith.all;
entity sequence is
port(
clk : in std_logic;
reset : in std_logic;
x: in std_logic;
z : out std_logic;
a : out std_logic;
b : out std_logic;
c : out std_logic;
d : out std_logic;
e : out std_logic;
f : out std_logic);
end sequence;
architecture behavioral of sequence is
type state_type is (Q0, Q1, Q2, Q3, Q4, Q5);
signal state, next_state : state_type;
begin
state_register: process (clk, reset)
begin
if (reset = '1') then --if reset is high, goto state Q0
state <= Q0;
elsif (clk'event and clk = '1') then --if not, and rising
state <= next_state; --edge, go to next state
end if;
end process;
next_state_func: process (x, state)
begin
case state is
when Q0 =>
if x = '0' then
next_state <= Q1;
else
next_state <= Q0;
end if;
when Q1 =>
if x = '0' then
next_state <= Q2;
else
next_state <= Q0;
end if;
when Q2 =>
if x = '1' then
next_state <= Q3;
else
next_state <= Q2;
end if;
when Q3 =>
if x ='0' then
next_state <= Q4;
else
next_state <= Q0;
end if;
when Q4 =>
if x = '1' then
next_state <= Q5;
else
next_state <= Q2;
end if;
when Q5 =>
if x = '1' then
next_state <= Q0;
else
next_state <= Q1;
end if;
end case;
end process;
-- This process controls the output of the sequence detector.
-- Each state has it's own output along with 'z' which indicates
-- the entire sequence 001011 has been detected.
output_func: process (x, state)
begin
case state is
when Q0 => z <= '0';
a <= '1';
b <= '0';
c <= '0';
d <= '0';
e <= '0';
f <= '0';
when Q1 => z <= '0';
a <= '0';
b <= '1';
c <= '0';
d <= '0';
e <= '0';
f <= '0';
when Q2 => z <= '0';
a <= '0';
b <= '0';
c <= '1';
d <= '0';
e <= '0';
f <= '0';
when Q3 => z <= '0';
a <= '0';
b <= '0';
c <= '0';
d <= '1';
e <= '0';
f <= '0';
when Q4 => z <= '0';
a <= '0';
b <= '0';
c <= '0';
d <= '0';
e <= '1';
f <= '0';
when Q5 => z <= '1';
a <= '0';
b <= '0';
c <= '0';
d <= '0';
e <= '0';
f <= '1';
end case;
end process;
end behavioral;
[1]:https://i.stack.imgur.com/pVwxL.jpg - 这是包含FSM状态图表的表。
答案 0 :(得分:0)
你的代码错了。看看output_func
过程;这是组合,只是解码当前状态,没有查看x
。 a
到f
输出是不必要的,只是对当前状态的6位解码 - 为什么?当前状态为z
时设置Q5
输出,这不是您想要的 - 整个过程是多余的。您需要在主FSM中设置z
,当前状态为Q5
,且 x
为1 - 即。在next_state <= Q0
转换。
关于你的实际问题 - 你不能用这个代码强制选择任何特定的F / F类型 - 合成器会做任何你想做的事情,这意味着它将在D类型中实现整个事物,因为JKs在过去的20年里已经过时了。 T类型可能也是如此。您需要重新开始,并假装您拥有T,D和JK的技术和库。将这些作为单独的实体编写,并重新编写代码以实例化这些组件,而不是让合成器推断它们。重新编写FSM以使用JK - 您给出的图表将向您展示如何使用JK。换句话说,导出每个F / F的J和K输入。 z
输出可以是D类型。你应该能够在某个地方适应T - 我已经把它作为锻炼给你了。