用vhdl语言设计当输入连续3个或3个以上1时输出才为1的状态机
以下是一个可能的解决方案:
library ieee;
use ieee.std_logic_1164.all;
entity state_machine is
port (
clk : in std_logic;
reset : in std_logic;
input : in std_logic;
output: out std_logic
);
end entity;
architecture behavioral of state_machine is
type state is (idle, count1, count2, count3);
signal current_state, next_state : state;
signal count : integer range 0 to 3 := 0;
begin
-- state transition process
process (clk, reset)
begin
if reset = '1' then
current_state <= idle;
count <= 0;
elsif rising_edge(clk) then
current_state <= next_state;
if input = '1' then
count <= count + 1;
else
count <= 0;
end if;
end if;
end process;
-- state output process
process (current_state, count)
begin
case current_state is
when idle =>
output <= '0';
if count >= 3 then
next_state <= count1;
else
next_state <= idle;
end if;
when count1 =>
output <= '0';
if count >= 3 then
next_state <= count2;
else
next_state <= idle;
end if;
when count2 =>
output <= '0';
if count >= 3 then
next_state <= count3;
else
next_state <= idle;
end if;
when count3 =>
output <= '1';
if count >= 3 then
next_state <= count3;
else
next_state <= idle;
end if;
end case;
end process;
end architecture;
这个状态机有4个状态:idle(空闲状态)、count1(计数1状态)、count2(计数2状态)和count3(计数3状态)。它有一个输入(input)和一个输出(output),以及时钟(clk)和复位(reset)信号。
在状态转换过程中,我们使用一个计数器(count)来记录连续1的数量。如果输入是0,则计数器重置为0。如果输入是1,则计数器加1。
在状态输出过程中,我们根据当前状态和计数器的值来确定输出。如果当前状态是idle、count1或count2,并且计数器的值大于等于3,则状态机进入下一个状态。否则,状态机回到idle状态。如果当前状态是count3,并且计数器的值大于等于3,则输出为1,否则输出为0
原文地址: http://www.cveoy.top/t/topic/daJV 著作权归作者所有。请勿转载和采集!