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我想为这个 + - 热门问题道歉，但我没有在任何地方找到 vhdl 的具体实现。我从头开始编写算法，但在数学实现方面遇到了问题。输出无效。什么都不重要，但只显示 1 个值。如果有人知道我需要做什么，如何解决它，将不胜感激任何帮助。
数学部分
library IEEE;
use IEEE.STD_LOGIC_1164.ALL;
use IEEE.NUMERIC_STD.ALL;
-- Uncomment the following library declaration if using
-- arithmetic functions with Signed or Unsigned values
--use IEEE.NUMERIC_STD.ALL;
-- Uncomment the following library declaration if instantiating
-- any Xilinx primitives in this code.
--library UNISIM;
--use UNISIM.VComponents.all;
entity massive is
port (
clk : in std_logic;
reset : in std_logic;
sinus : out std_logic_vector (15 downto 0));
end massive;
architecture Behavioral of massive is
type my_type is array (0 to 16) of signed (15 downto 0);
signal x : my_type;
signal y : my_type;
signal z : my_type;
signal j : my_type := ("1010111111001000", "0110011111000101", "0011011011010100", "0001101111010101", "0000110111111000",
"0000011011111110", "0000001101111111", "0000000111000000", "0000000011100000", "0000000001110000",
"0000000000111000", "0000000000011100", "0000000000001110", "0000000000000111", "0000000000000100",
"0000000000000010", "0000000000000001");
begin
process(clk)
begin
x(0) <= "0000010100000110";
y(0) <= "0000000000000000";
z(0) <= "0000000000000000";
if rising_edge(clk) then
if reset <= '1' then
For n in 0 to 15 loop
if (z(n) >= 0) then
x(n+1) <= x(n) - (y(n)/2**n);
y(n+1) <= y(n) + (x(n)/2**n);
z(n+1) <= z(n) + j(n);
else
x(n+1) <= x(n) +(y(n)/2**n);
y(n+1) <= y(n) -(x(n)/2**n);
z(n+1) <= z(n) - j(n);
end if;
end loop;
sinus <= std_logic_vector(y(16));
end if;
end if;
end process;
end Behavioral;
旋转部分
entity step_control is
generic (
first : integer := 0;
second : integer := 1;
third : integer := 2;
fourth : integer := 3;
);
Port ( clk : in STD_LOGIC;
Angle : out STD_LOGIC_VECTOR (12 downto 0);
quarter_in : out STD_LOGIC_VECTOR (1 downto 0));
end step_control;
architecture Behavioral of step_control is
signal Ang : std_logic_vector (12 downto 0) := (others => '0');
signal state : unsigned (1 downto 0) := to_unsigned(first,2);
signal count_ang : std_logic_vector (11 downto 0) := (others => '0');
begin
process (clk)
begin
if (rising_edge(clk)) then
case(state) is
when to_unsigned(first,2) => if (count_ang >= 3999) then --00
state <= to_unsigned(second,2);
count_ang <= "000000010000";
quarter_in <= "01";
Ang <= Ang - 16;
else
state <= to_unsigned(first,2);
quarter_in <= "00";
Ang <= Ang + 16;
count_ang <= count_ang + 16;
end if;
when to_unsigned(second,2) => if (count_ang >= 3999) then --01
state <= to_unsigned(third,2);
count_ang <= "000000010000";
quarter_in <= "10";
Ang <= Ang + 16;
else
state <= to_unsigned(second,2);
quarter_in <= "01";
Ang <= Ang - 16;
count_ang <= count_ang + 16;
end if;
when to_unsigned(third,2) => if (count_ang >= 3999) then
state <= to_unsigned(fourth,2);
count_ang <= "000000010000";
quarter_in <= "11";
Ang <= Ang - 16;
else
state <= to_unsigned(third,2);
quarter_in <= "10";
Ang <= Ang + 16;
count_ang <= count_ang + 16;
end if;
when to_unsigned(fourth,2) => if (count_ang >= 3999) then
state <= to_unsigned(first,2);
count_ang <= "000000010000";
quarter_in <= "00";
Ang <= Ang + 16;
else
state <= to_unsigned(fourth,2);
quarter_in <= "11";
Ang <= Ang - 16;
count_ang <= count_ang + 16;
end if;
when others => count_ang <= (others => '0');
end case;
end if;
end process;
Angle <= Ang;
end Behavioral;
和测试台(但我不知道，我有点在模块中询问。并且获得了我的“空”测试台)
ENTITY testmass IS
END testmass;
ARCHITECTURE behavior OF testmass IS
-- Component Declaration for the Unit Under Test (UUT)
COMPONENT massive
PORT(
clk : IN std_logic;
reset : IN std_logic;
sinus : OUT std_logic_vector(15 downto 0)
);
END COMPONENT;
--Inputs
signal clk : std_logic := '0';
signal reset : std_logic := '0';
--Outputs
signal sinus : std_logic_vector(15 downto 0);
-- Clock period definitions
constant clk_period : time := 10 ns;
BEGIN
-- Instantiate the Unit Under Test (UUT)
uut: massive PORT MAP (
clk => clk,
reset => reset,
sinus => sinus
);
-- Clock process definitions
clk_process :process
begin
clk <= '0';
wait for clk_period/2;
clk <= '1';
wait for clk_period/2;
end process;
-- Stimulus process
stim_proc: process
begin
-- hold reset state for 100 ns.
wait for 100 ns;
wait for clk_period*10;
-- insert stimulus here
wait;
end process;
END;
最佳答案
您的问题不是 Minimal, Complete and Verifiable example ，它是不可验证的:
Describe the problem. "It doesn't work" is not a problem statement. Tell us what the expected behavior should be. Tell us what the exact wording of the error message is, and which line of code is producing it. Put a brief summary of the problem in the title of your question.
输出无效。什么都不重要，只是显示 1 个值。
什么是唯一值？当有人试图复制您的问题时，我们看到的一件事是在实体大规模过程中对 z(n) 的每次评估的断言警告:
if (z(n) >= 0) then
这个问题对于 VHDL 信号来说很微妙。
您在流程中为信号赋值并期望它们立即可用。那不会发生。当任何进程尚未恢复并随后在当前模拟周期中暂停时，不会更新任何信号。
对于投影输出波形(队列)中的每个仿真时间，只有一个条目。后续分配(此处未发生)将导致仅最后一个值排队。
更重要的是 future 的值(value)在当前的模拟周期中不可用。
x、y 和 z 可以改为在流程中声明的变量:
architecture foo of massive is
-- not predefined before -2008:
function to_string (inp: signed) return string is
variable image_str: string (1 to inp'length);
alias input_str: signed (1 to inp'length) is inp;
begin
for i in input_str'range loop
image_str(i) := character'VALUE(std_ulogic'IMAGE(input_str(i)));
end loop;
return image_str;
end function;
begin
process (clk)
type my_type is array (0 to 16) of signed (15 downto 0);
variable x: my_type;
variable y: my_type;
variable z: my_type;
constant j: my_type := ("1010111111001000", "0110011111000101",
"0011011011010100", "0001101111010101",
"0000110111111000", "0000011011111110",
"0000001101111111", "0000000111000000",
"0000000011100000", "0000000001110000",
"0000000000111000", "0000000000011100",
"0000000000001110", "0000000000000111",
"0000000000000100", "0000000000000010",
"0000000000000001");
begin
x(0) := "0000010100000110";
y(0) := "0000000000000000";
z(0) := "0000000000000000";
if rising_edge(clk) then
if reset = '0' then -- reset not driven condition was <=
report "init values:" & LF & HT &
"x(0) = " & to_string(x(0)) & LF & HT &
"y(0) = " & to_string(y(0)) & LF & HT &
"z(0) = " & to_string(z(0));
for n in 0 to 15 loop
if z(n) >= 0 then
x(n + 1) := x(n) - y(n) / 2 ** n;
y(n + 1) := y(n) + x(n) / 2 ** n;
z(n + 1) := z(n) + j(n);
else
x(n + 1) := x(n) + y(n) / 2 ** n;
y(n + 1) := y(n) - x(n) / 2 ** n;
z(n + 1) := z(n) - j(n);
end if;
report "n = " & integer'image(n) & LF & HT &
"x(" & integer'image(n + 1) & ") = " &
to_string(x(n + 1)) & LF & HT &
"y(" & integer'image(n + 1) & ") = " &
to_string(y(n + 1)) & LF & HT &
"z(" & integer'image(n + 1) & ") = " &
to_string(z(n + 1));
end loop;
sinus <= std_logic_vector(y(16));
report "sinus = " & to_string(y(16));
end if;
end if;
end process;
end architecture foo;
添加报告语句以允许将值输出到模拟控制台。如果在连续分配给变量之间没有时间流逝，波形中的变量值就没有任何有用的意义。有些模拟器不会在波形转储中报告变量。
以上架构产生:
ghdl -a testmass.vhdl
ghdl -e testmass
ghdl -r testmass
testmass.vhdl:86:17:@5ns:(report note): init values:
x(0) = 0000010100000110
y(0) = 0000000000000000
z(0) = 0000000000000000
testmass.vhdl:100:21:@5ns:(report note): n = 0
x(1) = 0000010100000110
y(1) = 0000010100000110
z(1) = 1010111111001000
testmass.vhdl:100:21:@5ns:(report note): n = 1
x(2) = 0000011110001001
y(2) = 0000001010000011
z(2) = 0100100000000011
testmass.vhdl:100:21:@5ns:(report note): n = 2
x(3) = 0000011011101001
y(3) = 0000010001100101
z(3) = 0111111011010111
testmass.vhdl:100:21:@5ns:(report note): n = 3
x(4) = 0000011001011101
y(4) = 0000010101000010
z(4) = 1001101010101100
testmass.vhdl:100:21:@5ns:(report note): n = 4
x(5) = 0000011010110001
y(5) = 0000010011011101
z(5) = 1000110010110100
testmass.vhdl:100:21:@5ns:(report note): n = 5
x(6) = 0000011011010111
y(6) = 0000010010101000
z(6) = 1000010110110110
testmass.vhdl:100:21:@5ns:(report note): n = 6
x(7) = 0000011011101001
y(7) = 0000010010001101
z(7) = 1000001000110111
testmass.vhdl:100:21:@5ns:(report note): n = 7
x(8) = 0000011011110010
y(8) = 0000010010000000
z(8) = 1000000001110111
testmass.vhdl:100:21:@5ns:(report note): n = 8
x(9) = 0000011011110110
y(9) = 0000010001111010
z(9) = 0111111110010111
testmass.vhdl:100:21:@5ns:(report note): n = 9
x(10) = 0000011011110100
y(10) = 0000010001111101
z(10) = 1000000000000111
testmass.vhdl:100:21:@5ns:(report note): n = 10
x(11) = 0000011011110101
y(11) = 0000010001111100
z(11) = 0111111111001111
testmass.vhdl:100:21:@5ns:(report note): n = 11
x(12) = 0000011011110101
y(12) = 0000010001111100
z(12) = 0111111111101011
testmass.vhdl:100:21:@5ns:(report note): n = 12
x(13) = 0000011011110101
y(13) = 0000010001111100
z(13) = 0111111111111001
testmass.vhdl:100:21:@5ns:(report note): n = 13
x(14) = 0000011011110101
y(14) = 0000010001111100
z(14) = 1000000000000000
testmass.vhdl:100:21:@5ns:(report note): n = 14
x(15) = 0000011011110101
y(15) = 0000010001111100
z(15) = 0111111111111100
testmass.vhdl:100:21:@5ns:(report note): n = 15
x(16) = 0000011011110101
y(16) = 0000010001111100
z(16) = 0111111111111110
testmass.vhdl:109:17:@5ns:(report note): sinus = 0000010001111100
我们看到数组元素的值正在改变，而不是通过加法(或当 z(n) < 0 时减法)传播“X”，这是在上一个循环迭代中分配的。
另请注意，重置不会更改测试台中的值，并且在原始大规模过程中使用关系运算符“<=”对其值进行错误评估。
j 除了初始值外未分配，在上述架构中显示为常量。
我个人有点怀疑您是否可以在一个 10 ns 时钟内执行这 16 个链式加法或减法以及选择操作。
关于algorithm - 使用 cordic 算法生成正弦，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/46293416/
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