Hi everyone! I've some problems with a design, I have to Design an algorithmic state machine to implement a parallel series multiplier for eight-bit numbers encoded in SM (Sign-Magnitude). The multiplying is in a register with parallel output and the multiplier in a shift register of length m with serial output, the first bit to be the least significant. Each bit of multiplier, in his turn, multiplies by multiplying all, so that with operations multiplication is performed. I've the problem with the shift register with serial output, because I don´t know how to do it!! I'm learning VHDL right now, and I've never work with this hardware description language. Please if anyone can help me, I'd be very grateful. Thank you!!

--- Quote Start --- Hi everyone! I've some problems with a design, I have to Design an algorithmic state machine to implement a parallel series multiplier for eight-bit numbers encoded in SM (Sign-Magnitude). The multiplying is in a register with parallel output and the multiplier in a shift register of length m with serial output, the first bit to be the least significant. Each bit of multiplier, in his turn, multiplies by multiplying all, so that with operations multiplication is performed. I've the problem with the shift register with serial output, because I don´t know how to do it!! I'm learning VHDL right now, and I've never work with this hardware description language. Please if anyone can help me, I'd be very grateful. Thank you!! --- Quote End --- You need to show what have you done so far and where in design stage you want help. The 8 bit sign&magnitude can be any value from 0111 1111 to 1111 1111 when you multiply you do that per bit then add up as in primary school multiplication. You can ignore sign bit as you can conclude that at the end, for demo assume you have: 1) 111 1111 2) 111 1101 thus you multiply 1 by 111 1111 and you get 111 1111 then you multiply by zero which results in zero, shift one bit... xxxxxxx111 1111 xxxxxx000 0000 xxxxx111 1111 xxxx111 111 xx111 1111 x111 1111 111 1111 ------------------------ after 7 multiplications you add all to get result then decide sign bit

Thank you kaz, but how can I do that in VHDL?? I know how I have to do the operation, but I don't know how to create the register in VHDL. Right now, I have this: library IEEE; use IEEE.STD_LOGIC_1164.ALL; use IEEE.std_logic_unsigned.all; use IEEE.numeric_std.all; entity reg_des is port ( multiplicador: in std_logic; multiplicando: in std_logic_vector (7 downto 0); registro7: out std_logic_vector (7 downto 0); registro6: out std_logic_vector (7 downto 0); Q7: out std_logic; resultado: out bit_vector (15 downto 0); clk: in std_logic; reset: in std_logic); end reg_des; architecture behavioral of reg_des is begin reg_des : process (multiplicador, clk, reset) variable Q : std_logic_vector (8 downto 0); begin if reset = '1' then Q:= "000000000"; else if clk'event and clk = '1' then for i in 0 to 7 loop Q(i) := Q(i+1); if i = 0 then registro7(7) <= Q(i) and multiplicando(7); end if; if i = 1 then registro7(6) <= Q(i) and multiplicando(7); registro6 (7) <= Q(i-1) and multiplicando(6); end if; if i = 2 then registro7(5) <= Q(i) and multiplicando(7); registro6 (6) <= Q(i-1) and multiplicando(6); end if; if i = 3 then registro7(4) <= Q(i) and multiplicando(7); registro6 (5) <= Q(i-1) and multiplicando(6); end if; if i = 4 then registro7(3) <= Q(i) and multiplicando(7); registro6 (4) <= Q(i-1) and multiplicando(6); end if; if i = 5 then registro7(2) <= Q(i) and multiplicando(7); registro6 (3) <= Q(i-1) and multiplicando(6); end if; if i = 6 then registro7(1) <= Q(i) and multiplicando(7); registro6 (2) <= Q(i-1) and multiplicando(6); end if; if i = 7 then registro7(0) <= Q(i) and multiplicando(7); registro6 (1) <= Q(i-1) and multiplicando(6); end if; if i = 8 then registro6 (0) <= Q(i-1) and multiplicando(6); end if; end loop; Q(8):=multiplicador; Q7 <= Q(7); end if; end if; end process; end behavioral; ------------------------------------------------------ Is this the correct way to do it?? Thank you!

--- Quote Start --- You need to show what have you done so far and where in design stage you want help. The 8 bit sign&magnitude can be any value from 0111 1111 to 1111 1111 when you multiply you do that per bit then add up as in primary school multiplication. You can ignore sign bit as you can conclude that at the end, for demo assume you have: 1) 111 1111 2) 111 1101 thus you multiply 1 by 111 1111 and you get 111 1111 then you multiply by zero which results in zero, shift one bit... xxxxxxx111 1111 xxxxxx000 0000 xxxxx111 1111 xxxx111 111 xx111 1111 x111 1111 111 1111 ------------------------ after 7 multiplications you add all to get result then decide sign bit --- Quote End --- Thank you kaz, but how can I do that in VHDL?? I know how I have to do the operation, but I don't know how to create the register in VHDL. Right now, I have this: library IEEE; use IEEE.STD_LOGIC_1164.ALL; use IEEE.std_logic_unsigned.all; use IEEE.numeric_std.all; entity reg_des is port ( multiplicador: in std_logic; multiplicando: in std_logic_vector (7 downto 0); registro7: out std_logic_vector (7 downto 0); registro6: out std_logic_vector (7 downto 0); Q7: out std_logic; resultado: out bit_vector (15 downto 0); clk: in std_logic; reset: in std_logic); end reg_des; architecture behavioral of reg_des is begin reg_des : process (multiplicador, clk, reset) variable Q : std_logic_vector (8 downto 0); begin if reset = '1' then Q:= "000000000"; else if clk'event and clk = '1' then for i in 0 to 7 loop Q(i) := Q(i+1); if i = 0 then registro7(7) <= Q(i) and multiplicando(7); end if; if i = 1 then registro7(6) <= Q(i) and multiplicando(7); registro6 (7) <= Q(i-1) and multiplicando(6); end if; if i = 2 then registro7(5) <= Q(i) and multiplicando(7); registro6 (6) <= Q(i-1) and multiplicando(6); end if; if i = 3 then registro7(4) <= Q(i) and multiplicando(7); registro6 (5) <= Q(i-1) and multiplicando(6); end if; if i = 4 then registro7(3) <= Q(i) and multiplicando(7); registro6 (4) <= Q(i-1) and multiplicando(6); end if; if i = 5 then registro7(2) <= Q(i) and multiplicando(7); registro6 (3) <= Q(i-1) and multiplicando(6); end if; if i = 6 then registro7(1) <= Q(i) and multiplicando(7); registro6 (2) <= Q(i-1) and multiplicando(6); end if; if i = 7 then registro7(0) <= Q(i) and multiplicando(7); registro6 (1) <= Q(i-1) and multiplicando(6); end if; if i = 8 then registro6 (0) <= Q(i-1) and multiplicando(6); end if; end loop; Q(8):=multiplicador; Q7 <= Q(7); end if; end if; end process; end behavioral; ------------------------------------------------------ Is this the correct way to do it?? Thank you!

At least you got something... if it was me I will run a 7 states machine (or just a counter 0 to 6). Assuming y = A*B then I might go as below case count is when 0 => if A(0) = '1' then temp <= B; else temp <= "0000000"; end if; y <= y + temp; -- first of 7 additions, initialise y to zero when 1 => if A(1) = '1' then temp <= B; else temp <= "0000000"; end if; y <= y + temp&'0'; --increase weight by one digit ...etc you will need to declare proper types and manage bitwidth for addition.

--- Quote Start --- At least you got something... if it was me I will run a 7 states machine (or just a counter 0 to 6). Assuming y = A*B then I might go as below case count is when 0 => if A(0) = '1' then temp <= B; else temp <= "0000000"; end if; y <= y + temp; -- first of 7 additions, initialise y to zero when 1 => if A(1) = '1' then temp <= B; else temp <= "0000000"; end if; y <= y + temp&'0'; --increase weight by one digit ...etc you will need to declare proper types and manage bitwidth for addition. --- Quote End --- Sorry but I can´t understand what you want to do. Can you explain me?

Help multiplier vhdl | Altera Community

13 Replies

Altera_Forum

Honored Contributor

8 years ago

Im feeling pretty generous today. I hate seeing overly complicated case statements.

I think the following code does what you're trying to do:


library IEEE;
use IEEE.STD_LOGIC_1164.ALL;
use IEEE.numeric_std.all;
entity prueba2 is
port (
  A         : in  std_logic_vector( (7 downto 0);
  B         : in  unsigned (7 downto 0);
  inicio    : in  std_logic;
  clk       : in  std_logic;
  reset     : in  std_logic;
  q         : out unsigned (2 downto 0);
  resultado : out unsigned (15 downto 0)
);
end prueba2;
architecture behavioral of prueba2 is
signal cuenta : unsigned(2 downto 0);
signal y      : unsigned(15 downto 0);
begin
  multi : process (clk, reset)
  begin
    if reset = '1' then
      y  <= (others => '0');
    elsif rising_edge(clk) then
      if inicio = '1' then
        cuenta <= cuenta + 1;
      end if;
      if A /= x"00" then
        if A( to_integer(cuenta) ) = '1' then
          Y   <= Y + ( resize(B, y'length)*(2**to_integer(cuenta)) );
        end if;
      end if;
    end if;
  end process;
  q         <= cuenta;
  resultado <= y;
end architecture;

Altera_Forum

Honored Contributor

8 years ago

--- Quote Start ---

Im feeling pretty generous today. I hate seeing overly complicated case statements.

I think the following code does what you're trying to do:


library IEEE;
use IEEE.STD_LOGIC_1164.ALL;
use IEEE.numeric_std.all;
entity prueba2 is
port (
  A         : in  unsigned (7 downto 0);
  B         : in  unsigned (7 downto 0);
  inicio    : in  std_logic;
  clk       : in  std_logic;
  reset     : in  std_logic;
  q         : out unsigned (2 downto 0);
  resultado : out unsigned (15 downto 0)
);
end prueba2;
architecture behavioral of prueba2 is
signal cuenta : unsigned(2 downto 0);
signal y      : unsigned(15 downto 0);
begin
  multi : process (A, clk, reset)
  begin
    if reset = '1' then
      y  <= (others => '0');
    elsif rising_edge(clk) then
      if inicio = '1' then
        cuenta <= cuenta + 1;
      end if;
      if B /= x"00" then
        Y   <= Y + ( resize(B, 15)*(2**to_integer(cuenta)) );
      end if;
    end if;
  end process;
  q         <= cuenta;
  resultado <= y;
end architecture;

--- Quote End ---

well Tricky that might work but is too tricky for beginner, after all I can just say y = a*b

I assume it is exercise that gets more and more compact under pressure...great

Altera_Forum
Honored Contributor
8 years ago
--- Quote Start ---
well Tricky that might work but is too tricky for beginner, after all I can just say y = a*b
I assume it is exercise that gets more and more compact under pressure...great
--- Quote End ---

Thank you kaz and Tricky, finally before I read your answers I have solved it in the way that kaz as told me.
So I've understood it very well. Thank you so much for the help!!!!

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Help multiplier vhdl

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