I am trying to generate a random number for a little game and I tried to use the UNIFORM function from the package MATH_REAL.
I get a value when I test my code, but it never changes. I tried to make the code happen at every rising_edge but it didn't change anything. I couldn't find or think of an effective fix.
I think I know that problem is simply that my random number is generated only once, but I couldn't figure out a way to change this.
Here is what I implemented (my variation with the clock) :
rand: process (count1)
variable seed1, seed2 : positive;
variable rand : real;
variable int_rand : integer range 1 to 5;
begin
if rst_i='1' then
count1 <= 1;
elsif rising_edge(clk_i) then
UNIFORM(seed1, seed2, rand);
int_rand := INTEGER(TRUNC(rand*5.0));
count1 <= int_rand;
end if;
end process;
The value of count1 is then used in another process. The above code gives the same simulation results as simply :
rand: process (count1)
variable seed1, seed2 : positive;
variable rand : real;
variable int_rand : integer range 1 to 5;
begin
UNIFORM(seed1, seed2, rand);
int_rand := INTEGER(TRUNC(rand*5.0));
count1 <= int_rand;
end process;
Is there a simple way to ensure more that a random number is generated more than once?
When the purpose is to generate a random number on count1 for every rising edge of clk_i, then update the sensitivity list for the process to (assuming async reset by rst_i):
rand: process (rst_i, clk_i)
This will make the process reexecute at every change of rst_i or clk_i, where the sensitivity list of count1 reexecutes at change of count1 only.
Note that variable int_rand : integer range 1 to 5; should be range 0 to 4, since uniform result is > 0.0 and < 1.0.
An interesting behavior is, that a sensitivity list of rand: process (count1) as for the second process, without the if guarding generation on count1, will actually make the process reexecute several times start of simulation, until the generated count1 value is identical to the previous count1 value, in which case the is no change on count1 to trigger reexecution of the process, and then process is then not executed any more.
Related
I want an approximation of the Tanh function by saving the values in a LUT (by this I am doing a quantization). I want to choose the Number of entries in the LUT.
As an not-correct example, I imagine a code like
library IEEE;
use IEEE.STD_LOGIC_1164.ALL;
use IEEE.NUMERIC_STD.ALL;
use ieee.fixed_pkg.all;
entity tanh_lut is
generic (
MIN_RANGE: real := 0.0; -- Minimum value of x
MAX_RANGE: real := 5.0; -- Maximum value of x
DATA_RANGE_int: positive:= 8;
DATA_RANGE_frac: positive:= 8;
);
Port ( DIN : in sfixed(DATA_RANGE_int-1 downto -(DATA_RANGE_frac-1));
DOUT : out sfixed(DATA_RANGE_int-1 downto -(DATA_RANGE_frac-1))
end tanh_lut;
architecture Behavioral of tanh_lut is
begin
lut_gen: for i in 0 to LUT_SIZE-1 generate
constant x_val : real := MIN_RANGE + (MAX_RANGE - MIN_RANGE) * i / (LUT_SIZE-1);
constant x_val_next : real := MIN_RANGE + (MAX_RANGE - MIN_RANGE) * (i+1) / (LUT_SIZE-1);
constant y_val : real := tanh(x_val);
if DIN>=x_val_previous AND DIN<x_val then
DOUT <= to_sfixed(tanh(y_val),DOUT ) ;
END IF
end generate;
end Behavioral;
Per example, if I want 4 entries in the range 0 to 3, I want that it is synthesizing a code like:
if DIN>0 AND DIN<=1 then
DOUT <= to_sfixed(0, DOUT);
else DIN>1 AND DIN<=2 then
DOUT <= to_sfixed(0.76159415595, DOUT);
else DIN>2 AND DIN<=3 then
DOUT <= to_sfixed(0.96402758007, DOUT);
else DIN>3 AND DIN<=4 then
DOUT <= to_sfixed(0.99505475368, DOUT);
End if
Is there any way that a code like this or a code which implements the idea behind this is possible?
A simple LUT with addresses is not possible because the addresses are always integer and DIN is fixed point, e.g., 1.5
The other possibility would be two LUTs, one for mapping the Input to an address, another for mapping the address to the LUT entry, e.g., LUT1: 1.5=> address 5, LUT2: address 5 => 0.90. But by this I would double the amount of resources what I dont want
My requirements: things like the tanh(x) should not be synthesized, only the final value of tanh(x). It shoudl also be hardware efficient
It does not matter if you use a nested „if-elsif“ construct or if you use a new „if“ construct for each check.
So you can create a loop like this:
for i in 0 to c_number_of_checks-1 loop
if c_boundaries(i)<DIN and DIN<=c_boundaries(i+1) then
DOUT <= c_output_values(i);
end if;
end loop;
Of course you must provide the constants c_number_of_checks and c_boundaries, c_output_values. This can be done by:
constant c_number_of_checks : natural := 4;
type array_of_your_data_type is array (natural range <>) of your_data_type;
constant c_boundaries : array_of_your_data_type(c_number_of_checks downto 0) := init_c_boundaries(c_number_of_checks);
constant c_output_values : array_of_your_data_type(c_number_of_checks-1 downto 0) := init_c_output_values(c_number_of_checks);
This means you will need the functions init_c_boundaries, init_c_output_values, which create arrays of values, which can initialize the constant c_boundaries and c_output_values.
But this is not complicated (you can use from ieee.math_real the function TANH), as the functions need not to be synthesizable, as they are called only during compile time.
As you see, you will have some effort. So perhaps it is easier to follow the other suggestions. If you do so (value as address of a LUT) you should think about automatic ROM inference, which is provided by several tool chains and will give you a very efficient (small) hardware.
I'm currently working on writing a simple counter in VHDL, trying to genericize it as much as possible. Ideally I end up with a counter that can pause, count up/down, and take just two integer (min, max) values to determine the appropriate bus widths.
As far as I can tell, in order to get an integer of a given range, I just need to delcare
VARIABLE cnt: INTEGER RANGE min TO max := 0
Where min and max are defined as generics (both integers) in the entity. My understanding of this is that if min is 0, max is 5, for example, it will create an integer variable of 3 bits.
My problem is that I actually want to output this integer. So, naturally, I write
counterOut : OUT INTEGER RANGE min TO max
But this does not appear to be doing what I need. I'm generating a schematic block in Quartus Prime from this, and it creates a bus output from [min...max]. For example, if min = 0, max = 65, it outputs a 66 bit bus. Instead of the seven bit bus it should.
If I restricted the counter to unsigned values I might be able to just math out the output bus size, but I'd like to keep this as flexible as possible, and of course I'd like to know what I'm actually doing wrong and how to do it properly.
TL;DR: I want a VHDL entity to take generic min,max values, and generate an integer output bus of the required width to hold the range of values. How do?
If it matters, I'm using Quartus Prime Lite Edition V20.1.0 at the moment.
Note: I know I can use STD_LOGIC_VECTOR instead, but it is going to simulate significantly slower and is less easy to use than the integer type as far as I have read. I can provide more of my code if necessary, but it's really this one line that's the problem as far as I can tell.
I originally posted this on Stackexchange, but I think Stackoverflow might be a better place since it's more of a programming than a hardware problem.
EDIT: Complete code shown below
LIBRARY ieee;
USE ieee.std_logic_1164.all;
USE ieee.numeric_std.all;
USE ieee.std_logic_signed.all;
ENTITY Counter IS
GENERIC (modulo : INTEGER := 32;
min : INTEGER := 0;
max : INTEGER := 64);
PORT( pause : IN STD_LOGIC;
direction : IN STD_LOGIC; -- 1 is up, 0 is down
clk : IN STD_LOGIC;
counterOut : OUT INTEGER RANGE min TO max --RANGE 0 TO 32 -- THIS line is the one generating an incorrect output bus width
);
END ENTITY Counter;
-- or entity
ARCHITECTURE CounterArch OF Counter IS
BEGIN
PROCESS(direction, pause, clk)
VARIABLE cnt : INTEGER RANGE min TO max := 0;
VARIABLE dir : INTEGER;
BEGIN
IF direction = '1' THEN
dir := 1;
ELSE
dir := -1;
END IF;
IF clk'EVENT AND clk = '1' THEN
IF pause = '0'THEN
IF (cnt = modulo AND direction = '1') THEN
cnt := min; -- If we're counting up and hit modulo, reset to min value.
ELSIF (cnt = min AND direction = '0') THEN
cnt := modulo; --Counting down hit 0, go back to modulo.
ELSE
cnt := cnt + dir;
END IF;
END IF;
END IF;
counterOut <= cnt;
END PROCESS;
END ARCHITECTURE CounterArch;
I'm storing two tables in two signals. One table keeps the key (address) and the other keeps the value corresponding to the key. I need to compare an input to the key and, if they match, return the value stored.
The reason why I need this is for a dynamic lookup table for branch instruction prediction. In the fetch stage of a processor I get the input Instruction_Address and I return a branch_To_Address and a branch_Prediction. Initially I want to store 16 predictions/branch addresses and use a circular buffer ring to overwrite as needed.
I've been trying to use a FOR with a nested IF to search for the key inside keyTable.
The whole module seems to work fine, except when I compare two bit_vectors with the IF statement. I need this twice (one on read and another on write) and hence I need to "sweep" the keysTable so I can see if the address that is being looked up has an entry.
I noticed the error upon simulation, where the ELSE clause is being called always regardless of the keysTable having the right entries.
Verifiable example:
library IEEE;
use ieee.numeric_bit.all;
entity branch_prediction_table is
generic (
addrSize : NATURAL := 4;
tableSize : NATURAL := 4);
port (
clock : in bit;
input_addr: in bit_vector(addrSize-1 downto 0);
return_value : out bit );
end branch_prediction_table;
architecture branch_table of branch_prediction_table is
signal keysTable : bit_vector(addrSize*tableSize-1 downto 0) := ( others => '0');
signal valuesTable : bit_vector(tableSize*2-1 downto 0) := ( others => '0');
begin
tableProc: process(clock) is
variable valueFromTable : bit;
begin
if rising_edge(clock) then
search_table: for iR in (tableSize-1) to 0 loop
if (keysTable(addrSize*(iR+1)-1 downto addrSize*iR) = input_addr) then
valueFromTable := valuesTable((iR+1)*2-1);
EXIT search_table;
else
valueFromTable := '0';
end if;
end loop search_table;
return_value <= valueFromTable;
end if; -- rising_edge(clock)
end process tableProc;
end branch_table;
with verifiable testbench simulation TCL:
add wave -position insertpoint \
sim:/branch_prediction_table/addrSize \
sim:/branch_prediction_table/clock \
sim:/branch_prediction_table/input_addr \
sim:/branch_prediction_table/keysTable \
sim:/branch_prediction_table/return_value \
sim:/branch_prediction_table/tableSize \
sim:/branch_prediction_table/valuesTable
force -freeze sim:/branch_prediction_table/valuesTable 11111111 0
force -freeze sim:/branch_prediction_table/keysTable 1111101001100011 0
force -freeze sim:/branch_prediction_table/clock 0 0, 1 {5000 ps} -r {10 ns}
run 10 ns
force -freeze sim:/branch_prediction_table/input_addr 1010 0
run 20 ns
force -freeze sim:/branch_prediction_table/input_addr 1111 0
run 10 ns
and testbench simulation result showing that error is indeed in the IF:
I have tried converting them with to_integer(unsigned(bit_vector1)) = to_integer(unsigned(bit_vector2)) with no avail
As user1155120 pointed out:
The problem lies within search_table: for iR **in** (tableSize-1) to 0 loop
It should've been "down to" as L > R. Since I used "in" with L>R, that produces a null range and the for loop iteration is said to be complete.
(IEEE Std 1076-2008 5.2 Scalar types, "A range specifies a subset of values of a scalar type. A range is said to be a null range if the specified subset is empty. The range L to R is called an ascending range; if L > R, then the range is a null range. The range L downto R is called a descending range; if L < R, then the range is a null range.").
10.10 Loop statement "For the execution of a loop with a for iteration scheme, the discrete range is first evaluated. If the discrete range is a null range, the iteration scheme is said to be complete, ..."
Description:
I want to write vhdl code that finds the largest integer in the array A which is an array of 20 integers.
Question:
what should my algorithm look like, to input where the sequential statements are?
my vhdl code:
highnum: for i in 0 to 19 loop
i = 0;
i < 20;
i<= i + 1;
end loop highnum;
This does not need to be synthesizable but I dont know how to form this for loop a detailed example explaining how to would be appreciated.
Simply translating the C loop to VHDL, inside a VHDL clocked process, will work AND be synthesisable. It will generate a LOT of hardware because it has to generate the output in a single clock cycle, but that doesn't matter if you are just simulating it.
If that is too much hardware, then you have to implement it as a state machine with at least two states, Idle and Calculating, so that it performs only one loop iteration per clock cycle while Calculating, and returns to the Idle state when done.
First of all you should know how have you defined the array in vhdl.
Let me define an array for you.
type array_of_integer array(19 downto 0) of integer;
signal A : array_of_integer :=(others => 0);
signal max : integer;
-- Now above is the array in vhdl of integers all are initialized to value 0.
A(0) <= 1;
A(1) <= 2;
--
--
A(19)<= 19;
-- Now the for loop for calculating maximum
max <= A(0);
for i in 0 to 19 loop
if (A(i) > max) then
max <= A(i);
end if;
end loop;
-- Now If you have problems in understating that where to put which part of code .. in a ----vhdl entity format .. i.e process, ports, etc... you can reply !
Can someone please help me with the following:
Design a digital circuit, using VHDL, to keep track of time in the form of HH:MM:SS. The circuit should produce 6 separate four bit digital outputs (2 four bit outputs for the HH, 2 for the MM, 2 for the SS). The HH can just be a 2 digit number in the range 00 to 99 i.e. it’s not a clock, it just a counter for hours even though 99 hour tapes don’t exist. The time is to be displayed on the 6 right most 7 segment displays of the DE2. You have already designed a 7 segment decoder and driver as part of a previous lab, so that can be used to convert each 4 bit output into a 7 bit signal for each the 7-segment display. Don’t forget to set up the pin planer for these display (and all other signals)
The circuit should have the following single bit inputs: A Clock, an increment, a decrement and a reset. The increment/decrement inputs should cause the tape counter to add or subtract 1 second from the tape time on the next rising edge of the clock signal. If neither the increment or decrement inputs are present, the tape counter does not change. The reset is synchronous to the clock (to avoid glitches accidentally resetting it). The increment and decrement signals are all active high signals (i.e. a logic 1), the reset is active low (logic 0).
You tape counter should handle full hour, minute and second roll over, e.g. if the counter is showing 9:59:59, then the next increment should make it display 10:00:00 and vice versa when decrement is present.
Rather than solving your homework, I'd like to give you an idea. Most designers will tend to implement this clock using digit-by-digit rollover (some digits will rollover from 9-0, others from 5-0). I'd like to propose someting different.
The overall idea is: keep your time value in seconds as an integer. This will greatly facilitate the tasks of incrementing and decrementing. Then, you simply implement a conversion function that returns the number of hours, minutes, and seconds, given an integer number of seconds.
Your clock entity would look like this:
library ieee;
use ieee.std_logic_1164.all;
use work.clock_pkg.all;
entity clock is
port (
clock: in std_logic;
n_reset: in std_logic;
increment: in std_logic;
decrement: in std_logic;
hours: out natural range 0 to 99;
minutes: out natural range 0 to 59;
seconds: out natural range 0 to 59
);
end;
architecture rtl of clock is
signal time_in_seconds: natural range 0 to 359999;
begin
process (clock, n_reset) begin
if rising_edge(clock) then
if n_reset = '0' then
time_in_seconds <= 0;
elsif increment then
time_in_seconds <= time_in_seconds + 1;
elsif decrement then
time_in_seconds <= time_in_seconds - 1;
end if;
end if;
end process;
process (time_in_seconds) begin
(hours, minutes, seconds) <= seconds_to_time_type(time_in_seconds);
end process;
end;
As you can imagine, the workhorse of this solution is the seconds_to_time_type() function. You could implement it like this:
package clock_pkg is
type time_type is record
hours: natural range 0 to 99;
minutes, seconds: natural range 0 to 59;
end record;
function seconds_to_time_type(seconds: in natural) return time_type;
end;
package body clock_pkg is
function seconds_to_time_type(seconds: in natural) return time_type is
variable hh: natural range 0 to 99;
variable mm: natural range 0 to 119;
variable ss: natural range 0 to 119;
begin
hh := seconds / 3600;
mm := (seconds mod 3600) / 60;
ss := (seconds mod 3600) mod 60;
return (hh, mm, ss);
end;
end;
Now you have an entity that outputs separate integer values for hours, minutes, and seconds. Converting those values from integers to BCD, and showing those values on the displays is left as an exercise to the reader.
The typical way of implementing a counting clock is using binary coded decimal (BCD), where each digit consists of a separate n-bit counter, with a range as needed.
For example, in order to count seconds (from 0-59), you could use something like the following code:
process(clk, reset) begin
if(reset='1') then
second_tens <= (others=>'0');
second_ones <= (others=>'0');
elsif(rising_edge(clk)) then
if(count_en='1') then
if(second_ones = 9) then
second_ones <= (others=>'0');
if(second_tens = 5) then
second_tens <= (others=>'0');
-- Count up minutes.
else
second_tens <= second_tens + 1;
end if;
else
second_ones <= second_ones + 1;
end if;
end if;
end if;
end process;
Minutes and hours can be counted analogously.
You have skipped a step. You are trying to think about code with just a worded problem statement. First step is to design the hardware by drawing a block diagram. Break the problem down into pieces.
Initial partitioning might be Seconds, Minutes, and Hours. If you are counting in BCD, it you may wish to partition it further digit by digit. Work out what your hardware is supposed to do. Draw a picture. Write code that describes what is in the picture.
At the end of the day, your RTL block diagram is your HDL flow chart.