ONE clock period pulse based on trigger signal - triggers

Question

i am making a midi interface. UART works fine, it sends the 8 bit message along with a flag to a control unit. When the flag goes high, the unit will store the message in a register and make a clr_flag high in order to set the flag of UART low again. The problem is that i can not make this clr_flag one period long. I need it to be ONE period long, because this signal also controls a state machine that indicates what kind of message is being stored (note_on -> key_note -> velocity, for example).
My question here is, how can a signal (flag in this case) trigger a pulse just for one clk period? what i have now makes almost a pulse during a clock period, but i does it twice, because the flag has not become 0 yet. ive tried many ways and now i have this:
get_data:process(clk, flag)
begin
if reset = '1' then
midi <= (others => '0');
clr_flag <= '0';
control_flag <= '0';
elsif ((clk'event and clk='1') and flag = '1') then
midi <= data_in;
clr_flag <= '1';
control_flag <= '1';
elsif((clk'event and clk='0') and control_flag = '1') then
control_flag <= '0';
elsif((clk'event and clk='1') and control_flag = '0') then
clr_flag <= '0';
end if;
end process;
the problem with this double pulse or longer than one period pulse(before this, i had something that made clr_flag a two period clk pulse), is that the system will go though two states instead of one per flag.
so in short: when one signal goes high (independent of when it goes low), a pulse during one clock period should be generated.
thanks for your help.

Answer 1

The trick to making a single cycle pulse is realising that having made the pulse, you have to wait as long as the trigger input is high before getting back to the start. Essentially you are building a very simple state machine, but with only 2 states you can use a simple boolean to tell them apart.
Morten is correct about the need to adopt one of the standard patterns for a clocked process; I have chosen a different one that works equally well.
get_data:process(clk, reset)
variable idle : boolean;
begin
if reset = '1' then
idle := true;
elsif rising_edge(clk) then
clr_flag <= '0'; -- default action
if idle then
if flag = '1' then
clr_flag <= '1'; -- overrides default FOR THIS CYCLE ONLY
idle <= false;
end if;
else
if flag = '0' then
idle := true;
end if;
end if;
end if;
end process;

Answer 2

There are several issues to address in order to make the design for a one cycle
pulse using flip flops (registers).
First, the use of flip flops in hardware through VHDL constructions typically
follows a structure like:
process (clk, reset) is
begin
-- Clock
if rising_edge(clk) then
-- ... Flip flops to update at rising edge
end if;
-- Reset
if reset = '1' then
-- Flip flops to update at reset, which need not be all
end if;
end process;
So the get_data process should be updated accordingly, thus:
Sensitivity list should contain only clock (clk) and reset
The nested structure with if on event should be as above
Only rising edge of clk should be used, thus no check on clk = '0'
Making a one cycle pulse on clr_flag when flag goes high can be made with a
synchronous '0' to '1' detector on flag, using a version of flag that is
delayed a single cycle, called flag_ff below, and then checking for (flag =
''1) and (flag_ff = '0').
The resulting code may then look like:
get_data : process (clk, reset) is
begin
-- Clock
if rising_edge(clk) then
flag_ff <= flag; -- One cycle delayed version
clr_flag <= '0'; -- Default value with no clear
if (flag = '1') and (flag_ff = '0') then -- Detected flag going from '0' to '1'
midi <= data_in;
clr_flag <= '1'; -- Override default value making clr_flag asserted signle cycle
end if;
end if;
-- Reset
if reset = '1' then
midi <= (others => '0');
clr_flag <= '0';
-- No need to reset flag_ff, since that is updated during reset anyway
end if;
end process;

Answer 3

Synchronisation and Edge Detection for FSM
The Rise, Edge and Fall outputs will strobe for one cycle when those events are detected. Inputs and outputs are synchronised for use with a Finite State Machine.
entity SynchroniserBit is
generic
(
REG_SIZE: natural := 3 -- Default number of bits in sync register.
);
port
(
clock: in std_logic;
reset: in std_logic;
async_in: in std_logic := '0';
sync_out: out std_logic := '0';
rise_out: out std_logic := '0';
fall_out: out std_logic := '0';
edge_out: out std_logic := '0'
);
end;
architecture V1 of SynchroniserBit is
constant MSB: natural := REG_SIZE - 1;
signal sync_reg: std_logic_vector(MSB downto 0) := (others => '0');
alias sync_in: std_logic is sync_reg(MSB);
signal rise, fall, edge, previous_sync_in: std_logic := '0';
begin
assert(REG_SIZE >= 2) report "REG_SIZE should be >= 2." severity error;
process (clock, reset)
begin
if reset then
sync_reg <= (others => '0');
previous_sync_in <= '0';
rise_out <= '0';
fall_out <= '0';
edge_out <= '0';
sync_out <= '0';
elsif rising_edge(clock) then
sync_reg <= sync_reg(MSB - 1 downto 0) & async_in;
previous_sync_in <= sync_in;
rise_out <= rise;
fall_out <= fall;
edge_out <= edge;
sync_out <= sync_in;
end if;
end process;
rise <= not previous_sync_in and sync_in;
fall <= previous_sync_in and not sync_in;
edge <= previous_sync_in xor sync_in;
end;

Answer 4

Below is a way of creating a signal (flag2) that lasts exactly one clock period from a signal (flag1) that lasts at least one clock period.

Answer 5

I don't program in VHDL~ here is what I usually do for the same propose in Verilog:
always #(posedge clk or negedge rst) begin
if(~rst) flgD <= 1'b0;
else flgD <= flg;
end
assign trg = (flg^flgD)&flgD;

Answer 6

I am new to verilog and this is the sample code, which I used for triggering. Hope this serves your purpose. You can try same logic in VHDL.
module main(clk,busy,rd);
input clk,busy; // busy input condition
output rd; // trigger signal
reg rd,en;
always #(posedge clk)
begin
if(busy == 1)
begin
rd <= 0;
en <= 0;
end
else
begin
if (en == 0 )
begin
rd <= 1;
en <= 1;
end
else
rd <= 0;
end
end
endmodule

Answer 7

The below verilog code shall hold the value for the signals for one clock cycle exactly.
module PulseGen #(
parameter integer BUS_WIDTH = 32
)
(
input [BUS_WIDTH-1:0] i,
input clk,
output [BUS_WIDTH-1:0] o
);
reg [BUS_WIDTH-1:0] id_1 = 0 ;
reg [BUS_WIDTH-1:0] id_2 = 0 ;
always #(posedge clk)begin
id_1 <= i;
id_2 <= id_1;
end
assign o = (id_1 & ~id_2);

Answer 8

The way to achieve this is to create a debounce circuit. If you need a D flip-flop to change from 0 to 1, only for the first clock, just add an AND gate before its input like the image below:
So here you can see a D flip-flop and its debounce circuit.
P.S. Circuit created using this.

Answer 9

--------------------------------------------------------------------------------
--------------------------------------------------------------------------------
--input of minimum 1 clock pulse will give output of wanted length.
--load number 5 to PL input and you will get a 5 clock pulse no matter how long input is.
--------------------------------------------------------------------------------
--------------------------------------------------------------------------------
library ieee ;
use ieee.std_logic_1164.all ;
use ieee.std_logic_unsigned.all ;
entity fifth is
port (clk , resetN : in std_logic;
pdata : in integer range 0 to 5; --parallel data in. to choose how many clock the out pulse would be.
din : in std_logic;
dout : out std_logic
) ;
end fifth ;
architecture arc_fifth of fifth is
signal count : integer range 0 to 5;
signal pl : std_logic; --trigger detect output.
signal sample1 : std_logic;
signal sample2 : std_logic;
--trigger sync proccess.
begin
process(clk , resetN)
begin
if resetN = '0' then
sample1<='0';
sample2<='0';
elsif rising_edge(clk) then
sample1<=din;
sample2<=sample1;
end if;
end process;
pl <= sample1 and (not sample2); --trigger detect output. activate the counter.
--counter proccess.
process ( clk , resetN )
begin
if resetN = '0' then
count <= 0 ;
elsif rising_edge(clk) then
if pl='1' then
count<=pdata;
else
if count=0 then
count<=count;
else
count<=count-1;
end if;
end if;
end if ;
end process ;
dout<='1' when count>0 else '0';--output - get the wanted lenght pulse no matter how long is input
end arc_fifth ;