odd and even signals

[Patrick]

hello

I would like to know how to produce the bit signal with 2 baud signals
for DQPSK demodulation
I is the odd of the bit signal and Q is the even signal of bit signal

clk_bit is twice clk_baud

I have two signals I and Q and I want to generate out_bit with these
two signals I and Q

I made it like this but I think that it is not the right solution

Do you have suggestion about this design ?

signal I, Q : in std_logic;

ECH_IQ : process (init,clk_baud)
begin
if init='1' then
I_i <= '0';
Q_i <= '0';
elsif (clk_baud'event and clk_baud='1') then
I_i <= I;
Q_i <= Q;
end if;
end process ECH_IQ;

C1 <= I_i AND clk_baud; -- here i put the odd signal into C1
D1 <= Q_i AND (NOT clk_baud); -- and here the event signal into D1

-- I think it's not good to do a logic operator between a clock and a
signal

SIG_OUT_BIT : process (init,clk_bit)
begin
if init='1' then
out_bit <= '0';
elsif (clk_bit'event and clk_bit='1') then
out_bit <= D1 OR C1;
end if;
end process SIG_OUT_BIT;

 

[Jonathan Bromley]

On 21 Dec 2004 06:41:49 -0800, patrick.melet@dmradiocom.fr (Patrick)
wrote:

[description of something that is essentially a
Double Data Rate receiver]

clk_bit is twice clk_baud

How are these two clocks related? It matters very much.

-- I think it's not good to do a logic operator
between a clock and a signal

I think I agree, rather strongly - if the signal is
synchronous to that same clock.

The code you describe is a rather long-winded way to
describe a multiplexer; it's much easier to describe
like this:

process (init, clk_bit)
begin
if init = '1' then
out_bit <= '0';
elsif rising_edge (clk_bit) then
if clk_baud = '1' then
out_bit <= I_i;
else
out_bit <= Q_i;
end if;
end if;
end process;

However, this is not necessarily correct. It depends
on the timing relationship between clk_bit and clk_baud.
It may be that you know this, because (perhaps) you
derived clk_baud by dividing clk_bit by 2, or some
similar idea. If you *don't* know the relationship,
things get a great deal more difficult. Indeed, this is
the root of your concern about gating a clock with other
signals (which my code also does, albeit in a less
obvious way).
--
Jonathan Bromley, Consultant

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