2 use ieee.std_logic_1164.all;
3 use ieee.std_logic_arith.all;
4 use ieee.std_logic_unsigned.all;
5 use ieee.numeric_std.all;
12 qcount: out std_logic_vector (31 downto 0);
13 a_rise, a_fall, b_rise, b_fall, ab_event: out std_logic;
14 ab_error: out std_logic
18 architecture behavioral of qcounter is
27 subtype std_logic4 is std_logic_vector (3 downto 0);
28 signal a, b, a_prev, b_prev: std_logic;
29 signal count_prev: std_logic_vector (29 downto 0)
30 := "000000000000000000000000000000";
31 signal count: std_logic_vector (29 downto 0);
33 --ustaleni signalu a mezi tiky hodin
34 --reaguje na nabeznou hranu
42 --ustaleni signalu b mezi tiky hodin
43 --reaguje na nabeznou hranu
51 --prvni dva rady kombinacne
54 qcount(31 downto 2) <= count;
56 --k cemu tento prosess? jen pro prenos udalosti na piny?
57 comb_event: process (a_prev, b_prev, a, b) --proc je v sensitivity listu i stary stav?? jen kvuli nulovani?
65 if ((a xor a_prev) and (b xor b_prev)) = '1' then -- a i b se zmenily zaroven
66 -- forbidden double transition
69 a_rise <= (a xor a_prev) and a; --a nabezna
70 a_fall <= (a xor a_prev) and not a; --a sestupna
71 b_rise <= (b xor b_prev) and b; --b nabezna
72 b_fall <= (b xor b_prev) and not b; --b sestupna
73 ab_event <= (a xor a_prev) or (b xor b_prev); --a nebo b se zmenily
77 --prechod do 3. radu (binarne)
78 comb_count: process (a_prev, b_prev, a, b, count,count_prev) --proc jsou v sensitivity prev stavy?
80 if (a_prev = '0') and (b_prev = '1') and (a = '0') and (b = '0') then --posun dopredu
81 count <= count_prev + 1;
82 elsif (a_prev = '0') and (b_prev = '0') and (a = '0') and (b = '1') then --posun dozadu
83 count <= count_prev - 1;
89 --s nabeznou hranou hodin ulozime stare (prev) hodnoty, pripadne synchronni reset
90 --proc resuteje count_prev misto count?? neprojevi se tato zmena az pri pruchodu comb_count??
93 wait until clock'event and clock = '1';
95 count_prev <= "000000000000000000000000000000";
105 --proc pouzivan count_prev a neprirazuji rovnou do count?