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EEE20001

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Computer Science

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Oct 30, 2023

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Digital Electronics - Tutorial 27 of 46 Tutorial 8 Solution Q1) State sequence is: 1, 8, 4, 2, 9, 12, 6, 11, 5, 10, 13, 14, 15, 7, 3, 1 (repeat) or 0, 0 (repeat) To remove the 0-loop, add a 4-input NOR gate to detect “0000” and an XOR to force a ‘1’ into the SR. Q2) State seq – Old solution! Question changed to include E and C. D Q D Q D Q D Q Clock Idle 0 F1 0 F16 0 F16 1 1 5 6 1 1 1 else else else else 1 Idle L 1 F1L 1 F16 L 1 F161L 1 1 5 6 1 1 1 else else else else
Digital Electronics - Tutorial 28 of 46 Q3) State seq Idle 5c 10c 15c 20c 25c d5/ d5/ d5/ d10/ d10/ d10/ d20/ d20/ d20/ /r5 /r5 /p Inputs d5 = deposit 5c d10 = deposit 10c d20 = deposit 20c Actions r5 = refund 5c p = dispense product Inputs/Actions 30c /r5 Idle F1 F16 F16 1 1/0 5/1 6/0 1/0 1/0 1/0 Else/0 Else/0 Else/0 Else/0 1/0 T Q Controlling a T-FF
Digital Electronics - Tutorial 29 of 46 Q4) --------------------------------------------------------------------------------- -- 4-bit PRNS ---------------------------------------------------------------------------------- library IEEE; use IEEE.STD_LOGIC_1164.ALL; entity LFSR is Port ( clock : in STD_LOGIC; PRNS : out STD_LOGIC ); end entity LFSR; architecture Behavioral of LFSR is signal sr : std_logic_vector(3 downto 0); begin PRNS <= sr(0); sync: process( clock ) begin if rising_edge(clock) then if (sr = "0000") then -- force out of undesired loop sr <= "0001"; else -- implement PRNS using XOR feedback sr <= (sr(1) xor sr(0)) & sr(3 downto 1); end if; end if; ----------------------------------------------- -- OR ----------------------------------------------- -- sr(3)<= sr(1) xor sr(0); -- sr(2) <=sr(3); -- sr(1) <=sr(2); -- sr(0) <=sr(1); ----------------------------------------------- -- OR ----------------------------------------------- -- sr(3)<= sr(1) xor sr(0); -- sr(2 downto 0) <= sr(3 downto 1); end process sync; end architecture Behavioral;
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