第十章数孤统设计部例 10.1跑马灯设计 1.设计要求 控制8个LED进行花式显示,设计4中显示模式: ●S0:从左到右逐个点亮IFD; ●s1:从右到左逐个点亮LED ●S2从两边到中间逐个点亮LED ●S3:从中间到两边逐个点亮LED 4种模式循环切换,复位键(rst控制系统的运行和停止。 led clkx-时钟 clk rst一复位信号输入端 rst q[7…0 q-数据输出端 led的电路符号
第十章 数字系统设计范例 10.1 跑马灯设计 1.设计要求 控制8个LED进行花式显示，设计4中显示模式： ⚫S0:从左到右逐个点亮LED; ⚫S1:从右到左逐个点亮LED; ⚫S2:从两边到中间逐个点亮LED; ⚫S3:从中间到两边逐个点亮LED; 4种模式循环切换，复位键(rst)控制系统的运行和停止。 led的电路符号 clk q[7…0] clk –时钟 q – 数据输出端 led rst rst – 复位信号输入端
library ieee; use ieeestd_logic_1164.all use ieee std_logic_unsigned. all; entity led is port(elk:in std_ _ logic: rstin std _ logic; g: out std logic__vector(7 downto O)); dled architecture one ofled is type states is(0,s1,2,3;-定义四种模式 signal presentstates signal q1: std_logic_vector(7 downto O) signal countstd _ logic__vector(3 downto O) egn process(clk b egin ifrst=1 then 系统复位 present<=s0; q1 0000000; elsif clk event and clk=1 then
library ieee; use ieee.std_logic_1164.all; use ieee.std_logic_unsigned.all; entity led is port(clk:in std_logic; rst:in std_logic; q:out std_logic_vector(7 downto 0)); end led; architecture one of led is type states is (s0,s1,s2,s3); --定义四种模式 signal present:states; signal q1: std_logic_vector(7 downto 0); signal count:std_logic_vector(3 downto 0); begin process(clk,rst) begin if rst=‘1’ then --系统复位 present<=s0; q1<=“00000000”; elsif clk’event and clk=‘1’ then
case present 1s when s0=>ifq1=“00000then-S0模式,从左到右逐个点亮工ED 1<=·10000000 els if count=“0111then count<=“0000 q1<=“00000 present<=s1 else q1<=q1()&eq1(7 downto 1); count<=count+1 end s present<=s0 when s1=> if count=“0111”then-!1模式,从右到左逐个点亮LED count<=“0000 q1<=“10000 resent<=s else q1<=q1(6 downto O)&eq1(); count<=count+1; resent<=s1 end if:
case present is when s0 => if q1=“00000000” then --S0模式，从左到右逐个点亮LED q1<=“10000000”; else if count=“0111” then count<=“0000”; q1<=“00000001”; present<=s1; else q1<=q1(0)&q1(7 downto 1); count<=count+1; present<=s0; end if; end if; when s1 => if count=“0111” then --S1模式，从右到左逐个点亮LED count<=“0000”; q1<=“10000001”; present<=s2; else q1<=q1(6 downto 0)&q1(7); count<=count+1; present<=s1; end if;
when s2=> if count=“0111then-S2模式,从两边到中间逐个点亮LED count<=“0000 1<=“00011000 pr 3; else q1(7 downto 4)<=q1 ()& 1(7 downto 5) q1(3 downto 0)<=q1(2 downto O)&eq1(3) count<=count+1 present<=s2: end if: hens3=> if count=“0111′then-S3模式,从中间到两边逐个点亮LED count<=“000 1<=“1000000 present<=sO else q1(7 downto 4 <=q1(6 downto 4)&xq1(7 end if Present<=s t+ 9Z q1(3 downto O)<= (O)&eq1(3 downto 1); end case, dife end 9=g end one
when s2 => if count=“0111” then --S2模式，从两边到中间逐个点亮LED count<=“0000”; q1<=“00011000”; present<=s3; else q1(7 downto 4)<=q1(4)&q1(7 downto 5); q1(3 downto 0)<=q1(2 downto 0)&q1(3); count<=count+1; present<=s2; end if; when s3 => if count=“0111” then --S3模式，从中间到两边逐个点亮LED count<=“0000”; q1<=“10000000”; present<=s0; else q1(7 downto 4)<=q1(6 downto 4)&q1(7); q1(3 downto 0)<=q1(0)&q1(3 downto 1); count<=count+1; present<=s3; end if; end case; end if; end process; q<=q1; end one;
0. 19.9ns L厂L厂 B OOC 000000×10000000100000010000×0010000000X000010x00000 00000001 X000000 rst presen present, so resent. sI