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EDA技术与应用课后习题答案
SIGNAL Q : STD_LOGIC;
BEGIN
PR01: PROCESS(CLK0)
BEGIN
IF CLK ‘EVENT AND CLK=’1’
THEN Q<=NOT(CL OR Q);ELSE
END IF;
END PROCESS;
PR02: PROCESS(CLK0)
BEGIN
OUT1<=Q;
END PROCESS;
END ARCHITECTURE ONE;
END PROCESS;
4-5.给出1位全减器的VHDL描述。要求：
(1) 首先设计1位半减器，然后用例化语句将它们连接起来，图3-32中h_suber是半减器，diff是输出差，s_out是借位输出，sub_in是借位输入。
(2) 以1位全减器为基本硬件，构成串行借位的8位减法器，要求用例化语句来完成此项设计(减法运算是 x – y - sun_in = diffr)
4-5.答案
底层文件1：or2a.VHD实现或门操作
LIBRARY IEEE;
USE IEEE.STD_LOGIC_1164.ALL;
USE IEEE.STD_LOGIC_UNSIGNED.ALL;
ENTITY or2a IS
PORT(a,b:IN STD_LOGIC;
c:OUT STD_LOGIC);
END ENTITY or2a;
ARCHITECTURE one OF or2a IS
BEGIN
c <= a OR b;
END ARCHITECTURE one;
底层文件2：h_subber.VHD实现一位半减器
LIBRARY IEEE;
USE IEEE.STD_LOGIC_1164.ALL;
USE IEEE.STD_LOGIC_UNSIGNED.ALL;
ENTITY h_subber IS
PORT(x,y:IN STD_LOGIC;
diff,s_out::OUT STD_LOGIC);
END ENTITY h_subber;
ARCHITECTURE ONE OF h_subber IS
SIGNAL xyz: STD_LOGIC_VECTOR(1 DOWNTO 0);
BEGIN
xyz <= x & y;
PROCESS(xyz)
BEGIN
CASE xyz IS
WHEN "00" => diff<='0';s_out<='0';
WHEN "01" => diff<='1';s_out<='1';
WHEN "10" => diff<='1';s_out<='0';
WHEN "11" => diff<='0';s_out<='0';
WHEN OTHERS => NULL;
END CASE;
END PROCESS;
END ARCHITECTURE ONE;
顶层文件：f_subber.VHD实现一位全减器
LIBRARY IEEE;
USE IEEE.STD_LOGIC_1164.ALL;
USE IEEE.STD_LOGIC_UNSIGNED.ALL;
ENTITY f_subber IS
PORT(x,y,sub_in:IN STD_LOGIC;
diffr,sub_out:OUT STD_LOGIC);
END ENTITY f_subber;
ARCHITECTURE ONE OF f_subber IS
COMPONENT h_subber
PORT(x,y:IN STD_LOGIC;
diff,S_out:OUT STD_LOGIC);
END COMPONENT;
COMPONENT or2a
PORT(a,b:IN STD_LOGIC;
c:OUT STD_LOGIC);
END COMPONENT;
SIGNAL d,e,f: STD_LOGIC;
BEGIN
u1: h_subber PORT MAP(x=>x,y=>y,diff=>d,s_out=>e);
u2: h_subber PORT MAP(x=>d,y=>sub_in,diff=>diffr,s_out=>f);
u3: or2a PORT MAP(a=>f,b=>e,c=>sub_out);
END ARCHITECTURE ONE;
END ARCHITECTURE ART;
4-6.根据下图，写出顶层文件MX3256.VHD的VHDL设计文件。
4-6.答案
MAX3256顶层文件
LIBRARY IEEE;
USE IEEE.STD_LOGIC_1164.ALL;
USE IEEE.STD_LOGIC_UNSIGNED.ALL;
ENTITY MAX3256 IS
PORT (INA,INB,INCK: IN STD_LOGIC;
INC: IN STD_LOGIC;
E,OUT:OUT STD_LOGIC);
END ENTITY MAX3256;
ARCHITECTURE ONE OF MAX3256 IS
COMPONENT LK35 --调用LK35声明语句
PORT(A1,A2:IN STD_LOGIC;
CLK:IN STD_LOGIC;
Q1,Q2:OUT STD_LOGIC);
END COMPONENT;
COMPONENT D --调用D触发器声明语句
PORT(D,C:IN STD_LOGIC;
CLK:IN STD_LOGIC;
Q:OUT STD_LOGIC);
END COMPONENT;
COMPONENT MUX21--调用二选一选择器声明语句
PORT(B,A:IN STD_LOGIC;
S:IN STD_LOGIC;
C:OUT STD_LOGIC);
END COMPONENT;
SIGNAL AA,BB,CC,DD: STD_LOGIC;
BEGIN
u1: LK35 PORT MAP(A1=>INA,A2=>INB,CLK=INCK, Q1=>AA,Q2=>BB);
u2: D PORT MAP(D=>BB;CLK=>INCK,C=>INC,Q=>CC);
u3: LK35 PORT MAP (A1=>BB,A2=>CC,CLK=INCK, Q1=>DD,Q2=>OUT1);
u4: MUX21 PORT MAP (B=>AA,A=>DD,S=>BB,C=>E);
END ARCHITECTURE ONE;
设计含有异步清零和计数使能的16位二进制加减可控计数器。
4-7.答案：
LIBRARY IEEE;
USE IEEE.STD_LOGIC_1164.ALL;
USE IEEE.STD_LOGIC_UNSIGNED.ALL;
ENTITY CNT16 IS
PORT(CLK,RST,EN:IN STD_LOGIC;
CHOOSE:IN BIT;
SETDATA:BUFFER INTEGER RANCE 65535 DOWNTO 0;